JEE Basic Principles of Organic Chemistry Questions

The basicity of these molecules depends on the availability of the lone pair of electrons on the nitrogen atom. A more localized lone pair is more readily available for protonation and therefore results in higher basicity.

For Molecule (R):

Molecule (R) is a bridged bicyclic lactam in which the nitrogen atom is located at the bridgehead position.

According to Bredt's rule, a double bond cannot be formed at the bridgehead of a small bicyclic system because such a structure would introduce severe ring strain.

As a result, the usual amide resonance involving delocalization of the nitrogen lone pair into the carbonyl group cannot occur.

Since the lone pair remains localized on the nitrogen atom, it is readily available for protonation. Consequently, molecule (R) behaves similarly to an aliphatic tertiary amine and is the most basic among the three compounds.

For Molecule (P):

This is a saturated lactam in which the nitrogen lone pair undergoes normal amide resonance with the adjacent carbonyl group.

The resonance may be represented as

$$N-C=O \longleftrightarrow N^+=C-O^-.$$

Because the lone pair is significantly delocalized into the carbonyl system, its availability for protonation is greatly reduced, making molecule (P) weakly basic.

For Molecule (Q):

Molecule (Q) is an $$\alpha,\beta$$-unsaturated amide containing a carbon-carbon double bond conjugated with the carbonyl group.

The $$\pi$$ electrons of the alkene also participate in conjugation with the carbonyl group:

$$C=C-C=O \longleftrightarrow {}^+C-C=C-O^-.$$

This additional conjugation supplies electron density to the carbonyl oxygen, reducing the extent to which the nitrogen lone pair must delocalize into the carbonyl group.

Consequently, the nitrogen atom retains a greater amount of localized electron density than in molecule (P), making molecule (Q) more basic than molecule (P).

Therefore,

• (R) is the most basic because its lone pair remains localized.
• (Q) is more basic than (P) because cross-conjugation decreases the extent of amide resonance.

Hence, the correct order of basicity is

$$R > Q > P.$$

For every species, identify its general electronic feature and hybridisation before matching with the statements in LIST-II.

Case A: Carbocation
A carbocation carries a positive charge on carbon. The carbon is $$sp^2$$ hybridised and possesses an empty $$p$$-orbital that can accept a pair of electrons.
Hence, A corresponds to statement III.

Case B: Carbon-centred free radical
A carbon free radical has one unpaired electron on carbon. Depending on the surrounding atoms it may be $$sp^2$$ or $$sp^3$$ hybridised, but the defining feature is the single unpaired electron.
Hence, B corresponds to statement IV.

Case C: Nucleophile
A nucleophile is a species that donates (supplies) an electron pair to an electron-deficient centre in a reaction.
Hence, C corresponds to statement I.

Case D: Electrophile
An electrophile is electron-deficient; it seeks (receives) an electron pair from a nucleophile.
Hence, D corresponds to statement II.

Collecting the matches:
A - III, B - IV, C - I, D - II.

This is exactly the pairing shown in Option D. Therefore the correct answer is Option D.

Ethyl acetate is an organic solvent that is immiscible with water, so two separate layers form in a separating funnel.
The upper layer (lower density) is ethyl acetate, while the lower layer (higher density) is the aqueous solution of $$5 \text{ M } NaOH$$.

1. Behaviour of benzoic acid:
Benzoic acid is a carboxylic acid. In the presence of the strong base $$NaOH$$ it undergoes neutralisation: $$C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O$$.
Sodium benzoate $$\left(C_6H_5COONa\right)$$ is an ionic salt, highly soluble in water and almost insoluble in ethyl acetate. Hence benzoic acid transfers completely to the aqueous layer as its salt.

2. Behaviour of aniline:
Aniline is a weak base. In strongly basic medium ($$5 \text{ M } NaOH$$) it remains in the free-base (neutral) form $$C_6H_5NH_2$$ because there is no proton source to convert it to the water-soluble ammonium ion $$C_6H_5NH_3^+$$. Neutral aniline is only sparingly soluble in water but is quite soluble in an organic solvent such as ethyl acetate. Therefore aniline stays in the ethyl acetate layer.

3. Behaviour of chlorobenzene:
Chlorobenzene is a non-polar, neutral molecule. It does not react with $$NaOH$$ under these conditions and is insoluble in water. Thus chlorobenzene also remains in the ethyl acetate layer.

Final distribution:
• Aqueous (lower) layer : sodium benzoate (from benzoic acid) and excess $$NaOH$$.
• Organic (upper ethyl acetate) layer : chlorobenzene + aniline.

Hence the ethyl acetate layer contains chlorobenzene and aniline.
Correct Option D: chlorobenzene and aniline.

To determine which pairs have the first carbocation more stable than the second, we must compare the factors affecting carbocation stability in each case.

For Pair (A):

The first carbocation has its positive charge adjacent to a methoxy group $$(-OMe)$$. The oxygen atom possesses lone pairs that can be donated into the empty $$p$$-orbital of the carbocation through resonance, exhibiting a strong $$+M$$ effect. This resonance generates an oxonium ion structure in which the carbocation attains a complete octet, providing significant stabilization.

The second carbocation is a tertiary carbocation stabilized mainly by the $$+I$$ effect and hyperconjugation from the adjacent alkyl groups.

Since resonance stabilization that completes the octet is much stronger than hyperconjugation, the first carbocation is more stable than the second.

Hence, Pair (A) satisfies the given condition.

For Pair (B):

In the second carbocation, the nitro group $$(-NO_2)$$ is present at the para position and remains coplanar with the benzene ring. Therefore, it exerts its full $$-M$$ (resonance withdrawing) effect, strongly destabilizing the positive charge.

In the first carbocation, the nitro group is at the ortho position, while the carbocation-bearing carbon is attached to a bulky cyclohexyl group. Steric repulsion between these groups forces the nitro group out of the plane of the aromatic ring. As a result, resonance between the nitro group and the ring is inhibited, a phenomenon known as Steric Inhibition of Resonance (SIR).

Because the $$-M$$ effect is largely suppressed, only the weaker $$-I$$ effect operates, making the first carbocation comparatively more stable than the second.

Hence, Pair (B) also satisfies the given condition.

For Pair (C):

The first structure is an ordinary primary carbocation attached to a saturated cyclohexane ring.

The second structure is a primary carbocation adjacent to a cyclohexene system. The nearby double bond can stabilize the positive charge through homoallylic participation involving the $$\pi$$ electrons, giving additional stabilization.

Therefore, the second carbocation is more stable than the first, so this pair does not satisfy the required condition.

For Pair (D):

The first carbocation is a tert-butyl carbocation stabilized by hyperconjugation.

The second carbocation has a methoxy group adjacent to the positively charged carbon. The oxygen atom donates its lone pair through resonance ($$+M$$ effect), producing an octet-complete resonance structure and greatly stabilizing the carbocation.

Consequently, the second carbocation is much more stable than the first.

Therefore, this pair also does not satisfy the required condition.

Hence, the pairs in which the first carbocation is more stable than the second are (A) and (B) only.

Therefore, the correct answer is (A) & (B) only.

A nucleophile is a species that can donate a pair of electrons to an electrophilic centre. We examine each species in turn.

$$NH_3$$ (ammonia): Nitrogen carries a lone pair of electrons, making it a well-known nucleophile. ✓

$$PhSH$$ (thiophenol): Sulfur has two lone pairs and is large and polarisable, making thiophenol a strong nucleophile. ✓

$$(H_3C)_2S$$ (dimethyl sulfide): The sulfur atom has two lone pairs readily available for donation. ✓

$$H_2C{=}CH_2$$ (ethylene): While the $$\pi$$-bond can participate in electrophilic addition, ethylene lacks a lone pair on a specific atom. In the JEE convention it is not counted as a nucleophile. ✗

$${}^{\ominus}OH$$ (hydroxide ion): Carries a full negative charge and lone pairs on oxygen. One of the strongest common nucleophiles. ✓

$$H_3O^{\oplus}$$ (hydronium ion): Positively charged and electron-deficient; it is an acid/electrophile, not a nucleophile. ✗

$$(CH_3)_2CO$$ (acetone): Although the carbonyl oxygen has lone pairs, acetone is classified as an electrophile at its carbonyl carbon and is not counted as a nucleophile in this context. ✗

$$:C{=}NCH_3$$ (methyl isocyanide, $$CH_3NC$$): In an isocyanide the terminal carbon bears a lone pair, making it an excellent nucleophile (this property is exploited in the carbylamine reaction). ✓

The five nucleophiles are therefore: $$NH_3,\; PhSH,\; (H_3C)_2S,\; {}^{\ominus}OH,\; CH_3NC$$.

Hence, the correct answer is Option D (5).

The compound is named $$3,3\text{-dimethylhex}\!-\!1\text{-ene-4-yne}$$.

Step 1 : 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 : 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 : Assign the hybridisation of each carbon

Step 4 : 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 $$4,\;2,\;2$$.

We need to identify the correct statements about directing groups and activating/deactivating groups in electrophilic aromatic substitution.

Statement (A): $$-OCH_3$$ and $$-NHCOCH_3$$ are activating groups.

$$-OCH_3$$ (methoxy) is a strong activating group due to the lone pair donation from oxygen into the ring through resonance. $$-NHCOCH_3$$ (acetylamino) is a moderate activating group; the nitrogen lone pair donates electron density to the ring, though the carbonyl partially withdraws it. Overall, both are activating groups. Statement (A) is CORRECT.

Statement (B): $$-CN$$ and $$-OH$$ are meta-directing groups.

$$-CN$$ is an electron-withdrawing group (meta-directing), but $$-OH$$ is an electron-donating group that directs to ortho and para positions. Statement (B) is INCORRECT.

Statement (C): $$-CN$$ and $$-SO_3H$$ are meta-directing groups.

Both $$-CN$$ and $$-SO_3H$$ are electron-withdrawing groups that direct incoming electrophiles to the meta position. Statement (C) is CORRECT.

Statement (D): Activating groups act as ortho- and para-directing groups.

This is a well-established rule in electrophilic aromatic substitution. Activating groups donate electron density to the ring, preferentially stabilizing the carbocation intermediate at ortho and para positions. Statement (D) is CORRECT.

Statement (E): Halides are activating groups.

Halides ($$-F, -Cl, -Br, -I$$) are deactivating groups. Although they are ortho/para-directing (due to lone pair donation by resonance), the strong inductive electron-withdrawing effect makes the ring less reactive overall. Statement (E) is INCORRECT.

The correct statements are (A), (C), and (D).

The correct answer is Option 4: (A), (C) and (D) only.

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).

Step 1 - Recall the nature of hyperconjugation
Hyperconjugation involves delocalisation of the electrons of an $$\sigma$$-bond (generally C-H) with an adjacent empty p-orbital, a $$\pi$$-bond or a positively charged carbon centre. The overlap is present in the ground state of the molecule and does not require any external stimulus. Hence it is a permanent electronic effect, just like resonance.

Step 2 - Test Statement I
Statement I claims: “Hyperconjugation is not a permanent effect.”
But from Step 1 we know that hyperconjugation is permanent. Therefore, Statement I is $$\textbf{false}$$.

Step 3 - Understand the relation between alkyl substitution and hyperconjugation in carbocations
For a carbocation, each alkyl group attached to the positively charged carbon provides one or more $$\alpha$$-hydrogen atoms. Every $$\alpha$$-hydrogen can participate in hyperconjugation, generating a separate hyperconjugative structure. More such structures mean greater charge delocalisation and hence increased stability of the carbocation.

Step 4 - Test Statement II
Statement II says: “In general, the greater the number of alkyl groups attached to a positively charged C-atom, the greater is the hyperconjugation interaction and stabilisation of the cation.”
Step 3 confirms this: $$3^\circ$$ > $$2^\circ$$ > $$1^\circ$$ carbocations in stability because of increasing hyperconjugation. Hence, Statement II is $$\textbf{true}$$.

Step 5 - Choose the correct option
Statement I is false and Statement II is true. According to the option list, that corresponds to Option C.

Final answer: Option C.

The correct option is D (II, I).

Statement I compares the dipole moments of two aldehydes.
  • $$CH_3-CH=CH-CH=O$$ is an α,β-unsaturated (conjugated) aldehyde.
  • $$CH_3-CH_2-CH_2-CH=O$$ is a saturated aldehyde.

Dipole moment ( $$\mu$$ ) depends on the magnitude of charge separation and on the distance between the centres of positive and negative charge.
  • In the conjugated aldehyde the $$C=O$$ group is conjugated with the $$C=C$$ bond. By resonance the electron pair of the $$C=C$$ bond can shift towards the oxygen, giving forms such as
    $$CH_3-CH=CH-CH^{+}-O^{-}$$.
  • This resonance increases the positive character on the carbonyl carbon and the negative character on the oxygen, thereby increasing the $$C=O$$ dipole moment.
  • Any small dipole associated with the $$C=C$$ bond is approximately parallel to the carbonyl dipole, so the vectors add rather than cancel.

Therefore $$\mu\bigl(CH_3-CH=CH-CH=O\bigr) \gt \mu\bigl(CH_3-CH_2-CH_2-CH=O\bigr)$$ and Statement I is true.

Statement II compares the $$C_1-C_2$$ bond lengths.

• In $$CH_3-CH_2-CH_2-CH=O$$ both $$C_1$$ and $$C_2$$ are $$sp^3$$ hybridised. An $$sp^3-sp^3$$ single bond length is about $$1.54\ \text{Å}$$.
• In $$CH_3-CH=CH-CH=O$$, $$C_1$$ is $$sp^3$$ but $$C_2$$ is $$sp^2$$ (because it is part of the $$C=C$$ bond). An $$sp^3-sp^2$$ single bond has more $$s$$-character on one side and is therefore shorter (≈ $$1.50\ \text{Å}$$) than an $$sp^3-sp^3$$ bond.

Thus $$\text{bond length}(C_1-C_2)\bigl(CH_3-CH=CH-CH=O\bigr)$$ is smaller, not greater, than that in $$CH_3-CH_2-CH_2-CH=O$$. Statement II is false.

So, Statement I is true while Statement II is false. The correct choice is Option C.

We are asked to find the number of possible stereoisomers for 5-phenylpent-4-en-2-ol.

The structure of 5-phenylpent-4-en-2-ol can be represented as $$C_6H_5-CH=CH-CH_2-CH(OH)-CH_3$$; numbering from the OH end gives C1 as $$CH_3$$, C2 as $$CH(OH)$$, C3 as $$CH_2$$, while the C4=C5 double bond connects to a phenyl group at C5.

At C2 there is a chiral center bearing four different groups ($$-CH_3$$, $$-OH$$, $$-H$$, and $$-CH_2CH=CHC_6H_5$$), which results in 2 optical isomers (R and S).

Moreover, the C4=C5 double bond allows geometric isomerism because C5 carries phenyl ($$C_6H_5$$) and H, while C4 carries H and $$-CH_2CH(OH)CH_3$$, so E/Z isomerism is possible, giving 2 geometric isomers.

Combining these possibilities yields a total of 2 optical isomers times 2 geometric isomers, giving $$2 \times 2 = 4$$ stereoisomers.

The number of possible stereoisomers for 5-phenylpent-4-en-2-ol is 4.

We need to find the percentage of chlorine in an organic compound using the Carius method.

We start by finding the moles of AgCl.

The molar mass of AgCl is 108 + 35.5 = 143.5 g/mol.

The mass of AgCl is 143.5 mg, which is 0.1435 g.

Using these values, we calculate the moles of AgCl:

$$\frac{0.1435}{143.5} = 0.001$$ mol

Next, we determine the mass of chlorine obtained.

Each mole of AgCl contains one mole of Cl, so the moles of Cl are 0.001 mol.

The mass of Cl is then calculated as:

$$0.001 \times 35.5 = 0.0355$$ g, which is 35.5 mg.

Then we calculate the percentage of chlorine in the compound.

$$\% \text{Cl} = \frac{\text{Mass of Cl}}{\text{Mass of compound}} \times 100$$

Substituting the values gives:

$$= \frac{35.5}{180} \times 100 = 19.72\%$$

$$\approx 20\%$$

Therefore, the percentage of chlorine in the compound is 20%.

Molar mass=80, 90% C, 10% H. C: 0.9×80/12=6, H: 0.1×80/1=8. Formula: C₆H₈.

Degree of unsaturation = (2×6+2-8)/2 = (12+2-8)/2 = 3.

The answer is 3.

We need to evaluate two statements about oxidation states and $$p\pi - p\pi$$ bonding.

Analysis of Statement I:

"The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule."

This statement is incorrect. The oxidation state of an element is determined based on electronegativity, not electron gain enthalpy. In the concept of oxidation state, we assume that shared electron pairs are assigned entirely to the more electronegative atom. While electron gain enthalpy and electronegativity are related concepts, they are not the same:

- Electronegativity is the tendency of an atom to attract shared electrons in a chemical bond.

- Electron gain enthalpy is the energy change when a gaseous atom gains an electron to form an anion.

Oxidation states are assigned using electronegativity rules (e.g., in H$$_2$$O, oxygen is more electronegative, so both shared pairs are assigned to O, giving it an oxidation state of $$-2$$). Using electron gain enthalpy would not always give correct results because the two properties do not always follow the same trend (for example, fluorine has a lower electron gain enthalpy than chlorine, but fluorine is more electronegative).

Analysis of Statement II:

"$$p\pi - p\pi$$ bond formation is more prevalent in second period elements over other periods."

This statement is correct. Second period elements (such as C, N, O) have small atomic sizes, which allows their 2p orbitals to overlap laterally (sideways) very effectively to form strong $$\pi$$ bonds. For elements in higher periods (third period and beyond), the p orbitals are larger and more diffuse, leading to poor lateral overlap. This is why:

- Carbon readily forms $$C=C$$ and $$C\equiv C$$ bonds (with $$p\pi - p\pi$$ bonding)

- Nitrogen forms $$N=N$$ and $$N\equiv N$$ bonds

- Oxygen forms $$O=O$$ bonds

- But silicon prefers single bonds (Si-Si) rather than double bonds, and phosphorus and sulfur rarely form $$p\pi - p\pi$$ bonds

Elements in higher periods instead use $$d\pi - p\pi$$ bonding when they form multiple bonds.

Conclusion: Statement I is incorrect, and Statement II is correct.

The correct answer is Option (3): Statement I is incorrect but Statement II is correct.

We need to identify which processes have negative electron affinity (i.e., the process is endothermic when an electron is added).

Key Concept: Electron affinity (EA) is the energy change when an electron is added to a neutral gaseous atom. A negative electron affinity (in the convention where exothermic = positive EA) means the atom resists gaining an electron, i.e., energy must be supplied.

Elements with negative EA (endothermic electron gain) typically have either:

- Fully filled subshells (extra stability makes electron addition unfavourable)

- Half-filled subshells (extra stability due to exchange energy)

- Second electron additions (always endothermic due to electron-electron repulsion)

A. Be $$\rightarrow$$ Be$$^-$$:

Beryllium has configuration $$1s^2 2s^2$$ (completely filled 2s subshell). Adding an electron requires placing it in the higher-energy 2p orbital against the nuclear charge shielded by the filled 2s. This process is endothermic (negative EA). ✓

B. N $$\rightarrow$$ N$$^-$$:

Nitrogen has configuration $$1s^2 2s^2 2p^3$$ (half-filled 2p subshell). The half-filled 2p configuration is extra stable due to maximum exchange energy. Adding a fourth electron to the 2p shell would disrupt this stability and require pairing in an already occupied orbital. This process is endothermic (negative EA). ✓

C. O $$\rightarrow$$ O$$^{2-}$$:

This is a two-step process: O $$\rightarrow$$ O$$^-$$ $$\rightarrow$$ O$$^{2-}$$. While the first EA of oxygen is positive (exothermic), the second EA is always endothermic because adding an electron to an already negatively charged ion (O$$^-$$) requires overcoming strong electrostatic repulsion. The overall process of forming O$$^{2-}$$ is endothermic. ✓

D. Na $$\rightarrow$$ Na$$^-$$:

Sodium ($$1s^2 2s^2 2p^6 3s^1$$) can accept an electron to complete its 3s subshell ($$3s^2$$). This process is exothermic (positive EA), similar to how hydrogen gains an electron. ✗

E. Al $$\rightarrow$$ Al$$^-$$:

Aluminium ($$1s^2 2s^2 2p^6 3s^2 3p^1$$) can accept an electron into its 3p orbital. This process is exothermic (positive EA), as the 3p subshell is far from half-filled. ✗

Conclusion: Elements with negative electron affinity values are Be, N, and O (for the second electron addition). These correspond to A, B, and C.

The correct answer is Option 2: A, B and C only.

In bond-line notation, carbon atoms are represented by vertices or line ends and are not written explicitly. Hydrogen atoms attached to carbon are also omitted. Only heteroatoms and hydrogens attached to them are shown.

For $$HOCH(CN)_2$$, the central carbon is bonded to one $$-OH$$ group, two $$-CN$$ groups, and one hydrogen atom.

The central carbon should therefore be represented only by a vertex. The $$-OH$$ and $$-CN$$ groups must be shown explicitly, while the hydrogen attached to the central carbon remains hidden.

Options A, B, and C are incorrect because they explicitly show the central carbon and/or its attached hydrogen, which violates the conventions of bond-line notation.

Option D is correct because the central carbon is represented by a vertex, the $$-OH$$ and two $$-CN$$ groups are shown explicitly, and the hydrogen attached to the central carbon is omitted as required.

Hence, the correct answer is Option D.

Species: $$O^{2-}, F^-, Na^+, Mg^{2+}$$.

(A) All are isoelectronic: All have 10 electrons. $$O^{2-}$$ (8+2=10), $$F^-$$ (9+1=10), $$Na^+$$ (11-1=10), $$Mg^{2+}$$ (12-2=10). Correct. ✓

(B) All have the same nuclear charge: $$O$$ has 8 protons, $$F$$ has 9, $$Na$$ has 11, $$Mg$$ has 12. They have different nuclear charges. Incorrect. ✗

(C) $$O^{2-}$$ has the largest ionic radius: For isoelectronic species, the one with the smallest nuclear charge has the largest radius. $$O^{2-}$$ has the smallest nuclear charge (8), so it has the largest ionic radius. Correct. ✓

(D) $$Mg^{2+}$$ has the smallest ionic radius: $$Mg^{2+}$$ has the largest nuclear charge (12) among the isoelectronic species, so it has the smallest ionic radius. Correct. ✓

Correct statements: (A), (C), and (D).

The correct answer is Option (3): (A), (C) and (D) only.

We need to identify the species in which carbon has a sextet of electrons and can act as an electrophile.

Understand what a "sextet of electrons" means.

A carbon atom normally has 4 valence electrons and achieves an octet (8 electrons) by forming 4 bonds. If carbon has only a sextet (6 electrons), it is 2 electrons short of an octet, meaning it has an empty p-orbital.

Analyze each option.

Carbon free radical: Has 7 electrons (3 bonds + 1 unpaired electron). This is a septet, not a sextet. It is reactive but primarily acts as a radical, not an electrophile.

Carbanion: Has 8 electrons (3 bonds + 1 lone pair). Carbon has a complete octet and a negative charge. It acts as a nucleophile (electron donor), not an electrophile.

Carbocation: Carbon has only 6 electrons (3 bonds, no lone pair). It has an empty p-orbital, making it electron-deficient. Since it seeks electrons to complete its octet, it acts as an electrophile (electron acceptor). This matches both criteria: sextet of electrons and electrophilic character.

Pentavalent carbon: This would have 10 electrons around carbon, which is not a sextet.

The correct answer is Option (3): carbocation.

The incorrect statement regarding conformations of ethane:

- Ethane has infinite number of conformations: Correct - rotation around the C-C bond gives continuous conformations.

- The dihedral angle in staggered conformation is 60°: Correct - the H atoms are 60° apart in staggered form.

- Eclipsed conformation is the most stable conformation: Incorrect - the staggered conformation is the most stable. The eclipsed is the least stable.

- The conformations of ethane are interconvertible: Correct - they freely interconvert by rotation.

The incorrect statement is Option (3).

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Carbocation stability depends on the degree of substitution due to hyperconjugation and inductive effects from alkyl groups.

The order of carbocation stability is:

Tertiary > Secondary > Primary > Methyl

$$(CH_3)_3C^+ > (CH_3)_2CH^+ > CH_3CH_2^+ > CH_3^+$$

The correct answer is Option 3.

$$HO-CH_2-(CH_2)_3-CH_2-COCH_3$$: This is 7-hydroxyheptan-2-one.

IUPAC rules: ketone gets lowest locant when numbering. Numbering from ketone end: C1-C2(CO)-C3-C4-C5-C6-C7(OH). This gives 7-hydroxyheptan-2-one.

Statement I is correct. Statement II gives 2-oxoheptan-7-ol, which has the same locants but different principal characteristic group priority — ketone is preferred over alcohol, so the name should end in "-one" not "-ol". Hence Statement II is incorrect.

The answer is Option (1): Statement I is correct but Statement II is incorrect.

For IUPAC naming of substituted cyclohexanes:

- First identify all substituents on the cyclohexane ring.

- The compound has two methyl groups on the same carbon and an ethyl group on another carbon.

- Number the ring to give the lowest set of locants. The ethyl group and the two methyl groups should receive the lowest possible numbers.

- With dimethyl on one carbon and ethyl on another, numbering gives: the dimethyl carbon as position 1 and ethyl as position 3, yielding 3-Ethyl-1,1-dimethylcyclohexane.

The answer is 3-Ethyl-1,1-dimethylcyclohexane, which corresponds to Option (2).

1. Principal Group Priority: Carboxylic acid ($$-\text{COOH}$$) outranks aldehydes ($$-\text{CHO}$$), alcohols ($$-\text{OH}$$), and alkenes ($$\text{C}=\text{C}$$). The parent suffix is -oic acid.
2. Chain Numbering: Numbering begins at the carboxylic acid carbon ($$\text{C-1}$$) and extends along the longest continuous chain containing the double bond ($$\text{C}=\text{C}$$).
   • This gives a 7-carbon chain (heptane) with the double bond starting at position 6. Thus, the parent name is hept-6-enoic acid.
3. Substituents: * At C-2: $$-\text{CHO}$$ acting as a prefix is named formyl.
   • At C-4: $$-\text{OH}$$ acting as a prefix is named hydroxy.

Alphabetical Arrangement:

Combining the terms alphabetically (formyl before hydroxy):

2-formyl-4-hydroxyhept-6-enoic acid

Correct Option: B

Question: The interaction between a $$\pi$$ bond and a lone pair of electrons present on an adjacent atom is responsible for which effect?

Review the electronic effects in organic chemistry.

Hyperconjugation (Option 1): This involves the interaction between $$\sigma$$ bond electrons (particularly C-H bonds) and an adjacent empty or partially filled p-orbital or $$\pi$$ system. It does not involve lone pairs directly.

Inductive effect (Option 2): This is the transmission of charge through a chain of atoms via $$\sigma$$ bonds. It involves $$\sigma$$ bonds only, not $$\pi$$ bonds or lone pairs interacting with each other.

Electromeric effect (Option 3): This is a temporary effect that occurs when a reagent approaches a $$\pi$$ bond, causing complete transfer of the $$\pi$$ electron pair. It does not specifically involve lone pairs on adjacent atoms.

Resonance effect (Option 4): Resonance (also called the mesomeric effect) involves the delocalisation of electrons through $$\pi$$ bonds and lone pairs on adjacent atoms. When a lone pair on an atom adjacent to a $$\pi$$ bond participates in conjugation, it is the resonance effect. For example, in aniline ($$C_6H_5NH_2$$), the lone pair on N interacts with the $$\pi$$ system of the benzene ring.

The correct answer is Option 4: Resonance effect.

We need to identify the incorrect statement about the electromeric effect.

Option 1: "Electromeric effect dominates over inductive effect." This is correct. The electromeric effect, being a complete transfer of $$\pi$$-electrons, is generally stronger than the inductive effect, which is a partial displacement of $$\sigma$$-electrons.

Option 2: "The electromeric effect is a temporary effect." This is correct. Unlike the inductive effect (which is permanent), the electromeric effect occurs only when an attacking reagent approaches the molecule. It is a temporary polarization of $$\pi$$-bonds.

Option 3: "Hydrogen ion (H⁺) shows negative electromeric effect." This is INCORRECT. H⁺ is an electrophile (electron-deficient species). When H⁺ attacks a $$\pi$$-bond, the $$\pi$$-electrons shift toward the attacking reagent — this is a positive electromeric effect (+E), not negative. The negative electromeric effect ($$-E$$) occurs when the attacking reagent is a nucleophile, and the electron pair shifts away from the attacking species.

Option 4: "The organic compound shows electromeric effect in the presence of the reagent only." This is correct, consistent with the temporary nature of the effect.

The incorrect statement is Option (3).

We need the common name of benzene-1,2-diol.

Benzene-1,2-diol has two hydroxyl (-OH) groups on adjacent carbon atoms of the benzene ring (ortho positions).

- Benzene-1,2-diol (ortho) = Catechol (also called pyrocatechol)

- Benzene-1,3-diol (meta) = Resorcinol

- Benzene-1,4-diol (para) = Hydroquinone (quinol)

Note: o-Cresol is 2-methylphenol, not a diol.

The correct answer is Option (1): Catechol.

Key Concept: Organic compounds are broadly classified based on their carbon skeletons. Aliphatic compounds include both open-chain (acyclic) compounds and cyclic systems that mimic open-chain behavior (alicyclic). Cyclic compounds containing alternate single and double bonds that follow Hückel's rule are classified as aromatic.

Option [A]: Benzenoid aromatic. Benzenoid aromatic compounds contain at least one benzene ring ($$C_6H_6$$) structure characterized by alternating single and double bonds with a fully delocalized ring system of $$6\pi$$ electrons. Cyclohexene ($$C_6H_{10}$$) contains only one double bond and lacks a fully delocalized conjugated aromatic system. This is FALSE.

Option [B]: Benzenoid non-aromatic. The term benzenoid is explicitly reserved for structures containing benzene rings. Since cyclohexene does not feature a benzene ring network, it cannot be described as benzenoid. This is FALSE.

Option [C]: Acyclic. Acyclic compounds are open-chain organic compounds that do not contain any ring structure in their carbon backbone. Cyclohexene contains a closed six-membered carbon ring. This is FALSE.

Option [D]: Alicyclic. Alicyclic compounds (aliphatic cyclic) are organic compounds that are both aliphatic and cyclic. They contain one or more closed carbon rings, which can be saturated or unsaturated (containing double or triple bonds), but they do not possess aromatic character. Cyclohexene consists of a cyclic ring with one carbon-carbon double bond, matching this classification perfectly. This is TRUE.

Therefore, cyclohexene is classified as an alicyclic organic compound.

Answer: Option D — Alicyclic

The parent chain is pentane, which contains a continuous chain of five carbon atoms.

A phenyl group is attached to carbon 1 of the pentane chain. Therefore, numbering must begin from the end nearest the phenyl group.

The name also specifies the presence of two bromine substituents at carbon 2 and carbon 3.

In Option A, the bromine atoms are present at carbon 1 and carbon 3. Therefore, it represents $$1,3\text{-dibromo-1-phenylpentane}$$.

In Option B, the bromine atoms are present at carbon 3 and carbon 4. Therefore, it represents $$3,4\text{-dibromo-1-phenylpentane}$$.

In Option D, the bromine atoms are present at carbon 1 and carbon 2. Therefore, it represents $$1,2\text{-dibromo-1-phenylpentane}$$.

In Option C, the phenyl group is attached to carbon 1, while the bromine atoms are attached to carbon 2 and carbon 3. The remaining two carbons complete the pentane chain.

Thus, Option C correctly represents

$$2,3\text{-dibromo-1-phenylpentane}$$

Hence, the correct answer is Option C.

A)Both have the molecular formula $$(C_{3}H_{8}O)$$. In n-propanol, the $$(-OH)$$ group is at position 1, while in isopropanol, it is at position 2.

           Isopropanol

B)These compounds have the same molecular formula $$(C_4H_{10}O)$$ but differ in the distribution of carbon atoms on either side of the ether functional group $$(-O-)$$.

C)Both have the formula $$(C_{3}H_{6}O)$$, but they belong to different functional groups: propanone is a ketone and propanal is an aldehyde.

D)Both are isomers of pentane $$(C_{5}H_{12})$$ but differ in the structure of the carbon skeleton (branched chain vs. more highly branched chain).

Distillation is used to separate liquids having a large difference in boiling points. Chloroform and aniline have boiling points of approximately $$334\ K$$ and $$457\ K$$ respectively, making simple distillation suitable for their separation.

Therefore,

$$A \rightarrow (IV)$$

Fractional distillation is employed for separating mixtures of miscible liquids with closely spaced boiling points. A major industrial application is the separation of crude oil into its various fractions.

Therefore,

$$B \rightarrow (III)$$

Steam distillation is used for compounds that are steam volatile and immiscible with water. Aniline satisfies these conditions and can be separated from aniline-water mixtures by steam distillation.

Therefore,

$$C \rightarrow (II)$$

Distillation under reduced pressure is used for liquids having high boiling points that may decompose on heating at atmospheric pressure. Glycerol is separated from spent-lye using this technique because its boiling point is high and it undergoes decomposition upon prolonged heating.

Therefore,

$$D \rightarrow (I)$$

Hence, the correct matching is

$$A-(IV),\ B-(III),\ C-(II),\ D-(I)$$

According to rules of making resonance structures, the most stable resonance structure is one which will have no charge, next would be the structure in which electronegative atom gets a negative charge i.e. Oxygen must have negative charge. The least stable resonating structure would be the structure that has a positive charge on oxygen opposite to its tendency. Hence, III>II>I

Understanding resonance energy:

Many organic compounds can be represented by two or more Lewis structures (called resonance structures or canonical structures). The actual molecule is a resonance hybrid of all these structures.

The lowest energy resonance structure (also called the most stable contributing structure) is a hypothetical reference. The actual molecule, being a hybrid of all resonance structures, has a lower energy than any single resonance structure — including the most stable one.

Definition of resonance energy:

Resonance energy is defined as the difference in energy between the actual structure (resonance hybrid) and the most stable (lowest energy) resonance structure:

$$\text{Resonance Energy} = E_{\text{lowest energy resonance structure}} - E_{\text{actual structure (hybrid)}}$$

Since the actual structure is always more stable (lower energy) than any individual resonance structure, the resonance energy is always a positive quantity.

For example, benzene has a resonance energy of approximately 150 kJ/mol, meaning the actual benzene molecule is 150 kJ/mol more stable than the hypothetical cyclohexatriene (Kekulé) structure.

Among the given options:

Option A (electromeric energy): This refers to the temporary shift of electrons in a multiple bond under the influence of an attacking reagent. It is not related to the stability difference between resonance structures.

Option B (resonance energy): This is exactly the energy difference between the actual structure and the lowest energy resonance structure. This is correct.

Option C (ionization energy): This is the energy required to remove an electron from an isolated gaseous atom. Unrelated to resonance.

Option D (hyperconjugation energy): This refers to the stabilization due to the interaction of electrons in a $$\sigma$$ bond with an adjacent empty or partially filled p-orbital. While it contributes to stability, it is not defined as the difference between the actual and resonance structures.

The correct answer is Option B: resonance energy.

The Bromine substituted Carbon is a chiral carbon, hence it can have 2 configurations. Now, the double bond on left can either have cis or trans configuration and double bond on right could either have cis or trans configuration. Hence, total number of stereoisomers = 2*2*2 = 8

We need to identify the mechanism through which ionic reactions with organic compounds proceed. Heterolytic bond cleavage involves the breaking of a covalent bond such that both shared electrons go to one atom, forming a cation and an anion: $$ A:B \rightarrow A^+ + :B^- \quad \text{(or } A:^- + B^+\text{)} $$

In contrast, homolytic bond cleavage assigns one electron to each atom, generating free radicals: $$ A:B \rightarrow A\cdot + B\cdot $$

Ionic reactions involve the formation of ionic intermediates (carbocations, carbanions) and proceed through attack by nucleophiles or electrophiles. This requires heterolytic bond cleavage, where electron pairs move as a unit, creating charged species. Free radical mechanisms (from homolytic cleavage) produce neutral radicals, not ions. Therefore, ionic reactions proceed through heterolytic bond cleavage (B) only.

The correct answer is Option C: (B) only.

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Identify the functional group that shows negative resonance effect ($$-R$$ effect).

What is the negative resonance effect ($$-R$$ effect)?

A group shows $$-R$$ effect when it withdraws electron density from the conjugated system through resonance. These groups typically have a multiple bond to an electronegative atom (like $$C=O$$), which pulls electron density away from the attached system.

What is the positive resonance effect ($$+R$$ effect)?

Groups with lone pairs on the atom directly bonded to the conjugated system can donate electron density through resonance. Examples: $$-NH_2$$, $$-OH$$, $$-OR$$.

Analyzing each option:

$$-NH_2$$: Has a lone pair on N that can be donated to the ring. Shows +R effect.

$$-OH$$: Has lone pairs on O that can be donated. Shows +R effect.

$$-COOH$$: The carbonyl group ($$C=O$$) withdraws electron density from the conjugated system. The $$-COOH$$ group pulls electrons away through the $$\pi$$ system. Shows -R effect.

$$-OR$$: Has lone pairs on O that can be donated. Shows +R effect.

The correct answer is Option C: $$-COOH$$.

We need to find the total number of sigma ($$\sigma$$) and pi ($$\pi$$) bonds in 2-formylhex-4-enoic acid.

2-formylhex-4-enoic acid has a 6-carbon chain (hexanoic acid) with a formyl group ($$-CHO$$) at C2 and a double bond between C4 and C5.

Structure: $$CH_3-CH=CH-CH_2-CH(CHO)-COOH$$

Every single bond is 1$$\sigma$$ bond, and every double bond contains 1$$\sigma$$ + 1$$\pi$$ bond.

C-C sigma bonds in main chain: C1-C2, C2-C3, C3-C4, C4=C5($$\sigma$$), C5-C6 = 5

C2-CHO branch: 1

In $$-COOH$$: C=O($$\sigma$$) + C-OH + O-H = 3

In $$-CHO$$: C=O($$\sigma$$) + C-H = 2

C-H bonds: C2(1H) + C3(2H) + C4(1H) + C5(1H) + C6(3H) = 8

Total $$\sigma$$ bonds = 5 + 1 + 3 + 2 + 8 = 19

C4=C5: 1$$\pi$$ bond

C=O in $$-COOH$$: 1$$\pi$$ bond

C=O in $$-CHO$$: 1$$\pi$$ bond

Total $$\pi$$ bonds = 3

Total = 19 + 3 = 22

The total number of sigma and pi bonds is $$\boxed{22}$$.

We need to count the number of compounds containing sulphur as a heteroatom from: Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine.

Define heteroatom.

A heteroatom is any atom in an organic compound that is not carbon or hydrogen. In heterocyclic chemistry, it specifically refers to atoms like N, O, S that replace carbon in a ring.

Analyse each compound.

Furan: A five-membered heterocyclic ring containing oxygen as the heteroatom. No sulphur.

Thiophene: A five-membered heterocyclic ring containing sulphur as the heteroatom. Counted.

Pyridine: A six-membered heterocyclic ring containing nitrogen as the heteroatom. No sulphur.

Pyrrole: A five-membered heterocyclic ring containing nitrogen as the heteroatom. No sulphur.

Cysteine: An amino acid with the structure $$HSCH_2CH(NH_2)COOH$$. It contains a thiol group ($$-SH$$), so sulphur is present as a heteroatom. Counted.

Tyrosine: An amino acid with a phenol group ($$HO-C_6H_4-CH_2CH(NH_2)COOH$$). It contains oxygen and nitrogen but no sulphur.

Count.

Compounds containing sulphur as heteroatom: Thiophene and Cysteine = 2.

The answer is 2.

We need to find the total number of sigma and pi bonds in 2-oxohex-4-ynoic acid.

First, we draw its structure: 2-oxohex-4-ynoic acid is a 6-carbon chain with:

- Position 1: -COOH (carboxylic acid)

- Position 2: =O (oxo/keto group)

- Position 4-5: triple bond (ynoic = alkyne)

Thus, the structure is HOOC-CO-CH₂-C≡C-CH₃.

Next, we count the sigma bonds:

- O-H: 1 sigma

- C1=O (COOH): 1 sigma + 1 pi

- C1-O(H): 1 sigma

- C1-C2: 1 sigma

- C2=O: 1 sigma + 1 pi

- C2-C3: 1 sigma

- C3-H (×2): 2 sigma

- C3-C4: 1 sigma

- C4≡C5: 1 sigma + 2 pi

- C5-C6: 1 sigma

- C6-H (×3): 3 sigma

Summing these, the total number of sigma bonds is 1 + 1 + 1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 3 = 14, and the total number of pi bonds is 1 + 1 + 2 = 4.

Therefore, the total number of sigma and pi bonds is 14 + 4 = 18, so the answer is 18.

Find the number of optical isomers of 2-chlorobutane.

The structure of 2-chlorobutane can be represented as $$CH_3-CHCl-CH_2-CH_3$$.

In this molecule, carbon-2 is bonded to four different groups: $$-CH_3$$, $$-Cl$$, $$-H$$, and $$-CH_2CH_3$$, making C-2 a chiral center.

For a molecule with $$n$$ chiral centers and no internal plane of symmetry, the number of optical isomers is $$2^n$$. Here $$n = 1$$, so the number of optical isomers is $$2^1 = 2$$. These two isomers are enantiomers (R-2-chlorobutane and S-2-chlorobutane), which are non-superimposable mirror images of each other.

The answer is 2.

Key Concept: A chiral carbon atom (or asymmetric carbon) is a tetrahedral carbon atom ($$sp^3$$ hybridized) that is directly bonded to four entirely different atoms or groups of atoms.

Structure 1: Another molecule shown in the image top part (not completely spelled in text, let's analyze the given four textual lines and the structure image). Let's analyze the explicitly listed compounds sequentially:

Compound 1: $$CH_3-CH_2-CH(NO_2)-COOH$$ (2-Nitrobutanoic acid). Let's examine the carbon atoms in this chain. Carbon-2 is bonded to a methyl/ethyl-derived group ($$-CH_2-CH_3$$), a nitro group ($$-NO_2$$), a carboxylic acid group ($$-COOH$$), and a hydrogen atom ($$-H$$). Since all four groups attached to Carbon-2 are distinct, this carbon is a chiral center. This compound contains a chiral carbon.

Compound 2: $$CH_3-CH(I)-CH_2-NO_2$$ (2-Iodo-1-nitropropane). Let's look closely at Carbon-2. It is directly attached to a methyl group ($$-CH_3$$), an iodine atom ($$-I$$), a nitromethyl group ($$-CH_2-NO_2$$), and a hydrogen atom ($$-H$$). Because all four attached groups are different, Carbon-2 acts as a chiral center. This compound contains a chiral carbon.

Compound 3: $$CH_3-CH_2-CHBr-CH_2-CH_3$$ (3-Bromopentane). Let's examine Carbon-3, which holds the bromine atom. It is bonded to a bromine atom ($$-Br$$), a hydrogen atom ($$-H$$), and two identical ethyl groups ($$-CH_2-CH_3$$ on both sides). Because it does not have four different groups attached, it is achiral. This compound does not contain a chiral carbon.

Compound 4: $$CH_3-CH_2-CH(OH)-CH_2OH$$ (Butane-1,2-diol). Let's look closely at Carbon-2. It is bonded to an ethyl group ($$-CH_2-CH_3$$), a hydroxyl group ($$-OH$$), a hydroxymethyl group ($$-CH_2OH$$), and a hydrogen atom ($$-H$$). Since all four groups are different, Carbon-2 is a chiral center. This compound contains a chiral carbon.

Compound 5: The structure shown in the structural diagram image ($$H_3C-C(H)(I)-CH(I)-C_2H_5$$ or similar polyhalogenated alkane variant). Looking at the molecular graphic representation provided at the bottom: The left highlighted carbon is bonded to $$H_3C-$$, $$-H$$, $$-I$$, and the remaining bulky haloalkyl chain. Thus, it forms a chiral center. The adjacent carbon containing an iodine atom is bonded to $$-I$$, $$-H$$, an ethyl group ($$-C_2H_5$$), and the left haloalkyl portion. This also forms a second chiral center. This compound contains chiral carbons.

Counting the total number of compounds from the problem set that possess a chiral carbon center gives a total of 5 compounds.

Answer: 5

2-chlorobutane: CH$$_3$$CHClCH$$_2$$CH$$_3$$. Monochlorination gives C$$_4$$H$$_8$$Cl$$_2$$ isomers.

Possible positions for the second Cl:

1. C1: CH$$_2$$ClCHClCH$$_2$$CH$$_3$$ — C2 is chiral, optically active (2 enantiomers)

2. C2: CH$$_3$$CCl$$_2$$CH$$_2$$CH$$_3$$ — no chiral center, not optically active

3. C3: CH$$_3$$CHClCHClCH$$_3$$ — two chiral centers, can have (R,R), (S,S), and meso. The (R,R) and (S,S) are optically active = 2 enantiomers. Also (R,S) which is meso.

4. C4: CH$$_3$$CHClCH$$_2$$CH$$_2$$Cl — C2 is chiral, 2 enantiomers

Total optically active isomers: 2 + 0 + 2 + 2 = 6.

Therefore, the answer is $$\boxed{6}$$.

Step 1: Formation of Intermediate (A) and (B)

• Reactant: Propylbenzene ($$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}_3$$)
• Reagents: 1) $$\text{KMnO}_4\text{ - KOH, }\Delta$$   followed by 2) $$\text{H}_3\text{O}^+$$
• Reaction: Vigorous side-chain oxidation. Any alkyl group with at least one benzylic hydrogen is completely oxidized down to a carboxylic acid group attached directly to the benzene ring.
• Product (B): Benzoic acid ($$\text{C}_6\text{H}_5\text{COOH}$$)

Step 2: Formation of Major Product (C)

• Reactant: Benzoic acid ($$\text{C}_6\text{H}_5\text{COOH}$$)
• Reagents: $$\text{Br}_2\text{ / FeBr}_3$$ (Electrophilic Aromatic Substitution / Bromination)
• Reaction: The carboxylic acid group ($$-\text{COOH}$$) is a strongly deactivating and meta-directing group. Therefore, the bromine atom attacks the meta position.
• Product (C): m-Bromobenzoic acid (3-bromobenzoic acid)

Step 3: Counting the $$\pi$$-bonds in Product (C)

Let's count the $$\pi$$-bonds in the chemical structure of m-bromobenzoic acid:

1. Benzene Ring: Contains 3 $$\pi$$-bonds (alternate double bonds).
2. Carboxylic Acid Group ($$-\text{COOH}$$): Contains 1 $$\pi$$-bond (from the carbon-oxygen double bond, $$\text{C}=\text{O}$$).
3. Bromine Atom ($$-\text{Br}$$): Contains only $$\sigma$$-bonds (no $$\pi$$-bonds).

$$\text{Total }\pi\text{-bonds} = 3 + 1 = 4$$

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 $$\boxed{11}$$.

Step 1 - Index of hydrogen deficiency for $$P$$
For an open-chain compound $$C_9H_{18}O_2$$ the index is
$$\text{IHD}= \dfrac{2\times 9+2-18}{2}=1$$
So $$P$$ possesses one and only one π bond (no ring).

Step 2 - Nature of functional groups in $$P$$
$$P$$ decolorises bromine water ⇒ the single IHD corresponds to a $$C=C$$ bond.
$$P$$ also gives a positive iodoform test; without any carbonyl present (IHD already used up), this can only arise from a secondary alcohol of the type $$CH_3-C^*(OH)-$$. Hence both oxygen atoms of $$P$$ are present as non-carbonyl oxygens (alcohol/ether). In short, $$P$$ is an >C=C< containing di-oxygenated alcohol (or alcohol-ether).

Step 3 - Oxidative ozonolysis of the double bond
Ozonolysis followed by $$H_2O_2/H_2O$$ cleaves the $$C=C$$, converting each double-bond carbon into a -COOH group while leaving every pre-existing -OH intact. Thus both fragments, $$Q$$ and $$R$$, are $$\beta$$-hydroxy carboxylic acids.

Step 4 - Why $$Q$$ shows, but $$R$$ lacks, the iodoform test
The iodoform test in acids is positive only when the acid contains a $$CH_3CO-$$ or $$CH_3CH(OH)-$$ unit. Therefore the -OH in $$Q$$ is on the carbon that already bears a methyl group, giving $$CH_3-C(OH)(\ldots )-COOH$$; $$R$$ lacks this arrangement.

Step 5 - Oxidation with PCC followed by heating
(i) PCC oxidises the secondary -OH in each $$\beta$$-hydroxy acid to a carbonyl group, producing a $$\beta$$-keto acid.
(ii) A classic property of $$\beta$$-keto acids is thermal decarboxylation:
$$R-CO-CH_2-COOH \xrightarrow{\Delta} R-CO-CH_3 + CO_2$$
Thus $$Q$$ → $$S$$ and $$R$$ → $$T$$, where both $$S$$ and $$T$$ are methyl ketones.

Step 6 - Counting oxygen atoms in $$S$$ and $$T$$
A methyl ketone contains only one oxygen atom (the carbonyl oxygen).
Hence $$n_O(S)=1$$ and $$n_O(T)=1$$.

Step 7 - Required sum
$$n_O(S)+n_O(T)=1+1=2$$

Therefore, the sum of the number of oxygen atoms present in $$S$$ and $$T$$ is 2.

For one mole of a real gas, the Van der Waals equation is

$$\left(P+\frac{a}{V^2}\right)(V-b)=RT.$$

Expanding the equation,

$$PV-Pb+\frac{a}{V}-\frac{ab}{V^2}=RT.$$

Dividing throughout by (RT),

$$\frac{PV}{RT}-\frac{Pb}{RT}+\frac{a}{RTV}-\frac{ab}{RTV^2}=1.$$

Since the compressibility factor is defined as

$$Z=\frac{PV}{RT},$$

we obtain

$$Z-\frac{Pb}{RT}+\frac{a}{RTV}-\frac{ab}{RTV^2}=1.$$

At low pressures, the molar volume (V) is very large, so terms containing (\frac{1}{V^2}) can be neglected. Therefore,

$$Z\approx 1-\frac{a}{RTV}.$$

Point A corresponds to the low-pressure region where intermolecular attractive forces dominate, causing the gas to be more compressible than an ideal gas ((Z<1)). The deviation from ideal behaviour is represented by the attractive term (-\frac{a}{RTV}).

Hence, the compressibility factor at point A is

$$\boxed{Z=1-\frac{a}{RTV}}.$$

Therefore, the correct answer is

$$\boxed{1-\frac{a}{RTV}}.$$

The given Newman projections represent different conformations of butane viewed along the C₂-C₃ bond.

The stability of these conformations depends on torsional strain and steric strain.

• Torsional strain is maximum in eclipsed conformations and minimum in staggered conformations.

• Steric strain increases when the two methyl groups come closer to each other.

Option A represents the anti conformation. It is a staggered conformation in which the two methyl groups are separated by $$180^\circ$$. This arrangement minimizes both torsional strain and steric strain, making it the most stable conformation.

Option B represents the fully eclipsed conformation. The two methyl groups are directly eclipsing each other, resulting in maximum torsional and steric strain. Hence, it is the least stable conformation.

Option C represents the gauche conformation. Although it is staggered, the two methyl groups are only $$60^\circ$$ apart, leading to gauche interactions. Therefore, it is less stable than the anti conformation.

Option D represents an eclipsed conformation in which a methyl group eclipses a hydrogen atom. It possesses significant torsional strain and is less stable than the staggered conformations.

The stability order is:

$$\text{Anti} > \text{Gauche} > \text{Eclipsed} > \text{Fully Eclipsed}$$

Hence, the most stable conformation is the anti conformation.

$$\boxed{\text{Option A}}$$

To identify the correct matches, examine the type of isomerism exhibited by each pair.

A. Propanamine and N-Methylethanamine

Propanamine: $$CH_3CH_2CH_2NH_2$$

N-Methylethanamine: $$CH_3CH_2NHCH_3$$

Both compounds have the same molecular formula but differ in the distribution of alkyl groups on either side of the polyvalent nitrogen atom. Therefore, they are metamers.

$$A \rightarrow I$$

B. Hexan-2-one and Hexan-3-one

Hexan-2-one: $$CH_3COCH_2CH_2CH_2CH_3$$

Hexan-3-one: $$CH_3CH_2COCH_2CH_2CH_3$$

The functional group remains the same, but its position changes from carbon-2 to carbon-3. Therefore, they are positional isomers.

$$B \rightarrow II$$

C. Ethanamide and Hydroxyethanimine

Ethanamide: $$CH_3CONH_2$$

Hydroxyethanimine: $$CH_3C(OH)=NH$$

These structures differ by migration of a proton and shifting of a double bond. Therefore, they are tautomers.

$$C \rightarrow IV$$

D. o-Nitrophenol and p-Nitrophenol

In these compounds, the nitro group occupies different positions on the benzene ring relative to the hydroxyl group.

The change is from ortho-position to para-position, so they are positional isomers.

$$D \rightarrow II$$

Hence, the correct matching is:

$$A \rightarrow I,\quad B \rightarrow II,\quad C \rightarrow IV,\quad D \rightarrow II$$

Therefore, the correct answer is

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

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,

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

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.

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).

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).

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 $$6$$.

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.

The molecule contains:

$$-COOH$$ and $$C=O$$ functional groups.

According to IUPAC priority rules, the carboxylic acid has higher priority and is taken as the principal functional group.

Hence the suffix used is: $$\text{-oic acid}$$

The longest carbon chain containing the carboxylic acid group has:

$$6 \text{ carbon atoms}$$

Therefore the parent chain is: $$\text{hexanoic acid}$$

Numbering starts from the carboxylic acid carbon.

Thus:

$$C1 : COOH$$

$$C2 : CH \text{ carrying a methyl group}$$

$$C3 : CH_2$$

$$C4 : CH_2$$

$$C5 : C=O$$

$$C6 : CH_3$$

The substituents present are:

A methyl group at:

$$C2$$ and a ketone group represented by the prefix: $$\text{oxo-}$$ at: $$C5$$

Combining all parts of the name:

$${\text{2-methyl-5-oxohexanoic acid}}$$
Hence, Option A is correct

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 $$7$$.

For a compound to be aromatic, it must fulfill four criteria:

1. Cyclic: The structure must be a closed ring.
   $$\text{}$$
2. Planar: All atoms in the ring must be $$sp^2$$ hybridized so the p-orbitals can align.
   $$\text{}$$
3. Fully Conjugated: There must be a continuous system of alternating double bonds, positive charges, negative charges, or lone pairs around the entire ring.
   $$\text{}$$
4. Hückel's Electron Count: The conjugated loop must contain $$(4n + 2)\,\pi$$ electrons (where $$n = 0, 1, 2, 3...$$, giving allowed counts of $$2, 6, 10, 14...\,\pi$$ electrons)
   

• A (Cyclopentadienyl Anion): Cyclic, planar, fully conjugated. It has 2 double bonds ($$4\pi$$ $$e^-$$) + 1 negative charge lone pair ($$2\pi$$ $$e^-$$) = $$6\pi$$ electrons ($$n=1$$). Aromatic.
  $$\text{}$$
• B (Tropylium Cation): Cyclic, planar, fully conjugated. It has 3 double bonds ($$6\pi$$ $$e^-$$) + 1 empty p-orbital from the positive charge = $$6\pi$$ electrons ($$n=1$$). Aromatic.
  $$\text{}$$
• C (Naphthalene): Cyclic, planar, fully conjugated bicyclic system. It has 5 double bonds = $$10\pi$$ electrons ($$n=2$$). Aromatic.
  $$\text{}$$
• D (Cycloheptatrienyl Anion): It contains 3 double bonds ($$6\pi$$ $$e^-$$) + 1 negative charge lone pair ($$2\pi$$ $$e^-$$) = $$8\pi$$ electrons. This follows the $$4n$$ rule ($$n=2$$), making it Anti-aromatic (non-aromatic in reality as it tubs to lose planarity). Not Aromatic.

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: $$-SO_3H > -COCl > -CONH_2 > -CN$$.

1. Identify the Principal Functional Group:
2. The molecule contains an aldehyde ($$-CHO$$), a ketone ($$C=O$$), an alkene ($$C=C$$), and a nitro group $$(-NO_{2})$$. According to IUPAC priority rules, the aldehyde takes precedence and is assigned carbon C1.
3. Number the Principal Chain:
   Starting from the aldehyde carbon, the longest continuous chain containing the ketone, alkene, and nitro group has 7 carbons (hept-).
   • C1: Aldehyde carbon $$-al$$
   • C3: Ketone group (-oxo-)
   • C4: Alkene starts and has a methyl branch (-4-en-, 4-methyl-)
   • C6: Nitro group (6-nitro-)
4. Assemble the Name:
   Alphabetize the substituents: methyl, nitro, and oxo.
   • Substituents: 4-methyl, 6-nitro, 3-oxo.
   • Parent chain & suffix: hept-4-enal.

Full Name: 4-methyl-6-nitro-3-oxohept-4-enal (Option C)

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 final product is triphenylmethanol $$\mathrm{Ph_3C-OH}$$.

In the reaction of an ester with a Grignard reagent, the ester provides the carbonyl carbon and one substituent attached to it.

Methyl benzoate $$\mathrm{Ph-CO-OCH_3}$$ already contains one phenyl group.

Therefore, the remaining two phenyl groups must come from the Grignard reagent.

Hence, the Grignard reagent formed in step (i) must be phenylmagnesium bromide $$\mathrm{Ph-MgBr}$$.

The reactant required to form this Grignard reagent is bromobenzene $$\mathrm{Ph-Br}$$.

The first equivalent of $$\mathrm{Ph-MgBr}$$ converts the ester into benzophenone $$\mathrm{Ph-CO-Ph}$$.

The second equivalent attacks benzophenone to form the alkoxide $$\mathrm{Ph_3C-O^-}$$.

Acidic hydrolysis then gives triphenylmethanol $$\mathrm{Ph_3C-OH}$$.

Therefore, reactant $$\mathrm{A}$$ is bromobenzene $$\mathrm{Ph-Br}$$.

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.

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 minimum temperature at which a sealed flask will burst.

Volume = $$2 \text{ dm}^3 = 2 \times 10^{-3} \text{ m}^3$$

Mass of propane = 11 g

Burst pressure = 2 MPa = $$2 \times 10^6 \text{ Pa}$$

$$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$$

Calculate moles of propane

Molecular formula of propane: $$C_3H_8$$

Molar mass = $$3 \times 12 + 8 \times 1 = 44 \text{ g/mol}$$

$$n = \frac{11}{44} = 0.25 \text{ mol}$$

Apply the ideal gas equation

$$PV = nRT$$

$$T = \frac{PV}{nR}$$

Substitute values

$$T = \frac{2 \times 10^6 \times 2 \times 10^{-3}}{0.25 \times 8.3}$$

$$T = \frac{4000}{2.075}$$

$$T = 1927.7 \text{ K}$$

Convert to Celsius

$$T = 1927.7 - 273 = 1654.7 \approx 1655 \text{ °C}$$

The answer is 1655.

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.

We are given a two-step reaction where the yield of reaction I is 60% and the yield of reaction II is 50%.

The overall yield of a multi-step reaction is the product of the individual yields of each step.

Overall yield = Yield of Step I $$\times$$ Yield of Step II

$$\text{Overall yield} = \frac{60}{100} \times \frac{50}{100} = 0.6 \times 0.5 = 0.3$$

Converting to percentage:

$$\text{Overall yield} = 0.3 \times 100 = 30\%$$

Therefore, the overall yield of the complete reaction is $$\textbf{30}$$ %.

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 $$\textbf{42}$$ %.

At constant T, PV is constant. Thus, the PV graph is a straight line.

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.

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

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 tyre contains nitrogen gas at a fixed volume, so we can apply Gay-Lussac's Law, which states that at constant volume, $$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$.

The initial conditions are $$P_1 = 35$$ psi and $$T_1 = 27°C = 27 + 273 = 300$$ K. The tyre bursts when $$P_2 = 40$$ psi, and we need to find $$T_2$$.

Substituting into the equation: $$\frac{35}{300} = \frac{40}{T_2}$$

Solving for $$T_2$$: $$T_2 = \frac{40 \times 300}{35} = \frac{12000}{35} = 342.86 \text{ K}$$

Converting to Celsius: $$T_2 = 342.86 - 273 = 69.86°C \approx 70°C$$

The temperature at which the car tyre will burst is $$\mathbf{70}$$°C.

The de Broglie wavelength of an electron accelerated through a potential difference $$V$$ is given by:

$$\lambda = \frac{h}{\sqrt{2meV}}$$

Substituting the given values: $$h = 6.63 \times 10^{-34}$$ Js, $$m = 9.1 \times 10^{-31}$$ kg, $$e = 1.6 \times 10^{-19}$$ C, and $$V = 40{,}000$$ V:

$$2meV = 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 40000$$

$$= 2 \times 9.1 \times 1.6 \times 4 \times 10^{-31-19+4}$$

$$= 2 \times 58.24 \times 10^{-46} = 116.48 \times 10^{-46} = 1.1648 \times 10^{-44} \text{ kg}^2\text{m}^2\text{s}^{-2}$$

$$\sqrt{2meV} = \sqrt{1.1648 \times 10^{-44}} = 1.0793 \times 10^{-22} \text{ kg m s}^{-1}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{1.0793 \times 10^{-22}} = 6.14 \times 10^{-12} \text{ m}$$

So $$\lambda = X \times 10^{-12}$$ m, where $$X \approx 6$$.

The value of X is $$\mathbf{6}$$.

We are given that a proton and a $$Li^{3+}$$ nucleus are accelerated by the same potential. We need to find $$\frac{\lambda_{Li}}{\lambda_p}$$.

The de Broglie wavelength of a charged particle accelerated through potential $$V$$ is given by:

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

where $$h$$ is Planck's constant, $$m$$ is the mass, $$q$$ is the charge, and $$V$$ is the accelerating potential.

For the proton: $$\lambda_p = \frac{h}{\sqrt{2 m_p \cdot e \cdot V}}$$

For $$Li^{3+}$$: $$\lambda_{Li} = \frac{h}{\sqrt{2 m_{Li} \cdot 3e \cdot V}}$$

Here $$m_{Li} = 8.3 \, m_p$$ and the charge on $$Li^{3+}$$ is $$3e$$.

Taking the ratio:

$$\frac{\lambda_{Li}}{\lambda_p} = \frac{\sqrt{2 m_p \cdot e \cdot V}}{\sqrt{2 m_{Li} \cdot 3e \cdot V}} = \sqrt{\frac{m_p \cdot e}{m_{Li} \cdot 3e}}$$

$$= \sqrt{\frac{m_p}{8.3 \, m_p \times 3}} = \sqrt{\frac{1}{24.9}}$$

$$= \frac{1}{\sqrt{24.9}} = \frac{1}{4.99} \approx 0.2$$

We are told the ratio equals $$x \times 10^{-1}$$. So:

$$0.2 = x \times 10^{-1}$$

$$x = 2$$

So, the answer is $$2$$.

We have an LPG cylinder that contains gas initially at a pressure of $$P_1 = 300 \,\text{kPa} = 3 \times 10^5 \,\text{Pa}$$ and an initial temperature of $$t_1 = 27^\circ\text{C}$$. To work with gas-law equations we must always convert Celsius to Kelvin, because the Kelvin scale starts from absolute zero. The conversion formula is

$$T (\text{K}) = t (^\circ\text{C}) + 273.$$

Applying this conversion, the initial absolute temperature is

$$T_1 = 27 + 273 = 300 \,\text{K}.$$

The cylinder can safely withstand a maximum pressure of $$P_2 = 1.2 \times 10^6 \,\text{Pa}.$$ We are asked to find the temperature $$t_2$$ (in °C) at which the cylinder will reach this pressure and burst.

An LPG cylinder is a rigid container; therefore its volume remains constant during heating. For a fixed amount of an ideal gas kept at constant volume, pressure is directly proportional to absolute temperature. This relation is expressed by Gay-Lussac’s (or the Isochoric) gas law:

$$\frac{P}{T} = \text{constant} \qquad\Longrightarrow\qquad \frac{P_1}{T_1} = \frac{P_2}{T_2}.$$

Rearranging this formula to solve for the unknown absolute temperature $$T_2$$, we write

$$T_2 = \frac{P_2 \, T_1}{P_1}.$$

Now we substitute each known value carefully:

$$T_2 = \frac{1.2 \times 10^6 \,\text{Pa} \times 300 \,\text{K}}{3 \times 10^5 \,\text{Pa}}.$$

First, notice that the pascal (Pa) units cancel out. Next, perform the arithmetic with the numbers and powers of ten:

$$T_2 = \frac{1.2 \times 10^6}{3 \times 10^5} \times 300 \,\text{K}.$$

The fraction of the coefficients is $$\frac{1.2}{3} = 0.4$$, and the fraction of the powers of ten is $$\frac{10^6}{10^5} = 10^{6-5} = 10^1 = 10$$. Therefore

$$T_2 = 0.4 \times 10 \times 300 \,\text{K}.$$

Multiplying step by step, $$0.4 \times 10 = 4$$, and then $$4 \times 300 = 1200.$$ Hence

$$T_2 = 1200 \,\text{K}.$$

We now convert this absolute temperature back to the Celsius scale by reversing the earlier formula:

$$t_2 (^\circ\text{C}) = T_2 (\text{K}) - 273.$$

Substituting $$T_2 = 1200 \,\text{K}$$, we get

$$t_2 = 1200 - 273 = 927^\circ\text{C}.$$

This temperature is the minimum room temperature at which the cylinder pressure will reach $$1.2 \times 10^6 \,\text{Pa}$$ and consequently burst.

So, the answer is $$927^\circ\text{C}.$$

We are given that the empty (tare) mass of the cylinder is $$14.8\ \text{kg}.$$

When the cylinder is completely filled with LPG its total mass becomes $$29.0\ \text{kg}.$$

Hence the initial mass of LPG present is obtained by simple subtraction:

$$m_1 \;=\; 29.0\ \text{kg} - 14.8\ \text{kg} = 14.2\ \text{kg}.$$

At this stage the gauge shows an initial pressure

$$P_1 = 3.47\ \text{atm}.$$

During usage the mass of the cylinder (shell + remaining LPG) falls to $$23.0\ \text{kg}.$$

Therefore the mass of LPG that still remains is

$$m_2 \;=\; 23.0\ \text{kg} - 14.8\ \text{kg} = 8.2\ \text{kg}.$$

We treat LPG as an ideal gas. The ideal-gas equation is first stated:

$$PV = nRT,$$

where  $$n = \dfrac{m}{M}$$  is the number of moles, $$m$$ is the mass of gas, and $$M$$ its molar mass. For a rigid steel cylinder, the volume $$V$$ is constant. The problem says the ambient temperature is unchanged, so $$T$$ is also constant. Hence, for two different states of the same gas in the same cylinder, the ratio

$$\dfrac{P}{n} = \dfrac{RT}{V}$$

remains fixed. Substituting $$n = m/M$$, we get

$$P = \dfrac{RT}{VM}\,m.$$

Everything in the prefactor $$\dfrac{RT}{VM}$$ is constant, so pressure is directly proportional to mass when temperature and volume stay fixed:

$$\dfrac{P_1}{P_2} = \dfrac{m_1}{m_2}.$$

We now substitute the numerical values:

$$P_2 = P_1 \times \dfrac{m_2}{m_1} = 3.47\ \text{atm} \times \dfrac{8.2\ \text{kg}}{14.2\ \text{kg}}.$$

First compute the mass ratio:

$$\dfrac{8.2}{14.2} = 0.57746\;(\text{approximately}).$$

Then multiply by the initial pressure:

$$P_2 = 3.47 \times 0.57746 = 2.0038\ \text{atm}\;(\text{approximately}).$$

Rounding to the nearest whole number, we have

$$P_2 \approx 2\ \text{atm}.$$

So, the answer is $$2$$.

First, calculate the moles of each gas. The molar mass of methane ($$CH_4$$) is $$12 + 4 \times 1 = 16$$ g/mol, so moles of methane = $$\frac{6.4}{16} = 0.4$$ mol. The molar mass of carbon dioxide ($$CO_2$$) is $$12 + 2 \times 16 = 44$$ g/mol, so moles of carbon dioxide = $$\frac{8.8}{44} = 0.2$$ mol.

The total moles of gas in the mixture is $$n_{total} = 0.4 + 0.2 = 0.6$$ mol. The temperature in Kelvin is $$27 + 273 = 300$$ K.

Using the ideal gas law $$PV = nRT$$, the total pressure is $$P = \frac{nRT}{V} = \frac{0.6 \times 8.314 \times 300}{10} = \frac{1496.52}{10} = 149.652$$ kPa. Rounding to the nearest integer gives $$P \approx 150$$ kPa.

$$n_1 = \frac{2.8}{28} = 0.1\ mol$$

$$n_2 = \frac{0.2}{28} = 0.00714\ mol$$

Since the system is insulated,

$$n_1T_1 + n_2T_2 = (n_1+n_2)T_f$$

$$T_f = \frac{n_1T_1+n_2T_2}{n_1+n_2}$$

$$T_f = \frac{(0.1)(300)+(0.00714)(60)}{0.1+0.00714}$$

$$T_f = \frac{30+0.4284}{0.10714}$$

$$T_f = 284K$$

$$V_f = 1L + 2L = 3L = 3\times10^{-3}m^3$$

$$n_f = \frac{2.8}{28}+\frac{0.2}{28} = 0.10714\ mol$$

$$P_f = \frac{n_fRT_f}{V_f}$$

$$P_f = \frac{(0.10714)(8.31)(284)}{3\times10^{-3}}$$

$$P_f = 8.41\times10^4Pa$$

$$1bar = 10^5Pa$$

$$P_f = \frac{8.41\times10^4}{10^5}$$

$$P_f = 0.841bar$$

$$P_f = 84\times10^{-2}bar$$

$$\therefore x = 84$$

The starting alkene is:

$$\mathrm{cis\text{-}2\text{-}butene}$$

The reaction involves bromination:

$$\mathrm{Br_2/CCl_4}$$

Addition of bromine occurs through formation of a cyclic bromonium ion intermediate.

Hence, the reaction proceeds via:

$$\mathrm{anti\text{-}addition}$$

Stereochemical outcomes of anti-addition:

$$\mathrm{cis\text{-}alkene \rightarrow Racemic\ Mixture}$$

$$\mathrm{trans\text{-}alkene \rightarrow meso\ Compound}$$

Since the starting alkene is:

$$\mathrm{cis\text{-}2\text{-}butene}$$

the product formed is a racemic mixture of:

$$\mathrm{(2R,3R)\text{-}2,3\text{-}dibromobutane}$$

and

$$\mathrm{(2S,3S)\text{-}2,3\text{-}dibromobutane}$$

Thus, total stereoisomers formed:

$$\mathrm{= 2}$$

Final answer:

$$\mathrm{2}$$

We have been told that monochromatic radiation of wavelength $$\lambda = 500\ \text{nm}$$ (nanometres) is incident on a metal surface. The corresponding photon frequency is obtained from the fundamental relation between wavelength, frequency and the speed of light, namely $$c = \nu \lambda$$, where $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ is the speed of light.

Re-arranging, the frequency of the incident photons is

$$ \nu \;=\; \frac{c}{\lambda}. $$

Substituting $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ and $$\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5.0 \times 10^{-7}\ \text{m},$$ we get

$$ \nu \;=\; \frac{3.0 \times 10^{8}}{5.0 \times 10^{-7}} \;=\; 6.0 \times 10^{14}\ \text{Hz}. $$

The metal has a threshold (cut-off) frequency $$\nu_0 = 4.3 \times 10^{14}\ \text{Hz}$$. According to Einstein’s photoelectric equation, the maximum kinetic energy $$K_{\text{max}}$$ of an ejected electron is given by

$$ K_{\text{max}} = h(\nu - \nu_0), $$

where $$h = 6.63 \times 10^{-34}\ \text{J s}$$ is Planck’s constant.

First find the excess frequency above the threshold:

$$ \nu - \nu_0 = 6.0 \times 10^{14} - 4.3 \times 10^{14} = 1.7 \times 10^{14}\ \text{Hz}. $$

Now calculate the kinetic energy:

$$ K_{\text{max}} = h(\nu - \nu_0) = (6.63 \times 10^{-34}\ \text{J s})(1.7 \times 10^{14}\ \text{Hz}) = 1.1271 \times 10^{-19}\ \text{J}. $$

For an electron, this kinetic energy is related to its speed $$v$$ through the classical formula $$K = \tfrac12 m_e v^{2}$$, where the electron mass is $$m_e = 9.0 \times 10^{-31}\ \text{kg}$$. Stating the formula first:

$$ \frac12 m_e v^{2} = K_{\text{max}}. $$

Solving for $$v$$ gives

$$ v = \sqrt{\frac{2K_{\text{max}}}{m_e}}. $$

Substituting the numerical values:

$$ v = \sqrt{\frac{2(1.1271 \times 10^{-19})}{9.0 \times 10^{-31}}} = \sqrt{\frac{2.2542 \times 10^{-19}}{9.0 \times 10^{-31}}}. $$

Divide the mantissas and handle the powers of ten separately:

$$ \frac{2.2542}{9.0} = 0.250466\quad\text{and}\quad 10^{-19} \div 10^{-31} = 10^{12}. $$

Thus

$$ v = \sqrt{0.250466 \times 10^{12}} = \sqrt{2.50466 \times 10^{11}}. $$

Taking the square root:

$$ v \approx 5.0 \times 10^{5}\ \text{m s}^{-1}. $$

Because the question asks for the velocity expressed as “_________ $$\times 10^{5}$$ m s$$^{-1}$$ (nearest integer)”, we read off the factor as $$5$$.

So, the answer is $$5$$.

For a gas of molecular mass $$M$$ at a temperature $$T$$, the three characteristic molecular speeds are defined as follows:

1. Most Probable Speed ($$V_{\text{mp}}$$): The speed possessed by the maximum fraction of total molecules, representing the absolute peak of the curve.

   $$V_{\text{mp}} = \sqrt{\frac{2RT}{M}} \approx 1.414 \sqrt{\frac{RT}{M}}$$



2. Average Speed ($$V_{\text{av}}$$): The mathematical mean of the speeds of all individual molecules.

   $$V_{\text{av}} = \sqrt{\frac{8RT}{\pi M}} \approx 1.596 \sqrt{\frac{RT}{M}}$$



3. Root Mean Square Speed ($$V_{\text{rms}}$$): The square root of the mean of the squares of individual molecular speeds.

   $$V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \approx 1.732 \sqrt{\frac{RT}{M}}$$

Comparing the Speeds:

By comparing the numerical values under identical conditions of temperature and molar mass, we get a definitive increasing order of magnitude:

$$V_{\text{mp}} < V_{\text{av}} < V_{\text{rms}}$$

The precise ratio between these speeds is:

$$V_{\text{mp}} : V_{\text{av}} : V_{\text{rms}} = 1 : 1.128 : 1.224$$

Mapping to the Graph Axis:

On the horizontal "speed" axis of the Maxwell-Boltzmann distribution graph, values systematically increase from left to right:

• Label A sits at the leftmost position (lowest speed value), directly identifying the peak of the curve: $$A = V_{\text{mp}}$$
• Label B occupies the intermediate position on the curve: $$B = V_{\text{av}}$$
• Label C sits at the rightmost position (highest speed value), stretching down toward the high-speed tail: $$C = V_{\text{rms}}$$

Conclusion:

The values increase moving rightwards along the x-axis, matching the order of most probable, average, and root mean square speed values.

Answer: Option D — A = V$$_{mp}$$; B = V$$_{av}$$; C = V$$_{rms}$$

The ideal gas equation can be rewritten to express density by combining the basic gas law with molecular weight properties:

$$PV = nRT = \frac{m}{M}RT$$

Rearranging the equation to solve for density ($$d = \frac{m}{V}$$) gives:

$$P = \frac{m}{V}\frac{RT}{M} \implies P = \frac{dRT}{M}$$

Isolating the density ($$d$$), we get our working formula:

$$d = \left(\frac{M}{R}\right)\frac{P}{T}$$

Step-by-Step Analysis of Each Graph:

• Graph I: $$d$$ vs $$T$$ (at constant $$P$$)

  • From the formula, keeping $$P$$ constant means density is inversely proportional to temperature: $$d \propto \frac{1}{T}$$.
  • The graph of an inverse relationship is a rectangular hyperbola.
  • Status: Graph I is correct.



• Graph II: $$d$$ vs $$T$$ (at constant $$P$$)

  • As established above, $$d$$ and $$T$$ share an inverse relationship ($$d \propto \frac{1}{T}$$). As temperature increases, gas volume expands, causing the density to decrease.
  • Graph II depicts a linear increase with a positive slope ($$d \propto T$$), which is physically impossible for an ideal gas.
  • Status: Graph II is incorrect.



• Graph III: $$d$$ vs $$\frac{1}{T}$$ (at constant $$P$$)

  • Since $$d \propto \frac{1}{T}$$, plotting $$d$$ directly against the inverse of temperature ($$\frac{1}{T}$$) yields a linear relationship ($$y = mx$$).
  • The graph should be a straight line passing through the origin.
  • Status: Graph III is correct.



• Graph IV: $$d$$ vs $$P$$ (at constant $$T$$)

  • Keeping $$T$$ constant means density is directly proportional to pressure: $$d \propto P$$.
  • The graph should be a straight line passing through the origin with a positive slope.
  • Status: Graph IV is correct.

Conclusion:

Because density is inversely related to temperature at a constant pressure, Graph II is geometrically and scientifically flawed.

Answer: Option B — II

First, we recall the ideal-gas equation for any single component: $$P\,V = n\,R\,T$$ where $$P$$ is the pressure exerted by that component, $$V$$ is the volume it occupies, $$n$$ is the amount of substance in moles, $$R$$ is the universal gas constant, and $$T$$ is the absolute temperature.

Because the gases behave ideally and are contained in the same vessel, each gas exerts its own pressure as if the others were absent. This individual pressure is called the partial pressure. Mathematically, for any one gas, we have $$P_i = \dfrac{n_i R T}{V}.$$

We are told that one mole of $$H_2$$ exerts a partial pressure of $$2\;\text{atm}$$. Substituting $$n_{H_2}=1$$ into the formula gives

$$P_{H_2} \;=\; \dfrac{(1)\,R\,T}{V} \;=\; 2\;\text{atm}.$$

From this equality we can express the common factor $$\dfrac{R\,T}{V}$$ directly:

$$\dfrac{R\,T}{V} \;=\; 2\;\text{atm}.$$

Now, the mixture contains three different gases—$$H_2$$, He, and $$O_2$$—each present in one mole. Hence the total amount of substance is

$$n_{\text{total}} = 1 + 1 + 1 = 3\;\text{mol}.$$

The total pressure of the mixture, according to Dalton’s law, is the sum of the partial pressures. Using the ideal-gas expression for the whole mixture, we have

$$P_{\text{total}} = \dfrac{n_{\text{total}}\,R\,T}{V}.$$

Substituting $$n_{\text{total}} = 3$$ and our earlier result $$\dfrac{R\,T}{V} = 2\;\text{atm}$$, we obtain

$$P_{\text{total}} = 3 \times 2\;\text{atm} = 6\;\text{atm}.$$

Hence, the correct answer is Option A.

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

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