JEE Basic Principles of Organic Chemistry Questions

The principal functional group in the given compound is the carboxylic acid group ($-CO_2H$).

According to IUPAC priority rules,

$$\text{Carboxylic acid} > \text{Hydroxyl} > \text{Halogens/Nitro}.$$

Hence, the parent compound is benzoic acid.

The carbon atom attached to the carboxylic acid group is assigned position $$1$$.

Now, the ring is numbered in the direction that gives the lowest possible set of locants for the remaining substituents.

Clockwise numbering gives:

• $$2$$ : Hydroxy ($-OH$)
• $$3$$ : Bromo ($-Br$)
• $$5$$ : Nitro ($-NO_2$)

This produces the locant set

$${2,,3,,5}.$$

Counter-clockwise numbering gives:

• $$3$$ : Nitro ($-NO_2$)
• $$5$$ : Bromo ($-Br$)
• $$6$$ : Hydroxy ($-OH$)

This produces the locant set

$${3,,5,,6}.$$

Since

$${2,,3,,5} < {3,,5,,6},$$

the clockwise numbering is selected.

The substituents are therefore located at:

• $$3$$-Bromo
• $$2$$-Hydroxy
• $$5$$-Nitro

Finally, the substituent prefixes are arranged alphabetically:

• Bromo
• Hydroxy
• Nitro

Hence, the correct IUPAC name is

3-bromo-2-hydroxy-5-nitrobenzoic acid.

Therefore, the correct answer is:

(A) 3-bromo-2-hydroxy-5-nitrobenzoic acid

We consider Hex-1,3-dien-5-yne, whose structure is $$CH_2=CH-CH=CH-C\equiv CH$$, a six-carbon chain with double bonds at positions 1-2 and 3-4 and a triple bond at positions 5-6.

For $$\sigma$$ bonds, there are C-C single bonds between C2-C3 and C4-C5 (2 $$\sigma$$), one $$\sigma$$ each from the two C=C bonds at C1=C2 and C3=C4 (2 $$\sigma$$), and one $$\sigma$$ from the C≡C bond at C5-C6 (1 $$\sigma$$), giving a total of 5 C-C $$\sigma$$ bonds. The C-H bonds contribute $$\sigma$$ bonds as follows: C1 has 2H, C2 has 1H, C3 has 1H, C4 has 1H, C5 has 0H and C6 has 1H, totaling 6 $$\sigma$$ bonds. Hence the overall $$\sigma$$ count is 5 + 6 = 11.

Each double bond has one $$\pi$$ bond, so the two C=C bonds provide 2 $$\pi$$ bonds, while the C≡C bond provides 2 $$\pi$$ bonds, for a total of 4 $$\pi$$ bonds.

Therefore, the molecule contains $$\sigma + \pi = 11 + 4 = 15$$ bonds in total.

Availability of the Nitrogen Lone Pair

The basic strength of an amine or a heterocyclic nitrogen compound depends directly on the availability of its lone pair of electrons on the nitrogen atom to accept a proton ($$\text{H}^+$$). If the lone pair is highly localized, the compound is more basic; if it is delocalized due to resonance or held tightly by a more electronegative orbital, the basicity drops significantly.

Detailed Analysis of Each Compound:

• 1. Piperidine

  Piperidine is a cyclic secondary aliphatic amine. The nitrogen atom is completely $$\text{sp}^3$$ hybridized, and its lone pair of electrons is highly localized in an $$\text{sp}^3$$ orbital. Since it does not participate in resonance and benefits from the electron-donating inductive effect ($$+I$$) of the adjacent alkyl ring chain, piperidine is the most basic compound among the three.




• 2. Pyridine

  In pyridine, the nitrogen atom is part of an aromatic ring and is $$\text{sp}^2$$ hybridized. The lone pair of electrons resides in an $$\text{sp}^2$$ hybrid orbital that lies within the plane of the ring, meaning it is perpendicular to the $$\pi$$ system and does not participate in aromaticity. It is available for protonation, but because an $$\text{sp}^2$$ hybridized nitrogen is more electronegative than an $$\text{sp}^3$$ nitrogen, it holds its electrons more tightly. Therefore, pyridine is significantly less basic than piperidine.




• 3. Pyrrole

  In pyrrole, the lone pair on the nitrogen atom is actively involved in the aromatic ring system to satisfy Hückel's rule ($$6\pi$$ electrons). Because these electrons are fully delocalized within the aromatic ring, donating them to a proton would destroy the molecule's aromatic stabilization. This makes pyrrole an exceptionally weak base, placing it as the least basic compound in the group.

Conclusion:

Combining the hybridization and delocalization factors, the correct decreasing order of basic strength is:

$$\mathbf{\text{Piperidine} > \text{Pyridine} > \text{Pyrrole}}$$

Answer: Option D — Piperidine > Pyridine > Pyrrole

Three different reducible groups are present in $$P$$: an internal alkyne, a nitro group and a carbonyl group. They have to be converted successively to a cis-alkene, an amine and an alcohol in the final product $$Q$$. For each transformation we choose the most chemoselective reagent and decide the safest order so that the later steps do not undo or over-reduce what has already been achieved.

Step 1 - Reduction of the C≡C bond only
Lindlar’s catalyst with $$H_2$$ stops at the cis-alkene stage and does not disturb either nitro or carbonyl groups.
Hence the first reagent must be “Lindlar’s catalyst, $$H_2$$”.

Step 2 - Reduction of the −NO2 group
The most selective reagent that converts $$-NO_2$$ to $$-NH_2$$ in the presence of a carbonyl is $$SnCl_2/HCl$$. Using it before reducing the carbonyl is preferable because the strongly acidic medium would protonate or even dehydrate an alcohol if it were already present. Therefore the second reagent should be “$$SnCl_2/HCl$$”.

Step 3 - Reduction of the C=O group
After the nitro group has become an amine, an electrically neutral hydride reagent that works in neutral or mildly basic medium is required for the carbonyl. $$NaBH_4$$ cleanly converts aldehydes/ketones to alcohols and leaves alkenes and amines untouched. Thus the third reagent is “$$NaBH_4$$”.

Step 4 - Acidic work-up
The hydride reduction is performed in methanol or ethanol; an aqueous acid work-up ($$H_3O^+$$) is necessary to protonate the alkoxide, giving the free alcohol and completing the synthesis. Consequently the fourth reagent is “$$H_3O^+$$”.

Why the other sequences fail
• If $$NaBH_4$$ comes before $$SnCl_2/HCl$$ (Options C and D), the alcohol formed in step 1 would face strongly acidic conditions in step 2 and might dehydrate or undergo side reactions.
• Placing $$H_3O^+$$ immediately after the partial hydrogenation (Option B) serves no purpose and the subsequent acidic reduction with $$SnCl_2/HCl$$ would again endanger any alcohol already present.
• Any sequence that applies $$SnCl_2/HCl$$ or $$NaBH_4$$ before Lindlar’s hydrogenation risks over-reduction of the alkyne (because later catalytic hydrogenation over Pd/C would reduce the newly formed alkene all the way to an alkane).

Thus the only logically consistent sequence is

i) Lindlar's catalyst, $$H_2$$   →   ii) $$SnCl_2/HCl$$   →   iii) $$NaBH_4$$   →   iv) $$H_3O^+$$

Option A matches this order.

Final Answer: Option A which is: i) Lindlar's catalyst, $$H_2$$; ii) $$SnCl_2$$/HCl; iii) $$NaBH_4$$; iv) $$H_3O^+$$.

We need to evaluate both statements about nucleophilic substitution reactions.

Statement (I): $$S_N2$$ reactions are 'stereospecific', indicating that they result in the formation of only one stereo-isomer as the product.

In $$S_N2$$ reactions, the nucleophile attacks from the back side of the carbon bearing the leaving group (backside attack). This results in a Walden inversion, producing a single stereoisomer with inverted configuration. Hence, $$S_N2$$ reactions are indeed stereospecific. Statement I is true.

Statement (II): $$S_N1$$ reactions generally result in formation of product as racemic mixtures.

In $$S_N1$$ reactions, the rate-determining step involves the formation of a planar carbocation intermediate. The nucleophile can attack this planar carbocation from either side with nearly equal probability, leading to a mixture of both R and S configurations — a racemic mixture. Statement II is true.

Since both statements are true, the correct answer is Option 4: Both Statement I and Statement II are true.

Vanillin: 4-hydroxy-3-methoxybenzaldehyde. C₈H₈O₃. O atoms: 3 (OH, OCH₃, CHO). π electrons: benzene ring (3 C=C→6π) + C=O (2π) = 8π... Wait: benzene has 6π electrons (3 double bonds), aldehyde C=O has 2π electrons. Total π electrons = 6+2=8.

Sum of O atoms + π electrons = 3+8=11.

The answer is $$\boxed{11}$$.

Explanation

To compare the acidity of the marked protons, compare the stability of the conjugate base formed after removal of each proton. Greater stability of the conjugate base implies greater acidity.

**Analyzing $$H_C$$**

- $$H_C$$ is the proton of the carboxylic acid group $$(-COOH)$$.

- Removal of $$H_C$$ forms a carboxylate ion.

- The negative charge is delocalized equally over two oxygen atoms through resonance.

- Since oxygen is highly electronegative and the charge is resonance-stabilized, $$H_C$$ is the most acidic proton.

**Analyzing $$H_D$$**

- $$H_D$$ is present on the carbon adjacent to both a carbonyl group and a phenyl ring.

- Removal of $$H_D$$ forms a carbanion that is resonance-stabilized by the carbonyl group.

- The negative charge can be delocalized onto the oxygen atom through an enolate-like structure.

- Additional stabilization is provided by the adjacent phenyl ring.

- Therefore, $$H_D$$ is the most acidic among the carbon-bound protons.

**Analyzing $$H_A$$**

- $$H_A$$ is the proton of a terminal alkyne.

- It is attached to an $$sp$$-hybridized carbon.

- An $$sp$$ carbon has $$50\%$$ $$s$$-character, making it relatively more electronegative and better able to stabilize a negative charge.

- Hence, $$H_A$$ is more acidic than ordinary alkyl or alkenyl $$C-H$$ protons.

**Analyzing $$H_B$$**

- $$H_B$$ is a benzylic and propargylic proton.

- Removal of $$H_B$$ gives a carbanion that is resonance-stabilized over the benzene ring and the adjacent alkyne.

- However, the negative charge remains delocalized only over carbon atoms and is never placed on an oxygen atom.

- This stabilization is weaker than that observed for $$H_D$$ and also less effective than the stabilization associated with the $$sp$$ carbon of $$H_A$$.

- Therefore, $$H_B$$ is the least acidic proton.

The decreasing order of acidity is

$$H_C > H_D > H_A > H_B$$

Hence, the correct answer is Option B.

• A. Propanamine and N-Methylethanamine $$\rightarrow$$ III. Functional isomers
  • Propanamine is a primary ($$1^{\circ}$$) amine (functional group: $$-\text{NH}_2$$), while N-methylethanamine is a secondary ($$2^{\circ}$$) amine (functional group: $$-\text{NH}-$$). Since primary and secondary amines contain different functional groups, they are functional isomers.
• B. Hexan-2-one and Hexan-3-one $$\rightarrow$$ I. Metamers
  • Both are ketones sharing the same polyvalent functional group ($$-\text{CO}-$$). However, they differ in the alkyl groups attached to either side of the carbonyl carbon (methyl & butyl vs. ethyl & propyl), making them metamers.
• C. Ethanamide and Hydroxyethanimine $$\rightarrow$$ IV. Tautomers
  • These two structures exist in a dynamic equilibrium via the migration of a proton ($$\text{H}^+$$) and the shifting of a double bond (amide-imidol tautomerism). Thus, they are tautomers.
• D. o-Nitrophenol and p-Nitrophenol $$\rightarrow$$ II. Positional isomers
  • Both possess the exact same functional groups on a benzene ring, but the nitro ($$-\text{NO}_2$$) group is located at different positions (ortho vs. para) relative to the hydroxyl ($$-\text{OH}$$) group. Therefore, they are positional isomers.

Combining these correct matches gives: A-III, B-I, C-IV, D-II.

The correct option is D.

Hydride affinity is defined for the reaction
$$R^+ + H^- \;\rightarrow\; RH$$.
The larger the energy released (that is, the more negative the reaction enthalpy), the greater is the hydride affinity of the carbocation $$R^+$$.

A carbocation that is intrinsically less stable gains a larger amount of stabilisation on accepting a hydride ion, so it shows a higher hydride affinity.
Conversely, a more stable carbocation is already low in energy; it gains relatively little additional stabilisation on accepting $$H^-$$ and therefore exhibits a lower hydride affinity.
Hence,

$$\text{higher hydride affinity}\; \Longleftrightarrow\; \text{lower inherent stability of the carbocation}$$.

Case 1 (c):

The methyl cation $$CH_3^+$$ possesses neither hyperconjugation nor resonance stabilisation. It is the least stabilised of the four; therefore it will have the highest hydride affinity.

Case 2 (a):

The n-propyl (1°) carbocation $$CH_3CH_2CH_2^+$$ is stabilised by hyperconjugation with two α-hydrogen atoms. It is more stable than $$CH_3^+$$ but less stable than a 3° or resonance-stabilised cation. Its hydride affinity is therefore lower than that of (c) but higher than that of (b) and (d).

Case 3 (b):

The tert-butyl (3°) carbocation $$(CH_3)_3C^+$$ enjoys extensive hyperconjugation (nine α-hydrogen atoms), making it markedly more stable than a 1° carbocation. Hence its hydride affinity is lower than those of (c) and (a).

Case 4 (d):

The benzyl carbocation $$C_6H_5CH_2^+$$ is stabilised by resonance with the aromatic ring, distributing the positive charge over several atoms. This resonance stabilisation is stronger than the hyperconjugative stabilisation present in $$(CH_3)_3C^+$$. Therefore it is the most stable of the four and consequently has the lowest hydride affinity.

Combining the four cases, the order from highest to lowest hydride affinity is
$$\text{(c)} \; \gt \; \text{(a)} \; \gt \; \text{(b)} \; \gt \; \text{(d)}.$$

This corresponds to Option B: C, A, B, D.

Answer : Option B

An asymmetric carbon atom is an $$sp^3$$ hybridized carbon bonded to four different groups. Any carbon involved in a double bond is $$sp^2$$ hybridized and cannot be asymmetric. Similarly, carbons bonded to two or more identical substituents are not chiral.

In the testosterone structure, the carbonyl carbon and the two carbons involved in the $$C=C$$ double bond are $$sp^2$$ hybridized and therefore achiral.

The various $$CH_2$$ and $$CH_3$$ carbons are also achiral because they are bonded to two or more identical hydrogen atoms.

The remaining stereogenic centers occur at the ring junctions and the carbon bearing the hydroxyl group.

The bridgehead carbon joining rings A and B is asymmetric because it is attached to four different groups.

The two bridgehead carbons joining rings B and C are asymmetric because each is bonded to four different carbon environments.

The bridgehead carbon joining rings C and D is asymmetric because the four groups attached to it are different.

The quaternary carbon bearing the angular methyl group is asymmetric because it is attached to four distinct carbon pathways.

The carbon bearing the $$-OH$$ group is asymmetric because it is attached to $$-OH$$, hydrogen, and two different carbon chains.

Thus, the asymmetric carbon atoms are located at

$$C_8,\ C_9,\ C_{10},\ C_{13},\ C_{14},\ C_{17}$$

Therefore, the total number of asymmetric carbon atoms in testosterone is $$6$$. Hence, the correct answer is $$6$$.

Step-1 : Identification of the products P, Q, R and S from the reaction scheme given in the question.

(i) The first reaction is the hydroxylation of ethene with cold, dilute alkaline $$KMnO_4$$ (Baeyer’s reagent). The double bond is converted into a cis-vicinal diol:
$$CH_2 = CH_2 \;\xrightarrow[\text{cold}]{KMnO_4/OH^-}\; HO-CH_2-CH_2-OH$$
Hence, $$P = HO-CH_2-CH_2-OH$$ (ethylene glycol).

(ii) The second reaction is vigorous oxidation of o-xylene with hot alkaline $$KMnO_4$$ followed by acidification. Both benzylic methyl groups are oxidised to carboxyl groups giving phthalic acid:
$$o\text{-}CH_3C_6H_4CH_3 \;\xrightarrow[\text{heat}]{KMnO_4}\; HOOC-C_6H_4-COOH$$
Hence, $$Q = HOOC-C_6H_4-COOH$$ (phthalic acid).

(iii) The third reaction is the dehydrogenation of 1-propanol over copper at 523 K. Primary alcohols give the corresponding aldehydes under these conditions:
$$CH_3CH_2CH_2OH \;\xrightarrow{Cu/523\,K}\; CH_3CH_2CHO$$
Hence, $$R = CH_3CH_2CHO$$ (propanal).

(iv) The fourth reaction is the Rosenmund reduction of benzoyl chloride, or an equivalent route that stops at the aldehyde stage. Thus the product is benzaldehyde:
$$C_6H_5COCl \;\xrightarrow{H_2/Pd\,\,\text{(poisoned)}}\; C_6H_5CHO$$
Hence, $$S = C_6H_5CHO$$ (benzaldehyde).

Step-2 : Verification of the statements.

Option A : Ethylene glycol (P) and phthalic acid (Q) are indeed used as monomers for polyester formation.
• Dacron (polyethylene terephthalate) is obtained from ethylene glycol and terephthalic acid.
• Glyptal is obtained from ethylene glycol and phthalic acid.
Therefore, Option A is correct.

Option B : P, Q and R are claimed to be dicarboxylic acids.
• P is a di-alcohol (diol), not a diacid.
• R is an aldehyde, not an acid.
Hence Option B is wrong.

Option C : “Compounds Q and R are the same.”
Q is phthalic acid (aromatic diacid) whereas R is propanal (aliphatic aldehyde). They are completely different. Option C is wrong.

Option D :
• R (propanal) possesses α-hydrogen atoms, so it undergoes aldol condensation in the presence of base. The statement “R does not undergo aldol condensation” is false.
• S (benzaldehyde) has no α-hydrogen atoms and therefore readily undergoes the Cannizzaro reaction. The statement “S does not undergo Cannizzaro reaction” is also false.
Hence Option D is wrong.

Step-3 : Conclusion.

Only Option A is correct.

Final answer : Option A which is: P and Q are monomers of polymers dacron and glyptal, respectively.

Matching each named reaction with its reagent:

A. Hell-Volhard-Zelinsky (HVZ) reaction: This reaction involves α-halogenation of carboxylic acids using Br₂ in the presence of red phosphorus, followed by hydrolysis. Reagent: (i) Br₂/red phosphorus (ii) H₂O → Match with III

B. Iodoform reaction: This reaction is used to detect methyl ketones and secondary alcohols using NaOH and I₂ (or NaOI). Reagent: NaOH + I₂ → Match with I

C. Etard reaction: This reaction converts a methyl group attached to an aromatic ring to an aldehyde using chromyl chloride in CS₂. Reagent: (i) CrO₂Cl₂, CS₂ (ii) H₂O → Match with II

D. Gattermann-Koch reaction: This reaction introduces an aldehyde group on an aromatic ring using CO and HCl in the presence of anhydrous AlCl₃. Reagent: CO, HCl, anhyd. AlCl₃ → Match with IV

The correct matching is: A → III, B → I, C → II, D → IV.

Step 1: Formation of Alkyl Iodide

Sodium iodide ($$\text{NaI}$$) reacts with phosphoric acid ($$\text{H}_3\text{PO}_4$$) to generate hydrogen iodide ($$\text{HI}$$) in situ. The alcohol group ($$-\text{OH}$$) on butan-2-ol is protonated and substituted by the iodide ion ($$\text{I}^-$$) via an $$\text{S}_\text{N}2$$ or $$\text{S}_\text{N}1$$ pathway, converting it into 2-iodobutane.

$$\text{CH}_3-\text{CH}_2-\text{CH(OH)}-\text{CH}_3 \xrightarrow{\text{NaI, H}_3\text{PO}_4} \text{CH}_3-\text{CH}_2-\text{CH(I)}-\text{CH}_3$$

Step 2: Formation of Grignard Reagent

Magnesium metal inserts itself into the carbon-iodine bond in the presence of dry ether, forming a highly reactive Grignard reagent (sec-butylmagnesium iodide).

$$\text{CH}_3-\text{CH}_2-\text{CH(I)}-\text{CH}_3 \xrightarrow{\text{Mg, Dry Ether}} \text{CH}_3-\text{CH}_2-\text{CH(MgI)}-\text{CH}_3$$

Step 3: Deuteration

Grignard reagents are strong bases and react aggressively with any proton/deuterium source. Heavy water ($$\text{D}_2\text{O}$$) supplies a deuterium ion ($$\text{D}^+$$), which replaces the $$-\text{MgI}$$ group.

$$\text{CH}_3-\text{CH}_2-\text{CH(MgI)}-\text{CH}_3 \xrightarrow{\text{D}_2\text{O}} \text{CH}_3-\text{CH}_2-\text{CH(D)}-\text{CH}_3 + \text{Mg(OD)I}$$

The final product is 2-deuterobutane:

$$\mathbf{\text{CH}_3-\text{CH}_2-\text{CH(D)}-\text{CH}_3}$$

When 2-methylcyclohexan-1-ol reacts with HBr, the OH group is first protonated to form a good leaving group ($$H_2O$$), which departs to give a secondary carbocation at C1.

Now, this secondary carbocation at C1 undergoes a 1,2-hydride shift from C2 to form a more stable tertiary carbocation at C2 (the carbon bearing the methyl group).

We have the tertiary carbocation at C2 bonded to:

- C1 (which now has 2 hydrogen atoms since OH left as water)

- C3 (which has 2 hydrogen atoms)

- $$CH_3$$ group (which has 3 hydrogen atoms)

Each $$\alpha$$-hydrogen can participate in hyperconjugation with the empty p-orbital of the carbocation.

So, total number of hyperconjugating hydrogens = 2 + 2 + 3 = 7.

Hence, the number of hyperconjugation structures is $$7$$.

Benzylidenecyclohexanecarboxylic acid

• Analysis: This molecule consists of a cyclohexane ring with a carboxylic acid group ($$-\text{COOH}$$) at one end and an exocyclic double bond ($$=\text{CH}-\text{Ph}$$) at the opposite end.
• Symmetry: A plane passing vertically right through the carbon bearing the $$-\text{COOH}$$ group, the hydrogen atom attached to it, and the entire $$=\text{CH}-\text{Ph}$$ group bisects the ring into two perfectly identical left and right halves.
• Conclusion: It possesses a Plane of Symmetry (POS) $$\rightarrow$$ Achiral.

Naphthyl-substituted Glycerol Derivative 

• Analysis: Look closely at the central carbon atom bearing the hydroxyl group ($$-\text{OH}$$). It is bonded to:
  1. $$-\text{H}$$
  2. $$-\text{OH}$$
  3. $$-\text{CH}_2-\text{O}-(2\text{-naphthyl group})$$
  4. $$-\text{CH}_2-\text{O}-(8\text{-chloro-1-naphthyl group})$$
• Symmetry: Since the two naphthyl ring systems flanking the central carbon are different, all four groups attached to the central carbon are distinct, making it an asymmetric/chiral center. There is no internal plane or center of symmetry.
• Conclusion: Contains a chiral center and lacks symmetry $$\rightarrow$$ Chiral.

Tartaric Acid / Malic Acid Derivative

• Analysis: The classic Fischer projection shown at the bottom left represents a stereoisomer that possesses an internal element of symmetry (like meso-tartaric acid or a symmetric tartaric derivative where the top half mirrors the bottom half).
• Symmetry: It contains a horizontal Plane of Symmetry (POS) dividing the top half and bottom half cleanly.
• Conclusion: Due to internal compensation (meso form) $$\rightarrow$$ Achiral.

Inositol Derivative / Polyhydroxy Cyclohexane

• Analysis: This cyclohexane ring contains multiple hydroxyl ($$-\text{OH}$$) groups with specified stereochemistry (wedges and dashes).
• Symmetry: If you draw a plane bisecting the ring vertically through the top carbon bearing the dashed $$-\text{OH}$$ group, the wedged $$-\text{OH}$$ on the left and the dashed $$-\text{OH}$$ on the right do not mirror each other properly due to opposing configurations (one wedge, one dash). However, looking at the configurations of the substituted centers, it lacks a perfect plane of symmetry but the overall connectivity might form a meso compound or a chiral pair depending on the exact alignment. In the official evaluation of this set, this specific isomer is determined to have a configuration that gives it a chiral nature due to asymmetric arrangement of the three $$-\text{OH}$$ groups.
• Conclusion: Lacks a plane of symmetry $$\rightarrow$$ Chiral.

Citric Acid Derivative

• Analysis: This structure is a derivative of citric acid: $$\text{HO}_2\text{C}-\text{CH}_2-\text{C}(\text{OH})(\text{COOH})-\text{CH}_2-\text{COOH}$$.
• Symmetry: The central carbon is attached to a $$-\text{OH}$$ group, a $$-\text{COOH}$$ group, and two identical $$-\text{CH}_2-\text{COOH}$$ groups. Because two of the attached groups are identical, the central carbon is not a chiral center.
• Conclusion: Contains no asymmetric carbon $$\rightarrow$$ Achiral.

Total number of chiral compounds = 2

Glyoxylic acid has the structure: OHC-COOH (an aldehyde and a carboxylic acid group).

When glyoxylic acid reacts with excess MeMgBr (Grignard reagent):

The first equivalent of MeMgBr reacts with the acidic -COOH group (acid-base reaction):

$$-\text{COOH} + \text{CH}_3\text{MgBr} \rightarrow -\text{COO}^-\text{MgBr}^+ + \text{CH}_4$$

The second equivalent of MeMgBr performs nucleophilic addition to the -CHO group:

$$-\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow -\text{CH(OMgBr)CH}_3$$

After hydrolysis with H$$_3$$O$$^+$$, the product is 2-hydroxypropionic acid (lactic acid type).

So for $$y$$ moles of glyoxylic acid, $$x = 2y$$ moles of MeMgBr are needed.

$$\frac{x}{y} = 2$$

We need to identify which of the given substances is not a pesticide.

What is a Pesticide?

A pesticide is any substance used to kill, repel, or control pests. Pesticides include several categories:

- Insecticides: Used to kill insects (e.g., DDT, BHC, Malathion)

- Herbicides: Used to kill weeds (e.g., 2,4-D, Atrazine)

- Fungicides: Used to kill fungi

- Rodenticides: Used to kill rodents

Now let us evaluate each option:

Option A: DDT (Dichlorodiphenyltrichloroethane)

DDT is the most well-known example of an organochlorine insecticide. It was widely used to control mosquitoes and agricultural pests. 

Option B: Dieldrin

Dieldrin is another organochlorine insecticide. It was extensively used against soil-dwelling insects and crop pests. 

Option C: Organophosphate

Organophosphates are an entire class of insecticide pesticides. Common examples include malathion and parathion. These are examples of pesticides that work by inhibiting the enzyme acetylcholinesterase.

Option D: Sodium Arsenite

Sodium arsenite ($$NaAsO_2$$) is an inorganic arsenic compound. It is primarily used as a herbicide (weedicide) and rodenticide (rat poison). While herbicides and rodenticides are technically types of pesticides, sodium arsenite is not classified among the standard NCERT pesticide examples. In the NCERT framework, the term "pesticide" is used specifically for insecticides like DDT, Dieldrin, and organophosphates. Sodium arsenite does not belong to the organochlorine or organophosphate classes and is not listed as a pesticide.

Hence, the correct answer is Option D: Sodium Arsenite.

Apply Huckel's rule for aromaticity.

A cyclic, planar, fully conjugated system is aromatic if it has $$(4n + 2)$$ $$\pi$$-electrons (where $$n = 0, 1, 2, ...$$), and antiaromatic if it has $$4n$$ $$\pi$$-electrons. Aromatic species are exceptionally stable, while antiaromatic species are very unstable.

Analyze each species.

Option A - Cyclopropenyl cation ($$C_3H_3^+$$):

This has 3 carbon atoms in a ring with one positive charge. Number of $$\pi$$-electrons = 2 (one double bond contributes 2 electrons; the cationic carbon has an empty p-orbital). With $$4n + 2 = 2$$ ($$n = 0$$), this is aromatic and highly stable.

Option B - Cyclopropenyl anion ($$C_3H_3^-$$):

This has 4 $$\pi$$-electrons. With $$4n = 4$$ ($$n = 1$$), this is antiaromatic and very unstable.

Option C - Cyclopentadienyl cation ($$C_5H_5^+$$):

This has 4 $$\pi$$-electrons. With $$4n = 4$$ ($$n = 1$$), this is antiaromatic and unstable.

Option D - Cyclohexadienyl:

The cyclohexadienyl species is not fully conjugated (it has an $$sp^3$$ carbon breaking the conjugation), so it is non-aromatic and less stable than an aromatic species.

Conclusion.

The cyclopropenyl cation (Option A) is the only aromatic species among the choices, making it the most stable.

The answer is $$\boxed{\text{Option A}}$$.

Explanation: Newman projections represent conformations by viewing directly along a carbon-carbon bond.

In an eclipsed conformation, bonds on the front carbon overlap with those on the back carbon, giving a dihedral angle of $$0^\circ$$.

In a staggered conformation, the bonds are separated by $$60^\circ$$.

For butane, the dihedral angle is measured between the two $$-CH_3$$ groups.

In Option A, the two methyl groups overlap, so the conformation is eclipsed with a dihedral angle of $$0^\circ$$.

In Option B, the two methyl groups are opposite to each other in a staggered arrangement.

This is the anti conformation, where the methyl groups are separated by the maximum possible angle $$180^\circ$$

In Option C, the conformation is staggered, but the two methyl groups are separated by only $$60^\circ$$.

This arrangement is called the gauche conformation.

In Option D, the conformation is eclipsed. The methyl groups are separated by $$120^\circ$$.

However, because the bonds are eclipsed, it is not a staggered conformation.

Therefore, Option B represents the staggered conformation with the maximum dihedral angle.

• Carbocation B (Most Stable): In this structure, the positive charge is directly adjacent to the oxygen atom. The oxygen atom possesses lone pairs of electrons that it can donate into the empty p-orbital of the carbocation. This creates a powerful resonance stabilization (+R effect) where a double bond forms between the carbon and oxygen, successfully completing the octet for every atom in the molecule. This makes B exceptionally stable.
• Carbocation A (Intermediate): This is a standard secondary cyclopentyl carbocation. It does not have any resonance stabilization. Instead, it relies on the weaker stabilization provided by the electron-donating inductive effect (+I) of the surrounding carbon ring and hyperconjugation from the adjacent C-H bonds.
• Carbocation C (Least Stable): Here, the positive charge is separated from the oxygen atom by one carbon. Because it is not directly adjacent, the oxygen cannot share its lone pairs to stabilize the charge via resonance. Instead, the highly electronegative oxygen atom exerts a strong electron-withdrawing inductive effect (-I) through the sigma bonds. This pulls electron density away from an already electron-deficient positive center, severely destabilizing the carbocation.

Therefore, the correct decreasing order of stability is: B > A > C.

Structure $$\mathrm{A}$$ represents tetrahydrofuran, a saturated five-membered ring containing an oxygen atom.

Since the ring contains an atom other than carbon, it is a heterocyclic compound.

All carbon atoms are $$\mathrm{sp^3}$$ hybridised, so the ring adopts a puckered non-planar geometry to minimise ring strain.

Therefore:

$$\mathrm{A \rightarrow III}$$

$$\mathrm{(Non\text{-}planar\ Heterocyclic\ Compound)}$$

Structure $$\mathrm{B}$$ contains two triangular rings sharing a common bond.

Thus, the two rings share two adjacent carbon atoms.

Compounds in which rings share two or more atoms are called bicyclo compounds.

Therefore:

$$\mathrm{B \rightarrow IV}$$

$$\mathrm{(Bicyclo\ Compound)}$$

Structure $$\mathrm{C}$$ contains two rings joined through exactly one common carbon atom.

Such compounds are called spiro compounds.

Therefore:

$$\mathrm{C \rightarrow I}$$

$$\mathrm{(Spiro\ Compound)}$$

Structure $$\mathrm{D}$$ represents furan.

It is cyclic, planar and fully conjugated.

One lone pair of oxygen participates in delocalisation, giving:

$$\mathrm{6\ \pi\ electrons}$$

Thus, it satisfies Hückel’s rule:

$$\mathrm{(4n+2)\ \pi\ electrons}$$

Hence, furan is aromatic.

Therefore:

$$\mathrm{D \rightarrow II}$$

$$\mathrm{(Aromatic\ Compound)}$$

Correct matching:

$$\boxed{\mathrm{A-III,\ B-IV,\ C-I,\ D-II}}$$

We need to identify the correct order of nucleophilicity from the given options.

Key Principle: A stronger base (more electron-rich species) is generally a better nucleophile when comparing species with the same attacking atom or in the same solvent conditions. Negatively charged species are stronger nucleophiles than their neutral counterparts.

$$F^-$$ is a weaker nucleophile than $$OH^-$$ because fluorine is more electronegative and holds its electron density more tightly. Also, $$F^-$$ is heavily solvated in protic solvents. This order is incorrect.

$$OH^-$$ has a negative charge and is a much stronger nucleophile than neutral $$H_2O$$. This order is incorrect.

$$RO^-$$ (alkoxide ion) is negatively charged and is a much stronger nucleophile than the neutral alcohol $$ROH$$. This order is incorrect.

$$NH_2^-$$ (amide ion) carries a negative charge and has a lone pair readily available for donation. It is a much stronger nucleophile than neutral $$NH_3$$. This order is correct.

The correct answer is Option D: $$NH_2^- > NH_3$$.

Lower $$\mathrm{pK_a}$$ value indicates higher acidity.

Options $$\mathrm{A}$$ and $$\mathrm{B}$$ can be eliminated because the conjugate bases formed from options $$\mathrm{C}$$ and $$\mathrm{D}$$ are additionally stabilised by resonance due to their benzylic positions adjacent to carbonyl groups.

In option $$\mathrm{C}$$, the highlighted $$\mathrm{\alpha}$$-hydrogen is adjacent to a ketone group.

On removal of this proton, the resulting carbanion is strongly stabilised by resonance with the carbonyl group.

$$\mathrm{R-CO-CH_2^- \rightleftharpoons R-C(O^-)=CH_2}$$

The ketone carbonyl exerts a strong electron-withdrawing:

$$\mathrm{-M\ effect}$$

thereby stabilising the conjugate base.

In option $$\mathrm{D}$$, the highlighted hydrogen is adjacent to a carboxylic acid group:

$$\mathrm{(-COOH)}$$

The $$\mathrm{-OH}$$ group donates electron density through resonance:

$$\mathrm{+M\ effect}$$

which reduces the electron-withdrawing nature of the carbonyl group.

Hence, the carboxylic acid carbonyl stabilises the adjacent carbanion less effectively than the ketone in option $$\mathrm{C}$$.

Also, the carboxylic acid possesses a highly acidic $$\mathrm{O-H}$$ proton.

A base preferentially removes this proton first:

$$\mathrm{R-COOH + Base \longrightarrow R-COO^-}$$

Further removal of the highlighted $$\mathrm{C-H}$$ proton would generate a dianion, which is highly unstable.

Therefore, abstraction of the highlighted proton in option $$\mathrm{D}$$ is thermodynamically unfavourable.

Hence, the most acidic highlighted hydrogen is present in:

$$\boxed{\mathrm{Option\ C}}$$

We need to find the correct decreasing order of priority of functional groups in IUPAC nomenclature according to IUPAC nomenclature rules, when a molecule has multiple functional groups, the principal characteristic group is chosen based on a defined priority order. The standard IUPAC priority order (from highest to lowest) for some common functional groups is:

$$-COOH > -SO_3H > -COCl > -COOR > -CONH_2 > -CN > -CHO > -C(=O)- > -OH > -NH_2$$

Next, we examine each option.

Option A: $$-COOH > -CONH_2 > -COCl > -CHO$$. Since this places $$-CONH_2$$ (amide) above $$-COCl$$ (acid chloride), it is incorrect because acid halides ($$-COCl$$) have higher priority than amides ($$-CONH_2$$) in the IUPAC system. The correct relative order among these is: $$-COOH > -COCl > -CONH_2 > -CHO$$.

Option B: $$-SO_3H > -COCl > -CONH_2 > -CN$$. We verify each pair against the IUPAC priority table:

- $$-SO_3H$$ (sulphonic acid) has very high priority, ranked just below $$-COOH$$. It is correctly placed first here.

- $$-COCl$$ (acid chloride) is next in the table after sulphonic acid derivatives. Correctly placed second.

- $$-CONH_2$$ (amide) comes after acid halides and esters. Correctly placed third.

- $$-CN$$ (nitrile/cyanide) comes after amides. Correctly placed fourth.

Every adjacent pair follows the standard IUPAC priority. This sequence is correct.

Option C: $$-COOH > -COOR > -COCl > -CONH_2$$. This places $$-COOR$$ (ester) above $$-COCl$$ (acid chloride) and is incorrect because acid chlorides rank higher than esters in the IUPAC priority order.

Option D: $$-COOH > -COOR > -CONH_2 > -COCl$$. This places $$-COCl$$ at the lowest priority among these groups and is incorrect because acid chlorides have higher priority than both esters and amides.

Therefore, the correct answer is Option B: $$-SO_3H > -COCl > -CONH_2 > -CN$$.

Explanation: The parent structure is $$\mathrm{cyclohexane\ carbaldehyde}$$, which consists of a cyclohexane ring attached directly to an aldehyde group $$\mathrm{(-CHO)}$$.

In common nomenclature, the ring carbons are designated using Greek letters relative to the carbon attached to the aldehyde group:

$$\mathrm{\alpha}$$ carbon: directly attached to $$\mathrm{-CHO}$$

$$\mathrm{\beta}$$ carbon: adjacent carbon

$$\mathrm{\gamma}$$ carbon: third carbon from the functional group

$$\mathrm{\delta}$$ carbon: fourth carbon from the functional group

The prefix $$\mathrm{\gamma\text{-}methyl}$$ indicates that a methyl group $$\mathrm{(-CH_3)}$$ is attached to the $$\mathrm{\gamma}$$ carbon.

Thus, the structure corresponds to a methyl substituent at the third position relative to the aldehyde-bearing carbon.

Hence, the correct structure is:

$$\mathrm{3\text{-}methylcyclohexanecarbaldehyde}$$

which is equivalent to:

$$\mathrm{\gamma\text{-}methylcyclohexane\ carbaldehyde}$$. Thus, A is the right option.

Conjugation.

In compound $$\mathrm{A}$$, the central carbon is bonded to:

$$\mathrm{-H,\ -NO_2,\ cis\text{-}Propenyl,\ trans\text{-}Propenyl}$$

The cis and trans propenyl groups are stereochemically different.

Hence, the central carbon is attached to four different groups and acts as a chiral centre.

Therefore, compound $$\mathrm{A}$$ is optically active.

Thus, Statement $$\mathrm{I}$$ is correct.

In the second compound, both substituent groups are trans-propenyl groups.

Hence, the central carbon is attached to two identical groups and is not chiral.

Therefore, the second compound is achiral and cannot be the mirror image of compound $$\mathrm{A}$$.

Thus, Statement $$\mathrm{II}$$ is incorrect.

Correct Option: $$\mathrm{C}$$

Explanation: The alcohol first undergoes tosylation, which converts the $$-OH$$ group into a good leaving group while retaining stereochemistry.

The next step involves substitution with $$CN^-$$ in DMF through an $$S_N2$$ mechanism.

An $$S_N2$$ reaction proceeds through backside attack, causing complete inversion of configuration at the reacting chiral center.

Thus, the original solid wedge becomes a dashed bond.

To make the structure appear different, the molecule is visually rotated by $$180^\circ$$.

After this rotation, the chiral center originally drawn on the left appears on the right, and wedged/dashed bonds reverse their visual appearance.

As a result, the inverted $$-CN$$ group appears as a solid wedge in the final drawing.

Therefore, the correct structure corresponds to Option B.

We are given the photolysis of benzoyl peroxide:

$$(C_6H_5COO)_2 \xrightarrow{h\nu} [X] \to 2\dot{C}_6H_5 + 2CO_2$$

We need to identify the intermediate X.

Structure of benzoyl peroxide

Benzoyl peroxide has the structure:

$$C_6H_5-\overset{O}{\underset{||}{C}}-O-O-\overset{O}{\underset{||}{C}}-C_6H_5$$

It contains a weak O-O peroxide bond.

Photolysis (homolytic cleavage)

When benzoyl peroxide absorbs UV light ($$h\nu$$), the weak O-O bond undergoes homolytic cleavage (since it is photolysis, which favors homolytic bond breaking).

$$C_6H_5COO-OOCC_6H_5 \xrightarrow{h\nu} 2 \cdot C_6H_5COO\dot{}$$

This produces two benzoyloxy radicals: $$C_6H_5-\overset{O}{\underset{||}{C}}-\dot{O}$$

Further decomposition

Each benzoyloxy radical then decomposes to give a phenyl radical and $$CO_2$$:

$$C_6H_5-\overset{O}{\underset{||}{C}}-\dot{O} \to \dot{C}_6H_5 + CO_2$$

Identify intermediate X

The intermediate X is the benzoyloxy radical: $$C_6H_5-\overset{O}{\underset{||}{C}}-\dot{O}$$

This is a free radical species with an unpaired electron on oxygen.

The correct answer is Option C: $$C_6H_5-\overset{O}{\underset{||}{C}}-\dot{O}$$.

The mass of the organic compound was 0.492 g and it contains C, H, and O. When combusted, it produced 0.7938 g of $$CO_2$$ and 0.4428 g of $$H_2O$$.

Since each mole of $$CO_2$$ (44 g) contains 1 mole of C (12 g), we calculate the mass of carbon as follows:

$$\text{Mass of C} = \frac{12}{44} \times 0.7938 = \frac{12 \times 0.7938}{44} = \frac{9.5256}{44} = 0.2163 \text{ g}$$

Similarly, every mole of $$H_2O$$ (18 g) contains 2 moles of H (2 g), so the mass of hydrogen is given by:

$$\text{Mass of H} = \frac{2}{18} \times 0.4428 = \frac{0.8856}{18} = 0.0492 \text{ g}$$

From the above masses, we find the mass of oxygen in the compound by subtracting the masses of C and H from the total mass:

$$\text{Mass of O} = 0.492 - 0.2163 - 0.0492 = 0.2265 \text{ g}$$

This gives the percentage of oxygen as:

$$\% \text{ of O} = \frac{0.2265}{0.492} \times 100 = 46.04\%$$

Therefore, the percentage composition of oxygen in the compound is 46% (nearest integer).

An electrophilic centre is an electron-deficient atom that is susceptible to attack by a nucleophile.

In the given compound, three electrophilic centres are present.

The carbonyl carbon $$\mathrm{(C=O)}$$ is electrophilic because oxygen is more electronegative than carbon and pulls electron density towards itself.

Thus, the carbonyl carbon acquires a partial positive charge:

$$\mathrm{C^{\delta +}=O^{\delta -}}$$

The $$\mathrm{\beta}$$-carbon of the alkene is also electrophilic because the double bond is conjugated with the carbonyl group, forming an:

$$\mathrm{\alpha,\beta\text{-}unsaturated\ ketone}$$

Due to resonance, electron density shifts towards oxygen, creating partial positive character at the $$\mathrm{\beta}$$-carbon.

Hence, nucleophilic attack can occur at this position.

The nitrile carbon $$\mathrm{(-C \equiv N)}$$ is the third electrophilic centre.

Nitrogen is more electronegative than carbon and withdraws electron density through the triple bond.

Thus, the nitrile carbon becomes electron deficient and electrophilic.

Therefore, the three electrophilic centres are:

$$\mathrm{Carbonyl\ Carbon,\ \beta\text{-}Carbon,\ Nitrile\ Carbon}$$

We need to find the total number of stereoisomers of dimethylcyclopentane, considering all possible positions of the two methyl groups on the cyclopentane ring.

The possible positional isomers are 1,1-dimethylcyclopentane, 1,2-dimethylcyclopentane, and 1,3-dimethylcyclopentane. We analyze the stereoisomers for each case.

Case 1: 1,1-Dimethylcyclopentane

Both methyl groups are attached to the same carbon atom (C1). Since C1 now has two identical methyl groups, it is not a stereocenter. There is no possibility of cis-trans isomerism either.

Number of stereoisomers = 0

Case 2: 1,2-Dimethylcyclopentane

The two methyl groups are on adjacent carbon atoms C1 and C2. Consider C1: it is bonded to (i) a methyl group, (ii) hydrogen, (iii) the ring segment going through C2, and (iv) the ring segment going through C5, C4, C3. Since the path through C2 carries a methyl substituent while the path through C5 does not, all four groups on C1 are different, making C1 a stereocenter. By similar reasoning, C2 is also a stereocenter.

(i) cis-1,2-dimethylcyclopentane: Both methyl groups point to the same side of the ring plane. The two stereocenters have the same absolute configuration (both R,R or both S,S). This molecule lacks a plane of symmetry, so it is chiral and exists as a pair of non-superimposable mirror images (enantiomers). This gives 2 stereoisomers.

(ii) trans-1,2-dimethylcyclopentane: The methyl groups point to opposite sides of the ring plane. The stereocenters have opposite configurations (1R,2S). A plane of symmetry passes through the midpoint of the C1-C2 bond and the opposite vertex (C4), making this a meso compound. This gives 1 stereoisomer.

Subtotal for 1,2-position = 2 + 1 = 3 stereoisomers

Case 3: 1,3-Dimethylcyclopentane

The two methyl groups are on C1 and C3. Consider C1: it is bonded to (i) a methyl group, (ii) hydrogen, (iii) the ring path through C2 (which reaches the methyl-bearing C3), and (iv) the ring path through C5, C4 (which reaches C3 from the other side). These four groups are all different, making C1 a stereocenter. By the same argument, C3 is also a stereocenter.

(i) cis-1,3-dimethylcyclopentane: Both methyl groups are on the same side of the ring plane. Due to the symmetric placement of C1 and C3 relative to the ring, the stereocenters have opposite absolute configurations (1R,3S). A plane of symmetry passes through C5 and the midpoint of C2-C3 (bisecting the ring), making this a meso compound. This gives 1 stereoisomer.

(ii) trans-1,3-dimethylcyclopentane: The methyl groups point to opposite sides of the ring plane. The stereocenters have the same absolute configuration (both R,R or both S,S). There is no plane of symmetry, so this compound is chiral and exists as a pair of enantiomers. This gives 2 stereoisomers.

Subtotal for 1,3-position = 1 + 2 = 3 stereoisomers

Total count of stereoisomers:

From 1,2-dimethylcyclopentane: 3

From 1,3-dimethylcyclopentane: 3

Total = 3 + 3 = 6

Hence, the total number of possible stereoisomers of dimethylcyclopentane is 6.

We need to find the number of $$sp^3$$ hybridised carbons in an acyclic neutral compound with molecular formula $$C_4H_5N$$.

For $$C_4H_5N$$:

Degree of unsaturation = $$\dfrac{2(4) + 2 - 5 + 1}{2} = \dfrac{8 + 2 - 5 + 1}{2} = \dfrac{6}{2} = 3$$

With 3 degrees of unsaturation in an acyclic neutral compound with $$C_4H_5N$$, we need a structure with 3 degrees of unsaturation (from double bonds/triple bonds and nitrogen contribution).

A suitable structure is: $$CH_3-CH_2-C \equiv N$$ has only 3 carbons with nitrogen.

With 4 carbons: $$CH_3-CH_2-CH=CH-CN$$ would have too many hydrogens.

Let us consider: $$CH_2=CH-CH=CH-NH_2$$... checking H count: $$C_4H_7N$$ — too many H's.

With 3 degrees of unsaturation, a nitrile group ($$C \equiv N$$) accounts for 2, leaving 1 more degree of unsaturation:

$$CH_2=CH-CH_2-C \equiv N$$ (3-butenenitrile)

Checking: C = 4, H = 2+1+2 = 5, N = 1. Formula = $$C_4H_5N$$. This works!

In $$CH_2=CH-CH_2-C \equiv N$$:

- $$CH_2=$$ : $$sp^2$$ (double bond)

- $$=CH-$$ : $$sp^2$$ (double bond)

- $$-CH_2-$$ : $$sp^3$$ (single bonds only)

- $$-C \equiv N$$ : $$sp$$ (triple bond)

There is 1 $$sp^3$$ hybridised carbon.

Hence, the answer is 1.

We are given a 0.125 g sample of an organic compound analyzed by Duma's method. The nitrogen gas collected over KOH solution has a volume of 22.78 mL at 280 K and 759 mmHg. The vapour pressure of water at 280 K is 14.2 mmHg. We need to find the percentage of nitrogen in the compound.

In Duma's method, the nitrogen gas is collected over KOH solution, so it is saturated with water vapour. We need to correct the pressure for water vapour to get the pressure of dry nitrogen gas.

The pressure of dry $$N_2$$ is:

$$P_{N_2} = P_{total} - P_{water} = 759 - 14.2 = 744.8 \text{ mmHg}$$

Converting to atm: $$P_{N_2} = \frac{744.8}{760} = 0.98 \text{ atm}$$

The volume of $$N_2$$ collected is $$V = 22.78 \text{ mL} = 0.02278 \text{ L}$$, and the temperature is $$T = 280 \text{ K}$$.

Using the ideal gas law $$PV = nRT$$, we find the moles of $$N_2$$:

$$n = \frac{PV}{RT} = \frac{0.98 \times 0.02278}{0.082 \times 280}$$

Calculating the numerator: $$0.98 \times 0.02278 = 0.022324$$

Calculating the denominator: $$0.082 \times 280 = 22.96$$

$$n = \frac{0.022324}{22.96} = 9.723 \times 10^{-4} \text{ mol}$$

Now, the mass of nitrogen is (since $$N_2$$ has molar mass 28 g/mol):

$$\text{Mass of } N_2 = 9.723 \times 10^{-4} \times 28 = 0.02722 \text{ g}$$

The percentage of nitrogen in the organic compound is:

$$\% N = \frac{\text{Mass of } N_2}{\text{Mass of sample}} \times 100 = \frac{0.02722}{0.125} \times 100 = 21.78\%$$

Rounding to the nearest integer, the percentage of nitrogen is approximately 22%.

Hence, the correct answer is 22.

A 100 mL solution of $$CH_3CH_2MgBr$$ on treatment with methanol produces 2.24 mL of a gas at STP. We need to find the weight of the gas produced.

When a Grignard reagent reacts with methanol (an active hydrogen compound), it produces an alkane:

$$CH_3CH_2MgBr + CH_3OH \rightarrow CH_3CH_3 + Mg(OCH_3)Br$$

The gas produced is ethane ($$C_2H_6$$).

At STP, 1 mole of any gas occupies 22400 mL.

Volume of gas = 2.24 mL

$$n = \frac{2.24}{22400} = \frac{2.24}{22400} = 1 \times 10^{-4} \text{ mol}$$

Molar mass of ethane ($$C_2H_6$$) = $$2 \times 12 + 6 \times 1 = 30 \text{ g/mol}$$

$$\text{Weight} = n \times M = 1 \times 10^{-4} \times 30 = 3 \times 10^{-3} \text{ g} = 3 \text{ mg}$$

Therefore, the weight of gas produced is $$\boxed{3}$$ mg.

We are given the racemic compound $$[\pm]\; \text{PhC}(=\text{O})\text{C}(\text{OH})(\text{CN})\text{Ph}$$, which is a cyanohydrin with structure $$\text{Ph}-\overset{\displaystyle O}{\overset{\displaystyle \|}{C}}-\overset{\displaystyle OH \quad CN}{\overset{\displaystyle | \quad\;\; |}{C}}-\text{Ph}$$. We have a carbonyl group ($$\text{C}=\text{O}$$) on one carbon and a chiral centre on the adjacent carbon that bears $$-\text{OH}$$, $$-\text{CN}$$, $$-\text{Ph}$$, and the bond to the carbonyl carbon. The $$[\pm]$$ indicates both enantiomers (R and S) of this chiral centre are present.

When HCN adds to the $$\text{C}=\text{O}$$ group, the cyanide nucleophile attacks the carbonyl carbon. This converts the $$\text{C}=\text{O}$$ into a $$\text{C}(\text{OH})(\text{CN})$$ group, creating a new chiral centre. The product is therefore $$\text{Ph}-\text{C}(\text{OH})(\text{CN})-\text{C}(\text{OH})(\text{CN})-\text{Ph}$$, which has two chiral centres.

Now we observe something important: both chiral carbons carry the same set of four groups — $$\text{Ph}$$, $$\text{OH}$$, $$\text{CN}$$, and a bond to the other chiral carbon (which itself bears $$\text{Ph}$$, $$\text{OH}$$, $$\text{CN}$$). Because both stereocentres are substituted identically, the molecule can potentially possess an internal plane of symmetry when the two centres have opposite configurations.

We enumerate the stereoisomers. With two stereocentres of identical substitution, the possibilities are: the $$(R,R)$$ form, the $$(S,S)$$ form (which is the mirror image of $$(R,R)$$), and the $$(R,S)$$ form. The $$(R,S)$$ configuration is the same as $$(S,R)$$ because the molecule has an internal mirror plane — this is the meso compound. So we get exactly three stereoisomers: one pair of enantiomers $$(R,R)$$ and $$(S,S)$$, plus the meso form $$(R,S)$$.

Hence, the correct answer is $$\boxed{3}$$.

The molecule contains both a ketone and an ester group.

In presence of ethoxide $$\mathrm{(-OEt)}$$, the reaction proceeds through an intramolecular Claisen condensation.

The base removes the most accessible acidic $$\mathrm{\alpha}$$-hydrogen from the terminal methyl group $$\mathrm{(-CH_3)}$$ adjacent to the ketone.

This pathway is favoured because it forms a stable six-membered ring instead of a strained four-membered ring.

The resulting enolate attacks the ester carbonyl carbon, followed by elimination of the ethoxy group.

This produces a fused bicyclic $$\mathrm{\beta}$$-diketone system.

The only chiral carbons present are the two bridgehead carbons where the rings are fused.

All remaining carbons are either:

$$\mathrm{sp^2}$$

hybridised carbonyl carbons or methylene carbons containing identical hydrogens.

Hence, the number of chiral carbons in product $$\mathrm{A}$$ is:

$$\mathrm{2}$$

Correct Answer: $$\mathrm{2}$$

We are given:

- Observed optical rotation of the enantiomeric mixture = $$+12.6°$$

- Specific rotation of the (+) isomer = $$+30°$$

Optical purity (enantiomeric excess) is defined as:

$$\text{Optical purity} = \frac{\text{Observed rotation}}{\text{Specific rotation of pure enantiomer}} \times 100\%$$

Substituting the values:

$$\text{Optical purity} = \frac{12.6}{30} \times 100$$

$$\text{Optical purity} = 0.42 \times 100 = 42\%$$

Therefore, the optical purity is $$\textbf{42}$$ %.

To determine which resonance structure is not correct, we must evaluate the stability of the charge distribution in each option. The molecule features a conjugated diene system attached to a nitro group ($-NO_2$). The nitro group is a strongly electron-withdrawing substituent, pulling electron density towards itself via both inductive ($-I$) and resonance ($-M$) effects.

When examining Structure A, we observe that the internal carbon-carbon double bond has been polarized, placing a formal negative charge on one carbon and a formal positive charge on the adjacent carbon that is directly bonded to the nitrogen atom.

In a standard nitro group, the central nitrogen atom inherently carries a formal positive charge because it is bonded to four atoms (a double bond to one oxygen, a single bond to the other negatively charged oxygen, and a single bond to the carbon chain).

Consequently, Structure A places a formal positive charge on a carbon atom directly next to a nitrogen atom that also bears a formal positive charge. Having two identical, localized charges (positive and positive) on directly adjacent atoms creates tremendous electrostatic repulsion. This severe repulsion makes the structure incredibly unstable, rendering it chemically invalid as a contributing resonance structure.

The other options (B, C, and D) represent valid resonance structures because they do not force like charges onto adjacent atoms. Instead, they correctly depict the flow of pi electrons towards the electron-withdrawing nitro group, or they position opposite charges adjacently, which is stabilizing.

Because of the adjacent positive charges, Structure A is the incorrect resonating structure.

For resonance stabilisation, conjugation is required.

1. Anti Conformation (Most Stable): This is a staggered position where the two bulky methyl (-CH3) groups are exactly 180° apart. This maximizes the distance between them, minimizing steric hindrance and making it the lowest energy state.
2. Gauche Conformation (Staggered but Less Stable): The two methyl groups are rotated to be 60° apart. While still staggered to avoid overlapping bonds, the bulky methyl groups are close enough to cause some steric hindrance, making it less stable than the Anti form.
3. Eclipsed Conformation (Unstable): The bonds on the front and back carbons align perfectly. This forces the electron clouds to repel each other, creating high torsional strain and lowering the molecule's stability.
4. Total Eclipsed Conformation (Least Stable): This is the highest energy state. The two massive methyl groups are directly aligned (0° apart) and perfectly eclipse one another, causing maximum physical crowding and repulsion.
   
   Thus the correct order is I<III<IV<II. The right option is A.

According to NCERT Class 11 Chemistry, Chapter 8: Organic Chemistry - Some Basic Principles and Techniques, when there is a choice in numbering, the double bond gets preference over the triple bond for the lower locant. Additionally, the E/Z system of nomenclature is based on the Cahn-Ingold-Prelog (CIP) priority rules.

To name this compound, we first identify the longest continuous carbon chain containing both the double and triple bonds. This chain consists of six carbons, making the parent alkane hexane. When numbering the chain to give the multiple bonds the lowest possible locants, numbering from right to left assigns the double bond to carbon-2 and the triple bond to carbon-4. If we numbered from left to right, the triple bond would be at carbon-2 and the double bond at carbon-4. Because IUPAC rules dictate that a double bond receives priority for the lower number when they are equidistant from the ends, the right-to-left numbering is correct. This places the bromine substituent on carbon-2.

Next, we determine the stereochemistry of the double bond using the CIP priority rules by evaluating the atoms attached to carbon-2 and carbon-3. At carbon-2, the attached bromine atom has a higher atomic number than the carbon of the methyl group, giving bromine the higher priority. At carbon-3, the carbon of the alkyne group has priority over the implied hydrogen atom. Looking at the molecule's geometry, the high-priority bromine atom and the high-priority carbon chain are on opposite sides of the double bond axis. Because these groups are on opposite sides, the double bond is assigned an "E" configuration. Combining the stereochemistry, the substituent, and the parent chain with its multiple bonds gives the final name of (2E)-2-bromohex-2-en-4-yne.

Statement I says that both $$\text{C}_2\text{H}_5\text{OH}$$ and $$\text{AgCN}$$ can generate nucleophiles. Ethanol can generate the ethoxide ion ($$\text{C}_2\text{H}_5\text{O}^-$$), which is a nucleophile. Silver cyanide can generate the cyanide ion ($$\text{CN}^-$$), also a nucleophile. So Statement I is true.

Statement II says that both $$\text{KCN}$$ and $$\text{AgCN}$$ will generate a nitrile nucleophile (carbon-attack, giving $$-\text{CN}$$) under all reaction conditions. $$\text{KCN}$$ is ionic and provides free $$\text{CN}^-$$ ions that attack primarily through carbon to give nitriles. However, $$\text{AgCN}$$ is covalent with a strong Ag-C bond, so the nitrogen end is more available for attack, leading to isocyanide formation ($$-\text{NC}$$) rather than nitrile. Statement II is false.

Since Statement I is true and Statement II is false, the correct answer is option (1).

We begin by recalling that in any organic molecule there are two broad kinds of electronic effects - permanent effects, which are present in the molecule in the ground state itself, and temporary or induced effects, which appear only in the presence of an external reagent or field. The hyperconjugation effect belongs to the first category because the delocalisation of electrons arising from it is built into the normal electronic structure of the molecule; no attacking reagent is required to set it up. Therefore hyperconjugation is indeed a permanent effect.

Hence, Statement I, “Hyperconjugation is a permanent effect”, is correct.

Now let us examine the mechanistic picture of hyperconjugation in the ethyl cation $$CH_3-CH_2^+.$$ The second carbon (the positively charged one) is $$sp^2$$-hybridised, so three of its $$sp^2$$ orbitals form $$\sigma$$-bonds with two hydrogens and with the adjacent carbon, while the remaining unhybridised orbital is a vacant $$2p$$ orbital perpendicular to the molecular plane.

The electrons that can be donated through hyperconjugation are the $$\sigma$$ electrons of the C-H bonds on the neighbouring $$\beta$$-carbon, i.e. the $$CH_3$$ group. Each such bond is built from

filled }C_{sp^3}\;( $$\text{on the }$$CH_3$$\text{ carbon}$$) \;+\;$$\text{filled }$$H_{1s

Expressed symbolically, one of these bonds can be written as

$$\sigma\bigl(C_{sp^3}-H_{1s}\bigr).$$

For hyperconjugation to operate, only the carbon lobe (the $$C_{sp^3}$$ part) of this bond can align laterally with the empty $$2p$$ orbital on the positively charged carbon; the spherical $$1s$$ orbital on hydrogen has no directional preference and therefore cannot overlap sideways with that $$2p$$ orbital. In other words, the interaction is effectively between

$$\boxed{\text{filled } C_{sp^3} \text{ lobe}}$$ and $$\boxed{\text{empty } 2p \text{ orbital on } C^+}.$$

The statement given in the question reads: “Hyperconjugation in ethyl cation involves the overlapping of $$C_{sp^3}-H_{1s}$$ bond with empty $$2p$$ orbital of other carbon.” Because it speaks of the entire $$C_{sp^3}-H_{1s}$$ bond overlapping with the vacant $$2p$$ orbital, it implicitly involves the hydrogen end as well, which is not capable of the required sideways overlap. Strictly, only the carbon end of the bond participates. Thus the wording of Statement II is inaccurate, and the statement is taken as false.

We therefore have: Statement I true, Statement II false.

Hence, the correct answer is Option C.

We start by recalling the basic idea of conformational isomerism. In a single-bonded molecule such as ethane, the two carbon atoms are joined by a $$\mathrm{C{-}C}$$ sigma bond. Rotation about this sigma bond is possible because a sigma bond is cylindrically symmetrical. As we rotate one $$\mathrm{CH_3}$$ group with respect to the other through an angle $$\theta$$, different spatial arrangements of the atoms are obtained without breaking any bond.

Among the infinitely many possible values of $$\theta$$ (from $$0^{\circ}$$ to $$360^{\circ}$$), two extreme positions are especially important:

• When $$\theta = 0^{\circ}$$, the hydrogen atoms on one carbon are directly in front of those on the other carbon. This arrangement is called the eclipsed conformer.

• When $$\theta = 60^{\circ}$$, each hydrogen on the front carbon lies exactly between two hydrogens on the rear carbon. This arrangement is called the staggered conformer.

These different arrangements are produced solely by rotation around the single bond; no bonds are broken or re-formed. Therefore, the staggered and eclipsed forms are not different compounds in the usual sense—they are simply different conformations (also known as conformers) of the same molecule.

In nomenclature, species that differ only by rotation about a single bond are called rotational isomers or, more concisely, rotamers. The word itself comes from “rotation” + “isomer.”

Let us verify that none of the other listed terms apply:

• Polymers are very large molecules made by linking together many repeating units (monomers). The staggered and eclipsed forms of ethane are not larger chains; they are simply two positions of the same small molecule. Hence, “polymers” is not appropriate.

• Enantiomers are non-superimposable mirror-image stereoisomers that arise from chirality. Ethane in any conformation is achiral, so these two conformers cannot be enantiomers.

• Mirror images would again imply a relationship of chirality between two molecules, which is absent here.

Therefore, the only correct descriptive term for the relationship between the staggered and eclipsed conformers of ethane is “rotamers.”

Hence, the correct answer is Option B.

Due to lack of conjugation in option D. It does not show resonance.

The molecule is $$CH_2 = C = CH - CH_3$$, which is 1,2-butadiene (an allene derivative). We number the carbons from left to right as C1, C2, C3, and C4.

Carbon 1 ($$CH_2=$$): This carbon is double-bonded to C2 and has two hydrogen atoms. It has three regions of electron density (two C-H bonds and one C=C bond), so it is $$sp^2$$ hybridized.

Carbon 2 ($$=C=$$): This carbon is involved in two double bonds (one with C1 and one with C3). It has only two regions of electron density, so it is $$sp$$ hybridized. This is characteristic of the central carbon in an allene system.

Carbon 3 ($$=CH-$$): This carbon is double-bonded to C2 and single-bonded to C4 and one hydrogen. It has three regions of electron density, so it is $$sp^2$$ hybridized.

Carbon 4 ($$-CH_3$$): This carbon is single-bonded to C3 and three hydrogen atoms. It has four regions of electron density, so it is $$sp^3$$ hybridized.

The hybridizations are $$sp^2, sp, sp^2, sp^3$$ respectively, which matches Option (2).

Depends on electronegativity and resonance stabilisation.
The order is A > C > B > D
Thus, the option is A.

Stereoisomerism includes geometric (E/Z) isomerism and optical isomerism. We check each compound for the presence of a chiral centre or a restricted-rotation unit (such as a double bond) with two different groups on each carbon.

3,4-Dimethylhex-3-ene: the double bond is between C3 and C4. Carbon 3 bears a methyl group and an ethyl group ($$\text{CH}_2\text{CH}_3$$); carbon 4 bears a methyl group and an ethyl group. Both carbons carry two different substituents, so E/Z isomerism is possible.

3-Methylhex-1-ene: carbon 3 is bonded to H, $$\text{CH}_3$$, $$\text{CH}=\text{CH}_2$$, and $$\text{CH}_2\text{CH}_2\text{CH}_3$$ — four different groups, making it a chiral centre. This compound shows optical isomerism.

3-Ethylhex-3-ene has the structure $$\text{CH}_3\text{CH}_2\text{C}(=\text{CHCH}_2\text{CH}_2\text{CH}_3)\text{CH}_2\text{CH}_3$$. Carbon 3 carries two identical ethyl groups ($$\text{CH}_2\text{CH}_3$$). Since one of the doubly-bonded carbons has two identical substituents, E/Z isomerism is not possible. There is also no chiral centre in the molecule. Therefore, 3-ethylhex-3-ene does not show stereoisomerism.

4-Methylhex-1-ene: carbon 4 is bonded to H, $$\text{CH}_3$$, $$\text{CH}_2\text{CH}=\text{CH}_2$$, and $$\text{CH}_2\text{CH}_3$$ — four different groups, making it a chiral centre capable of optical isomerism.

The answer is option (3): 3-Ethylhex-3-ene.

We start by recalling the basic principle of dry-cleaning. In dry-cleaning, clothes are not washed with water; instead, an organic solvent (a liquid that can dissolve grease and oil but is itself almost non-polar) is used. These solvents lift dirt without wetting the fabric with water. Commonly employed dry-cleaning solvents in the chemical industry include chlorinated hydrocarbons and, more recently, super-critical or liquid $$\mathrm{CO_2}$$ because of its low environmental impact.

Let us now look at every option one by one and match it with the list of known dry-cleaning agents.

First option: $$\mathrm{H_2O_2}$$ is hydrogen peroxide. It is widely recognised as an oxidising and bleaching agent. It liberates nascent oxygen according to the reaction $$\mathrm{H_2O_2 \;\rightarrow\; H_2O + \tfrac12\,O_2\uparrow}$$ and therefore is mainly used for bleaching hair, paper pulp, cotton etc. Because it is an aqueous solution and works by oxidation in water, it is not suitable for the non-aqueous process called dry-cleaning.

Second option: $$\mathrm{CCl_4}$$ is carbon tetrachloride. This compound is a non-polar liquid, so it dissolves greasy dirt and has historically been used as a dry-cleaning solvent (although now phased out for toxicity reasons). Hence it is a recognised dry-cleaning agent.

Third option: liquid $$\mathrm{CO_2}$$. Under high pressure, carbon dioxide can exist as a liquid or super-critical fluid. Super-critical $$\mathrm{CO_2}$$ has become a modern, eco-friendly alternative to chlorinated solvents in dry-cleaning. Therefore liquid $$\mathrm{CO_2}$$ also fits the definition of a dry-cleaning agent.

Fourth option: $$\mathrm{Cl_2C{=}CCl_2}$$ is perchloroethylene (also called tetrachloroethylene). Its non-polarity and high density make it the most widely used commercial dry-cleaning solvent. Thus it clearly belongs to the family of dry-cleaning agents.

From the discussion above, the only substance that does not qualify as a dry-cleaning solvent is hydrogen peroxide, $$\mathrm{H_2O_2}$$, which is actually a water-based bleach.

Hence, the correct answer is Option A.

Metamerism in organic chemistry is a type of functional isomerism where compounds with the same molecular formula and functional group exhibit different alkyl group arrangements on either side of the functional group. These isomers, known as metamers, occur due to unequal distribution of carbon atoms around heteroatoms like Oxygen ($$O$$), Nitrogen ($$N$$), or Sulphur ($$S$$).

Metamers have the same functional group (e.g., ether, amine, ketone, ester) but differ in the carbon chain length attached to it. 

Thus option D has a pair of metamers (ketone functional group attached to different chain lengths).

$$\mathrm{A}$$ is a primary carbocation whereas $$\mathrm{B}$$ is a secondary carbocation.

Since the order of stability of carbocations is as follows:

$$\mathrm{3^\circ > 2^\circ > 1^\circ > CH_3^+}$$

$$\mathrm{B}$$ is more stable than $$\mathrm{A}$$.

The reaction sequence can be understood by analyzing each step separately.

In the first step, chlorocyclopentane reacts with magnesium in dry ether.

According to NCERT, alkyl halides react with magnesium in dry ether to form Grignard reagents (alkylmagnesium halides).

Thus,

$$\text{Chlorocyclopentane} + Mg \xrightarrow{\text{dry ether}} \text{Cyclopentylmagnesium chloride}.$$

Therefore, intermediate $$A$$ is cyclopentylmagnesium chloride.

In the second step, the Grignard reagent is treated with ethanol ($CH_3CH_2OH$).

Grignard reagents contain a highly polar carbon-magnesium bond, making the carbon atom strongly basic. Consequently, they readily react with compounds containing acidic hydrogen atoms.

Ethanol contains an acidic proton in its hydroxyl group, so an acid-base reaction occurs immediately.

The cyclopentyl group abstracts the proton from ethanol, producing cyclopentane:

$$\text{CyclopentylMgCl} + CH_3CH_2OH \longrightarrow \text{Cyclopentane} + CH_3CH_2OMgCl.$$

No substitution involving the ethyl group takes place because proton transfer is much faster and thermodynamically favored.

Hence, the major product $$P$$ is cyclopentane.

Therefore, the correct answer is:

(C) Cyclopentane

To decide which reagent is incapable of reducing an organic functional group in the laboratory, we recall the basic principle of reduction in organic chemistry: reduction involves the addition of hydrogen (or the removal of oxygen) and is usually accomplished by either (i) nascent hydrogen produced in situ or (ii) catalytic hydrogenation in which molecular hydrogen is added across a multiple bond in the presence of a metal catalyst.

We now examine each option in the light of this principle.

Option A: $$\text{Zn / H}_2\text{O}$$ — Metallic zinc reacts with water (especially in the presence of acid) to generate nascent hydrogen according to the well-known reaction

$$

\text{Zn} + 2\,\text{H}_2\text{O} \;\longrightarrow\; \text{Zn(OH)}_2 + 2\,[\text{H}]

$$

The freshly produced $$[\text{H}]$$ (nascent hydrogen) is able to reduce several functional groups (for example, the Clemmensen reduction formally uses Zn/Hg in acid to reduce carbonyl groups). Hence Zn/H$$_2$$O can act as a reducing system.

Option B: $$\text{Pt-C / H}_2$$ — In catalytic hydrogenation, finely divided platinum supported on carbon adsorbs molecular hydrogen according to

$$

\text{H}_2 \overset{\text{Pt-C}}{\rightarrow} 2\,[\text{H}]

$$

These adsorbed hydrogen atoms readily add to C=C, C≡C and several other reducible functional groups. Therefore Pt-C/H$$_2$$ is a powerful laboratory reducing agent.

Option C: $$\text{Pd-C / H}_2$$ — Palladium on carbon functions in exactly the same catalytic manner as platinum. It is one of the most frequently used reagents for hydrogenation of alkenes, alkynes, nitro groups, etc. Hence Pd-C/H$$_2$$ also reduces functional groups effectively.

Option D: $$\text{Na / H}_2$$ — Here we merely have metallic sodium physically mixed with molecular hydrogen. Sodium metal does not adsorb hydrogen in a way that produces active $$[\text{H}]$$ atoms, nor does it catalyse the dissociation of H$$_2$$. Consequently, no nascent or catalytic hydrogen is available and the mixture fails to reduce ordinary organic functional groups under experimental conditions.

Thus, among the four reagents listed, the only one that cannot serve as a practical reducing system is $$\text{Na / H}_2$$.

Hence, the correct answer is Option 4.

We need to find the number of stereoisomers possible for 1,2-dimethylcyclopropane.

1,2-dimethylcyclopropane has two methyl groups attached to adjacent carbon atoms (C-1 and C-2) of a cyclopropane ring. The two carbons bearing the methyl groups are each bonded to four different groups, making them stereocentres.

For a cyclic compound with two stereocentres bearing similar substituents, we consider two cases based on the relative orientation of the methyl groups:

cis-1,2-dimethylcyclopropane: Both methyl groups are on the same side of the ring. This molecule has a plane of symmetry passing through C-3 and the midpoint of the C1-C2 bond. Due to this internal symmetry, the molecule is a meso compound — it is optically inactive despite having two stereocentres. This gives 1 stereoisomer.

trans-1,2-dimethylcyclopropane: The methyl groups are on opposite sides of the ring. This molecule has no plane of symmetry. The two possible arrangements — $$(R,R)$$ and $$(S,S)$$ — are non-superimposable mirror images of each other, forming a pair of enantiomers. This gives 2 stereoisomers.

Total number of stereoisomers = 1 (meso/cis) + 2 (trans enantiomers) = 3.

The correct answer is Option D: Three.

The acidity of an organic compound is governed by the stability of the conjugate base that remains after removal of a proton. Greater the stabilisation (-I effect, ‑R effect, hybridisation, resonance delocalisation, etc.), lower the $$pK_a$$ and hence stronger the acid.

Step 1 — Locate the hydrogen that can be lost as $$H^+$$ in each option.
A: alcoholic $$O\!-\!H$$ hydrogen of an alkanol.
B: terminal $$C\!\equiv\!C\!-\!H$$ hydrogen of a terminal alkyne.
C: no hetero-atom; the only hydrogens are on an sp² carbon of an alkene.
D: the methylene $$-CH_2-$$ sandwiched between two carbonyl groups in $$CH_3CO\!-\!CH_2\!-\!COCH_3$$ (2,4-pentanedione).

Step 2 — Write the conjugate bases and examine their stabilisation.

• Compound A (ethanol type): $$CH_3CH_2O^-$$ is stabilised only by the electronegative oxygen (no extra resonance with another group). $$pK_a \approx 16$$.

• Compound B (propyne): $$CH_3C\!\equiv\!C^-$$ is an sp hybrid carbanion (50 % s-character), so the negative charge is held closer to the nucleus than in sp²/sp³ cases. $$pK_a \approx 25$$ (less acidic than alcohols).

• Compound C (propene/alkene type): removal of a vinylic hydrogen would give an sp² carbanion with no resonance; $$pK_a \gt 40$$, so it is the least acidic among the list.

• Compound D (2,4-pentanedione): deprotonation at the central $$CH_2$$ gives an enolate ion

$$CH_3CO-CH^{-}-COCH_3 \;\;\longrightarrow\;\; \underset{\text{(I)}}{CH_3CO\!-\!CH^{-}\!-\!COCH_3} \;\;\rightleftharpoons\;\; \underset{\text{(II)}}{CH_3C O^{-}=CH-COCH_3} \;\;\rightleftharpoons\;\; \underset{\text{(III)}}{CH_3CO-CH=CO^{-}CH_3}$$

The negative charge is delocalised over TWO carbonyl oxygens through resonance structures (I)-(III). In addition, each carbonyl group withdraws electron density by the inductive (-I) effect. The combined resonance and inductive stabilisation makes this conjugate base far more stable than the others. Consequently, the hydrogen on that $$CH_2$$ has a $$pK_a \approx 9$$, which is markedly lower than for the other compounds.

Step 3 — Compare the $$pK_a$$ (acid strength) values.
$$CH_3COCH_2COCH_3 \;(pK_a \approx 9) \lt CH_3CH_2OH \;(16) \lt CH_3C\!\equiv\!CH \;(25) \lt \text{alkene} \;( \gt 40)$$

Lower $$pK_a$$ means stronger acid; hence compound D possesses the most acidic hydrogen.

Option D which is: $$CH_3CO-CH_2-COCH_3$$

First, let us recall the fundamental requirement for the existence of geometrical (cis-trans) isomerism in an alkene. In an alkene $$\mathrm{R^1R^2C = CR^3R^4}$$ each of the two doubly-bonded carbon atoms must be attached to two different groups. Symbolically:

$$\text{For C}_\alpha:\; R^1 \neq R^2 \quad\text{and}\quad \text{For C}_\beta:\; R^3 \neq R^4$$

If either carbon carries two identical substituents, interchange of positions of the groups around the $$\pi$$-bond no longer produces a distinct compound, and geometrical isomerism becomes impossible. Now we apply this criterion to every option one by one, writing out the full structure each time.

Option A : 2-methylpent-2-ene

The parent pent-2-ene is $$\mathrm{CH_3-CH = CH-CH_2-CH_3}$$. Introducing the 2-methyl group on the second carbon gives

$$\mathrm{CH_3-C(CH_3)=CH-CH_2-CH_3}$$

At the doubly-bonded carbon on the left (the second carbon) the attached groups are $$\mathrm{CH_3}$$ and another $$\mathrm{CH_3}$$ (identical) while the other carbon of the double bond has $$\mathrm{H}$$ and $$\mathrm{CH_2-CH_2-CH_3}$$. Because the first carbon of the $$\mathrm{C=C}$$ bears two identical $$\mathrm{CH_3}$$ groups, we have

$$R^1 = R^2 = \mathrm{CH_3}$$

which violates $$R^1 \neq R^2$$, so no geometrical isomerism is possible here.

Option B : 4-methylpent-2-ene

First write pent-2-ene again: $$\mathrm{CH_3-CH = CH-CH_2-CH_3}$$. Now place a methyl group on carbon 4 (counting from the left end):

$$\mathrm{CH_3-CH = CH-CH(CH_3)-CH_3}$$

Let us examine the two double-bonded carbons separately.

• For the left carbon of the double bond (carbon 2) the two single-bonded substituents are

$$R^1 = \mathrm{CH_3}\quad\text{and}\quad R^2 = \mathrm{H}$$

Clearly, $$R^1 \neq R^2$$.

• For the right carbon of the double bond (carbon 3) the two single-bonded substituents are

$$R^3 = \mathrm{H}\quad\text{and}\quad R^4 = \mathrm{CH(CH_3)-CH_3}$$

Again, $$R^3 \neq R^4$$.

Since both carbons meet the inequality condition, two distinct spatial arrangements (cis and trans, or Z and E) are possible. Therefore geometrical isomerism is exhibited by 4-methylpent-2-ene.

Option C : 4-methylpent-1-ene

Its structure is $$\mathrm{CH_2 = CH-CH_2-CH(CH_3)-CH_3}$$. The leftmost carbon of the double bond is $$\mathrm{CH_2}$$ and hence has two identical hydrogens:

$$R^1 = R^2 = \mathrm{H}$$

Because $$R^1 = R^2$$, the necessary condition fails and geometrical isomerism is not possible.

Option D : 2-methylpent-1-ene

The structure is $$\mathrm{CH_2 = C(CH_3)-CH_2-CH_2-CH_3}$$. The terminal $$\mathrm{CH_2}$$ of the double bond again carries two identical hydrogens:

$$R^1 = R^2 = \mathrm{H}$$

With identical groups on one carbon, geometrical isomerism is not feasible.

Among all the given compounds, only 4-methylpent-2-ene satisfies the requirement that each carbon of the $$\mathrm{C=C}$$ bond bears two different substituents. Hence it alone can exist as a pair of geometrical isomers.

Hence, the correct answer is Option B.

1. Carboxylic Acid Hydrogen (b)

• Location: Part of the -COOH group attached to the aliphatic ring.
• Analysis: When H_b is removed, the negative charge is delocalized over two highly electronegative oxygen atoms via resonance. Generally, carboxylic acids are significantly more acidic than phenols and alkynes because of this equivalent resonance stabilization.
• Conclusion: H_b is the most acidic hydrogen.

2. Phenolic Hydrogens (c and d)

Both H_c and H_d belong to phenolic -OH groups attached directly to the benzene ring. When they are removed, the negative charge is stabilized by resonance into the aromatic benzene ring.

To determine which one is more acidic between c and d, we must analyze the electronic effects of the strongly electron-withdrawing nitro group (-NO2):

• Hydrogen c: * The -OH group holding H_c is located ortho (adjacent) to the -NO2 group.
  • The nitro group exerts both a strong negative inductive effect (-I) and a strong negative resonance effect (-M) at the ortho position. This dramatically pulls electron density away, stabilizing the resulting phenoxide ion exceptionally well.
• Hydrogen d:
  • The -OH group holding H_d is located meta to the -NO2 group.
  • Resonance effects (-M) do not operate at the meta position; the nitro group can only stabilize this conjugate base through its weaker, distance-dependent negative inductive effect (-I).
• Conclusion: Because the ortho position receives both -M and -I stabilization while the meta position only receives -I, H_c is more acidic than H_d (c > d).

3. Terminal Alkyne Hydrogen (a)

• Location: Attached to an sp-hybridized carbon atom (C triple bond C).
• Analysis: While an sp-hybridized carbon is relatively electronegative for a carbon atom (holding onto a negative charge better than sp2 or sp3 carbons), carbon is still fundamentally far less electronegative than oxygen. Consequently, a terminal alkyne is much less acidic than both carboxylic acids and phenols.
• Conclusion: H_a is the least acidic hydrogen.

Final Order of Acidity

Combining all our findings from strongest to weakest:

b > c > d > a

To determine the correct IUPAC name of the given compound, we follow the standard IUPAC nomenclature rules for substituted aromatic compounds.

The molecule contains a benzene ring substituted with four functional groups:

• $$-CHO$$ (aldehyde)
• $$-CH_2OH$$ (hydroxymethyl)
• $$-NH_2$$ (amino)
• $$-NO_2$$ (nitro)

Among these, the aldehyde group has the highest priority according to IUPAC nomenclature rules:

$$\text{Aldehyde} > \text{Alcohol} > \text{Amine} > \text{Nitro}.$$

Therefore, the aldehyde group is selected as the principal functional group, and the parent compound is benzaldehyde.

The carbon atom attached to the $$-CHO$$ group is assigned position $$1$$. The ring can then be numbered in either direction to obtain the lowest possible set of locants for the remaining substituents.

If numbered clockwise, the substituent positions are

$${3,,4,,6}.$$

If numbered counterclockwise, the substituent positions are

$${2,,4,,5}.$$

According to the lowest set of locants rule, the set

$${2,,4,,5}$$

is preferred because it gives the lower number at the first point of difference.

Thus, the substituent positions are:

• $$C1$$ : $$-CHO$$
• $$C2$$ : $$-NO_2$$ (nitro)
• $$C4$$ : $$-CH_2OH$$ (hydroxymethyl)
• $$C5$$ : $$-NH_2$$ (amino)

The substituent prefixes are then arranged alphabetically:

• Amino
• Hydroxymethyl
• Nitro

Combining the substituents with their locants and the parent name gives the IUPAC name:

5-amino-4-hydroxymethyl-2-nitrobenzaldehyde.

Hence, the correct answer is

$$\boxed{\text{5-amino-4-hydroxymethyl-2-nitrobenzaldehyde}}.$$

To determine which compound exhibits geometrical isomerism, we first recall the necessary condition for its existence. For a molecule containing a carbon-carbon double bond to show geometrical (cis-trans or E/Z) isomerism, each carbon of the double bond must be attached to two different substituents. If either carbon is bonded to two identical groups, geometrical isomerism is not possible.

In 4-chloromethylenecyclohexane, the exocyclic double-bonded carbon is attached to two identical hydrogen atoms $$\left(=CH_2\right)$$. Since one carbon of the double bond has identical substituents, the compound cannot exhibit geometrical isomerism.

In 3-methyl-1-chloromethylenecyclohexane, the exocyclic carbon of the double bond is attached to one hydrogen atom and one chlorine atom $$\left(=CHCl\right)$$, which are different substituents. The ring carbon involved in the double bond is connected to two non-equivalent paths around the ring because of the methyl substituent at the $$3$$-position. As both double-bonded carbons have two different groups attached, this compound can exhibit geometrical isomerism.

In 5-methyl-1-chloromethylenecyclohexane, the arrangement is symmetrical, making the two paths around the ring equivalent. Consequently, the ring carbon of the double bond is effectively attached to identical groups, so the compound does not exhibit geometrical isomerism.

Similarly, in 1-chloromethylenecyclohexane, the ring carbon involved in the double bond is attached to two identical sides of the unsubstituted cyclohexane ring. Therefore, it cannot exhibit geometrical isomerism.

Hence, among the given compounds, only 3-methyl-1-chloromethylenecyclohexane (Option B) satisfies the conditions required for geometrical isomerism and can exist as E/Z isomers.

1. Electronegativity and Hybridization:

   As the s-character of a carbon atom increases ($$\text{sp}^3 = 25\%$$, $$\text{sp}^2 = 33.3\%$$, $$\text{sp} = 50\%$$), its electronegativity increases. A more electronegative atom holds its lone pair of electrons more tightly, making it more stable and less willing to donate them to a proton (less basic). Therefore, basicity follows the order:

   $$\text{sp}^3\text{-hybridized C}^- > \text{sp}^2\text{-hybridized C}^- > \text{sp-hybridized C}^-$$

2. Inductive Effect ($$+I$$):

   Alkyl groups are electron-donating groups via the $$+I$$ effect. In $$\text{sp}^3$$ carbanions, the presence of electron-donating methyl groups increases electron density on the negative carbon, increasing its tendency to donate electrons and making it highly basic.

Detailed Stability and Basicity Breakdown:

• $$(v)\text{ }\text{CN}^-$$ (Cyanide ion): The negative charge resides on a highly electronegative nitrogen atom, and the carbon is $$\text{sp}$$-hybridized. This is exceptionally stable, making it the weakest base.
• $$(iii)\text{ }\text{HC}\equiv\text{C}^-$$ (Acetylide ion): The negative charge is on an $$\text{sp}$$-hybridized carbon ($$50\%$$ s-character). It holds the charge tightly, making it a weak base.
• $$(ii)\text{ }\text{CH}_2=\text{CH}^-$$ (Vinyl carbanion): The negative charge is on an $$\text{sp}^2$$-hybridized carbon ($$33.3\%$$ s-character). It is more basic than $$\text{sp}$$ but less basic than $$\text{sp}^3$$.
• $$(iv)\text{ }\text{CH}_3^-$$ (Methyl carbanion): An $$\text{sp}^3$$-hybridized carbanion ($$25\%$$ s-character) with no additional alkyl groups destabilizing it. It is strongly basic.
• $$(i)\text{ }(\text{CH}_3)_3\text{C}^-$$ (tert-Butyl carbanion): An $$\text{sp}^3$$-hybridized carbanion flanked by three electron-donating methyl groups ($$+I$$ effect). This drastically destabilizes the negative charge, making it the most unstable intermediate and the strongest base.

Conclusion:

Combining these factors gives the following increasing order of basicity (weak to strong):

$$\mathbf{(v) < (iii) < (ii) < (iv) < (i)}$$

Answer: Option C

To decide whether a reaction gives a racemic mixture, check two points: (i) does the reaction create at least one chiral centre, and (ii) if it does, are the two possible enantiomers formed in equal amounts? A 50 : 50 pair of enantiomers is a racemic mixture.

Below each option is examined on the above lines.

Option A: $$CH_3CH(OH)CH_3 \xrightarrow{SOCl_2} CH_3CH(Cl)CH_3$$

The conversion of a secondary alcohol to the chloride with $$SOCl_2$$ (no pyridine) follows the $$S_{N}i$$ mechanism.
 • A chlorosulfite intermediate is formed, which is achiral and can rotate freely about the C-O bond.
 • When $$Cl^-$$ makes the internal (front-side) attack, an equal amount of each enantiomer of 2-chlorobutane is produced.
Therefore the product is a racemic mixture.

Option B: $$cis\!-\!CH_3CH\!=\!CHCH_3 \xrightarrow{Br_2/CCl_4} BrCH_2CHBrCH_2CH_3$$

Addition of $$Br_2$$ to an alkene proceeds by anti addition through a cyclic bromonium ion.
For the cis alkene the two new stereocentres (C-2 and C-3) end up with opposite configurations in the two products: $$RS$$ and $$SR$$. These are mirror images formed in 1 : 1 ratio, so the product is racemic.

Option C: $$CH_3CH_2CH=CH_2 \xrightarrow{HBr} CH_3CH_2CH(Br)CH_3$$

The electrophilic addition is Markovnikov: H⁺ adds to the terminal carbon giving a planar secondary carbocation $$CH_3CH_2\!-\!\overset{+}{C}HCH_3$$. The nucleophile $$Br^-$$ attacks this planar centre equally from either face, producing ($$R$$)- and ($$S$$)-2-bromobutane in equal amounts. Hence a racemic mixture is obtained.

Option D: $$trans\!-\!CH_3CH\!=\!CHCH_3 \xrightarrow{Br_2/CCl_4} CH_3CH(Br)CH(Br)CH_3$$

Again the reaction goes through anti addition of bromine. For the trans alkene, anti addition puts the two bromine atoms on opposite sides of the former double bond in such a way that the product, 2,3-dibromobutane, possesses an internal mirror plane:

$$\displaystyle CH_3-CH(Br)-CH(Br)-CH_3$$

This single compound has the absolute configuration $$2R,3S$$ (or equivalently $$2S,3R$$); because the molecule contains a plane of symmetry, it is a meso form and is achiral. Only one achiral product is formed, so no racemic mixture appears.

Thus, the only reaction that does not produce a racemic product is:

Option D which is: addition of $$Br_2$$ to trans-2-butene giving meso-2,3-dibromobutane.

A carbon atom is considered chiral if it satisfies two conditions:

1. It must be $$sp^3$$ hybridized (bonded to four atoms or groups via single bonds).
2. It must be directly attached to four entirely different chemical groups.

Any carbon atom involved in a double bond ($$sp^2$$ hybridized, such as those in the aromatic quinoline ring or the side phenyl/methyl groups) cannot be chiral.

Step-by-Step Structural Identification:

Let's break down the molecule into its main structural units to locate the chiral centers:

• 1. The Linker Carbon (The Benzylic/Alcoholic Position):

  Look at the central carbon atom attached to the hydroxyl ($$-\text{OH}$$) group, a methyl ($$-\text{CH}_3$$) group, the aromatic quinoline ring, and the adjacent quinuclidine ring system. This carbon is single-bonded to four completely distinct structural components, making it a chiral center (1).




• 2. The Quinuclidine (Bicyclic) Cage System:

  In the bicyclic nitrogen-containing ring system, we look at the bridgehead and substituted carbons:

  • The Nitrogen atom: Though it has a lone pair and three different bonds, it is not a carbon atom.
  • The Bridgehead Carbon (opposite to N): This carbon connects three separate carbon pathways. Two pathways match identically in ring symmetry, making this specific bridgehead carbon achiral.
  • The Carbon attached to the linker: This carbon links the cage to the alcohol unit. Its four attachments are unique, making it a chiral center (2).
  • The Carbon with the external aromatic ring substituent: The carbon atom on the upper part of the cage holding the secondary aromatic ring ($$-\text{C}_6\text{H}_4\text{CH}_3$$ group) is attached to four unique paths, making it a chiral center (3).
  • Two adjacent bridge carbons: Due to the asymmetric substitutions introduced on the cage rings, two additional skeletal carbon vertices within this bridged bicyclic network lose their internal plane of symmetry and are bonded to four unique groupings, contributing two more chiral centers (4 and 5).



• 3. Aromatic Rings:

  All carbon atoms within the quinoline bicyclic ring system and the terminal toluene ring are $$sp^2$$ hybridized (forming aromatic double bonds). None of these can be chiral centers.

Conclusion:

Counting the asymmetric positions carefully, we locate exactly 1 chiral carbon at the central alcohol linkage and 4 chiral carbons embedded within the rigid, asymmetric quinuclidine core matrix.

Answer: 5

Chloramphenicol is an antibiotic whose IUPAC name is 2,2-dichloro-$$N$$-[(1$$R$$,2$$R$$)-2-hydroxy-1-(hydroxymethyl)-2-(4-nitrophenyl)ethyl]acetamide.

Its structural formula is:

$$O_2N - C_6H_4 - CH(OH) - CH(NH - CO - CHCl_2) - CH_2OH$$

To find chiral carbons, we look for $$sp^3$$ hybridised carbon atoms bonded to four different groups.

Carbon 1: The carbon bearing the $$-OH$$ group, i.e. $$-CH(OH)-$$. Its four substituents are: (i) $$-C_6H_4(NO_2)$$ (the para-nitrophenyl ring), (ii) $$-OH$$, (iii) $$-H$$, (iv) the adjacent carbon bearing the amide group. All four groups are different, so this carbon is chiral.

Carbon 2: The carbon bearing the amide group, i.e. $$-CH(NHCO-CHCl_2)-$$. Its four substituents are: (i) $$-NHCO-CHCl_2$$, (ii) $$-CH_2OH$$, (iii) $$-H$$, (iv) the adjacent carbon bearing $$-OH$$. All four groups are different, so this carbon is also chiral.

Other carbons in the molecule are either part of the aromatic ring ($$sp^2$$), the carbonyl carbon ($$sp^2$$), the $$-CHCl_2$$ carbon (two identical Cl substituents, so not chiral), or the $$-CH_2OH$$ carbon (two identical H atoms, so not chiral).

Therefore, the number of chiral carbons in chloramphenicol is $$2$$.

The correct answer is $$2$$.

First we recall the definition of an $$\text{sp}^2$$-hybridised carbon. A carbon atom is $$\text{sp}^2$$-hybridised when it forms three $$\sigma$$-bonds and one unhybridised $$p$$-orbital participates in a $$\pi$$-bond; this happens in every carbon that is part of a double bond ($$C=C$$ or $$C=O$$) and in every carbon atom of an aromatic ring such as benzene.

Now we write the full structural formula of Aspartame (methyl L-α-aspartyl-L-phenylalaninate). In condensed form the molecule can be expressed as

$$$\mathrm{CH_3OOC\!-\!CH(NH_2)\!-\!CH_2\!-\!CO\!-\!NH\!-\!CH(CH_2\!-\!C_6H_5)\!-\!COOCH_3}$$$

For clarity we expand it, separating every carbon so that we can inspect their hybridisation:

$$$ \begin{aligned} &\text{(i) } \mathrm{COOCH_3} &&\longrightarrow &&\underline{\mathrm{O=}}\boxed{\mathrm{C}}\!-\!\mathrm{O}\!-\!\mathrm{CH_3} \\ &\text{(ii) } \mathrm{-CH(NH_2)-} &&\longrightarrow &&\boxed{\mathrm{C}}\!-\!\mathrm{H},\;\mathrm{NH_2} \\ &\text{(iii) } \mathrm{-CH_2-} &&\longrightarrow &&\boxed{\mathrm{C}}\!-\!\mathrm{H_2} \\ &\text{(iv) } \mathrm{-CO-} &&\longrightarrow &&\underline{\mathrm{O=}}\boxed{\mathrm{C}}\!-\!\mathrm{NH-} \\ &\text{(v) } \mathrm{-CH(}\mathrm{CH_2\!-\!C_6H_5}\mathrm{)-} &&\longrightarrow &&\boxed{\mathrm{C}}\!-\!\mathrm{H},\;\mathrm{CH_2},\;\mathrm{C_6H_5} \\ &\text{(vi) } \mathrm{-COOCH_3} &&\longrightarrow &&\underline{\mathrm{O=}}\boxed{\mathrm{C}}\!-\!\mathrm{O}\!-\!\mathrm{CH_3} \end{aligned} $$$

The phenyl ring $$\mathrm{C_6H_5}$$ is an ordinary benzene ring, so every one of its six carbons is $$\text{sp}^2$$-hybridised.

We now count the $$\text{sp}^2$$ carbons one by one.

1. In the benzene (phenyl) ring there are $$6$$ carbons. All $$6$$ are $$\text{sp}^2$$.

2. Each carbonyl group $$\mathrm{C=O}$$ contributes one $$\text{sp}^2$$ carbon because the carbon in $$\mathrm{C=O}$$ is involved in one double bond. Aspartame possesses three separate carbonyls:   (a) the ester carbonyl on the extreme left,   (b) the amide carbonyl in the middle, and   (c) the second ester carbonyl on the right. Therefore the number of $$\text{sp}^2$$ carbons from carbonyls is $$3$$.

3. All remaining carbons (those in $$\mathrm{CH_3}$$, $$\mathrm{CH_2}$$ and $$\mathrm{CH}$$ groups that are not part of a double bond or an aromatic ring) are $$\text{sp}^3$$, so they do not contribute to our count.

Adding the contributions, we have

$$$\text{Total sp}^2 \text{ carbons} \;=\; 6 \;(\text{benzene}) \;+\; 3 \;(\text{carbonyls}) \;=\; 9.$$$

So, the answer is $$9$$.

Double bond holds more priority hence the numbering would start from carbon nearest to it.

Sum of locants must be the smallest so we start numbering from Para and go right from it.

IUPAC rules state start numbering from alphabetical order if all other priorities are equal

CH3SO3-: The negative charge is highly delocalized over three highly electronegative oxygen atoms via resonance. Because it is incredibly stable, it is an exceptionally weak base and a very poor nucleophile (often acting as an excellent leaving group).
CH3COO- : The negative charge is delocalized over two oxygen atoms. It is more basic than mesylate, making it a moderate nucleophile
OH- : The negative charge is localized entirely on a single oxygen atom with no resonance stabilization. Therefore, it is a very strong base and a strong nucleophile.
H2O : Water is a neutral molecule. Neutral nucleophiles are significantly less nucleophilic than their conjugate anionic counterparts OH-. 

The numbering starts from the rightmost carbon, since COOH comes above in the priority order, the prefix -OH group is used as suffix of the Carboxylic Acid will be taken. Hence, as the longest carbon chain is of 5 members including -COOH. The IUPAC name would be 3-Hydroxy-4-methylpentanoic acid

The angle would be 120 + 29 = 159. H-H' = 29 and C-H' to C-H = 120.

We begin by recalling that the different conformations of an acyclic (open-chain) molecule arise from rotation about single (σ) bonds. The relative stability of any two such conformers is decided by the types of strains or interactions that become more or less severe as the bond rotates.

Four commonly discussed factors are:

1. Steric interactions - also called van der Waals repulsions. When two bulky groups come too close, their electron clouds repel. Increased repulsion raises potential energy, so minimising steric crowding stabilises a conformation.

2. Torsional strain - extra energy due to eclipsing of bonds. It is quantified by the Fourier form $$E_{\text{torsion}} = \frac{V}{2}\bigl(1 - \cos 3\theta\bigr),$$ where $$\theta$$ is the dihedral angle. A staggered arrangement ($$\theta = 60^\circ,\,180^\circ$$) lowers this energy, whereas an eclipsed arrangement ($$\theta = 0^\circ$$) raises it. Thus torsional strain definitely governs conformational stability.

3. Electrostatic forces of interaction - favourable dipole-dipole alignments (antiparallel) or unfavourable ones (parallel) can make one conformer lower or higher in energy. Therefore these intermolecular-style intramolecular effects also contribute.

4. Angle strain - this is the strain that results when bond angles deviate from the ideal tetrahedral value $$109.5^\circ$$ for an $$sp^3$$ carbon. However, in an acyclic molecule the C-C-C bond angles are virtually free to assume their ideal values; simple rotation about a σ-bond does not force them to compress or expand. Angle strain is a concept that becomes important mainly in cyclic systems (small rings such as cyclopropane or cyclobutane) where the ring geometry locks the angles at non-ideal values.

So, among the four listed factors, steric interactions, torsional strain, and electrostatic interactions all influence how much potential energy a conformer possesses, whereas angle strain does not come into play for an open chain.

Hence, the correct answer is Option B.

• $$\text{--CO}_2\text{Et (Ester)} \rightarrow \text{--CH}_2\text{OH (Primary Alcohol)}$$ : Reduced
• $$\text{--CO}_2\text{H (Carboxylic Acid)} \rightarrow \text{--CO}_2\text{H (Carboxylic Acid)}$$ : Protected / Unchanged
• $$\text{--CN (Nitrile)} \rightarrow \text{--CHO (Aldehyde)}$$ : Reduced selectively to an aldehyde

Detailed Evaluation of Each Option:

• Option A: (i) $$\text{NaBH}_4$$, (ii) Raney $$\text{Ni/H}_2$$, (iii) $$\text{H}_3\text{O}^+$$

  • $$\text{NaBH}_4$$ is a mild reducing agent that generally does not reduce esters, carboxylic acids, or nitriles under standard conditions. Thus, step (i) will fail to convert the ester to an alcohol.
  • Raney $$\text{Ni/H}_2$$ reduces nitrile ($$-\text{CN}$$) groups completely down to primary amines ($$-\text{CH}_2\text{NH}_2$$), not aldehydes ($$-\text{CHO}$$).
  • Result: Incorrect transformation products.



• Option B: (i) $$\text{LiAlH}_4$$, (ii) $$\text{H}_3\text{O}^+$$

  • $$\text{LiAlH}_4$$ is an exceptionally strong, non-selective reducing agent. It will reduce the ester ($$-\text{CO}_2\text{Et}$$) to a primary alcohol, but it will also simultaneously reduce the carboxylic acid ($$-\text{CO}_2\text{H}$$) to a primary alcohol and the nitrile ($$-\text{CN}$$) group to a primary amine ($$-\text{CH}_2\text{NH}_2$$).
  • Result: Destroys the carboxylic acid and creates an amine instead of an aldehyde.



• Option C: (i) $$\text{B}_2\text{H}_6$$, (ii) DIBAL-H, (iii) $$\text{H}_3\text{O}^+$$

  • Diborane ($$\text{B}_2\text{H}_6$$) is highly chemoselective for the reduction of carboxylic acids over esters and nitriles. It would rapidly reduce the functional $$-\text{CO}_2\text{H}$$ group to an alcohol, leaving the ester untouched. This is the exact opposite of the target compound requirements.
  • Result: Reduces the wrong acidic functional group.



• Option D: (i) $$\text{B}_2\text{H}_6$$, (ii) $$\text{SnCl}_2/\text{HCl}$$, (iii) $$\text{H}_3\text{O}^+$$

  • Step (i): When coordinated under specific, controlled stoichiometric conditions where the carboxylic acid group forms a stable acyloxyborane intermediate, $$\text{B}_2\text{H}_6$$ selectively permits the targeted reduction of the ester group to a primary alcohol ($$-\text{CH}_2\text{OH}$$) while leaving the structural carboxylic acid framework preserved after mild workup.
  • Step (ii) & (iii): The combination of $$\text{SnCl}_2/\text{HCl}$$ followed by acid hydrolysis ($$$\text{H}_3\text{O}^+$$) is the classic Stephen Reduction. This sequence selectively converts the nitrile group ($$-\text{CN}$$) into an iminium salt intermediate, which is subsequently hydrolyzed cleanly to form the target aldehyde ($$-\text{CHO}$$) group without affecting the rest of the molecule.
  • Result: Perfectly yields the desired product.

Conclusion:

Only the sequence in Option D offers the necessary functional group selectivity to reduce the ester and transform the nitrile group while keeping the carboxylic acid functional group intact.

Answer: Option D — (i) $$\text{B}_2\text{H}_6$$, (ii) $$\text{SnCl}_2/\text{HCl}$$, (iii) $$\text{H}_3\text{O}^+$$

For geometric isomerism, each carbon of a $$C=C$$ bond must be bonded with two different groups. The carbon of the double bond $$I$$ is bonded with 2 methyl groups. Thus, geometric isomerism is not possible at the double bond $$I$$

The second carbon (C2) is attached to $$-OH$$, $$-CO_2H$$, $$H$$, and the C3 group.

(1) $$-OH$$: Oxygen (atomic number 8) is the highest.

(2) $$-C3$$ group: This carbon is attached to $$(Cl, C, H)$$. Since Chlorine (17) has a higher atomic number than Oxygen (8), this group beats the carboxylic acid.

(3) $$-CO_2H$$: This carbon is attached to $$(O, O, O)$$.

(4) $$-H$$: Hydrogen is always the lowest.

The sequence $$1 \rightarrow 2 \rightarrow 3$$ ($$OH \rightarrow C3 \rightarrow CO_2H$$) moves clockwise. In a Fischer projection, if the lowest priority group ($$H$$) is on a horizontal bond, we reverse the result. Clockwise becomes S.

The third carbon (C3) is attached to $$-Cl$$, the C2 group, $$-CH_3$$, and $$H$$.

(1) $$-Cl$$: Chlorine (17) is the highest.

(2) $$-C2$$ group: This carbon is attached to $$(O, C, H)$$. Oxygen beats the hydrogens of the methyl group.

(3) $$-CH_3$$: Carbon attached to $$(H, H, H)$$.

(4) $$-H$$: Lowest priority.

The sequence $$1 \rightarrow 2 \rightarrow 3$$ ($$Cl \rightarrow C2 \rightarrow CH_3$$) moves counter-clockwise. Since $$H$$ is on a horizontal bond, we reverse the result. Counter-clockwise becomes R.

First, let us recall that conventional dry-cleaning of clothes has long been carried out with the solvent tetrachloroethylene, commonly called perchloroethylene or simply “perc.” However, environmental chemists have shown that $$\text{C}_2\text{Cl}_4$$ is a volatile chlorinated hydrocarbon; when it evaporates it adds harmful organic vapours to the atmosphere and its residues in soil and groundwater are toxic to living organisms. Therefore, modern industry looks for substitute fluids that clean garments equally well but release far fewer pollutants.

With this goal in mind, researchers examined several other small molecules.

We have the option $$SO_2$$ (sulphur dioxide). Although $$SO_2$$ is not chlorinated, it is still an acidic oxide; when it dissolves in atmospheric moisture it forms sulphurous and sulphuric acids, leading directly to acid rain. Hence using $$SO_2$$ would aggravate air-quality problems rather than solve them.

Next we look at $$NO_2$$ (nitrogen dioxide). This brown gas is an important component of urban smog; photochemical reactions involving $$NO_2$$ produce tropospheric ozone, another major pollutant. So employing $$NO_2$$ in dry cleaning would likewise worsen air pollution.

The usual industrial solvent, $$C_2Cl_4$$ (tetrachloroethylene), produces chlorinated emissions and can persist in the environment for years, contaminating soil and aquifers. Its vapours are also suspected carcinogens. Therefore, continuing with $$C_2Cl_4$$ does not constitute a “better” environmental strategy.

That leaves $$CO_2$$ (carbon dioxide). In everyday discussion $$CO_2$$ is linked to the greenhouse effect, but here it is used in a closed dry-cleaning machine under high pressure, where it exists as liquid or supercritical $$CO_2$$. After the cleaning cycle the gas can be re-compressed and recycled inside the machine, so there is practically no release to the atmosphere. Because $$CO_2$$ is non-toxic, non-flammable, inexpensive, and already abundant in the air, its controlled reuse in such equipment avoids introducing new harmful chemicals and thus represents the most environmentally benign alternative among the four substances listed.

Therefore, choosing carbon dioxide for dry cleaning is the superior method to control environmental pollution.

Hence, the correct answer is Option B.

First, we recall a very important general fact about Grignard reagents. A Grignard reagent has the general formula $$R{-}MgX$$, where $$R$$ is an alkyl group and $$X$$ is a halogen. Because the carbon-magnesium bond is highly polar, the carbon atom behaves as a strong carbanion. Therefore, a Grignard reagent is extremely reactive toward any source of acidic hydrogen (denoted by $$H^+$$), such as water, alcohols, phenols, carboxylic acids, and even terminal alkynes.

The standard reaction that expresses this reactivity is written first in its most general form:

$$R{-}MgX \;+\; H{-}Y \;\rightarrow\; R{-}H \;+\; MgX{-}Y$$

Here, $$H{-}Y$$ is any compound supplying a proton. The key observation is that the $$R$$ group from the Grignard reagent simply abstracts (removes) the proton. As a result, the hydrocarbon $$R{-}H$$ is liberated, often as a gas if it is small and volatile, while the magnesium-containing residue $$MgX{-}Y$$ remains in solution or as a solid.

Now we apply this general idea to the specific substances given in the problem. The Grignard reagent supplied is $$CH_3MgBr$$. The proton source present is methanol, which we write as $$CH_3OH$$. Placing these into the general reaction pattern, we write:

$$CH_3MgBr \;+\; CH_3OH \;\rightarrow\; CH_3H \;+\; MgBr(OCH_3)$$

At this stage we notice that the formula $$CH_3H$$ can be simplified. Joining the carbon atom with three hydrogens from the methyl group and the single hydrogen abstracted from methanol gives the simplest alkane, methane, whose correct molecular formula is $$CH_4$$. Therefore we rewrite the equation more neatly as:

$$CH_3MgBr \;+\; CH_3OH \;\rightarrow\; CH_4 \;+\; Mg(OCH_3)Br$$

Because methane is a very small, non-polar molecule, it is a gas under ordinary laboratory conditions. Consequently, when the mixture is warmed, methane escapes as a gas.

We can underline the reasoning in words: the methyl group $$CH_3{-}$$ from the Grignard reagent accepts a proton $$H^+$$ from methanol, directly forming methane $$CH_4$$. No other gaseous hydrocarbon (such as ethane or propane) can appear, because only one carbon atom is present in the fragment leaving as the hydrocarbon.

Therefore, the gas evolved is methane, which corresponds to Option A.

Hence, the correct answer is Option A.

First, let us recall the basic classification of synthetic detergents. Detergents are broadly placed into three categories based on the electrical charge that appears on the part of the molecule which goes into water when the detergent is dissolved.

We have

$$\text{1. Anionic detergents}$$ — their hydrophilic end carries a negative charge (usually $$\;{\large -SO_3^-}\;$$ or $$\;{\large -SO_4^-}\;$$).
$$\text{2. Cationic detergents}$$ — their hydrophilic end carries a positive charge (most commonly the quaternary ammonium ion $$\;{\large R_4N^+}\;$$).
$$\text{3. Non-ionic detergents}$$ — their hydrophilic end is neutral (generally an alcohol, ester, or ether chain).

Now we shall look at each option one by one and identify the nature of the hydrophilic (water-loving) part.

Option A : Cetyltrimethyl ammonium bromide
The name shows a quaternary ammonium group $$\;{\large (CH_3)_3N^+}$$ with a bromide counter-ion $$\;Br^-$$. Because the positive charge resides on nitrogen, the detergent ion produced in water is $$\;{\large (CH_3)_3N^+R}\;$$. A detergent whose head bears a positive charge is, by definition, cationic, not anionic.

Option B : Glyceryl oleate
This substance is an ester formed from glycerol and oleic acid. No ionic group is present; the molecule remains neutral in aqueous solution. Therefore it belongs to the non-ionic class, not to the anionic class.

Option C : Hexadecyltrimethyl ammonium bromide
Again we notice the quaternary ammonium ion $$\;{\large (CH_3)_3N^+}$$ paired with bromide. Exactly like option A, it is a cationic detergent.

Option D : Sodium lauryl sulphate
The structural formula of lauryl sulphate ion may be written as

$$\text{Lauryl group} - SO_4^- \;=\; CH_3(CH_2)_{11}OSO_3^-$$

In aqueous solution the compound dissociates as

$$\underbrace{CH_3(CH_2)_{11}OSO_3^-}_{\text{detergent ion carrying a negative charge}} \;+\; \underbrace{Na^+}_{\text{sodium counter-ion}}$$

The hydrophilic end clearly possesses a negative charge (the sulphate group $$-SO_4^-$$). Hence sodium lauryl sulphate is an anionic detergent.

Collecting our observations, only option D furnishes an ion with a negative charge in solution, satisfying the definition of an anionic detergent. Therefore the correct choice is option D.

Hence, the correct answer is Option D.

To determine the optically inactive compound, we need to recall that optical inactivity occurs when a compound does not rotate the plane of polarized light. This typically happens if the molecule is achiral, meaning it either lacks a chiral center or possesses a plane of symmetry. A chiral center is a carbon atom bonded to four different groups. Let's analyze each option step by step.

Starting with Option A: 2-chloropropanal. The structure is $$CH_{3}-CHCl-CHO$$. The carbon at position 2 is bonded to four groups: hydrogen (H), chlorine (Cl), methyl group ($$CH_{3}$$), and aldehyde group ($$CHO$$). Since $$CH_{3}$$ and $$CHO$$ are different, and all four groups are distinct (H, Cl, $$CH_{3}$$, $$CHO$$), this carbon is a chiral center. The molecule has no plane of symmetry, so it is chiral and optically active.

Next, Option B: 2-chloro-2-methylbutane. The structure is $$CH_{3}-C(Cl)(CH_{3})-CH_{2}-CH_{3}$$, which can be written as:

Moving to Option C: 2-chlorobutane. The structure is $$CH_{3}-CHCl-CH_{2}-CH_{3}$$. The carbon at position 2 is bonded to H, Cl, $$CH_{3}$$, and $$CH_{2}CH_{3}$$. These groups are all different: methyl ($$CH_{3}$$) and ethyl ($$CH_{2}CH_{3}$$) have different sizes and compositions. Thus, carbon-2 is a chiral center, and the molecule lacks a plane of symmetry, making it chiral and optically active.

Finally, Option D: 2-chloropentane. The structure is $$CH_{3}-CHCl-CH_{2}-CH_{2}-CH_{3}$$. The carbon at position 2 is bonded to H, Cl, $$CH_{3}$$, and propyl group ($$CH_{2}CH_{2}CH_{3}$$). All four groups are distinct (H, Cl, $$CH_{3}$$, $$CH_{2}CH_{2}CH_{3}$$), so carbon-2 is a chiral center. The molecule has no plane of symmetry, rendering it chiral and optically active.

Therefore, the only optically inactive compound is Option B, 2-chloro-2-methylbutane, due to its plane of symmetry and absence of a chiral center.

Hence, the correct answer is Option B.

First, we recall the definition of positional (position) isomerism. Two compounds are called positional isomers when (i) both possess the same molecular formula, (ii) both contain the same functional group, and (iii) the only difference between the two is the position of that functional group or of a multiple bond on the same carbon skeleton.

Now we inspect the two substances given in option 1. The first molecule in that pair is written as $$CH_3CH_2CH_2OH$$. Counting all atoms, we have

$$C:3,\; H:8,\; O:1 \;\; \Longrightarrow \;\; \text{molecular formula } C_3H_8O.$$

The second molecule in the same pair is written as $$CH_3CH(OH)CH_3$$. Again, counting every atom explicitly, we get

$$C:3,\; H:8,\; O:1 \;\; \Longrightarrow \;\; \text{molecular formula } C_3H_8O.$$

So both substances definitely satisfy the first requirement of having an identical molecular formula. Next we note that both contain the hydroxyl functional group $$-OH$$, so the functional group is also common. The only remaining point to check is the position of this $$-OH$$ group on the parent carbon chain.

In $$CH_3CH_2CH_2OH$$ the $$-OH$$ is attached to the terminal (number 1) carbon, giving the name propan-1-ol. In $$CH_3CH(OH)CH_3$$ the $$-OH$$ is bonded to the middle (number 2) carbon, giving the name propan-2-ol. Therefore the two differ solely in the location of the same functional group on an unchanged three-carbon skeleton; this is exactly the hallmark of positional isomerism.

For completeness, we compare this situation with the remaining options. In each of the other three pairs, either the molecular formula is different (violating condition (i)), or the carbon skeleton is rearranged (giving chain isomers rather than positional isomers), or the functional groups themselves are different (giving functional isomers). Hence none of those remaining pairs satisfies all three criteria listed at the outset.

Because only the pair in option 1 fulfils every one of the requirements for positional isomerism, we can confidently conclude that this is the correct choice.

Hence, the correct answer is Option 1.

Option A: B is more stable than A because B is a tertiary ($$3^\circ$$) radical stabilized by more hyperconjugation, whereas A is only a secondary ($$2^\circ$$) radical.

Option B: B is more stable than A because the cyclopropyl radical (A) suffers from high angle strain and has greater s-character in its radical orbital, making it highly unstable compared to the standard cyclohexyl radical (B).

Option C: B is more stable than A because B is an allylic radical stabilized by resonance, whereas A is a vinylic radical where the unpaired electron sits directly on a $$sp^2$$ carbon, making it exceptionally unstable.

Option D: A is more stable than B because the triphenylmethyl radical ($$\text{Ph}_3\text{C}^\bullet$$) is extensively stabilized by resonance across three aromatic rings, which vastly outperforms the simple hyperconjugation stabilizing the tert-butyl radical ($$(\text{CH}_3)_3\text{C}^\bullet$$).

The longest continuous carbon chain contains 6 carbon atoms, identifying the parent alkane as hexane. 

The molecule has two alkyl branches: an ethyl group and a methyl group.

Numbering from either end results in the same locant set (3, 4). According to IUPAC rules, when locants are identical from both sides, the lower number is assigned to the substituent that comes first alphabetically.

Ethyl (E) precedes Methyl (M). Thus, Ethyl is assigned position 3 and Methyl is assigned position 4.

IUPAC Name: 3-ethyl-4-methylhexane

To determine which acid does not exhibit optical isomerism, we must recall that optical isomerism arises when a molecule has a chiral center, typically a carbon atom with four different substituents, and the molecule lacks a plane of symmetry. A chiral molecule is non-superimposable on its mirror image. Let's examine each option step by step.

Starting with option A: Lactic acid. Its molecular formula is $$CH_{3}CH(OH)COOH$$. The central carbon atom (the one attached to the OH group) is bonded to four different groups: a hydrogen atom (H), a hydroxyl group (OH), a methyl group ($$CH_{3}$$), and a carboxyl group ($$COOH$$). Since all four groups are distinct, this carbon is a chiral center. Therefore, lactic acid exhibits optical isomerism and exists as two enantiomers.

Next, option B: Tartaric acid. Its molecular formula is $$HOOCCH(OH)CH(OH)COOH$$. This molecule has two carbon atoms, each bonded to an OH group. Each of these carbons is attached to four different groups: H, OH, COOH, and the other $$CH(OH)COOH$$ group. Thus, both carbons are chiral centers. Tartaric acid exists in three stereoisomeric forms: the d-form, the l-form (which are enantiomers and optically active), and the meso-form (which has a plane of symmetry and is optically inactive due to internal compensation). However, since tartaric acid can exist in optically active forms (the d- and l-enantiomers), it does exhibit optical isomerism.

Now, option C: Maleic acid. Maleic acid is the cis-isomer of butenedioic acid, with the formula $$HOOCCH=CHCOOH$$. The structure can be represented as:

$$\begin{array}{c} \text{H} & & \text{COOH} \\ & \backslash & \\ & C = C \\ & / & \\ \text{COOH} & & \text{H} \end{array}$$

In maleic acid, the double bond between the two carbon atoms restricts rotation. Each carbon in the double bond is sp² hybridized and bonded to three atoms: one hydrogen (H), one carboxyl group (COOH), and the other carbon via the double bond. Since neither carbon has four substituents, there are no chiral centers. Additionally, the molecule is planar and has a plane of symmetry that bisects the double bond and aligns the two identical halves, making it achiral. Therefore, maleic acid does not exhibit optical isomerism.

Finally, option D: α-amino acids. These have the general formula $$R-CH(NH_{2})COOH$$, where the α-carbon is bonded to four groups: a hydrogen atom (H), an amino group ($$NH_{2}$$), a carboxyl group ($$COOH$$), and a side chain (R). For most α-amino acids, R is different from H, $$NH_{2}$$, and $$COOH$$, making the α-carbon a chiral center. For example, alanine (R = $$CH_{3}$$) has a chiral center and exhibits optical isomerism. However, glycine is an exception where R = H, so the α-carbon has two identical H atoms and is not chiral. Despite glycine not exhibiting optical isomerism, the class of α-amino acids as a whole includes compounds that do exhibit optical isomerism (like alanine). Therefore, this option does not universally lack optical isomerism.

Comparing all options, lactic acid (A) and tartaric acid (B) exhibit optical isomerism due to chiral centers. α-amino acids (D) include some that exhibit optical isomerism. Maleic acid (C) has no chiral center and is achiral, so it does not exhibit optical isomerism.

Hence, the correct answer is Option C.

Option A is a tertiary ($$3^\circ$$) allylic carbocation. The positive charge sits on a carbon attached to three other carbons, and it is directly adjacent to a conjugated double-bond system. Furthermore, the double bond is in conjugation with a phenyl ($$\text{Ph}$$) ring, which allows extensive delocalization of the positive charge throughout the aromatic ring via resonance. Hence, it is the most stable.

We are given that $$(-)-1$$-chloro-1-phenylethane dissolves in toluene and racemises slowly when a small amount of $$SbCl_5$$ is added.

First, let us understand the structure. 1-Chloro-1-phenylethane is:

$$C_6H_5 - CHCl - CH_3$$

The carbon bearing the chlorine atom (C-1) has four different groups: $$-C_6H_5$$, $$-CH_3$$, $$-Cl$$, and $$-H$$. This makes it a chiral centre, and the compound is optically active (it is given as the ($$-$$) enantiomer).

Now, $$SbCl_5$$ is a strong Lewis acid. It can accept the lone pair from the chlorine atom of the organic substrate:

$$C_6H_5CHClCH_3 + SbCl_5 \rightarrow C_6H_5CH^+CH_3 + SbCl_6^-$$

Here, $$SbCl_5$$ abstracts the $$Cl^-$$ from the chiral carbon, generating a carbocation at that carbon.

The resulting carbocation $$C_6H_5CH^+CH_3$$ is a secondary benzylic carbocation. The positive charge is stabilised by resonance with the phenyl ring. Crucially, this carbocation is $$sp^2$$ hybridised and therefore has a planar geometry (trigonal planar around the positively charged carbon).

Because the carbocation is planar, when the $$Cl^-$$ (from $$SbCl_6^-$$) attacks back, it can approach from either face of the plane with equal probability:

(i) Attack from one face regenerates the original $$(-)$$ enantiomer.

(ii) Attack from the opposite face generates the $$(+)$$ enantiomer.

Over time, a 50:50 mixture of both enantiomers (a racemic mixture) is formed, and the optical rotation drops to zero. This is the observed racemisation.

The key intermediate responsible for racemisation is the planar carbocation.

Note: A free radical (Option B) would require homolytic cleavage, but $$SbCl_5$$ is a Lewis acid that promotes heterolytic (ionic) cleavage. A carbanion (Option C) would require removal of $$H^+$$, not $$Cl^-$$. A carbene (Option D) would require loss of $$HCl$$, which is not the mechanism here.

The correct answer is Option A: Carbocation.

(III) Benzyl Carbocation ($$C_6H_5-\overset{+}{C}H_2$$)

The positive charge on the carbon atom is in conjugation with the $$\pi$$-electron system of the benzene ring. The charge is delocalized to the ortho and para positions of the ring. It has strong $$+M$$ (Resonance). It has 4 resonating structures in total. This extensive delocalization makes it highly stable.

(I) Allyl Carbocation ($$CH_2=CH-\overset{+}{C}H_2$$)

The positive charge is in conjugation with a single carbon-carbon double bond. It also has $$+M$$ (Resonance) and has 2 resonating structures.

While it is resonance-stabilized, the degree of delocalization is lower than that in the benzyl carbocation.

(II) n-Propyl Carbocation ($$CH_3-CH_2-\overset{+}{C}H_2$$)

This is a primary ($$1^\circ$$) alkyl carbocation. There is no resonance possible here. Hyperconjugation and $$+I$$ effects are present here from the ethyl group. It has 2 $$\alpha$$-hydrogens. Since resonance is a much stronger stabilizing factor than hyperconjugation, this is the least stable carbocation among the three.

Stability: Benzyl (III) > Allyl (I) > n-Propyl (II)

Maleic acid and fumaric acid are two different arrangements of the same molecular formula $$C_4H_4O_4$$ (the structure of $$HOOC - CH = CH - COOH$$).

Because the C=C double bond cannot freely rotate, the two $$-COOH$$ groups can be located either on the same side of the double bond (cis) or on opposite sides (trans).

This difference arises purely from restricted rotation about the double bond, not from any change in carbon chain, functional group, or tautomeric shift. Hence they are classified as geometrical (cis-trans) isomers.

Therefore, the correct choice is:
Option D which is: Geometrical isomers

Resonance refers to the delocalisation of $$\pi$$-electrons (and lone-pair electrons that can convert to $$\pi$$-electrons) over two or more adjacent, coplanar, conjugated atoms. A molecule or ion will show resonance only if such a continuous conjugated system exists.
The key checkpoints are:
- At least one $$\pi$$-bond or an adjacent lone pair in an orbital that can overlap sideways.
- Continuous overlap of $$p$$ orbitals over three or more atoms.

Now examine every option.

Option A Dimethyl ether, $$CH_3-O-CH_3$$
The central oxygen possesses two lone pairs, but both adjacent atoms are sp3-hybridised carbons that contain only $$\sigma$$ bonds. No adjacent empty or partially filled $$p$$ orbital is present, so the lone pairs on oxygen cannot delocalise. Hence no alternative Lewis (resonance) structures are possible. Therefore dimethyl ether cannot be represented by resonance structures.

Option B Nitrate anion, $$NO_3^-$$
The nitrogen is $$sp^2$$-hybridised and each oxygen carries either a $$\pi$$ bond or a lone pair that can participate in $$p$$-orbital overlap. This gives three equivalent Lewis structures obtained by shifting the $$N$$=O double bond and negative charge around the three oxygens. Hence nitrate ion definitely exhibits resonance.

Option C Carboxylate anion, $$RCOO^-$$
A carboxylate ion has two oxygen atoms bonded to the same carbon; the negative charge and the $$\pi$$ bond can be interchanged between the two oxygens, giving two equivalent resonance structures. So the carboxylate anion is resonating.

Option D Toluene, $$C_6H_5CH_3$$
The benzene ring itself contains a conjugated $$\pi$$ system with six electrons shared over six carbon atoms, represented by two Kekulé resonance forms (and additional Dewar forms). Thus toluene also possesses resonance.

Among the four, the only species lacking any conjugated $$\pi$$ system is dimethyl ether. Hence it cannot be represented by resonance structures.

Final answer: Option A which is: Dimethyl ether

The given structure, represents a three-carbon chain (propane) in which an O atom bridges the first and second carbon atoms, forming a three-membered cyclic ether (an epoxide).

Step 1 - Select the parent chain.
The longest continuous carbon chain contains three carbon atoms, so the parent hydrocarbon is propane.

Step 2 - Identify and locate the functional group.
An oxygen atom is bonded to C-1 and C-2, creating an $$\ce{C-O-C}$$ bridge. According to IUPAC, such a bridge on adjacent carbons is named as an “epoxy” substituent. Its locants are written as “1,2-epoxy”.

Step 3 - Number the parent chain.
Number the propane skeleton from the end that gives the lowest possible set of locants to the epoxy group. Either direction gives 1 and 2, so we keep the conventional left-to-right order.

Step 4 - Assemble the name.
Prefix (position, type of bridge): $$1,2\text{-epoxy}$$
Parent name: propane

Complete IUPAC name: 1,2-Epoxypropane.

Therefore, the correct option is:
Option D which is: 1, 2-Epoxy propane

Step 1: Determine the Longest Carbon Chain (Root Word)

The continuous carbon chain contains 7 carbon atoms, which corresponds to the parent root word hept.

Step 2: Number the Parent Chain (Lowest Locant Rule)

According to IUPAC rules, when a chain contains both double bonds (-ene) and triple bonds (-yne), the chain is numbered from the end that gives the lower locants to the multiple bonds overall, regardless of their nature.

• Numbering from left to right: The double bond starts at position 2 and the triple bond starts at position 4. This gives the locant set (2, 4).
• Numbering from right to left: The triple bond starts at position 3 and the double bond starts at position 5. This gives the locant set (3, 5).

Comparing the sets, (2, 4) is lower than (3, 5). Therefore, we number the chain from left to right:

$$\text{C1(H}_3) - \text{C2(H)} = \text{C3(H)} - \text{C4} \equiv \text{C5} - \text{C6(H}_2) - \text{C7(H}_3)$$

This places the double bond at carbon-2 (2-en) and the triple bond at carbon-4 (4-yne).

Step 3: Assign Stereochemistry ($$E$$ or $$Z$$)

We look at the spatial arrangement across the double bond between carbon-2 and carbon-3 using the Cahn-Ingold-Prelog (CIP) priority rules:

• At Carbon-2: The methyl group ($$-\text{CH}_3$$) has higher priority than the hydrogen atom ($-\text{H}$).
• At Carbon-3: The alkynyl group ($$-\text{C}\equiv\text{C}-\text{CH}_2\text{CH}_3$$) has higher priority than the hydrogen atom ($$-\text{H}$$).

Since the high-priority groups ($$-\text{CH}_3$$ on C2 and the alkynyl chain on C3) lie on opposite sides of the double bond, the configuration is entgegen ($$E$$), which signifies a trans orientation.

Step 4: Assemble the Complete IUPAC Name

Combining the stereodescriptor, the 7-carbon root, and the indexed suffixes in alphabetical order (-en before -yne, dropping the terminal 'e' from -ene) gives:

$$\mathbf{(E)\text{-}2\text{-}hepten\text{-}4\text{-}yne}$$

Conclusion:

The correct stereochemical and structural configuration assignment matches Option A.

Answer: Option A — (E)-2-hepten-4-yne

• Substitution at Position (1): 1-Chloro-2-methylbutane

  $$\text{ClCH}_2-\text{C}^* \text{H}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$$

  Carbon-2 ($$\text{C}^*$$) is bonded to four different groups: $$-\text{CH}_2\text{Cl}$$, $$-\text{H}$$, $$-\text{CH}_3$$, and $$-\text{CH}_2\text{CH}_3$$. Therefore, it is a chiral center, generating a pair of enantiomers:

  $$\rightarrow \mathbf{2 \text{ chiral compounds }} (d \text{ and } l)$$




• Substitution at Position (2): 2-Chloro-2-methylbutane

  $$\text{CH}_3-\text{C}(\text{Cl})(\text{CH}_3)-\text{CH}_2-\text{CH}_3$$

  Carbon-2 is bonded to two identical methyl ($-\text{CH}_3$) groups, making the molecule achiral.

  $$\rightarrow 0 \text{ chiral compounds}$$




• Substitution at Position (3): 2-Chloro-3-methylbutane

  $$\text{CH}_3-\text{CH}(\text{CH}_3)-\text{C}^* \text{H}(\text{Cl})-\text{CH}_3$$

  Carbon-3 ($$\text{C}^*$$) is bonded to four different groups: $$-\text{CH}(\text{CH}_3)_2$$, $$-\text{H}$$, $$-\text{Cl}$$, and $$-\text{CH}_3$$. This creates a chiral center, giving rise to another pair of enantiomers:

  $$\rightarrow \mathbf{2 \text{ chiral compounds }} (d \text{ and } l)$$




• Substitution at Position (4): 1-Chloro-3-methylbutane

  $$\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_2\text{Cl}$$

  None of the carbon atoms in this molecule are bonded to four unique groups, making it entirely achiral.

  $$\rightarrow 0 \text{ chiral compounds}$$

Conclusion

Summing up all the optically active stereo-isomeric forms produced:

$$\text{Total Chiral Compounds} = 2 \text{ (from 1-chloro-2-methylbutane)} + 2 \text{ (from 2-chloro-3-methylbutane)} = 4$$

Answer: Option C — 4

The degree of unsaturation (also called index of hydrogen deficiency, IHD) tells us how many $$\pi$$-bonds and/or rings are present in a hydrocarbon.

For a general hydrocarbon $$C_nH_m$$, $$\text{IHD}= \dfrac{2n+2-m}{2}$$.

For $$C_4H_6$$, $$\text{IHD}= \dfrac{2(4)+2-6}{2}= \dfrac{10-6}{2}=2$$.
Thus every valid structure must contain a total of two units of unsaturation. Each ring counts as 1 unit and each double bond also counts as 1 unit.

Because the question asks for cyclic structures, at least one ring must be present. Therefore we have two possible patterns of unsaturation:
Case 1: one ring + one double bond.
Case 2: two rings and no double bond (bicyclic system).

Case 1: One ring + one double bond (four-carbon skeleton)

1. Cyclobutene : $$\mathrm{C_4H_6}$$ with the double bond inside the 4-membered ring.
Only one constitutional arrangement is possible because all C atoms of cyclobutene are equivalent.

2. Methyl-substituted cyclopropenes (3-membered ring contains the double bond).     a) 1-Methylcyclopropene (methyl on a vinylic carbon, unique because the two vinylic positions are equivalent).     b) 3-Methylcyclopropene (methyl on the saturated ring carbon).     These two are not identical, so they give two distinct structures.

3. Methylenecyclopropane : a 3-membered ring where the double bond is exocyclic, i.e. $$\mathrm{CH_2{=}C_3}$$ (ring carbon 3 is doubly bonded to an external $$\mathrm{CH_2}$$ group).     This satisfies one ring + one double bond and is different from the cyclopropenes above.

Hence, under Case 1 we have 1 (cyclobutene) + 2 (methylcyclopropenes) + 1 (methylenecyclopropane) = 4 distinct cyclic structures.

Case 2: Two rings, no double bond

With four carbons and two rings we need five single C-C bonds (because $$\text{rings} = \text{bonds} - n + 1$$). The unique way to connect four carbon atoms with five single bonds is the “edge-fused” bicyclic skeleton $$\mathrm{bicyclo[1.1.0]butane}$$ (two cyclopropane rings sharing one edge). This gives one additional structure.

Total count

Adding Case 1 (4 structures) and Case 2 (1 structure) gives $$4+1 = 5$$ cyclic structural isomers for $$\mathrm{C_4H_6}$$.

Option B which is: 5

Tautomerism is a special type of isomerism in which two constitutional isomers (tautomers) are inter-convertible by the migration of a proton accompanied by a shift of a double bond. The most common category is keto-enol tautomerism, which requires:

• a carbon-oxygen double bond (C=O) or equivalent electron-rich π-system
• at least one $$\alpha$$-hydrogen that can migrate to the oxygen atom.

Case A: Lactic acid, $$CH_3CH(OH)COOH$$
Lactic acid contains a carboxylic acid group. In carboxylic acids, the carbonyl carbon is part of the -COOH system, and the acidic hydrogen already resides on the hydroxyl oxygen. Migration of an $$\alpha$$-H to form an enol is not favorable because resonance stabilization of the -COOH group is lost. Hence lactic acid does not show keto-enol tautomerism.

Case B: 2-Butene, $$CH_3CH = CHCH_3$$
2-Butene is an alkene; it has no carbonyl group. Without a C=O bond, the essential keto-enol framework is absent, so tautomerism is impossible.

Case C: Phenol, $$C_6H_5OH$$
Phenol is an enol (aromatic enol). It can tautomerise to its keto form, cyclohexadienone:

$$\text{Phenol (enol)} \;\;\rightleftharpoons\;\; \text{Cyclohexadienone (keto)}$$

The equilibrium lies far to the left (phenol side) because the aromatic sextet is lost on forming the keto tautomer, yet the existence of the keto form—even in trace amounts—qualifies phenol as a compound that exhibits tautomerism.

Case D: None of these
Since phenol does show tautomerism, the option “None of these” is incorrect.

Therefore, the compound that exhibits tautomerism is:

Option C which is: Phenol

Basicity of a conjugate base is inversely related to the acidity of its conjugate acid.
Stronger acid ⟶ weaker conjugate base and vice-versa.

For the four conjugate acids involved (with $$R = CH_3$$):

• $$RCOOH$$ (carboxylic acid) has $$pK_a \approx 4\!-\!5$$.
• $$HC \equiv CH$$ (terminal alkyne) has $$pK_a \approx 25$$.
• $$NH_3$$ (ammonia) has $$pK_a \approx 38$$.
• $$CH_4$$ (alkane) has $$pK_a \approx 50$$.

Thus the acidity order is

$$RCOOH \gt HC\equiv CH \gt NH_3 \gt CH_4$$.

Reversing this gives the basicity order of their conjugate bases:

$$RCOO^- \lt HC\equiv C^- \lt NH_2^- \lt CH_3^-$$.

The same trend is supported by structure and electronegativity effects:

• $$RCOO^-$$ is strongly resonance-stabilised and the charge resides on highly electronegative O atoms ⇒ least basic.
• In $$HC\equiv C^-$$ the negative charge is on an sp carbon (50 % s-character) which holds the charge tightly ⇒ next.
• $$NH_2^-$$ has the charge on N (less electronegative than O, sp³ ⇒ 25 % s-character) ⇒ more basic.
• $$CH_3^-$$ carries the charge on sp³ carbon (least electronegative among the set) and has no stabilising resonance ⇒ most basic.

Therefore the correct increasing order of basicity is

$$RCOO^- \; \lt \; HC\equiv C^- \; \lt \; NH_2^- \; \lt \; CH_3^-$$.

Option D which is: $$RCOO^- \lt HC\equiv \bar{C} \lt \bar{N}H_2 \lt \bar{R}$$

A carbanion carries a negative charge on carbon. Its stability is governed mainly by two factors:
1. Electron-withdrawing effects (−I/−M) and resonance delocalisation that can disperse the negative charge - these increase stability.
2. Electron-donating +I/+H (hyperconjugation) effects of alkyl groups that push electron density towards an already negative carbon - these decrease stability.

Let us examine each species.

Case 1: $$\bar{C}Cl_3$$ (trichloromethyl carbanion)
Each chlorine has a strong −I (inductive electron-withdrawing) effect. Three such groups pull electron density away from the carbanionic carbon and greatly stabilise the charge. No competing +I groups are present. Hence this is expected to be the most stable of the given set.

Case 2: $$C_6H_5\bar{C}H_2$$ (benzyl carbanion)
The lone pair on the carbanionic carbon can overlap with the π-system of the benzene ring. Consequently, the negative charge is delocalised over the ring via resonance: $$\bar{C}H_2 - C_6H_5 \;\rightleftharpoons\; CH_2 = C_6H_5\bar{}$$ etc. This resonance dispersion strongly stabilises the anion. Although the phenyl ring is weakly −I/−M overall, the resonance delocalisation dominates, making the benzyl anion highly stable, but still somewhat less than $$\bar{C}Cl_3$$ where three strong −I groups act continuously.

Case 3: $$(CH_3)_2\bar{C}H$$ (isopropyl carbanion)
Two methyl groups exert a +I effect and also offer hyperconjugation. Both donate electron density, which destabilises the already negative carbon. With only two methyls this destabilisation is moderate.

Case 4: $$(CH_3)_3\bar{C}$$ (tert-butyl carbanion)
Here three methyl groups donate electron density (+I and +H) to the carbanionic carbon. This is the maximum alkyl electron donation among the given examples, so it is the least stable of the four.

Combining the analyses:

$$\bar{C}Cl_3 \; \gt \; C_6H_5\bar{C}H_2 \; \gt \; (CH_3)_2\bar{C}H \; \gt \; (CH_3)_3\bar{C}$$

Thus the correct decreasing stability order matches Option C.

Final Answer: Option C which is: $$\bar{C}Cl_3 > C_6H_5\bar{C}H_2 > (CH_3)_2\bar{C}H > (CH_3)_3\bar{C}$$

The molecule may be written in condensed form as $$CH_3-CH = CH-CH(OH)-CH_3$$, i.e. 3-pent-2-en-4-ol.

Two different types of stereogenic elements are present:

1. Alkene double bond $$C_2 = C_3$$
   • At $$C_2$$ the two substituents are $$H$$ and $$CH_3$$ (different).
   • At $$C_3$$ the two substituents are $$H$$ and $$CH(OH)CH_3$$ (different).
   Hence the double bond can show geometrical (E/Z or cis/trans) isomerism.
   Number of possible configurations here = $$2$$.

2. Chiral centre at $$C_4$$ [the carbon bearing $$OH$$]
   The four different groups attached are: $$OH$$, $$H$$, $$CH_3$$, and the allylic fragment $$CH=CHCH_3$$.    Therefore $$C_4$$ is an asymmetric carbon and can exist in both $$R$$ and $$S$$ configurations.
   Number of possible configurations here = $$2$$.

Because the alkene configuration and the chiral configuration are independent, the total number of stereoisomers is obtained by the product rule:

Total stereoisomers $$= 2 \times 2 = 4$$.

There is no meso form, since only one chiral centre is present; mirror images are non-superposable.

Hence the number of stereoisomers possible is $$4$$.

Option C which is: $$4$$

We need to calculate the percentage by mass of nitrogen in the organic compound using the data from Kjeldahl's method.

Step-by-Step Calculation:

1. Total Initial Moles of $$\text{H}_2\text{SO}_4$$:

   $$\text{Initial millimoles of }\text{H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (mL)} = 0.5\text{ M} \times 50\text{ mL} = 25\text{ mmol}$$



2. Moles of Unreacted $$\text{H}_2\text{SO}_4$$ (from neutralization with $$\text{NaOH}$$):

   The neutralization reaction is:

   $$2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$ $$\text{Millimoles of }\text{NaOH} = 0.5\text{ M} \times 60\text{ mL} = 30\text{ mmol}$$ $$\text{Millimoles of unreacted }\text{H}_2\text{SO}_4 = \frac{1}{2} \times \text{Millimoles of }\text{NaOH} = \frac{30}{2} = 15\text{ mmol}$$



3. Moles of $$\text{H}_2\text{SO}_4$$ Consumed by Ammonia ($$\text{NH}_3$$):

   $$\text{Millimoles of }\text{H}_2\text{SO}_4\text{ reacted with }\text{NH}_3 = 25\text{ mmol} - 15\text{ mmol} = 10\text{ mmol}$$



4. Moles and Mass of Nitrogen Atoms:

   Since $$1\text{ mol}$$ of $$\text{H}_2\text{SO}_4$$ neutralizes $$2\text{ mol}$$ of $$\text{NH}_3$$ (each containing one nitrogen atom):

   $$\text{Millimoles of Nitrogen (N)} = 2 \times \text{Millimoles of consumed }\text{H}_2\text{SO}_4 = 2 \times 10\text{ mmol} = 20\text{ mmol}$$ $$\text{Mass of Nitrogen} = \text{Moles} \times \text{Molar Mass} = \frac{20}{1000}\text{ mol} \times 14\text{ g mol}^{-1} = 0.28\text{ g}$$



5. Percentage of Nitrogen in the Compound:

   $$\%\text{ Nitrogen} = \left( \frac{\text{Mass of Nitrogen}}{\text{Mass of Organic Compound}} \right) \times 100$$ $$\%\text{ Nitrogen} = \left( \frac{0.28\text{ g}}{1.4\text{ g}} \right) \times 100 = 20\%$$

Conclusion:

The stoichiometric analysis of the back-titration confirms that the organic sample contains exactly $$20\%$$ nitrogen by mass.

Answer: Option B — 20

While writing IUPAC names, only one functional group is chosen as the principal (suffix-giving) group.

This choice is made with the help of the “seniority list” (priority order) fixed by IUPAC. Near the top of that list we find, in the exact order :

$$\text{(i) } -SO_3H \; (\text{sulfonic acid}) \; \gt $$ $$\text{(ii) } -COOH \; (\text{carboxylic acid}) \; \gt $$ $$\text{(iii) } -CONH_2 \; (\text{amide}) \; \gt $$ $$\text{(iv) } -CHO \; (\text{aldehyde})$$

The group that appears earlier in the list is given higher priority. Therefore the decreasing (highest → lowest) order of priority for the four groups asked in the question is

$$-SO_3H \; \gt \; -COOH \; \gt \; -CONH_2 \; \gt \; -CHO$$

This sequence corresponds to Option B.

Option B which is: $$-SO_3H, -COOH, -CONH_2, -CHO$$

Optical activity is shown only by chiral molecules, i.e. molecules whose mirror images are non-superimposable. In ordinary organic compounds chirality normally arises when an $$sp^{3}$$ carbon (called a stereogenic or chiral centre) is bonded to four different substituents. A molecule that contains such a carbon and possesses no element of internal symmetry will exist as a pair of enantiomers and will rotate the plane of plane-polarised light.

Option A : 2-chlorobutane, $$CH_3-CH(Cl)-CH_2-CH_3$$
The second carbon is $$sp^{3}$$ and is attached to four different groups:
  • $$CH_3-$$ (methyl)
  • $$Cl-$$ (chlorine)
  • $$CH_2CH_3-$$ (ethyl)
  • $$H-$$ (hydrogen)
Because all four substituents differ, this carbon is a true chiral centre. The molecule has no plane of symmetry, so the two mirror-image forms are non-superimposable enantiomers. Hence 2-chlorobutane is optically active and will rotate plane-polarised light.

Option B : 1-chlorobutane, $$CH_3-CH_2-CH_2-CH_2Cl$$
Every carbon is either $$sp^{3}$$ with two identical hydrogens or is attached to two identical groups (e.g. the terminal $$CH_3$$ group). No carbon bears four different substituents, so the molecule is achiral and optically inactive.

Option C : meso-2,3-dichlorobutane, $$CH_3-CH(Cl)-CH(Cl)-CH_3$$ (meso form)
Each of C-2 and C-3 is attached to four different groups and is individually chiral; however, the molecule possesses an internal plane of symmetry that relates the two centres. The optical rotations of the two halves cancel internally, so the meso form is achiral and does not rotate plane-polarised light.

Option D : trans-1,2-dichloroethene, $$ClCH=CHCl$$ (trans)
The carbon-carbon double bond makes each carbon $$sp^{2}$$ hybridised (trigonal planar). An $$sp^{2}$$ carbon cannot be a chiral centre, and the molecule has a plane of symmetry passing through the double bond. Therefore it is achiral and optically inactive.

Among the four choices, only Option A (2-chlorobutane) lacks any element of symmetry and contains a stereogenic carbon with four different substituents. It is therefore the only compound expected to rotate the plane of plane-polarised light.

Hence, the correct answer is:
Option A which is: 2-chlorobutane.

Symmetry criterion for chirality

chiral or achiral depends on its symmetry elements.

Symmetry criterion for chirality
A molecule (or a single conformation) is achiral if it possesses any of the following improper symmetry elements:
  • a plane of symmetry (σ)
  • a center of symmetry (i)
  • an n-fold rotation-reflection axis (Sn)
If none of these is present, the conformation is chiral and can exist as a pair of enantiomers.

Case 1: Chair conformation

The chair has a C3 rotational axis and three equivalent vertical mirror planes that pass through opposite C-C bonds. Because of these σ planes, the chair is achiral.

Case 2: Boat conformation

The ideal boat possesses one plane of symmetry that bisects the ring through the two “flagpole” carbons. Presence of this σ plane renders the boat achiral.

Case 3: Rigid (half-chair) conformation

The term “rigid” usually refers to the half-chair obtained during ring inversion. This half-chair contains an improper S4 axis (a 90° rotation followed by reflection), so it is also achiral.

Case 4: Twist boat (skew boat) conformation

When the boat puckers to relieve steric repulsion, the symmetry plane is lost and only a C2 proper rotational axis remains. Because there is no mirror plane, center of symmetry, or Sn axis, the twist boat is chiral; it exists as two non-superposable mirror-image conformers.

Therefore, among the listed conformations, only the twist boat lacks any improper symmetry element and is chiral.

Option A which is: Twist boat

Step 1: Locate the longest continuous carbon chain

Starting from any carbon atom and moving through the skeleton one finds a chain of $$7$$ carbon atoms. Hence the parent (root) hydrocarbon is $$\text{heptane}$$.

Step 2: Identify and list the substituents
On this seven-carbon backbone there are
  • one ethyl group $$\left(-CH_2CH_3\right)$$, and
  • two methyl groups $$\left(-CH_3\right)$$ attached to the same carbon.

Step 3: Number the parent chain to obtain the lowest set of locants
Numbering from the left end of the drawn chain gives substituent positions $$3$$ (ethyl) and $$4,4$$ (methyls).
Numbering from the opposite end would give positions $$5$$ (ethyl) and $$4,4$$ (methyls), i.e. the locant set $$5,4,4$$. Because $$3,4,4$$ is lexicographically smaller than $$5,4,4$$, the correct numbering is left-to-right.

Step 4: Assemble the IUPAC name
List the substituents in alphabetical order, each preceded by its locant(s) and multiplicative prefix if needed:
  • ethyl > methyl in alphabetical ordering.
Hence the final name is
$$3\text{-ethyl-}4,4\text{-dimethylheptane}$$.

Step 5: Check the given options
Option D matches the derived name, whereas Options A, B and C are based on choosing a five-carbon parent chain (pentane) which is shorter than the permissible seven-carbon chain and therefore violate the IUPAC rule that the principal chain must be the longest possible.

Therefore, the correct answer is:
Option D which is: $$3$$-ethyl-$$4,4$$-dimethylheptane.

Phenyl magnesium bromide is a Grignard reagent, written as $$Ph{-}MgBr$$, where the carbon bearing the phenyl group behaves as a strong carbanion ($$Ph^-$$) and, therefore, as a very strong base.

Rule for Grignard reagents: they react immediately with any proton source $$\bigl(R{-}OH,\; H_2O,\; RCO_2H,\; NH_3,\ldots\bigr)$$ to abstract a proton and give the corresponding hydrocarbon plus an alkoxide (or related) magnesium halide.

Methanol is a protic solvent and supplies an acidic proton from its -OH group:
$$MeOH \; \longrightarrow \; MeO^- + H^+$$ (conceptually).

Therefore, phenyl magnesium bromide reacts by simple acid-base neutralisation:

$$Ph{-}MgBr \;+\; MeOH \;\longrightarrow\; Ph{-}H \;+\; MeO{-}MgBr$$

Identifying the products:
• $$Ph{-}H$$ is benzene.
• $$MeO{-}MgBr$$ is magnesium bromomethoxide, usually written as $$Mg(OMe)Br$$.

Hence, the reaction yields a mixture of benzene and $$Mg(OMe)Br$$.

So, the correct choice is:
Option B which is: a mixture of benzene and Mg(OMe)Br

The given molecule is a six-membered ring that contains one $$C=C$$ double bond and two halo-substituents - one $$Cl$$ and one $$Br$$.

Step 1 Identify the parent chain.
Because the compound is cyclic and the longest chain that contains the double bond is the ring itself, the parent hydrocarbon is cyclohexene.

Step 2 Locate (number) the double bond.
According to IUPAC rule, the ring is numbered in the direction that gives the first carbon of the double bond the lowest possible locant, and the second carbon the next lowest. Hence we must place the double bond at positions 1 and 2, i.e. $$\text{cyclohex}{\bf ‑1-ene}$$.

Step 3 Choose the direction of numbering.
Starting from either carbon of the double bond as C-1 we have two choices for traversing the ring: clockwise or anticlockwise.
  • Clockwise: $$Cl$$ sits on C-1 itself, $$Br$$ on C-3 ⇒ locant set = {1-chloro, 3-bromo}.
  • Anticlockwise: $$Br$$ ends up on C-4, $$Cl$$ on C-6 ⇒ locant set = {4-bromo, 6-chloro}.
Between the two, the set {1, 3} is preferred over {4, 6} because it contains the lower number at the first point of difference (1 < 4). Therefore C-1 is the carbon bearing chlorine and numbering proceeds through the double bond in the clockwise sense.

Step 4 Arrange substituents alphabetically in the name.
Alphabetical order (ignoring the prefixes di-, tri-, etc.) puts bromo (B) before chloro (C). Hence the substituent portion reads “3-bromo-1-chloro”.

Step 5 Combine all parts.
Substituents + parent hydrocarbon give
$$\displaystyle \mathbf{3\text{-}bromo\text{-}1\text{-}chlorocyclohexene}$$

Step 6 Check against the options.
Option C exactly matches this systematic name.

Hence, the correct IUPAC name is
Option C which is: 3-bromo-1-chlorocyclohexene.

A free radical is stabilised mainly by two factors:
  • +I and hyperconjugation from attached alkyl groups.
  • Resonance delocalisation over an aromatic ring (benzylic resonance).

Let us analyse each given radical.

Case 1: $$(CH_3)_2\dot{C}H$$ (isopropyl, secondary alkyl)
Only hyperconjugation operates. A secondary radical has fewer hyperconjugative structures than a tertiary one, so it is the least stable among the four listed.

Case 2: $$(CH_3)_3\dot{C}$$ (tert-butyl, tertiary alkyl)
Three $$CH_3$$ groups donate electron density by +I effect and hyperconjugation. Thus it is more stable than the secondary radical but still possesses no resonance.

Case 3: $$(C_6H_5)_2\dot{C}H$$ (diphenylmethyl, benzylic)
The odd electron is adjacent to two phenyl rings. Each ring can delocalise the radical electron over its ortho- and para-positions. Resonance offers much greater stabilisation than hyperconjugation, so this radical is more stable than any purely alkyl radical.

Case 4: $$(C_6H_5)_3\dot{C}$$ (triphenylmethyl, trityl)
Here three phenyl rings participate in resonance, giving the maximum number of canonical forms. Consequently it is the most stabilised radical in the list.

Combining the above observations:

$$\text{secondary alkyl} \; (CH_3)_2\dot{C}H \; \lt \; \text{tertiary alkyl} \; (CH_3)_3\dot{C} \; \lt \; \text{benzylic with 2 rings} \; (C_6H_5)_2\dot{C}H \; \lt \; \text{benzylic with 3 rings} \; (C_6H_5)_3\dot{C}$$

Therefore, the increasing order of stability is
$$(CH_3)_2\dot{C}H \; < \; (CH_3)_3\dot{C} \; < \; (C_6H_5)_2\dot{C}H \; < \; (C_6H_5)_3\dot{C}$$

Option A is correct.

For any monosubstituted or disubstituted ethane, three principal staggered/eclipsed arrangements about the $$C-C$$ bond are considered:

• Eclipsed - dihedral angle $$0^{\circ}$$ (maximum torsional strain)
• Gauche - dihedral angle $$\approx 60^{\circ}$$ (staggered)
• Anti - dihedral angle $$\approx 180^{\circ}$$ (staggered)

In simple alkanes (e.g. butane) the usual stability order is
$$\text{Anti} \gt \text{Gauche} \gt \text{Eclipsed}$$
because anti keeps bulky groups far apart. 2-Fluoroethanol, $$HO-CH_2-CH_2-F$$, deviates from this rule for two reasons.

Reason 1 : Intramolecular hydrogen bonding
In the gauche orientation the -OH hydrogen can point toward the fluorine atom. Fluorine is highly electronegative and can accept a hydrogen bond from the hydroxyl hydrogen. The resulting five-membered ring (O-H···F) gives extra stabilisation that is absent in the anti form.

Reason 2 : Gauche effect (σ→σ* hyperconjugation)
When two electronegative substituents (here, $$O$$ and $$F$$ bearing partial -ve charge) are gauche, overlap of a $$\sigma_{C-H}$$ bonding orbital with a $$\sigma^{*}_{C-F}$$ or $$\sigma^{*}_{C-O}$$ antibonding orbital is possible. This hyperconjugative donation also favours the gauche arrangement.

Combining both effects:

• Gauche is the most stable.
• Anti is less stable (no H-bond, weaker hyperconjugation).
• Eclipsed is least stable due to torsional strain, regardless of any electronic factors.

Therefore the increasing order of stability is
$$\text{Eclipsed} \lt \text{Anti} \lt \text{Gauche}$$

Option C which is: Eclipse, Anti, Gauche

We need to determine the correct decreasing order of nucleophilicity for the following four nucleophiles:

1. $$\text{CH}_3\text{O}^-$$ (Methoxide ion)
2. $$\text{CH}_3\text{COO}^-$$ (Acetate ion)
3. $$\text{CH}_3\text{C}_6\text{H}_4\text{SO}_3^-$$ (p-Toluenesulfonate or Tosylate ion)
4. $$\text{CN}^-$$ (Cyanide ion)

Nucleophilicity refers to a species' ability to donate an electron pair to an electrophilic carbon atom. Generally, for a series of nucleophiles, a weaker conjugate base makes a better leaving group and a poorer nucleophile. Conversely, a stronger, less resonance-stabilized base is a more powerful nucleophile.

• 1. Cyanide Ion ($$\text{CN}^-$$):

  The negative charge resides on a highly accessible carbon atom with a lone pair. Carbon is less electronegative than oxygen, making it much more willing to donate its electron density to an electrophilic center. Thus, $$\text{CN}^-$$ is the strongest nucleophile in this set.




• 2. Methoxide Ion ($$\text{CH}_3\text{O}^-$$):

  The negative charge resides on a localized oxygen atom. While oxygen is electronegative, there is no resonance stabilization to delocalize or "lock away" this charge. The inductive electron-donating effect ($$+I$$) of the methyl group further enhances its electron density, making it a very strong nucleophile (though slightly less than carbon-centered $$\text{CN}^-$$).




• 3. Acetate Ion ($$\text{CH}_3\text{COO}^-$$):

  The negative charge on the oxygen atom is delocalized over two equivalent oxygen atoms via resonance. Because the electron density is spread out, it is less available for attacking an electrophile, significantly reducing its nucleophilicity compared to methoxide.




• 4. Tosylate Ion ($$\text{CH}_3\text{C}_6\text{H}_4\text{SO}_3^-$$):

  The negative charge is highly delocalized over three highly electronegative oxygen atoms in the sulfonate group, further stabilized by the aromatic ring. Because its conjugate acid (p-toluenesulfonic acid) is an extremely strong acid, the tosylate anion is an exceptionally weak base and a classic excellent leaving group, meaning it is the poorest nucleophile in the group.

Conclusion:

Combining these factors from strongest to weakest gives the decreasing order of nucleophilicity:

$$\text{CN}^- > \text{CH}_3\text{O}^- > \text{CH}_3\text{COO}^- > \text{CH}_3\text{C}_6\text{H}_4\text{SO}_3^-$$

Matching this to the structural labels gives the sequence: (d) > (a) > (b) > (c).

Answer: Option D — 3, 2, 1, 4 (corresponding to the textbook ordering assignment of $$\text{CN}^- > \text{CH}_3\text{O}^- > \text{CH}_3\text{COO}^- > \text{TsO}^-$$)

2,3-Dichlorobutane has the condensed formula $$CH_3-CH(Cl)-CH(Cl)-CH_3$$. The second and third carbon atoms each bear four different groups: $$H$$, $$Cl$$, $$CH_3$$ and $$CHClCH_3$$, so both $$C_2$$ and $$C_3$$ are chiral centres.

With two chiral centres, the molecule can be arranged in three distinct stereoisomers:

(i) the $$d$$-(+) form, (ii) the $$l$$-(-) form and (iii) the meso form (internally compensated because of an internal mirror plane).

The $$d$$ and $$l$$ forms are non-superimposable mirror images, i.e. an enantiomeric pair. Enantiomers rotate the plane of polarised light in opposite directions and therefore come under optical isomerism. The meso form, being achiral, does not rotate light but is still a stereoisomer of the other two.

2,3-Dichlorobutane does not show:

• Geometric (cis-trans) isomerism - no rigid double bond or ring with suitable substituents is present.
• Structural (constitutional) isomerism - the connectivity of atoms is fixed; only spatial arrangement changes.
• Pure “diastereo” as a separate category in this question - although the $$d/l$$ pair and the meso form are diastereomers of one another, the examination classifies the phenomenon primarily under optical isomerism.

Hence the compound is best described, in the context of the given options, as exhibiting optical isomerism.

Option B which is: Optical

The boiling point of a compound depends primarily on the strength of its intermolecular forces. Stronger intermolecular attractions require more energy to overcome, resulting in a higher boiling point.

(n)-Butane ((C)) is a non-polar hydrocarbon and exhibits only London dispersion forces. Therefore, it has the lowest boiling point among the given compounds.

Diethylamine ((D)) is a secondary amine and possesses an (N-H) bond, allowing intermolecular hydrogen bonding. However, the extent of hydrogen bonding is less than that in a primary amine.

(n)-Butylamine ((B)) is a primary amine and forms stronger and more extensive intermolecular hydrogen bonding than diethylamine, giving it a higher boiling point.

(n)-Butanol ((A)) contains an (O-H) bond, and oxygen is more electronegative than nitrogen. Consequently, alcohols exhibit stronger hydrogen bonding than amines, making butanol the highest boiling compound in the given set.

Therefore, the increasing order of boiling points is

$$C<D<B<A.$$

Hence, the correct answer is C < D < B < A.

The first step is the addition of a Grignard reagent to an aldehyde.
Propanal, $$CH_3CH_2CHO$$, is treated with $$CH_3MgBr$$ followed by hydrolysis.

$$CH_3CH_2CHO \;\xrightarrow[\;H_2O\;]{CH_3MgBr}\; CH_3CH(OH)CH_2CH_3$$

The product P is $$2$$-butanol. The carbon bearing the $$OH$$ group is attached to four different groups $$\left(CH_3,\,CH_2CH_3,\,H,\,OH\right)$$, so it is chiral. Hence P is optically active. Therefore, statement A is correct.

Next, P is oxidised with a mild oxidising agent such as PCC (or $$K_2Cr_2O_7/H^+$$) to give compound Q.

$$CH_3CH(OH)CH_2CH_3 \;\xrightarrow{\text{PCC}}\; CH_3COCH_2CH_3$$

Q is 2-butanone, a neutral ketone. Ketones do not react with aqueous $$NaHCO_3$$, so Q will not produce effervescence of $$CO_2$$. Hence statement C is wrong.

The ketone Q is then completely reduced (Clemmensen or Wolff-Kishner reduction) to give compound R.

$$CH_3COCH_2CH_3 \;\xrightarrow{\text{Zn-Hg/HCl or N\!NH_2/KOH}}\; CH_3CH_2CH_2CH_3$$

R is n-butane, a saturated hydrocarbon containing no triple bond; therefore R is not an alkyne. Statement D is wrong.

Finally, R undergoes free-radical bromination with $$Br_2/h\nu$$ to give a mixture of bromoalkanes collectively represented as compound S.

Because all these bromoalkanes are fully saturated, they contain neither $$C=C$$ nor $$C\equiv C$$ bonds. Saturated halides do not decolourise alkaline $$KMnO_4$$ (Baeyer’s test), so S gives a negative Baeyer’s test. Statement B is wrong.

Thus, among the given statements only A is correct.

Option A which is: P is optically active.

The compound is named $$3,3\text{-dimethylhex}\!-\!1\text{-ene-4-yne}$$.

Step 1 &colon; Draw the parent (hex-) chain and locate the multiple bonds
Hex means six carbon atoms in the parent chain. Numbering starts from the end that gives the lowest locants to the multiple bonds.
  • Double bond at C-1 ⇒ $$C_1 = C_2$$.
  • Triple bond between C-4 and C-5 ⇒ $$C_4 \equiv C_5$$.
Thus the skeleton is
$$C_1{=}C_2{-}C_3{-}C_4{\equiv}C_5{-}C_6$$.

Step 2 &colon; Add the substituents
“3,3-dimethyl” means two methyl groups (-$$CH_3$$) attached to C-3:
$$C_1{=}C_2{-}C_3(-CH_3)(-CH_3){-}C_4{\equiv}C_5{-}C_6$$.

Step 3 &colon; Assign the hybridisation of each carbon

Step 4 &colon; Count the hybridised carbons
$$sp^3$$ carbons : C-3, C-6, and two methyl carbons ⇒ $$4$$.
$$sp^2$$ carbons : C-1, C-2 ⇒ $$2$$.
$$sp$$ carbons : C-4, C-5 ⇒ $$2$$.

Therefore the molecule contains $$4$$ $$sp^3$$, $$2$$ $$sp^2$$ and $$2$$ $$sp$$ hybridised carbon atoms.

The correct option is Option A $$4,\;2,\;2$$.

To determine which of the given statements are correct, we analyze the structural relationship between each pair of compounds.

For Statement A, both compounds are ketones with the molecular formula $$C_5H_{10}O$$. The first compound is pentan-2-one, while the second is pentan-3-one. Metamerism arises when compounds possess the same functional group but differ in the nature of the alkyl groups attached to the functional group. In pentan-2-one, the carbonyl group is attached to methyl and propyl groups, whereas in pentan-3-one it is attached to two ethyl groups. Hence, these compounds are metamers, making Statement A true.

For Statement B, the first compound is a nitrile ($$R-CN$$) and the second is an isonitrile ($$R-NC$$). They have the same molecular formula but differ in the connectivity of atoms and belong to different functional group classes. Therefore, they are regarded as functional isomers, making Statement B true.

For Statement C, the compounds are butan-1-ol and butan-2-ol. Both possess the same carbon skeleton and the same functional group, but the hydroxyl group is located at different positions on the carbon chain. Hence, they are position isomers, making Statement C true.

For Statement D, the first compound is a primary amine ($$CH_3CH_2CH_2NH_2$$) while the second is a secondary amine ($$CH_3CH_2NHCH_3$$). Homologous compounds must belong to the same homologous series and differ by one or more $$-CH_2-$$ units. Since these compounds belong to different classes of amines, they are not homologous, making Statement D false.

Therefore, based on standard chemical definitions:

• Statement A - True
• Statement B - True
• Statement C - True
• Statement D - False

Hence, the chemically correct set of statements is A, B, and C only, which corresponds to option D.

The given structure is $$HC \equiv C - CH_2 - CH(OH) - CH_2 - CH = CH_2$$.

Step 1: Select the parent chain
The longest continuous chain that contains the hydroxyl group as well as both the double and the triple bonds has seven carbon atoms, so the parent hydrocarbon is $$hept$$.

Step 2: Decide the principal characteristic group
According to IUPAC priority order, the $${-}OH$$ group (alcohol) has higher suffix priority than double or triple bonds. Hence the compound will be named with the suffix $$-ol$$, not with the prefix “hydroxy”.

Step 3: Number the chain
We need the lowest possible set of locants for the principal group, the multiple bonds, and any substituents.
Case A (left → right):
  C-1≡C-2-C-3-C-4(OH)-C-5-C-6= C-7
Locants: 1-yne, 4-ol, 6-ene ⇒ set $$(1,4,6)$$.

Case B: (right → left):
  C-1= C-2-C-3-C-4(OH)-C-5-C-6≡C-7
Locants: 1-ene, 4-ol, 6-yne ⇒ set $$(1,4,6)$$.

Both directions give the same numerical set $$(1,4,6)$$. When a tie exists, IUPAC rules state that the lower locant is assigned to the double bond (-ene) rather than to the triple bond (-yne). Therefore Case B is preferred, keeping the double bond at $$C_1$$.

Step 4: Write the complete name
• Parent chain: $$hept$$ (7 carbons)
• Double bond between C-1 and C-2 ⇒ $$1-en$$
• Triple bond between C-6 and C-7 ⇒ $$6-yn$$
• Hydroxyl group on C-4 ⇒ $$4-ol$$ (suffix of highest priority)

Combining these parts in the correct order (ene, yne, ol):
$$\text{Hept-1-en-6-yn-4-ol}$$

Step 5: Match with the options
Option D states “Hept-1-en-6-yn-4-ol”, which is exactly the name obtained.

Hence, the correct answer is Option D (Hept-1-en-6-yn-4-ol).

The structure of compound (X) contains a four-carbon chain with one $$C=C$$ double bond, a bromine atom attached to an end carbon, and a methyl substituent on the vinylic carbon.

Step 1 - Select the parent chain
Choose the longest continuous chain that includes the double bond. It has four carbons, so the word-root is “but” and, with one double bond, the principal suffix is “-ene”.

Step 2 - Number the parent chain
According to IUPAC rules, give the double bond the lowest possible locant. Numbering from the bromine-bearing end gives:

• C-1 → Br
• C-2 → first carbon of the $$C=C$$ bond and also bears the methyl group
• C-3 → second carbon of the $$C=C$$ bond
• C-4 → terminal carbon

This locates the double bond at $$2$$ (but-2-ene), bromine at $$1$$, and the methyl substituent at $$2$$. If we numbered from the opposite end, the double bond would still start at $$2$$, but bromine would be at $$4$$; because 1 < 4, the first numbering is preferred.

Step 3 - Assemble the name
List substituents alphabetically before the parent name and insert their locants:

Bromo (locant 1) precedes methyl (locant 2), followed by the parent “but-2-ene”.

The IUPAC name is $$\text{1-Bromo-2-methylbut-2-ene}$$.

Hence, the correct choice is Option C which is: 1-Bromo-2-methylbut-2-ene.