JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 23

Answer: (b) Propene

Explanation:
1. Hyperconjugation needs a C–H bond on an sp³ carbon attached to an sp² or sp carbon (or to a carbocation/radical centre).
2. Propene has CH₃ attached to a C=C, so it shows hyperconjugation. Ethyne, phenol and ethanol have no such α C–H bonds next to a multiple bond.

Teacher's Note:
1. Count α-hydrogens next to an unsaturated carbon.
2. Hyperconjugation is also called no-bond resonance.

Answer: (a) CH₃CH₂CH₂⁺

Explanation:
1. The n-propyl cation is primary with only two α C–H bonds for hyperconjugation and little +I support.
2. Isopropyl is secondary, t-butyl is tertiary and the benzylic cation is stabilised by the ring, so n-propyl is the least stable.

Teacher's Note:
1. Order: benzylic/tertiary > secondary > primary.
2. More α-hydrogens means more hyperconjugation.

Answer: (d) Propene

Explanation:
1. Stability of an alkene by hyperconjugation depends on the number of α C–H bonds on the carbons of the double bond.
2. Ethene has 0, propene 3, but-1-ene 2 and 3-methylbut-1-ene only 1, so propene has the most.

Teacher's Note:
1. Count the α-hydrogens.
2. This is a simple counting rule, not the only factor in real alkene stability.

Answer: (c) pent-1-ene < cis-pent-2-ene < trans-pent-2-ene < 2-methylbut-2-ene

Explanation:
1. The more alkyl groups on the double bond carbons, the more stable the alkene: pent-1-ene (mono) < pent-2-ene (di) < 2-methylbut-2-ene (tri).
2. Trans isomers are more stable than cis because there is less steric strain.

Teacher's Note:
1. Substitution increases stability.
2. Trans > cis for disubstituted alkenes.

Answer: (b) migration of an active (α) hydrogen atom

Explanation:
1. Tautomers are constitutional isomers in rapid equilibrium that differ in the position of a proton (active hydrogen) and a double bond.

Teacher's Note:
1. Tautomerism needs a mobile hydrogen.
2. It is different from resonance, where only electrons move.

Answer: (a) Cyclohexa-2,4-dien-1-one

Explanation:
1. Enolisation of cyclohexa-2,4-dienone gives phenol, which is aromatic and very stable, so the equilibrium lies almost entirely on the enol side.
2. The others form non-aromatic enols whose content is small (acetone about 0.0001%) or moderate (1,3-dicarbonyls stabilised by hydrogen bonding).

Teacher's Note:
1. Aromatic enol means complete enolisation.
2. Compare the stability of the enol form.

Answer: (d) Tautomerism

Explanation:
1. Urea can shift an N–H hydrogen to oxygen, giving the isourea form HN=C(OH)NH₂, so it exhibits keto-enol-like (amide–imidol) tautomerism.

Teacher's Note:
1. Amides and ureas can show amide–imidol tautomerism.
2. Tautomerism involves transfer of a proton.

Answer: (c) 2,2-Dimethylpropanal ((CH₃)₃C–CHO)

Explanation:
1. Tautomerism needs a hydrogen on the α-carbon.
2. (CH₃)₃C–CHO has no α-hydrogen, so it cannot enolise. Propanone and nitromethane have α-H, and the enol C₆H₅CH=CHOH converts to the aldehyde.

Teacher's Note:
1. No α-H means no keto-enol tautomerism.
2. Nitro compounds with α-H show nitro–aci-nitro tautomerism.

Answer: (b) Nitrobenzene

Explanation:
1. Nitroalkanes with an α-hydrogen exist partly in the acidic aci-nitro form and dissolve in alkali as salts.
2. Nitrobenzene has no α-hydrogen (the carbon attached to NO₂ is aromatic and has no H to lose in this way), so it does not dissolve.

Teacher's Note:
1. α-H is needed for aci-nitro tautomerism.
2. Nitroalkane solubility in NaOH tests for α-H.

Answer: (a) Malonaldehyde (OHC–CH₂–CHO)

Explanation:
1. Enol content depends on how well the enol is stabilised by conjugation and intramolecular hydrogen bonding.
2. The enol of malonaldehyde is a fully conjugated hydrogen-bonded ring with no alkyl substituents to cause strain, and it is the major form; diethyl malonate has the least enol content.

Teacher's Note:
1. Compare: aldehyde > ketone > ester carbonyls in enolising ability.
2. Diesters enolise the least.

Answer: (d) Benzylamine

Explanation:
1. In benzylamine the lone pair is not conjugated with the ring, so it is far more basic than aniline.
2. Diphenylamine and triphenylamine have the lone pair delocalised into two or three rings, and 4-nitroaniline has an electron-withdrawing NO₂.

Teacher's Note:
1. Delocalised lone pair means weaker base.
2. Benzylamine behaves like an aliphatic amine.

Answer: (c) FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH

Explanation:
1. The −I effect of the halogen falls in the order F > Cl > Br, so the acid strength follows the same order.
2. Acetic acid, with no halogen, is the weakest.

Teacher's Note:
1. More electronegative halogen means stronger acid.
2. Electronegativity: F > Cl > Br.

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

Explanation:
1. s-character of the carbanion carbon: (a) sp, 50%; (b) and (c) sp², 33%; (d) sp³, 25%.
2. Phenyl anion (b) is slightly more stable than the alkyl vinyl anion (c) because the aryl ring is more electron-withdrawing. Order: a > b > c > d.

Teacher's Note:
1. Higher s-character means more stability.
2. Alkyl groups destabilise carbanions.

Answer: (a) a > b > c

Explanation:
1. The more stable the carbanion, the weaker it is as a base.
2. Stability (s-character) is c > b > a, so basic strength is a > b > c.

Teacher's Note:
1. Stability and basicity run in opposite directions.
2. Conjugate acid strength: ethyne > ethene > ethane.

Answer: (d) Phenol

Explanation:
1. Phenol loses H⁺ to give the phenoxide ion, whose negative charge is delocalised over the ring.
2. In alcohols the negative charge of the alkoxide stays on oxygen and alkyl groups destabilise it, so alcohols are far weaker acids.

Teacher's Note:
1. Resonance stabilises phenoxide.
2. Phenol pKₐ about 10; alcohols about 16–18.

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

Explanation:
1. Approximate pKₐ values: picric acid 0.4, acetic acid 4.76, phenol 10, cyclohexanol about 16.
2. So the decreasing order of acidity is III > II > IV > I.

Teacher's Note:
1. Three nitro groups make phenol as strong as a mineral acid.
2. Carboxylic acids are stronger than phenols.

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

Explanation:
1. A para NO₂ group acts by both −I and −M effects, a meta NO₂ by −I only, so 4-nitrophenol (pKₐ 7.15) is stronger than 3-nitrophenol (8.4).
2. Phenol (10.0) is stronger than 4-methylphenol (10.3), whose CH₃ group donates electrons.

Teacher's Note:
1. −M works from ortho and para positions only.
2. +I groups reduce acidity.

Answer: (a) H₃O⁺

Explanation:
1. BH₃ (6 electrons on boron), NO₂⁺ and Cl⁺ are electron-deficient and accept electron pairs in reactions with nucleophiles.
2. In H₃O⁺ oxygen has a full octet; it acts as a proton donor (an acid) rather than as an electrophile at oxygen, so it is the odd one out here.

Teacher's Note:
1. Electrophiles have incomplete octets or a positive centre that accepts electrons.
2. H₃O⁺ is better described as an acid.

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

Explanation:
1. Approximate pKₐ values: propanoic 4.87, acetic 4.76, phenylacetic 4.31, acetoacetic 3.6.
2. So pKₐ decreases in the order II > I > III > IV (the stronger acid has the lower pKₐ).

Teacher's Note:
1. The C=O group (−I) strengthens acetoacetic acid most.
2. The phenyl group is −I, so phenylacetic acid is stronger than acetic acid.

Answer: (c) Benzylamine

Explanation:
1. Benzylamine's lone pair is localised on nitrogen.
2. In aniline it is delocalised into the ring (and more so with NO₂), and in acetanilide it is also delocalised onto the carbonyl oxygen.

Teacher's Note:
1. Amides are very weakly basic.
2. Alkyl-type amines are strongest.

Answer: (b) a, b and c

Explanation:
1. (a) correct: oxygen lone pair donates to the cation (+M).
2. (b) correct: secondary is more stable than primary.
3. (c) correct: allyl cation is resonance-stabilised.
4. (d) wrong: a vinyl cation (sp carbon, empty p orbital in the plane) is less stable than ethyl.

Teacher's Note:
1. Check each statement separately.
2. Resonance gives the biggest stabilisation.

Answer: (a) C-2

Explanation:
1. The hydrogens on C-2 (α to the carbonyl) are acidic because the resulting enolate is resonance-stabilised.
2. The aldehydic H on C-1 and hydrogens farther away (C-3, C-4) are not activated.

Teacher's Note:
1. Only α-hydrogens are acidic in carbonyl compounds.
2. Enolate resonance gives the stabilisation.

Answer: (d) 3-Acetylpentane-2,4-dione ((CH₃CO)₃CH)

Explanation:
1. The C–H flanked by three carbonyl groups gives an anion delocalised over three C=O groups.
2. Pentane-2,4-dione has two such groups, and acetone and ethanal only one.

Teacher's Note:
1. More carbonyl groups on the same carbon means more acidic C–H.
2. Resonance stabilisation of the conjugate base decides acidity.

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

Explanation:
1. Piperidine is a secondary aliphatic amine (sp³ N, free lone pair) and the strongest base (pKₐ of conjugate acid about 11).
2. Pyridine has an sp² nitrogen (pKₐ 5.2), aniline's lone pair is delocalised (4.6), and pyrrole's lone pair is part of the aromatic sextet (essentially non-basic).

Teacher's Note:
1. A lone pair that takes part in aromaticity is unavailable for protonation.
2. sp³ N is more basic than sp² N.

Answer: (b) a < b < c

Explanation:
1. Bond length increases as the s-character of the bonded carbons decreases.
2. Central bond of butadiene: sp²–sp² (shortest, with some double-bond character); propene: sp³–sp²; ethane: sp³–sp³ (longest).

Teacher's Note:
1. More s-character means a shorter bond.
2. Conjugation also shortens the central bond of butadiene.