JEE Chemistry Hydrocarbons MCQs Set 30

Answer: (a)

Explanation:
1. The white precipitate (AgBr) with alcoholic \(\ce{AgNO3}\) shows a reactive (benzylic) bromine, so (A) has a \(\ce{-CH2Br}\) group: options (b), with Br on the ring, is excluded.
2. \(\ce{C8H6O4}\) is phthalic acid (a benzene dicarboxylic acid) which forms an anhydride only when the two \(\ce{-COOH}\) groups are ortho.
3. So the two side-chain carbons must be ortho: 2-methylbenzyl bromide, option (d). The meta (a) and para (c) isomers give isophthalic and terephthalic acids, which do not form anhydrides.

Teacher's Note:
1. Benzylic halide plus ortho substitution: only ortho-phthalic acid forms a cyclic anhydride.

Answer: (c) (ii), (iii) and (iv)

Explanation:
1. Aromatic requires a planar, fully conjugated cyclic system with \((4n+2)\,\pi\) electrons.
2. (i) cyclopentane is saturated, so it is not aromatic.
3. (ii) thiophene: 4 \(\pi\) electrons from two C=C plus 2 from the sulphur lone pair = 6, aromatic. (iii) naphthalene has 10 \(\pi\) electrons, aromatic. (iv) pyrrole: 4 + 2 from nitrogen lone pair = 6, aromatic.
4. So (ii), (iii) and (iv) are aromatic.

Teacher's Note:
1. Count the pi electrons, including a lone pair that sits in the ring p-orbital system.

Answer: (b) (i), (ii) and (iii)

Explanation:
1. Geometrical isomerism needs two different groups on each carbon of the double bond.
2. (i) F, Cl on one carbon and H, D on the other: both carbons carry two different groups, so cis/trans exists.
3. (ii) F, Cl on each carbon: different groups on each carbon, so cis and trans exist.
4. (iii) CH3 and C2H5 on each carbon: exists. (iv) one carbon has two CH3 groups, so no geometrical isomerism.

Teacher's Note:
1. Check each doubly bonded carbon separately for two different substituents.

Answer: (c) (i) and (iii)

Explanation:
1. Cyclopropenyl cation: planar, 3 p-orbitals with 2 \(\pi\) electrons (\(4n+2\), n = 0): aromatic.
2. Cyclooctatetraene has 8 \(\pi\) electrons and is tub-shaped, not planar: non-aromatic.
3. Biphenyl contains two benzene rings: aromatic.
4. Cyclopropenyl anion has 4 \(\pi\) electrons (\(4n\)): anti-aromatic, not aromatic.
5. So (i) and (iii) are aromatic.

Teacher's Note:
1. Apply Huckel's (4n + 2) rule to each ring; the ring must also be planar and fully conjugated.

Answer: (c) both methylenecyclohexane and 1-methylcyclohexene

Explanation:
1. Markovnikov addition puts \(\ce{H+}\) on the carbon that already holds more hydrogens, giving the more stable carbocation.
2. Methylenecyclohexane \(\ce{CH2=C6H10}\) is protonated on \(\ce{CH2}\) to give the tertiary 1-methylcyclohexyl cation. 1-Methylcyclohexene is protonated on ring C-2, giving the same tertiary cation.
3. Chloride attack gives 1-chloro-1-methylcyclohexane in both cases. 3-Methylcyclohexene would first give a secondary cation, so it does not give this product directly.

Teacher's Note:
1. Both alkenes lead to the same tertiary carbocation, 1-methylcyclohexyl.

Answer: (d)

Explanation:
1. Ozonolysis cuts the ring C=C and puts a C=O on each carbon of it. The product has an aldehyde end (\(\ce{CHO}\)) and a methyl ketone end.
2. Joining the two carbonyl carbons back through the chain gives a ring where one alkene carbon carries CH3 (it became the ketone) and the other carries H (the aldehyde), with the second CH3 on the carbon opposite the double bond: 1,4-dimethylcyclopentene.
3. (a) 3,3-dimethylcyclopentene and (b) 3,4-dimethylcyclopentene give aldehyde-aldehyde products; (d) 1,2-dimethylcyclopentene gives the diketone heptane-2,6-dione.

Teacher's Note:
1. Reverse the ozonolysis: join the two carbonyl carbons to re-form the ring double bond.

Answer: (a)

Explanation:
1. Benzene (6 \(\pi\) electrons), naphthalene (10) and thiophene (6, counting the sulphur lone pair) satisfy Huckel's rule and are planar and fully conjugated.
2. Cyclopentadiene has an \(sp^3\) carbon in the ring, so conjugation is not continuous around the ring: it is non-aromatic.

Teacher's Note:
1. A ring with an sp3 carbon cannot be fully conjugated.

Answer: (d)

Explanation:
1. Hot alkaline \(\ce{KMnO4}\) oxidises the whole side chain to \(\ce{-COOH}\): B is benzoic acid.
2. The \(\ce{-COOH}\) group is deactivating and meta-directing, so \(\ce{Br2/FeCl3}\) gives C = 3-bromobenzoic acid.
3. Fischer esterification with ethanol gives D = ethyl 3-bromobenzoate, option (d).

Teacher's Note:
1. Side-chain oxidation gives benzoic acid, whose COOH group directs bromine to the meta position.

Answer: (b)

Explanation:
1. Staggered conformers are more stable than eclipsed ones, so (b) and (c) are ruled out.
2. Of the two staggered forms, (a) has the two OH groups anti (180 degrees apart) and (d) has them gauche (60 degrees).
3. In ethylene glycol the gauche conformer is favoured because the two OH groups can form an intramolecular hydrogen bond, so (d) is the most stable.

Teacher's Note:
1. Gauche is more stable than anti here because of intramolecular hydrogen bonding.

Answer: (a) III > I > II

Explanation:
1. The nitro group withdraws electrons by -M and -I effects and stabilises the phenoxide ion, so 4-nitrophenol is the strongest acid.
2. The methyl group donates electrons (+I, hyperconjugation) and destabilises the phenoxide, making p-cresol weaker than phenol.
3. Order: III > I > II.

Teacher's Note:
1. Electron-withdrawing groups strengthen phenols as acids; electron-donating groups weaken them.

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

Explanation:
1. Anisole: the \(\ce{-OCH3}\) group donates strongly by resonance (+M), so it is the most reactive.
2. Toluene: \(\ce{-CH3}\) donates weakly (+I and hyperconjugation), more reactive than benzene.
3. \(\ce{-CF3}\) withdraws strongly by -I effect and deactivates the ring, so it is the least reactive.
4. Increasing reactivity: (trifluoromethyl)benzene, benzene, toluene, anisole, i.e. IV, I, II, III.

Teacher's Note:
1. Activating groups speed up electrophilic substitution; deactivating groups slow it down.

Answer: (b) 8

Explanation:
1. 3-Methylpentane has four kinds of hydrogen: the two terminal \(\ce{CH3}\) groups (C-1/C-5), the two \(\ce{CH2}\) groups (C-2/C-4), the \(\ce{CH}\) (C-3) and the \(\ce{CH3}\) on C-3.
2. Cl on C-1: 1-chloro-3-methylpentane, one chiral centre: 2 stereoisomers. Cl on C-2: 2-chloro-3-methylpentane, two chiral centres: 4 stereoisomers.
3. Cl on C-3: 3-chloro-3-methylpentane, achiral: 1. Cl on the methyl group: 3-(chloromethyl)pentane, achiral: 1.
4. Total \(2+4+1+1=8\).

Teacher's Note:
1. Count the chiral centres of each monochloro product to include stereoisomers.

Answer: (d) 0

Explanation:
1. Each \(\ce{C=C}\) is cleaved into two \(\ce{C=O}\) groups (aldehydes, as the carbons carry H).
2. The ends give \(\ce{CH3CHO}\) (two molecules). The two inner \(\ce{CH(CH3)}\) carbons each end up flanked by two \(\ce{CHO}\) groups: \(\ce{OHC-CH(CH3)-CHO}\) (two molecules).
3. Both acetaldehyde and methylpropanedial have no chiral carbon (the latter has two identical \(\ce{CHO}\) groups), so no product is optically active.

Teacher's Note:
1. Write each fragment and look for a carbon with four different groups.

Answer: (c) 3

Explanation:
1. o-Xylene is a resonance hybrid of two Kekule structures, so ozonolysis is considered for both of them.
2. Kekule structure with the two methyl-bearing carbons double bonded: cleavage gives two molecules of methylglyoxal \(\ce{CH3CO-CHO}\) and one of glyoxal \(\ce{OHC-CHO}\).
3. The other Kekule structure (the methyl-bearing carbons joined by a single bond): cleavage gives biacetyl \(\ce{CH3CO-COCH3}\) and two molecules of glyoxal.
4. Different products overall: glyoxal, biacetyl and methylglyoxal: 3.

Teacher's Note:
1. Consider both Kekule structures; the distinct products are glyoxal, biacetyl and methylglyoxal.

Answer: (b) 3

Explanation:
1. Dehydration by E1 (Saytzeff) removes H from C-1 or C-3.
2. Removing H from C-1 gives but-1-ene; removing H from C-3 gives but-2-ene, which exists as cis and trans isomers.
3. Total products: 1-butene, cis-2-butene and trans-2-butene = 3.

Teacher's Note:
1. Remember the cis and trans forms of but-2-ene when counting products.

Answer: (b) 4

Explanation:
1. Count the different kinds of hydrogen: (1) the two equivalent \(\ce{CH3}\) groups, (2) the two equivalent tertiary C-H (C-1/C-2), (3) the ring \(\ce{CH2}\) groups next to them (C-3/C-5), (4) the ring \(\ce{CH2}\) opposite them (C-4).
2. Each kind gives one structural isomer: \(\ce{CH2Br}\) derivative, 1-bromo-1,2-dimethyl, 3-bromo-1,2-dimethyl and 4-bromo-1,2-dimethylcyclopentane.
3. Total = 4.

Teacher's Note:
1. Group equivalent hydrogens by symmetry; each group gives one structural isomer.