JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 32

Answer: (b) (ii) < (i) < (iii)

Explanation:
1. The para methoxy group donates electrons by the +M effect, which destabilises the carboxylate ion, so (ii) is weaker than benzoic acid.
2. The ortho isomer is stronger than benzoic acid because of the ortho effect and the –I effect of oxygen at a close distance.
3. Increasing acidity: p-methoxy < benzoic acid < o-methoxy.

Teacher's Note:
1. For para substituents, +M donors weaken benzoic acid and –M or –I groups strengthen it.
2. Ortho substituents usually raise the acidity.

Answer: (a) the anion formed after loss of the first H⁺ of oxalic acid is more stable because of the stronger –I effect of the nearby –COOH group

Explanation:
1. In oxalic acid the second –COOH is directly attached to the carboxylate carbon, so its –I effect acts at the shortest possible distance.
2. This spreads the negative charge and stabilises the monoanion, giving a larger Ka1.
3. In malonic acid a CH2 group separates the two COOH groups and weakens the effect.

Teacher's Note:
1. The –I effect falls off rapidly with distance.
2. Molecular mass has no bearing on acid strength.

Answer: (d) phenoxide ion is stabilised through delocalisation of the negative charge

Explanation:
1. In the phenoxide ion the negative charge on oxygen is delocalised into the benzene ring (ortho and para positions) by resonance.
2. In ethoxide the charge is localised on oxygen and is further intensified by the +I effect of the ethyl group.
3. The better stabilised anion means a more acidic parent compound.

Teacher's Note:
1. Acid strength is judged by the stability of the conjugate base.
2. A stable conjugate base is a weak base.

Answer: (a) TTTT

Explanation:
1. S1: true, pKa 8.40 is lower than 9.98 because –I stabilises the phenoxide.
2. S2: true, hydrogen bonding in the ortho isomer holds the OH hydrogen and lowers its acidity slightly (7.23 against 7.15).
3. S3: true, the –M effect of NO2 delocalises the negative charge only from the ortho and para positions.
4. S4: true, a group at the meta position cannot conjugate with the OH-bearing carbon.
5. All four are true: TTTT.

Teacher's Note:
1. The –M effect operates only from the ortho and para positions.
2. Intramolecular hydrogen bonding reduces ionisation.

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

Explanation:
1. Two nitro groups (ortho and para) give the greatest stabilisation of the phenoxide: pKa about 4.1.
2. One para nitro group gives pKa 7.15, phenol 9.98 and the electron donating p-cresol 10.2.
3. Order: III > IV > I > II.

Teacher's Note:
1. More –M/–I groups at ortho and para positions mean more acidity.
2. A +I group such as methyl lowers the acidity.

Answer: (b) \( \ce{HC#CNa + H2O -> HC#CH + NaOH} \)

Explanation:
1. An acid displaces a weaker acid from its salt, never a stronger one.
2. (a) Formic acid is stronger than acetic acid, so the reaction would go the other way.
3. (b) Water (pKa 15.7) is a far stronger acid than ethyne (pKa about 25), so sodium acetylide is protonated by water: feasible.
4. (c) p-Nitrophenol is stronger than phenol, so the reaction is not feasible.
5. (d) Benzenesulfonic acid is a strong acid, much stronger than benzoic acid, so the reaction does not occur.

Teacher's Note:
1. The conjugate base of the weaker acid is the stronger base.
2. The order of acid strength is RSO3H > RCOOH > H2CO3 > phenol > water > alcohol > ethyne.

Answer: (d) by both the Lewis and Lowry-Bronsted concepts

Explanation:
1. An amine accepts a proton through the lone pair on nitrogen, which is Lowry-Bronsted basicity.
2. The same lone pair can be donated to any electron deficient species such as BF3, which is Lewis basicity.
3. Amines contain no OH group, so the Arrhenius concept does not apply directly.

Teacher's Note:
1. Every Lewis base with a lone pair can also accept H+.
2. The Arrhenius concept needs OH– to be released.

Answer: (b) all of these

Explanation:
1. Nitrogen is less electronegative than oxygen, so its lone pair is held less tightly and is donated more easily.
2. Hence amines are more basic than alcohols and ethers; in esters the oxygen lone pair is also delocalised into C=O, making esters even weaker bases.

Teacher's Note:
1. Basicity decreases across a period as electronegativity increases.
2. Delocalisation of the lone pair reduces basicity.

Answer: (a) \( \ce{(C2H5)2NH} \) > \( \ce{(C2H5)3N} \) > \( \ce{C2H5NH2} \) > \( \ce{NH3} \)

Explanation:
1. For ethyl amines the +I effect is stronger, so the pKb values are 3.00 for the secondary amine, 3.25 for the tertiary amine, 3.29 for the primary amine and 4.75 for ammonia.
2. Hence the order is 2° > 3° > 1° > NH3.

Teacher's Note:
1. In water, the best balance of +I effect, solvation and steric hindrance is reached in the secondary amine.
2. All alkylamines are stronger bases than NH3.

Answer: (d) \( \ce{CH3CH2^-} \) > \( \ce{CH2=CH^-} \) > \( \ce{NH2^-} \) > \( \ce{HC#C^-} \) > \( \ce{HO^-} \) > \( \ce{F^-} \)

Explanation:
1. The stronger the base, the weaker its conjugate acid, so rank the anions by the pKa of the conjugate acid.
2. Approximate pKa values: ethane 50, ethene 44, ammonia 35, ethyne 25, water 15.7, hydrogen fluoride 3.2.
3. Hence the basic strength falls in the order ethyl anion > vinyl anion > amide ion > ethynide ion > hydroxide > fluoride.

Teacher's Note:
1. Rank by the pKa of the conjugate acid.
2. An sp carbanion is more stable (less basic) than an sp2 one, which is more stable than an sp3 one.

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

Explanation:
1. Protonation of acetamidine gives an amidinium ion in which the positive charge is shared equally by both nitrogen atoms, so it is the strongest base (conjugate acid pKa about 12.4).
2. Dimethylamine (pKa of conjugate acid 10.7) is slightly stronger than ethylamine (10.6).
3. Acetamide is a very weak base because its nitrogen lone pair is delocalised into C=O.
4. Order: I > III > II > IV.

Teacher's Note:
1. Resonance stabilisation of the conjugate acid raises basicity.
2. Amides are the weakest bases of these because of lone pair delocalisation.

Answer: (c) guanidine, \( \ce{H2N-C(=NH)-NH2} \)

Explanation:
1. Protonation of guanidine gives the guanidinium ion, whose positive charge is shared equally among three nitrogen atoms by resonance.
2. This stabilisation makes guanidine one of the strongest neutral organic bases (conjugate acid pKa about 13.6).
3. The alkylamines have pKa near 10.7 and triphenylamine is almost non-basic.

Teacher's Note:
1. The more stabilised the conjugate acid, the stronger the base.
2. Guanidine is as strong a base as many hydroxides.

Answer: (b) piperidine

Explanation:
1. Piperidine is a saturated secondary amine with a localised sp3 nitrogen lone pair (conjugate acid pKa about 11).
2. Pyridine's lone pair is in an sp2 orbital (pKa 5.2), aniline's is delocalised into the ring (4.6), and in pyrrole the lone pair is part of the aromatic sextet (essentially non-basic).

Teacher's Note:
1. sp3 nitrogen is more basic than sp2 nitrogen.
2. A lone pair that is part of an aromatic sextet is not available for protonation.

Answer: (a) 2 < 5 < 1 < 3 < 4

Explanation:
1. Electron withdrawing groups lower basicity and donating groups raise it.
2. p-Nitro (–M and –I) is the strongest withdrawing arrangement (conjugate acid pKa 1.0); m-nitro acts by –I only (2.5).
3. Aniline is 4.6, p-toluidine 5.1 (+I, hyperconjugation) and p-anisidine 5.3 (+M).
4. Increasing basicity: 2 < 5 < 1 < 3 < 4.

Teacher's Note:
1. The para nitro group conjugates with the amino group, so it reduces basicity more than the meta nitro group.
2. +M donors in the para position increase basicity.

Answer: (a) N-ethylbenzylamine, \( \ce{C6H5CH2NHC2H5} \)

Explanation:
1. N-Ethylbenzylamine is a secondary alkyl amine: a CH2 group separates the ring from nitrogen, so the lone pair is localised.
2. In N-propylaniline the lone pair is delocalised into the ring.
3. In the two amides the lone pair is delocalised into C=O, making them very weak bases.

Teacher's Note:
1. Look for an amine nitrogen with no attached aryl or acyl group.
2. Amides are the weakest bases.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Assertion: pKa of p-nitrophenol is 7.15 and of o-nitrophenol 7.23, so the para isomer is the stronger acid. True.
2. Reason: in the ortho isomer the OH hydrogen is held by a hydrogen bond to the nitro group, so it is released less easily. True, and it explains the assertion.

Teacher's Note:
1. Intramolecular hydrogen bonding is possible only in the ortho isomer.
2. Both isomers get –M stabilisation of the phenoxide ion.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Assertion: benzoic acid (pKa 4.2) is stronger than carbonic acid and liberates CO2 from NaHCO3, while phenol (pKa 10) does not. True.
2. Reason: benzoic acid is stronger than phenol. True, and it is the reason for the different behaviour.

Teacher's Note:
1. NaOH cannot distinguish them because both dissolve in it.
2. NaHCO3 reacts only with acids stronger than carbonic acid.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A tertiary carbocation has nine alpha hydrogens for hyperconjugation and three +I alkyl groups, so it is more stable than a primary one and is formed more easily.
2. Both statements are true and the reason explains the assertion.

Teacher's Note:
1. Stability order: 3° > 2° > 1° > methyl.
2. Stable intermediates are formed faster.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The unpaired electron in the allyl radical is delocalised over two carbon atoms through the adjacent pi bond.
2. This delocalisation stabilises it more than an alkyl radical, which has only hyperconjugation.

Teacher's Note:
1. Resonance stabilisation is greater than hyperconjugation.
2. Allylic C–H bonds are weak for this reason.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In R–C≡O+ both carbon and oxygen have complete octets; in R–C+=O the carbon has only six electrons.
2. Structures with all octets complete are more stable, so both statements are true and the reason explains the assertion.

Teacher's Note:
1. A structure with fewer charges and complete octets contributes more.
2. Acylium ions are stabilised by this triple bond structure.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Assertion is true: in an amide the nitrogen lone pair is delocalised into the carbonyl group and is not available for protonation, whereas in an amine it is.
2. Reason is a true statement but it is not the explanation: the difference arises from resonance in the amide, not from nitrogen being less electronegative than oxygen.

Teacher's Note:
1. Amide resonance is the key factor.
2. Amides are practically neutral in water.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion: pKa values are 2.98 for salicylic acid against 4.08, 4.58 and 4.2 for the others. True.
2. The real reason is that the salicylate ion is stabilised by an intramolecular hydrogen bond between the phenolic OH and the carboxylate, not by steric inhibition of resonance. The reason is false.

Teacher's Note:
1. Chelation by hydrogen bonding stabilises the conjugate base.
2. Both the ortho effect and hydrogen bonding contribute.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Assertion is true because of resonance, which makes both C–O bonds identical (each with bond order 1.5).
2. Reason is also true as a general statement, but it does not explain why the two bonds are equal, so it is not the correct explanation.

Teacher's Note:
1. Resonance averages the bond lengths.
2. Acetate C–O length is about 127 pm, between 123 pm (C=O) and 134 pm (C–O).

Answer: (d) \( \ce{-N^+(CH3)3} \)

Explanation:
1. Groups carrying a full positive charge on the atom attached to carbon are far stronger –I groups than neutral groups such as –F.
2. Among the cationic groups the quaternary ammonium group –N+(CH3)3 is the strongest in the standard –I series: –N+R3 > –S+R2 > –NH3+ > –NO2 > –CN > –F.

Teacher's Note:
1. Positive charge on the attached atom greatly increases the –I effect.
2. Neutral –F is weaker than the charged groups.