General Organic Chemistry MCQ for JEE Main Revision

Step 1: Determine Principal Functional Group Between the carboxylic acid ($\ce{-COOH}$) and the aldehyde ($\ce{-CHO}$), the carboxylic acid has higher priority. The parent chain must start from the acid group.

Step 2: Numbering and Chain

1. Numbering begins at $\ce{-COOH}$.
2. The longest chain containing the double bond and the other carbonyl has 6 carbons $\rightarrow$ Hex-3-enoic acid.

Step 3: Substituents

• C2: Methyl group.
• C5: Methyl group.
• C6: Carbonyl group (aldehyde carbon) $\rightarrow$ Named as "oxo" when part of the chain.

Step 4: Final IUPAC Name 2,5-Dimethyl-6-oxohex-3-enoic acid.

$\boxed{\text{Answer: (c) 2,5-Dimethyl-6-oxo-hex-3-enoic acid}}$

Resonance Stability Principles $\boxed{\text{Answer: (b)}}$

Step 1: Matching Effects to Mechanisms

• (A) Aniline resonance: Electrons move into the ring. This is $+R$ effect. (A-IV)
• (B) Cyclohexene + $\ce{H^+}$: Double bond electrons migrate to the attacking reagent ($\ce{H^+}$). This is $+E$ effect. (B-III)
• (C) Cyclohexene + $\ce{CN^-}$: Double bond electrons move away from the attacking reagent ($\ce{CN^-}$). This is $-E$ effect. (C-I)
• (D) Nitrobenzene resonance: Electrons move toward the substituent. This is $-R$ effect. (D-II)

Step 2: Correct Match (A)-(IV), (B)-(III), (C)-(I), (D)-(II).

$\boxed{\text{Answer: (a)}}$

Strategy: Electron density on the benzene ring is increased by electron-donating groups ($+M/+I$) and decreased by electron-withdrawing groups ($-M/-I$).

Step 1: Analyze the substituents

• (IV): The nitrogen of the amide/amine is directly attached to the ring and bears a negative charge (anion). This is a very strong $+M$ effect, making the ring extremely electron-rich.
• (I): Standard saturated ring fused to benzene. No strong M-effects.
• (III): Carbonyl group ($-M$) attached to the ring withdraws electron density.
• (II): Two carbonyl groups ($-M$) attached to the ring withdraw density even more effectively.

Step 2: Determine Order Decreasing density: IV > I > III > II

$\boxed{\text{Answer: (Option C) IV > I > III > II}}$

Strategy: Major resonating contributors are determined by: (1) Completeness of octets, (2) Greater number of covalent bonds, (3) Lack of charge separation, and (4) Negative charge on more electronegative atoms.

Step 1: Analyze each set

• (P): (II) is major because both C and N have complete octets and there are more bonds compared to (I).
• (Q): (II) is major because oxygen is more electronegative than carbon and carries the negative charge more stably than the carbanion in (I).
• (R): (II) is major because all atoms (including Carbon and Nitrogen) have complete octets, whereas in (I) the carbocation has an incomplete octet.
• (S): (II) is major because it has no charge separation on the ring and all atoms have complete octets.

Step 2: Conclusion The major contributors are II, II, II, II.

$\boxed{\text{Answer: (Option C) II, II, II, II}}$

Strategy: Stability of ions is determined by resonance, inductive effects, and charge distribution.

Step 1: Analyze Option (C) Anion (1) is the phenoxide ion ($C_6H_5O^-$). Anion (2) is the benzoate ion ($C_6H_5COO^-$). Benzoate is much more stable than phenoxide because the negative charge is delocalized over two equivalent oxygen atoms in the carboxylate group, whereas in phenoxide it is delocalized into the less electronegative carbon atoms of the ring.

Step 2: Verify other options

• (A): Cation on carbonyl carbon vs vinylic. (B): Pyridinium derivative vs O-anion. (D): Formate vs Trifluoroacetate. Option (C) correctly states Phenoxide < Benzoate.

$\boxed{\text{Answer: (Option C)}}$

Strategy: Hyperconjugation requires (1) a vacant p-orbital (carbocation), a partially filled p-orbital (radical), or a $\pi^*$ orbital (alkene), and (2) adjacent $C-H$ $\sigma$-bonds that are not constrained by geometry (Bredt's Rule).

Step 1: Analyze the species

• (A): Tertiary radical at a bridgehead-like or constrained position with availabe $\alpha$-H.
• (B): Triphenylmethyl radical. Stabilized purely by resonance, no $\alpha$-H available for hyperconjugation.
• (C): Radical at a bridgehead. According to Bredt's Rule, hyperconjugation is inhibited because the required planar transition state for the double bond cannot be achieved.
• (D): Allylic radical system where $\alpha$-hydrogens are available on the adjacent carbons of the rings.

Step 2: Conclusion Hyperconjugation is most clearly and effectively observed in the systems with unconstrained $\alpha$-H like in Option (D).

$\boxed{\text{Answer: (Option D)}}$

Check the Acidity Lab in the Organic Chemistry Hub page under revision tools on www.canvasclasses.in to visualise such comparisons.

Step 1: Comparing Acidic Strength Acidity depends on the stability of the conjugate base.

• Phenol: Resonace stabilized phenoxide ion.
• Sulfonic Acid ($\ce{R-SO3H}$): Highly stabilized by resonance with three oxygens.
• Carboxylic Acid ($\ce{R-COOH}$): Stabilized by resonance with two oxygens.
• Terminal Alkyne ($\ce{R-C \equiv C-H}$): The negative charge on $sp$-hybridized carbon is less stable than resonance-stabilized oxygens.

Step 2: Identifying the Least Acidic Terminal alkynes ($pK_a \approx 25$) are significantly less acidic than carboxylic acids ($pK_a \approx 5$), phenols ($pK_a \approx 10$), or sulfonic acids ($pK_a < 0$).

$\boxed{\text{Answer: (a) Compound D }}$

Step 1: Acidity of Active Methylene Protons Deprotonation occurs most readily where the conjugate base is most stable.

• Structure A: 2,4-pentanedione. The active methylene hydrogens are flanked by two methyl ketones. The $-I$ and $-R$ effects of two keto groups stabilize the anion.
• Structure B: Dimethyl malonate. Esters are less electron-withdrawing than ketones due to the $+R$ effect of the alkoxy group.
• Structure C: Ethyl acetoacetate. One ketone and one ester.

Step 2: Conclusion Structure A has two ketone groups, which provide the strongest stabilization to the conjugate base compared to esters.

$\boxed{\text{Answer: (a) A only}}$

Step 1: Comparing Nitrogen Basicity Basicity depends on the availability of the nitrogen lone pair.

• (C) Pyrrolidine: $sp^3$ nitrogen, localized lone pair. Most basic.
• (D) Imidazole: One nitrogen lone pair is localized ($sp^2$), the other is aromatic. Still fairly basic as the $sp^2$ nitrogen can accept a proton.
• (A) Pyridine: $sp^2$ nitrogen lone pair is localized but in a more electronegative orbital than $sp^3$.
• (B) Pyrrole: lone pair is involved in aromaticity (6π system). Least basic.

Step 2: Conclusion Order: (C) > (D) > (A) > (B).

$\boxed{\text{Answer: (d) (B) < (A) < (D) < (C)}}$

pKr+ provides a quantitative measure of carbocation stability. A HIGHER (more positive) value indicates GREATER stability.

1. Tropylium Cation (C): It is aromatic ($6\pi$ electrons, $4n+2$ where $n=1$), giving it exceptional stability.
2. Triphenylmethyl Cation (A): The positive charge is delocalized over 3 phenyl rings.
3. Diphenylmethyl Cation (B): The charge is delocalized over 2 phenyl rings.
4. Sec-butyl Cation (D): Only stabilized by inductive effect and hyperconjugation.

Step 2: Stability Order Stability decreases in the order: Tropylium > Triphenylmethyl > Diphenylmethyl > Secondary alkyl.

$\boxed{\text{Answer: (a) C > A > B > D}}$

Identifying Most Stable Carbocation Carbocation stability is determined by resonance, inductive effect, and hyperconjugation. $\boxed{\text{Answer: (c)}}$

Hydride Affinity and Carbocation Stability Hydride affinity is the tendency of a carbocation to accept a hydride ion. It is inversely proportional to the stability of the carbocation. The more stable the carbocation, the lower its affinity for hydride.

• (D): Tricyclopropylmethyl cation (Exceptionally stable due to Bent orbital overlap). Lowest hydride affinity.
• (B): Triphenylmethyl cation (Stable due to resonance with three rings).
• (A): Allylic + tertiary stabilized cation.
• (C): tert-butyl cation (Least stable among the four).

Step 2: Order of Affinity Affinity order: (C) > (A) > (B) > (D).

$\boxed{\text{Answer: (b)}}$

Step 1: Analyzing Acidic Hydrogens in Ketones Acidity depends on the resonance stabilization of the conjugate base (enolate ion). Step 2: Determining Maximum Acidity The enolate formed in structure (d) is stabilized by resonance with two adjacent $\ce{C=O}$ groups, making these hydrogens highly acidic ($pK_a \approx 9$).

$\boxed{\text{Answer: (d)}}$

Identifying Nucleophiles Nucleophiles are species that have a lone pair of electrons or a $\pi$ bond that can be donated to an electrophile. Evaluating Other Species

• $\ce{H3O^+}$: Electrophile (positive charge, no lone pair to donate).
• $\ce{(CH3)_2CO}$ and $\ce{>C=NCH3}$: Electrophilic at the C-atom. Total count = 5.

$\boxed{\text{Answer: (c) 5}}$

Linear conjugation (in C) is more stable than cross-conjugation (in B). Linear systems allow for continuous, uninterrupted delocalization of electrons across the entire molecule, maximizing resonance energy. In contrast, cross-conjugated systems have broken paths where the conjugation is split, leading to less efficient delocalization and lower stability.

Keto-Enol Tautomerism:

The enol content of a carbonyl compound depends on the stability of the enol form. Factors that stabilize the enol form include:

1. Conjugation with other double bonds
2. Intramolecular hydrogen bonding
3. Aromaticity (if applicable)

Analysis of Options:

(a) Cyclohexane-1,3-dione: The two carbonyl groups are separated by one CH₂ group, so the enol form has extended conjugation and hydrogen bonding making enol form particularly stable.

(b) Cyclohexane-1,3,5-trione: This compound has three carbonyl groups alternating with CH₂ groups. The enol form is aromatic and has maximum conjugation and multiple intramolecular hydrogen bonds, making it the most stable enol form. Aromaticity alone is the biggest differentiator here.

(c) Cyclohexane-1,4-dione: Has two carbonyl groups with two CH₂ groups between them. The enol form has no conjugation and no hydrogen bonding due to both carbonyl groups positioned diagonally across.

(d) Cyclohexane-1,2,3-trione: Three adjacent carbonyl groups. The middle carbonyl group can't participate in tautomerism. Also three carbonyl groups repel each other strongly due to being on same side. The enol form has limited stabilization.

Conclusion: Cyclohexane-1,3,5-trione (option b) shows the highest enol content due to maximum stabilization through aromaticity and multiple intramolecular hydrogen bonding sites.

$\boxed{\text{Answer: (b)}}$

Conditions for Geometrical Isomerism: For a compound to show geometrical (cis-trans or E/Z) isomerism:

1. Must have a C=C double bond (restricted rotation)
2. Each carbon of the double bond must have two different groups
3. The double bond should allow for different spatial arrangements

$\boxed{\text{Answer: (c)}}$

Step 1: Identify Stereocenters Step 2: Calculate Total Isomers Since the molecule is unsymmetrical, the total number of stereoisomers is $2^n$, where $n$ is the number of stereocenters ($n = 3$). $\text{Total Isomers} = 2^3 = 8$

$\boxed{\text{Answer: (a) 8}}$

Tropolone has a total of 8 π electrons, but only 6 π electrons within the ring are involved in aromaticity. The first statement is correct that it has aromatic character and 8 π electrons.

Second statement is incorrect because the pi electrons of the carbonyl group are not involved in aromaticity.

Step 1: Analyzing Squaric Acid Reaction Squaric acid reacts with 2 equivalents of $\ce{OH-}$ to form the squarate dianion (Compound B).

• (A) Aromaticity: The squarate dianion is a 2π-electron system (resonance hybrid) which is aromatic and very stable. (True)
• (C) Tautomerism: Squaric acid (Compound A) exists in equilibrium with its tautomeric forms. (True)
• (D) Bond Lengths: Due to aromaticity and resonance, all $C-C$ bond lengths in the square dianion are identical. (True)
• Statement (B) says completion is very slow, which is typically false for acid-base neutralizations of such sharp acids.

$\boxed{\text{Answer: (d) (A), (C) and (D) only}}$

Strategy: A compound is aromatic if it is cyclic, planar, fully conjugated, and has $4n+2$ $\pi$ electrons. It is non-aromatic if it fails any of the first three criteria (usually conjugation is broken by an $sp^3$ atom). Step 1: Analyze each compound

• (a): Aziridinium cation-like system. The top nitrogen is $sp^3$ and has no lone pair/p-orbital for conjugation. The system is not fully conjugated. Therefore, it is non-aromatic.
• (b): Pyrylium cation. $6 \pi$ electrons, fully conjugated. Aromatic.
• (c): Cyclopentadienyl anion. $6 \pi$ electrons, fully conjugated. Aromatic.
• (d): Furan-like system (1,3-dioxole derivative). $6 \pi$ electrons (including lone pairs from oxygens), fully conjugated. Aromatic.

Step 2: Conclusion Compound (a) is non-aromatic.

$\boxed{\text{Answer: (Option a)}}$