JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 63

Answer: (c) sp > sp\(^2\) > sp\(^3\)

Explanation:
1. Electronegativity of carbon increases with the percentage of s-character: sp (50%) > sp\(^2\) (33%) > sp\(^3\) (25%).
2. The more electronegative carbon withdraws electrons more strongly, so the -I effect follows the same order.

Teacher's Note:
1. More s-character means more electronegative carbon.

Answer: (d) 4 > 1 > 3 > 2

Explanation:
1. Electron-donating CH3 increases electron density on N, so 4-methylaniline is the strongest base.
2. Electron-withdrawing groups reduce basicity: the NO2 group (-M, -I) lowers it more than CHO, so 4-nitroaniline is weakest.

Teacher's Note:
1. EDG on the ring increases and EWG decreases the basicity of aniline.

Answer: (a) 8

Explanation:
1. A carbanion carbon has three bond pairs and one lone pair.
2. Valence shell electrons = 6 + 2 = 8.

Teacher's Note:
1. Bond pairs plus lone pair give the octet.

Answer: (a) \( \ce{-NH3+} \) (in anilinium ion)

Explanation:
1. Aniline, phenol and chlorobenzene have lone pairs on the atom attached to the ring which are delocalised into it (+M effect).
2. In \( \ce{-NH3+} \) the nitrogen has no lone pair, so it cannot donate by resonance.

Teacher's Note:
1. A lone pair or a pi bond is needed for resonance.

Answer: (d) Benzyl more stable than allyl; ethyl more stable than vinyl

Explanation:
1. The positive charge in the benzyl cation is delocalised over the whole ring (more resonance forms) so it is more stable than the allyl cation.
2. The vinyl cation has the positive carbon in an sp orbital (more electronegative), so it is less stable than the ethyl cation.

Teacher's Note:
1. More delocalisation means more stable; sp carbocations are unstable.

Answer: (c) Cyclopentadienyl anion

Explanation:
1. The cyclopentadienyl anion is planar and cyclic with 6 pi electrons (4n + 2, n = 1): aromatic.
2. The cation has 4 pi electrons (anti-aromatic), cyclobutadiene has 4, and the cyclohexadienyl anion has an sp\(^3\) carbon so it is not fully conjugated.

Teacher's Note:
1. The anion has a lone pair in a p orbital that joins the ring conjugation.

Answer: (b) \( \ce{NH2-C(=NH)-NH2} \) (guanidine)

Explanation:
1. Protonation of guanidine gives the guanidinium ion, with the charge delocalised over three nitrogen atoms.
2. This large resonance stabilisation makes it a much stronger base than dimethylamine, aniline or urea.

Teacher's Note:
1. The conjugate acid with maximum resonance stabilisation indicates the strongest base.

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

Explanation:
1. Alkyl groups have a +I effect and destabilise the carboxylate ion.
2. The more alkyl carbons, the weaker the acid; formic acid has none, so it is strongest.

Teacher's Note:
1. Larger alkyl group means weaker acid.

Answer: (a) Cyclopropene

Explanation:
1. Cyclopropene has an sp\(^3\) carbon in the ring, so the ring is not fully conjugated.
2. Tropylium (6 pi), cyclopropenyl cation (2 pi) and naphthalene (10 pi) are aromatic.

Teacher's Note:
1. A sp\(^3\) ring atom breaks conjugation.

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

Explanation:
1. In cyclohexylamine the C-N bond is a pure single bond (longest).
2. In aniline the lone pair delocalisation gives the C-N some double-bond character, and the para CHO group increases this delocalisation further (4-aminobenzaldehyde, shorter).
3. The C=N bond of \( \ce{CH2=NH} \) is the shortest.

Teacher's Note:
1. More double-bond character means a shorter bond.

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

Explanation:
1. Two chlorines (II) give a stronger total -I effect than one (I).
2. Cl one carbon farther away (III) acts more weakly than in I, but still more than no Cl (IV).

Teacher's Note:
1. More and closer EWG groups mean stronger acid.