JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 22

Answer: (a) SO₃

Explanation:
1. An electrophile is electron-deficient and accepts an electron pair.
2. In SO3 sulphur is bonded to three oxygens and has no lone pair or negative charge, so it is electron-poor and attacks electron-rich sites.
3. RNH2, nitrate ion and ROH all have lone pairs to donate, so they are nucleophiles.

Teacher's Note:
1. Electrophiles are electron-pair acceptors (Lewis acids).
2. Species with a lone pair or a negative charge are normally nucleophiles.

Answer: (d) NH₂⁻

Explanation:
1. A nucleophile donates a pair of electrons.
2. The amide ion NH₂⁻ carries a negative charge and lone pairs, so it is a nucleophile.
3. NO₂⁺ (cation) and :CCl₂ (carbene, electron-deficient) are electrophiles, and CH₃ free radical (one unpaired electron) is a free radical.

Teacher's Note:
1. Check for lone pairs and negative charge.
2. Cations and electron-deficient carbenes are electrophiles.

Answer: (c) Carbocation

Explanation:
1. A carbocation has three bonds and an empty p orbital, so carbon has 3 pairs = 6 electrons.
2. A carbanion has 8 electrons (3 bond pairs and a lone pair) and a free radical has 7.

Teacher's Note:
1. Carbocation: sextet. Carbanion: octet. Radical: 7 electrons.
2. Count the bonding and non-bonding electrons around carbon.

Answer: (b) base as well as a nucleophile

Explanation:
1. A carbanion has a lone pair on carbon and a negative charge.
2. It can accept a proton (acts as a base) and can attack electron-poor centres (acts as a nucleophile).

Teacher's Note:
1. Lone-pair carriers are both bases and nucleophiles.
2. Carbanions are conjugate bases of C–H acids.

Answer: (a) ⁻CH₂NO₂

Explanation:
1. Alkyl groups (+I) intensify the negative charge, so CH₃⁻ is more stable than ethyl, isopropyl or sec-butyl carbanions.
2. The nitro group withdraws electrons by −I and −M effects and delocalises the charge, so ⁻CH₂NO₂ is much more stable than CH₃⁻.

Teacher's Note:
1. Electron-withdrawing groups stabilise carbanions; alkyl groups destabilise them.
2. Order of carbanion stability: methyl > 1° > 2° > 3°.

Answer: (d) −NH₃⁺ > −NO₂ > −CN

Explanation:
1. The group with a full positive charge, −NH₃⁺, withdraws electrons most strongly.
2. Between the neutral groups, −NO₂ (sp² N, two oxygens) is stronger than −CN.

Teacher's Note:
1. A formal positive charge gives the strongest −I effect.
2. Learn the order: −NH₃⁺ > −NO₂ > −CN > −COOH > −F.

Answer: (c) HC≡C⁻

Explanation:
1. Stability of a carbanion increases with the s-character of the carbon bearing the lone pair.
2. HC≡C⁻ is sp-hybridised (50% s), vinyl anions are sp² (33%), alkyl anions sp³ (25%).

Teacher's Note:
1. More s-character means the lone pair is held closer to the nucleus.
2. Order: sp > sp² > sp³.

Answer: (b) III > II > I

Explanation:
1. The lone pair of III is in an sp orbital (50% s), of II in an sp² orbital (33% s) and of I in an sp³ orbital (25% s).
2. More s-character gives greater stability: III > II > I.

Teacher's Note:
1. Relate stability to s-character.
2. The same order explains why ethyne is more acidic than ethene and ethane.

Answer: (a) (CH₃)₂CHNH₂

Explanation:
1. The more alkyl groups (+I) on or near nitrogen, the more available the lone pair for protonation.
2. Of the primary amines, isopropylamine has the largest, most branched alkyl group, so it is the strongest base here; ammonia is the weakest.

Teacher's Note:
1. Alkyl groups push electrons towards nitrogen.
2. Basic strength: iPrNH₂ > EtNH₂ > MeNH₂ > NH₃ for these.

Answer: (d) CH₃COOH

Explanation:
1. Electron-withdrawing F and Cl stabilise the carboxylate ion and make the acid stronger (lower pKₐ).
2. HCOOH has no +I group. Acetic acid has the +I methyl group, which destabilises the carboxylate, so it is the weakest and has the highest pKₐ.

Teacher's Note:
1. Higher pKₐ means a weaker acid.
2. +I groups weaken carboxylic acids.

Answer: (c) Trifluoroacetic acid

Explanation:
1. Three fluorine atoms exert a very strong −I effect and stabilise the carboxylate ion greatly.
2. This outweighs the single Cl in chloroacetic acid and the weaker −I of OCH₃.

Teacher's Note:
1. Count and compare the −I strength of the substituents.
2. F is more electronegative than Cl and O.

Answer: (b) −CH₃

Explanation:
1. Alkyl groups push electrons towards the carbon they are attached to (+I effect).
2. −OH, −OCH₃ and −Cl are more electronegative than carbon and show a −I effect.

Teacher's Note:
1. Alkyl groups are weakly electron-releasing by induction.
2. −OH and −OCH₃ show +M but −I.

Answer: (a) CH₃CH₂CCl₂COOH

Explanation:
1. Ionisation depends on the stability of the carboxylate ion.
2. Two Cl atoms on the carbon next to COOH (the α-carbon) give the strongest −I effect, and the effect falls off quickly with distance, so 2,2-dichlorobutanoic acid is the strongest acid.

Teacher's Note:
1. More halogens and closer halogens increase acid strength.
2. The inductive effect decreases rapidly with distance.

Answer: (d) CH₃CH₂CHClCOOH

Explanation:
1. Chlorine on the α-carbon (2-chlorobutanoic acid) exerts a strong −I effect on the carboxylate.
2. In 4-chlorobutanoic acid the Cl is far away, and acetic and formic acids have no Cl, so 2-chlorobutanoic acid is strongest.

Teacher's Note:
1. Position of the halogen matters: α is best.
2. Compare pKₐ values: about 2.8 for 2-chlorobutanoic acid, 4.5 for 4-chlorobutanoic acid, 4.76 for acetic acid and 3.75 for formic acid.

Answer: (c) HCOOH

Explanation:
1. Alkyl groups release electrons and weaken the acid.
2. Formic acid has an H in place of CH₃, so no +I effect: pKₐ 3.75 against 4.76 for acetic acid.
3. Propanoic, butanoic and 2-methylpropanoic acids have larger alkyl groups and are weaker than acetic acid.

Teacher's Note:
1. More alkyl substitution means weaker acid.
2. HCOOH is the strongest simple aliphatic monocarboxylic acid.

Answer: (b) CCl₃COOH

Explanation:
1. The lowest pKₐ means the strongest acid.
2. Three Cl atoms on the α-carbon give the strongest −I effect, so trichloroacetic acid is the strongest acid in the list.

Teacher's Note:
1. Strongest acid has the lowest pKₐ.
2. Trichloroacetic acid has pKₐ about 0.7.

Answer: (a) 4-Methoxybenzyl cation (p-CH₃O–C₆H₄–CH₂⁺)

Explanation:
1. The charge on the benzylic carbon is delocalised into the ring.
2. A p-OCH₃ group donates electron density by its strong +M effect and stabilises the cation most; Cl and NO₂ withdraw electrons, and benzyl has no extra stabilisation.

Teacher's Note:
1. +M donor groups at the para position stabilise benzylic cations.
2. −M and −I groups destabilise them.

Answer: (d) 4-Nitrobenzoic acid

Explanation:
1. Carboxylic acids are stronger than phenol.
2. A p-NO₂ group withdraws electrons by −I and −M, stabilising the benzoate ion, so 4-nitrobenzoic acid is stronger than benzoic acid, whereas p-CH₃ weakens the acid.

Teacher's Note:
1. Electron-withdrawing groups raise the acidity of benzoic acid.
2. Electron-donating groups lower it.

Answer: (c) Benzene

Explanation:
1. Benzene has a fully delocalised aromatic π system with resonance energy of about 150 kJ/mol.
2. Cyclohexane has no π bond, and cyclohexene has an isolated one, and the diene has only limited conjugation.

Teacher's Note:
1. Aromaticity gives special stabilisation.
2. Resonance energy is the extra stability from delocalisation.

Answer: (b) CH₂=CH–Cl (vinyl chloride)

Explanation:
1. In vinyl chloride the lone pair on Cl is delocalised into the C=C (+M), giving the C–Cl bond partial double-bond character.
2. Also the carbon is sp²-hybridised, so the bond is shorter than the sp³ C–Cl bonds in the others.

Teacher's Note:
1. Resonance shortens a single bond.
2. Vinyl and aryl halides have short, strong C–X bonds and are unreactive in substitution.

Answer: (a) d < b < c < a

Explanation:
1. The ethyl cation has no resonance, so it is the least stable: d is lowest.
2. Benzyl (b) is resonance-stabilised; p-methyl (c) adds +I and hyperconjugation; p-methoxy (a) adds a strong +M effect and is the best.
3. So d < b < c < a.

Teacher's Note:
1. Resonance beats simple inductive effects.
2. Strong +M donors stabilise most.

Answer: (d) Benzylamine (C₆H₅CH₂NH₂)

Explanation:
1. In benzylamine the NH₂ group is separated from the ring by CH₂, so the lone pair is not delocalised and it is available like an alkylamine.
2. In the anilines the lone pair is delocalised into the ring, and nitro groups reduce basicity further.

Teacher's Note:
1. Aliphatic amines are stronger bases than aromatic amines.
2. Delocalisation of the lone pair reduces basicity.

Answer: (c) IV > I > III > II

Explanation:
1. The +I/hyperconjugation of CH₃ makes 4-methylaniline more basic than aniline.
2. −M/−I groups (CHO, NO₂) in the para position lower basicity, and NO₂ is the stronger acceptor, so the aldehyde is less deactivating.
3. Order: IV > I > III > II.

Teacher's Note:
1. Electron donors raise basicity of anilines; acceptors lower it.
2. Compare the strength of the acceptor groups.

Answer: (b) 4-Nitrobenzyl carbanion (p-O₂N–C₆H₄–CH₂⁻)

Explanation:
1. A para nitro group delocalises the negative charge of the benzylic carbanion onto the oxygens of NO₂ (−M and −I).
2. The methoxy group destabilises the carbanion, and in the 2-phenylethyl carbanion the charge is not conjugated with the ring.

Teacher's Note:
1. Electron-withdrawing groups stabilise carbanions.
2. Conjugation with a −M group gives the greatest stabilisation.

Answer: (a) Cycloheptatrienyl (tropylium) cation

Explanation:
1. The tropylium ion is a planar, cyclic, fully conjugated ion with 6 π electrons (Hückel 4n+2 with n = 1), so it is aromatic and exceptionally stable.
2. The cyclopentadienyl cation has 4 π electrons (antiaromatic) and benzyl and allyl cations are only resonance-stabilised.

Teacher's Note:
1. Aromatic cations are the most stable.
2. Tropylium bromide is an ionic solid.