JEE Chemistry Organic Chemistry MCQs Set 62

Answer: (a) maltase

Explanation:
1. Maltose is a disaccharide of two glucose units.
2. The enzyme maltase hydrolyses it into two molecules of glucose.
3. Zymase converts glucose to ethanol, diastase converts starch to maltose, and sucrase (invertase) hydrolyses sucrose.

Teacher's Note:
1. Enzymes are named after their substrates: maltase acts on maltose.
2. Remember the order starch, then maltose, then glucose, then ethanol.

Answer: (c) nylon 6,6

Explanation:
1. Nylon 6,6 is a polyamide with \(-\mathrm{CO-NH-}\) groups, which form strong hydrogen bonds between chains.
2. Natural rubber, Teflon and polystyrene have only weak van der Waals forces.

Teacher's Note:
1. Polyamides and polyesters are fibres because of strong intermolecular forces.
2. Elastomers like rubber have the weakest forces.

Answer: (d) benzonitrile

Explanation:
1. \(\mathrm{PCl_5}\) is a dehydrating agent.
2. It removes a molecule of water from the primary amide: \(\ce{C6H5CONH2 ->[PCl5] C6H5CN}\).

Teacher's Note:
1. Dehydration of an amide with \(\mathrm{P_2O_5}\), \(\mathrm{PCl_5}\) or \(\mathrm{SOCl_2}\) gives a nitrile.
2. Aniline needs the Hofmann bromamide reaction instead.

Answer: (d) ethanal, \(\mathrm{CH_3CHO}\)

Explanation:
1. Aldehydes are more reactive than ketones, since a ketone has two electron releasing alkyl groups and more steric hindrance.
2. Among aldehydes, the smaller the alkyl group, the lower the steric hindrance and the electron release.
3. Ethanal is therefore the most reactive.

Teacher's Note:
1. Reactivity order: formaldehyde > other aldehydes > ketones.
2. Steric hindrance and the +I effect both lower the reactivity.

Answer: (a) \(\mathrm{CH_3^-} > \mathrm{NH_2^-} > \mathrm{OH^-} > \mathrm{F^-}\)

Explanation:
1. These are all anions of the second period, with the attacking atom being C, N, O and F.
2. Moving left to right, electronegativity increases, so the electrons are held more tightly and the nucleophilicity falls.
3. Order: \(\mathrm{CH_3^-} > \mathrm{NH_2^-} > \mathrm{OH^-} > \mathrm{F^-}\).

Teacher's Note:
1. Across a period, nucleophilicity decreases with increasing electronegativity.
2. This is the same order as the basicity of the species.

Answer: (c) N-(4-chlorophenyl)acetamide

Explanation:
1. The Cl group is ortho-para directing, and the para position is favoured by sterics, so A is 4-chloronitrobenzene.
2. Sn/HCl reduces the nitro group to an amino group, giving 4-chloroaniline.
3. \(\mathrm{CH_3COCl}\) acetylates the \(-\mathrm{NH_2}\) group to give N-(4-chlorophenyl)acetamide.

Teacher's Note:
1. Halogens deactivate the ring but still direct incoming groups to the ortho and para positions.
2. Acetylation of an amine gives an amide and not an ester.

Answer: (b) free radical substitution reaction

Explanation:
1. NBS with peroxides or light supplies a low concentration of \(\mathrm{Br_2}\) and \(\mathrm{Br}\) radicals.
2. The bromine radical abstracts the allylic hydrogen to form a resonance stabilised allyl radical, which gives \(\mathrm{CH_2=CH-CH_2Br}\).
3. This is a free radical substitution at the allylic position.

Teacher's Note:
1. NBS is the reagent for allylic and benzylic bromination.
2. The double bond remains intact because the bromine concentration is very low.

Answer: (a) azoxybenzene

Explanation:
1. In alkaline medium nitrobenzene undergoes bimolecular reduction, and the mild reducing agent glucose stops the reduction at the azoxy stage.
2. The product is azoxybenzene, \(\mathrm{C_6H_5-N=N^+(O^-)-C_6H_5}\).
3. Zn/NaOH gives azobenzene, and excess Zn/NaOH gives hydrazobenzene.

Teacher's Note:
1. In neutral medium the reduction gives phenylhydroxylamine; in strongly acidic medium aniline.
2. In alkaline medium the products depend on the reducing agent strength.

Answer: (d) 1-bromopropane in DMSO

Explanation:
1. \(\mathrm{S_N2}\) is fastest for unhindered primary halides.
2. Polar aprotic solvents such as DMSO and DMF do not solvate the anion strongly, which makes the nucleophile more reactive.
3. 1-Bromopropane is primary and DMSO is polar aprotic, while the tertiary halides prefer \(\mathrm{S_N1}\) or elimination.

Teacher's Note:
1. S\(_N\)2 reactivity: methyl > primary > secondary >> tertiary.
2. Protic solvents (water, alcohols) hydrogen bond to the nucleophile and slow \(\mathrm{S_N2}\).

Answer: (c) 4

Explanation:
1. The functional groups are the ketone (C=O), the alkene (C=C), the alcohol (\(-\mathrm{OH}\)) and the alkyne (\(-\mathrm{C{\equiv}C-}\)).
2. These are four different kinds of functional groups.

Teacher's Note:
1. Count each type of functional group once, whatever the number of times it appears.
2. The alkyl framework of the steroid is not a functional group.

Answer: (d) all (1), (2) and (3)

Explanation:
1. (1) Both are \(\mathrm{C_3H_6O}\), one a ketone and one an aldehyde, so they are functional isomers.
2. (2) Both are \(\mathrm{C_5H_{10}O_2}\) acids with different carbon chains (branched and straight), so they are chain isomers.
3. (3) Both are esters of the same formula with different groups on the two sides of the \(-\mathrm{COO-}\) group, so they are metamers. Hence all three are correct.

Teacher's Note:
1. Metamerism needs the same functional group with different alkyl or aryl groups on its two sides.
2. Chain isomers differ in the carbon skeleton only.

Answer: (c) \(\mathrm{CH_3-CH^+-C_6H_5}\)

Explanation:
1. (a) is a primary cation with an electron-withdrawing carbonyl group, so it is very unstable.
2. (b) is stabilised by 9 hyperconjugating C-H bonds.
3. (c) is secondary and benzylic, with the charge delocalised over the benzene ring (several resonance structures), which stabilises it more than hyperconjugation or allylic resonance in (d).
4. So (c) is the most stable.

Teacher's Note:
1. Resonance stabilisation over a benzene ring is greater than that over a vinyl group.
2. Order: benzylic > allylic > tertiary > secondary > primary.

Answer: (b) cyclopentadienyl anion

Explanation:
1. Aromatic species must be cyclic, planar, fully conjugated, and have \((4n+2)\) \(\pi\) electrons.
2. The cyclopentadienyl anion has 6 \(\pi\) electrons (n = 1) and is aromatic.
3. Cyclobutadiene (4 \(\pi\) e\(^-\)) and the cycloheptatrienyl anion (8 \(\pi\) e\(^-\)) are anti-aromatic, and cyclohexa-1,4-diene is not conjugated.

Teacher's Note:
1. Apply Huckel's rule \((4n+2)\) to the number of \(\pi\) electrons.
2. The cycloheptatrienyl cation (tropylium) is aromatic, but the anion is not.

Answer: (c) toluene (methylbenzene)

Explanation:
1. Hyperconjugation depends on the number of \(\alpha\) C-H bonds next to the \(\pi\) system.
2. Styrene has none, vinylcyclohexane has 1, toluene has 3 and in 3,3-dimethylbut-1-ene the carbon next to the double bond is quaternary with no \(\alpha\) hydrogen.
3. Toluene has the most \(\alpha\) hydrogens (3).

Teacher's Note:
1. Count only the C-H bonds on the carbon directly attached to the unsaturated carbon.
2. A quaternary carbon next to a double bond cannot show hyperconjugation.

Answer: (c) more extent of hydrogen bonding

Explanation:
1. Glycerol has three OH groups and glycol (ethane-1,2-diol) has two.
2. The larger number of OH groups gives more extensive intermolecular hydrogen bonding in glycerol, which makes it more viscous.

Teacher's Note:
1. Viscosity rises with the strength and number of intermolecular hydrogen bonds.
2. Molecular weight alone does not explain viscosity.