JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 25

Answer: (c) trans-But-2-ene

Explanation:
1. Heat of hydrogenation measures how much energy the alkene loses on going to butane; the more stable the alkene, the smaller it is.
2. trans-But-2-ene (about 115 kJ/mol) is the most stable of the butenes, cis-But-2-ene is about 120 kJ/mol and but-1-ene about 127 kJ/mol.
3. Buta-1,3-diene takes up two moles of H₂ per mole and releases about 239 kJ/mol.

Teacher's Note:
1. A more substituted, trans alkene is more stable and has a smaller heat of hydrogenation.

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

Explanation:
1. pKₐ values: 4-nitrophenol 7.15, 3-nitrophenol 8.4, phenol 10.0, 4-methylphenol 10.3.
2. The para NO₂ group acts by −M and −I effects, the meta NO₂ by −I only, and CH₃ donates electrons.

Teacher's Note:
1. Para-nitro is stronger than meta-nitro because of the −M effect.

Answer: (a) (4n + 2) π electrons in a planar, cyclic, fully conjugated system

Explanation:
1. Hückel's rule: aromatic stability needs a planar, cyclic, fully conjugated ring with (4n + 2) π electrons, as in benzene (6), naphthalene (10) and the tropylium cation (6).

Teacher's Note:
1. 4n π electrons gives anti-aromaticity.

Answer: (d) CD₃–CO–CD₃

Explanation:
1. Acetone enolises in either direction; each enolisation exchanges one H for D, and repeated enolisation in D₂O replaces all six α-hydrogens, giving CD₃COCD₃ (the product returns to the keto form).

Teacher's Note:
1. Exchange of α-hydrogens in D₂O is a test for enolisable hydrogens.

Answer: (c) a carbocation

Explanation:
1. SbCl₅ pulls out Cl⁻ and forms the planar, achiral benzylic carbocation C₆H₅CH⁺CH₃. The chloride re-attacks from either face, giving both enantiomers equally (racemisation).

Teacher's Note:
1. A planar carbocation intermediate causes loss of optical activity.

Answer: (b) Benzylamine (C₆H₅CH₂NH₂)

Explanation:
1. Benzylamine has a localised lone pair on nitrogen, while in the three anilines it is delocalised into the ring and made less available by NO₂ groups.

Teacher's Note:
1. Aliphatic amines are stronger bases than aromatic amines.

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

Explanation:
1. Protonation of the amidine gives a resonance-stabilised amidinium ion (pKₐ about 12.4), so it is the strongest base. Dimethylamine (10.7) is stronger than ethylamine (10.6) and the amide is essentially neutral.

Teacher's Note:
1. Resonance-stabilised cations make strong bases.

Answer: (d) sp², sp², sp, sp

Explanation:
1. The carbon of =CH₂ and the carbon of =CH– each have a double bond (sp²); the nitrile carbon (C≡N) and the nitrogen have a triple bond (sp). Left to right the hybridisations are sp², sp², sp, sp.

Teacher's Note:
1. A double bond means sp², a triple bond means sp.

Answer: (c) CH₃Cl

Explanation:
1. CCl₄ is symmetrical with zero dipole moment. The measured values are CH₃Cl 1.87 D, CH₂Cl₂ 1.60 D and CHCl₃ 1.04 D, so CH₃Cl is highest.

Teacher's Note:
1. Dipole moments of the chloromethanes fall as more Cl atoms are added, as the bond dipoles partly cancel.

Answer: (b) II > III > I

Explanation:
1. The rate of nucleophilic addition falls with increasing steric hindrance and electron donation to the carbonyl carbon.

Teacher's Note:
1. Ethanal has one small alkyl group and an H, propanone has two methyl groups, and benzophenone has two bulky, electron-releasing phenyl groups, so the order is II > III > I.

Answer: (a) concentrated H₃PO₄

Explanation:
1. The dehydration needs a strong acid whose conjugate base is a poor nucleophile. H₃PO₄ (and H₂SO₄) supply H⁺ without a competing nucleophile, whereas HCl and HBr would give the alkyl halide by substitution.

Teacher's Note:
1. Use a non-nucleophilic acid for dehydration.

Answer: (d) aq. NaOH + CHCl₃

Explanation:
1. Chloroform and aqueous alkali generate the electrophile dichlorocarbene, :CCl₂, which attacks the ortho position of the phenoxide ion; hydrolysis then gives the CHO group.

Teacher's Note:
1. Chloroform with base gives dichlorocarbene.

Answer: (c) :CCl₂ (dichlorocarbene)

Explanation:
1. CHCl₃ loses H⁺ to OH⁻ giving ⁻CCl₃, which loses Cl⁻ to give the neutral, electron-deficient carbene :CCl₂, the attacking electrophile.

Teacher's Note:
1. Carbenes have six electrons and are electrophilic.

Answer: (b) σ–p

Explanation:
1. In hyperconjugation the electrons of a C–H σ bond overlap partially with an adjacent empty or partially filled p orbital (or a π orbital).

Teacher's Note:
1. Hyperconjugation is also called σ–p conjugation.

Answer: (a) Anhydrous AlCl₃, because it is covalent and its vacant orbital is filled by the oxygen of the ether

Explanation:
1. Anhydrous AlCl₃ is an electron-deficient Lewis acid that dissolves by forming a coordinate complex with the ether oxygen. In the hydrate, the aluminium already holds water, so it is ionic and does not dissolve in the non-polar ether.

Teacher's Note:
1. Like dissolves like: covalent in a covalent solvent.

Answer: (d) 3-Hydroxypropanoic acid forms a lactide on heating

Explanation:
1. Lactide is a cyclic diester formed by two molecules of an α-hydroxy acid. 3-Hydroxypropanoic acid is a β-hydroxy acid and loses water to give propenoic (acrylic) acid. The other three statements are correct (decarboxylation of a malonic acid, cyclic anhydride formation, decarboxylation of a β-keto acid).

Teacher's Note:
1. Alpha gives lactide; beta gives an unsaturated acid; gamma and delta give lactones.

Answer: (c) it easily gives ethyl radicals at low temperature, which start the chlorine radical chain

Explanation:
1. The Pb–C bonds are weak and break homolytically at a low temperature, producing ethyl radicals. These abstract Cl from Cl₂ to give Cl•, which starts the chain: Cl+ CH₄ → HCl + CH₃•.

Teacher's Note:
1. Radical initiators allow chain reactions at lower temperatures.

Answer: (b) 1.15

Explanation:
1. Neopentane has 12 equivalent primary H and ethane has 6. Per-hydrogen reactivity ratio = (2.3 ÷ 12) ÷ (1 ÷ 6) = 0.1917 ÷ 0.1667 = 1.15.

Teacher's Note:
1. Divide the product ratio by the number of equivalent hydrogens before comparing.

Answer: (a) both propagation steps are exothermic only for HBr

Explanation:
1. For HBr, step 1 (Bradds to the alkene, ΔH = −25.1 kJ/mol) and step 2 (H abstraction from HBr, ΔH = −50.2 kJ/mol) are both exothermic, so the chain propagates. For HCl, step 2 is endothermic (+12.6), and for HI, step 1 is endothermic (+46), so the chain breaks.

Teacher's Note:
1. A radical chain needs both propagation steps to be favourable.

Answer: (d) But-2-ene, because it forms a more stable secondary carbocation

Explanation:
1. Protonation of the alkene is rate-determining and gives a carbocation: ethene gives the unstable primary ethyl cation, while but-2-ene gives the secondary sec-butyl cation, which is more stable.

Teacher's Note:
1. A more stable carbocation intermediate means a faster electrophilic addition.