JEE Basic Principles of Organic Chemistry Questions

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.

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.

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.

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.

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

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.

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).

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.

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.

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

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.

In the given structure Numbering would be start from leftmost carbon hence methyl at 5 , oxo at 2 and since functional group is acetic acid hence name would be , 5-methyl-2-oxo-hexane-6-oic acid

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 $$7$$.

aromatic compounds  should have cyclic planar structure with (4n+2)pi electrons .

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

• 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.

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: $$NH_2^- > NH_3$$.

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

CH3COOH is primarily functional groups, so numbering would be start from there. C attached to primarily functional group would be rank 1 as per IUPAC nomenclature hence methyl would be at 4 nitro at 6 ,oxo at 3 and double bond at 4 with total 6 carbon hence name would be 4-methyl-6 nitro-3-oxhept-4-enal

Longest 

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).

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 $$\boxed{3}$$.

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).

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:

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:

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

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,

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

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

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

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

$$T_f = 284K$$

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

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

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

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

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

$$1bar = 10^5Pa$$

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

$$P_f = 0.841bar$$

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

$$\therefore x = 84$$

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 $$5$$.

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.

We have an initial spherical balloon of radius $$r_1 = 3\;\text{cm}$$ which contains helium gas at a pressure $$P_1 = 48 \times 10^{-3}\;\text{bar}$$. A second balloon of radius $$r_2 = 12\;\text{cm}$$ is filled with the same amount of helium gas and is kept at the same temperature. For an ideal gas the relation $$PV = nRT$$ holds. Because the temperature $$T$$ and the amount of gas $$n$$ remain unchanged, the product $$PV$$ is constant, so

$$P_1 V_1 = P_2 V_2.$$

The volume of a sphere is given by the formula $$V = \dfrac{4}{3}\pi r^{3}$$. Substituting the two radii, we write

$$V_1 = \dfrac{4}{3}\pi r_1^{3}, \qquad V_2 = \dfrac{4}{3}\pi r_2^{3}.$$

Dividing the second volume by the first, the common factor $$\dfrac{4}{3}\pi$$ cancels:

$$\frac{V_2}{V_1} \;=\; \frac{r_2^{3}}{r_1^{3}}.$$

Now we substitute the numerical radii:

$$\frac{V_2}{V_1} = \left(\frac{12}{3}\right)^{3} = 4^{3} = 64.$$

Hence $$V_2 = 64\,V_1.$$ Putting this result in $$P_1 V_1 = P_2 V_2$$, we have

$$P_1 V_1 = P_2 (64\,V_1).$$

The volume factor $$V_1$$ now cancels from both sides, yielding

$$P_2 = \frac{P_1}{64}.$$

Substituting $$P_1 = 48 \times 10^{-3}\;\text{bar}$$, we get

$$P_2 = \frac{48 \times 10^{-3}}{64}\;\text{bar}.$$

We first carry out the division of the numerical coefficients. Writing the fraction explicitly,

$$\frac{48}{64} = \frac{48 \div 16}{64 \div 16} = \frac{3}{4} = 0.75.$$

Therefore,

$$P_2 = 0.75 \times 10^{-3}\;\text{bar}.$$

The answer is required in the form “$$\dots \times 10^{-6}\;\text{bar}$$”. To convert $$0.75 \times 10^{-3}$$ into a $$\times 10^{-6}$$ expression, we shift the factor $$10^{-3}$$ three powers lower:

$$0.75 \times 10^{-3} \;=\; 0.75 \times 10^{3} \times 10^{-6}.$$\; (since $$10^{-3} = 10^{3} \times 10^{-6}$$)

Now, $$0.75 \times 10^{3} = 0.75 \times 1000 = 750.$$ Hence

$$P_2 = 750 \times 10^{-6}\;\text{bar}.$$

So, the answer is $$750$$.

For photoelectric emission we use Einstein’s photoelectric equation, which states

$$E_{\text{photon}} = \phi \;+\; K.E.$$

where $$E_{\text{photon}}$$ is the energy of the incident photon, $$\phi$$ is the work-function of the metal, and $$K.E.$$ is the kinetic energy of the emitted electron.

First we calculate the photon energy from its wavelength. The energy-wavelength relation is

$$E_{\text{photon}} \;=\; \dfrac{h\,c}{\lambda}.$$

We have the numerical values

$$h = 6.63 \times 10^{-34}\,\text{J s}, \qquad c = 3.00 \times 10^{8}\,\text{m s}^{-1}, \qquad \lambda = 300\,\text{nm} = 300 \times 10^{-9}\,\text{m}.$$

Substituting these in the formula gives

$$E_{\text{photon}} = \dfrac{(6.63 \times 10^{-34})\,(3.00 \times 10^{8})}{300 \times 10^{-9}}\;\text{J}.$$

Multiplying the numerators,

$$6.63 \times 3.00 = 19.89, \quad\text{and}\quad 10^{-34}\times 10^{8}=10^{-26},$$

so the numerator becomes

$$19.89 \times 10^{-26}\;\text{J m}.$$

For the denominator, write

$$300 \times 10^{-9} = (3.00 \times 10^{2}) \times 10^{-9} = 3.00 \times 10^{-7}\;\text{m}.$$

Now divide:

$$E_{\text{photon}} = \dfrac{19.89 \times 10^{-26}}{3.00 \times 10^{-7}}\;\text{J} = \left(\dfrac{19.89}{3.00}\right) \times 10^{-26+7}\;\text{J}.$$

Calculating the numeric and exponent parts separately,

$$\dfrac{19.89}{3.00} = 6.63, \qquad -26 + 7 = -19,$$

hence

$$E_{\text{photon}} = 6.63 \times 10^{-19}\;\text{J}.$$

The work-function of sodium is given as

$$\phi = 4.41 \times 10^{-19}\;\text{J}.$$

Using Einstein’s equation, the kinetic energy of the photoelectrons is

$$K.E. = E_{\text{photon}} - \phi = (6.63 \times 10^{-19}) - (4.41 \times 10^{-19})\;\text{J}.$$

Subtracting the coefficients while keeping the common power of ten,

$$6.63 - 4.41 = 2.22,$$

so

$$K.E. = 2.22 \times 10^{-19}\;\text{J}.$$

Finally, the problem asks for the answer in units of $$\times 10^{-21}\;\text{J}.$$ To rewrite the kinetic energy accordingly, note that

$$2.22 \times 10^{-19}\;\text{J} = 2.22 \times 10^{2} \times 10^{-21}\;\text{J} = 222 \times 10^{-21}\;\text{J}.$$

So, the answer is $$222$$.

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

Its structural formula is:

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

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

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

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

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

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

The correct answer is $$2$$.

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

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

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

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

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

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

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

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

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

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

Adding the contributions, we have

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

So, the answer is $$9$$.

We recall that the compression factor is defined as $$Z=\dfrac{pV}{RT}.$$

For the gases in the problem the experimentally observed equation of state neglects the attraction term and is written as $$p=\dfrac{RT}{V-b},$$ where $$b$$ is the van der Waals excluded-volume constant.

We first manipulate this equation to obtain $$V$$ in terms of $$p$$.

Starting with $$p(V-b)=RT,$$ we multiply out to get $$pV-pb=RT.$$

Now we isolate $$V$$. Adding $$pb$$ to both sides and dividing by $$p$$, we obtain $$V=b+\dfrac{RT}{p}.$$

We substitute this expression for $$V$$ in the definition of the compressibility factor: $$Z=\dfrac{pV}{RT}= \dfrac{p\left(b+\dfrac{RT}{p}\right)}{RT}.$$

We separate the two terms in the numerator: $$Z=\dfrac{pb}{RT}+\dfrac{p}{RT}\cdot \dfrac{RT}{p}.$$

The second term simplifies immediately to 1, giving the very convenient linear relation $$Z=1+\dfrac{b}{RT}\,p.$$

Thus, at the fixed temperature $$T$$ of the experiment, the plot of $$Z$$ versus $$p$$ is a straight line with slope $$\text{slope}= \dfrac{b}{RT}.$$

Because $$R$$ and $$T$$ are common to all the gases, the steepness of the line is governed solely by the magnitude of $$b$$. The larger the value of $$b$$, the steeper the increase of $$Z$$ with pressure.

The van der Waals constant $$b$$ reflects the effective molecular size; for the noble gases it increases with atomic size in the order $$\text{Ne} < \text{Ar} < \text{Kr} < \text{Xe}.$$

Therefore, xenon (Xe) has the largest $$b$$ and consequently the largest value of $$\dfrac{b}{RT}$$, giving the greatest slope in the $$Z$$ vs. $$p$$ plot.

Hence, the correct answer is Option D.

For real gases we use the relation $$PV = Z\,nRT$$, where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the number of moles, $$T$$ is the temperature, $$R$$ is the universal gas constant and $$Z$$ is the compressibility factor.

Both gases A and B are taken at the same temperature and contain an equal number of moles, so $$n$$, $$R$$ and $$T$$ are common to both. Hence we can write

$$P_A V_A = Z_A nRT \quad\text{and}\quad P_B V_B = Z_B nRT.$$

Dividing the first equation by the second we get

$$\dfrac{P_A V_A}{P_B V_B} = \dfrac{Z_A}{Z_B}.$$

The problem tells us that the volume of gas A is twice that of gas B, so $$V_A = 2V_B.$$ Substituting this into the ratio gives

$$\dfrac{P_A (2V_B)}{P_B V_B} = \dfrac{Z_A}{Z_B}.$$

Simplifying the left side, $$V_B$$ cancels:

$$\dfrac{2P_A}{P_B} = \dfrac{Z_A}{Z_B}.$$

We are also told that the compressibility factor of gas A is three times that of gas B, so $$Z_A = 3Z_B.$$ Substituting this value yields

$$\dfrac{2P_A}{P_B} = \dfrac{3Z_B}{Z_B} = 3.$$

Now solving for the ratio of the pressures, we have

$$2P_A = 3P_B.$$

This is exactly the relation asked for between the two pressures. Hence, the correct answer is Option A.

We begin by writing the van der Waals equation. For n moles of a real gas the equation is stated as

$$\left(P+\frac{a n^{2}}{V^{2}}\right)\,(V-nb)=nRT.$$

Here $$a$$ corrects the pressure term for intermolecular attractions, while $$b$$ corrects the volume term for the finite size of the molecules. A larger $$a$$ therefore signifies stronger attractions (the gas can be squeezed more easily), and a larger $$b$$ signifies a larger “excluded volume’’ per mole (each mole physically occupies more space).

To compare the four gases we put $$n=1$$ (one mole):

$$\left(P+\frac{a}{V^{2}}\right)\,(V-b)=RT. \quad -(1)$$

Throughout the comparison we imagine all gases kept at the same temperature $$T$$ and pressure $$P$$ so that the right-hand side, $$RT,$$ is the same for every gas.

Comparison of gases A and C (volume occupied)

For gas A $$a_A = 642.32\;\text{kPa dm}^6\text{ mol}^{-1},\qquad b_A = 0.05196\;\text{dm}^3\text{ mol}^{-1}.$$

For gas C $$a_C = 431.91\;\text{kPa dm}^6\text{ mol}^{-1},\qquad b_C = 0.05196\;\text{dm}^3\text{ mol}^{-1}.$$

Both gases have exactly the same value of $$b,$$ but gas A possesses a larger value of $$a$$ than gas C. From equation (1) we see that for a fixed $$P$$ the factor $$\displaystyle\left(P+\frac{a}{V^{2}}\right)$$ becomes larger when $$a$$ is larger. To keep the product equal to the constant $$RT,$$ the other factor $$(V-b)$$ must then become smaller. Therefore,

$$a_A > a_C\quad\Longrightarrow\quad (V_A-b) < (V_C-b)\quad\Longrightarrow\quad V_A < V_C.$$

Thus one mole of gas C occupies more volume than one mole of gas A under identical conditions.

Comparison of gases B and D (ease of compressibility)

For gas B $$a_B = 155.21,\; b_B = 0.04136.$$

For gas D $$a_D = 155.21,\; b_D = 0.4382.$$

Both gases have the same value of $$a,$$ so the attraction effect is identical. Their difference lies in $$b$$: gas D has a much larger excluded volume than gas B. To see how this affects compressibility, we rewrite equation (1) in the form of the compressibility factor $$Z$$:

$$Z \;=\; \frac{PV}{RT}\;=\;\frac{V}{V-b} \;-\; \frac{a}{RTV}. \quad -(2)$$

The first term $$\dfrac{V}{V-b}$$ (the repulsive correction) becomes larger as $$b$$ becomes larger, making $$Z$$ increase above 1. A larger $$Z$$ indicates that the gas resists compression (its volume is already “puffed up’’ by repulsions). Consequently, a gas with the larger $$b$$ is less compressible.

Because $$b_D \gg b_B,$$ we have

$$Z_D > Z_B \quad\Longrightarrow\quad \text{gas D is less compressible than gas B}.$$

Equivalently, gas B is more compressible than gas D.

Putting both parts together

We have shown that

(i) Gas C occupies more volume than gas A, and

(ii) Gas B is more compressible than gas D.

The option that states both facts correctly is Option C.

Hence, the correct answer is Option C.

We recall the van der Waals relation between the critical temperature $$T_c$$ and the constants $$a$$ and $$b$$:

$$T_c=\frac{8a}{27Rb}$$

Here $$R$$ is the universal gas constant. Since $$R$$ is the same for all gases being compared, the factor $$\dfrac{8}{27R}$$ is common and need not be calculated explicitly. Thus, for the purpose of comparison, the critical temperature is directly proportional to the ratio $$\dfrac{a}{b}$$.

So, to find which gas has the highest $$T_c$$, we need to compute $$\dfrac{a}{b}$$ for each gas. The given data are:

Argon (Ar): $$a = 1.3\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 3.2\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Neon (Ne): $$a = 0.2\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 1.7\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Krypton (Kr): $$a = 5.1\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 1.0\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Xenon (Xe): $$a = 4.1\;\text{atm dm}^6\text{ mol}^{-2}, \; b = 5.0\times10^{-2}\;\text{dm}^3\text{ mol}^{-1}$$

Now we evaluate the ratios one by one.

For Ar: $$\frac{a}{b}=\frac{1.3}{3.2\times10^{-2}}=\frac{1.3}{0.032}=40.625$$

For Ne: $$\frac{a}{b}=\frac{0.2}{1.7\times10^{-2}}=\frac{0.2}{0.017}=11.7647$$

For Kr: $$\frac{a}{b}=\frac{5.1}{1.0\times10^{-2}}=\frac{5.1}{0.01}=510$$

For Xe: $$\frac{a}{b}=\frac{4.1}{5.0\times10^{-2}}=\frac{4.1}{0.05}=82$$

We observe that:

$$\frac{a}{b}\;(\text{Ne}) < \frac{a}{b}\;(\text{Ar}) < \frac{a}{b}\;(\text{Xe}) < \frac{a}{b}\;(\text{Kr})$$

Since $$T_c \propto \dfrac{a}{b}$$, the gas with the largest $$\dfrac{a}{b}$$ value will possess the highest critical temperature. The highest ratio is clearly for Krypton (Kr) with a value of 510.

Hence, the correct answer is Option A.

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

We have to select that compound which fulfils both of the following conditions:

(i) it must react with the Grignard reagent $$\mathrm{C_2H_5MgBr}$$ (ethyl-magnesium bromide), and

(ii) it must decolourize an aqueous bromine solution (bromine water).

First we recall the general behaviour of a Grignard reagent. Ethyl-magnesium bromide is represented as $$\mathrm{C_2H_5^- \,\,Mg^{2+}Br^-}$$, so the carbon attached to magnesium behaves as a very strong base and nucleophile. Therefore, any compound containing an acidic hydrogen (-OH, -NH, -SH, etc.) instantly reacts with a Grignard reagent according to

$$\mathrm{R{-}H \;+\; C_2H_5MgBr \;\longrightarrow\; R^-\,MgBr^+ \;+\; C_2H_6}\,,$$

where $$\mathrm{R{-}H}$$ denotes the acidic compound and the ethyl group becomes $$\mathrm{C_2H_6}$$ (ethane).

Now we recall the test with bromine water. Aqueous $$\mathrm{Br_2}$$ is ruddy brown; it gets decolourised by

  • compounds containing an electron-rich double bond $$\mathrm{C=C}$$ (alkenes, enols), and

  • compounds whose aromatic ring is strongly activated, for example phenol, which undergoes electrophilic bromination to give 2,4,6-tribromophenol:

$$\mathrm{C_6H_5OH + 3\,Br_2(aq) \;\longrightarrow\; C_6H_2Br_3OH + 3\,HBr}$$

and the reddish colour of $$\mathrm{Br_2}$$ disappears.

Let us check each option qualitatively:

• Option A carries a $$\mathrm{C\equiv N}$$ and a carbonyl oxygen. It does react with a Grignard reagent (nitrile and carbonyl are both reactive), but neither group can decolourise bromine water because there is no strongly activated aromatic ring nor an alkenic double bond able to add bromine in water. So Option A fails condition (ii).

• Option B contains a nitrile $$\mathrm{-C\!\equiv N}$$ and an ester $$\mathrm{-CO_2CH_3}$$. Again, it reacts with the Grignard reagent, but an ester/nitrile does not affect bromine water. Hence Option B also fails condition (ii).

• Option C shows an ether $$\mathrm{-OCH_3}$$ and a carbon-carbon double bond $$\mathrm{CH=CH_2}$$. The double bond definitely decolourises bromine water, satisfying condition (ii). However, simple alkenes do not possess an acidic hydrogen and therefore do not react with ethyl-magnesium bromide. Thus Option C fails condition (i).

• Option D bears a hydroxyl group $$\mathrm{-OH}$$, and from the context of typical JEE questions this -OH is almost always on an aromatic ring (phenol). We analyse it quantitatively:

Reaction with Grignard reagent

Applying the general acid-base formula written earlier, we let $$\mathrm{ArOH}$$ represent phenol:

$$\mathrm{ArOH + C_2H_5MgBr \;\longrightarrow\; ArO^-\,MgBr^+ + C_2H_6}.$$

The hydrogen of the hydroxyl group is sufficiently acidic (pKₐ ≈ 10), so it is abstracted, giving magnesium phenoxide and ethane gas. Condition (i) is thus satisfied.

Decolourisation of bromine water

The aromatic ring of phenol is strongly activated by the $$\mathrm{-OH}$$ group. When bromine water is added, electrophilic substitution occurs at the ortho and para positions:

$$\mathrm{ArOH + 3\,Br_2 (aq) \;\longrightarrow\; Ar(Br)_3OH + 3\,HBr},$$

and the brown colour of $$\mathrm{Br_2}$$ vanishes. Hence condition (ii) is also satisfied.

Therefore only the compound possessing the hydroxyl (phenolic) group fulfils both required properties.

Hence, the correct answer is Option 4.

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

We have to compare the acid strengths of the four molecules $$CH(CN)_3$$, $$CHI_3$$, $$CHBr_3$$ and $$CHCl_3$$. In every case the acidic hydrogen is attached to the same kind of carbon framework, namely a $$-CHX_3$$ type group where the three substituents X are either $$CN$$, $$I$$, $$Br$$ or $$Cl$$. So the deciding factor is how strongly each group X withdraws electron density from the $$C-H$$ bond. A greater electron-withdrawing effect stabilises the conjugate base more, which in turn gives a stronger acid.

First, we recall the relationship between acid strength and the stability of its conjugate base. The acid-base equilibrium can be written as

$$\text{HA}\rightleftharpoons \text{H}^+ + \text{A}^-.$$

The equilibrium constant for this reaction is the acid-dissociation constant $$K_a$$, given by the formula

$$K_a=\dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}.$$

When the conjugate base $$\text{A}^-$$ is stabilised by any effect (inductive, resonance, hyperconjugation, etc.), the numerator becomes effectively larger, the value of $$K_a$$ increases and the acid is stronger (p$$K_a=-\log K_a$$ becomes smaller).

Now we analyse the nature of the groups:

• The $$CN$$ group exerts a very powerful −I (inductive) effect because the nitrogen is highly electronegative and is doubly bonded to carbon. Moreover, the $$CN$$ group also has a −M (electron-withdrawing resonance) effect because the $$\pi$$ electrons can delocalise toward the nitrogen. Both effects strongly pull electron density away from the central carbon, greatly stabilising the conjugate base $$C(CN)_3^-$$.

• The halogens $$I$$, $$Br$$ and $$Cl$$ also show a −I effect; however, their strengths decrease in the order $$Cl > Br > I$$ because inductive withdrawal becomes weaker as the atom becomes larger and less electronegative.

• None of the halogens can offer an additional strong −M effect toward a saturated carbon the way the $$CN$$ group does in this molecule, so their total withdrawing capacity is much less than that of $$CN$$.

So, to arrange the groups in order of total electron withdrawal we have

$$CN \;\; \gg \;\; Cl \; > \; Br \; > \; I.$$

Consequently, the conjugate base of $$CH(CN)_3$$ is the most stabilised, giving the largest $$K_a$$ and the smallest p$$K_a$$ value among the four acids. In other words, $$CH(CN)_3$$ is the strongest acid.

Therefore the order of acid strength is

$$CH(CN)_3 \; \text{(strongest)} \; > \; CHCl_3 \; > \; CHBr_3 \; > \; CHI_3 \; \text{(weakest)}.$$

Among the given options the molecule $$CH(CN)_3$$ corresponds to Option A.

Hence, the correct answer is Option A.

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

Four commonly discussed factors are:

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

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

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

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

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

Hence, the correct answer is Option B.