JEE Chemistry Hydrocarbons MCQs Set 33

Answer: (d) N-(4-nitrophenyl)benzamide (nitration of the N-phenyl ring, para to nitrogen)

Explanation:
1. The ring attached to nitrogen is activated by the NH group (lone pair donation, ortho-para directing), while the benzoyl ring is deactivated by the C=O group.
2. Nitration therefore occurs on the N-phenyl ring, at the less hindered para position.

Teacher's Note:
1. Anilide nitrogen donates electrons to its own ring.
2. Acyl groups on a ring deactivate it.

Answer: (b) deactivates the ring towards electrophilic substitution

Explanation:
1. The nitro group withdraws electrons (-I and -M), so the ring is less nucleophilic and reacts more slowly with electrophiles.
2. It actually activates the ring towards nucleophilic aromatic substitution, so the last statement is wrong.

Teacher's Note:
1. Electron-withdrawing groups deactivate EAS and activate nucleophilic substitution.
2. Nitrobenzene is not basic.

Answer: (c) N = 6 and P = 4

Explanation:
1. Four constitutional isomers form; two of them (1-chloro-2-methylbutane and 2-chloro-3-methylbutane) are chiral and each exists as two enantiomers, so N = 6.
2. Enantiomers have identical boiling points and cannot be separated by fractional distillation, so there are only four fractions.

Teacher's Note:
1. Distillation separates only compounds with different physical properties.
2. Enantiomers have identical boiling points.

Answer: (a) CH₃-CHBr-OCH₃

Explanation:
1. The methoxy group stabilises the adjacent carbocation by +M effect (oxocarbenium ion), so H⁺ adds to the CH₂ carbon.
2. Bromide attacks the oxocarbenium carbon giving CH₃CHBrOCH₃; under anhydrous conditions it is not hydrolysed further.

Teacher's Note:
1. Vinyl ethers give Markovnikov addition with strong stabilisation of the cation by oxygen.
2. Water would hydrolyse the alpha-bromo ether to the aldehyde.

Answer: (d) the first transition state (formation of the sigma complex)

Explanation:
1. The path is reactants, first transition state, sigma complex (intermediate), second transition state, product.
2. The first step has the larger activation energy because aromaticity is lost, so the first transition state is the highest point on the profile.

Teacher's Note:
1. The slow step has the highest transition state.
2. An intermediate is a minimum between two transition states.

Answer: (b) one of the propagation steps is endothermic in both cases

Explanation:
1. The chain needs X· to add to the alkene and the resulting radical to take H from HX. For HCl the H-Cl bond is so strong that hydrogen abstraction by the carbon radical is endothermic. For HI the addition of I· to the alkene is endothermic (weak C-I bond).
2. So the chain cannot propagate properly with HCl or HI; only HBr has both steps favourable.

Teacher's Note:
1. Both steps of the chain must be exothermic or nearly so.
2. H-Cl is strong, C-I is weak.

Answer: (d) II > I > III

Explanation:
1. In (I) and (II) both groups activate the same positions (2, 4 and 6) and reinforce each other. The NH₂ group is a stronger activator than OCH₃, so (II) is more reactive than (I).
2. In (III) the nitro group deactivates the ring, which opposes the methoxy group, so (III) is the least reactive.

Teacher's Note:
1. When two groups reinforce, the ring is more reactive.
2. -NH₂ is more strongly activating than -OR.

Answer: (b) 4

Explanation:
1. Monochlorination gives four constitutional isomers: 1-chloro-2-methylbutane (chiral), 2-chloro-2-methylbutane (achiral), 2-chloro-3-methylbutane (chiral) and 1-chloro-3-methylbutane (achiral).
2. Each chiral isomer exists as a pair of enantiomers, so there are 2 x 2 = 4 chiral compounds (verified with RDKit).

Teacher's Note:
1. Count stereocentres in each constitutional isomer.
2. Each chiral compound with one stereocentre is a pair of enantiomers.

Answer: (b) 8

Explanation:
1. 3-Methylpentane gives four constitutional monochloro isomers: 1-chloro-3-methylpentane (chiral, 2 enantiomers), 2-chloro-3-methylpentane (two stereocentres, 4 stereoisomers), 3-chloro-3-methylpentane (1) and 3-(chloromethyl)pentane (1).
2. Total 2 + 4 + 1 + 1 = 8 (checked with RDKit).

Teacher's Note:
1. Count constitutional isomers first, then stereoisomers.
2. n stereocentres give up to 2ⁿ stereoisomers.

Answer: (d) toluene

Explanation:
1. Friedel-Crafts reactions fail on strongly deactivated rings (COOH, NO₂) and on rings carrying a basic NH₂ group, which complexes with AlCl₃.
2. Toluene has an activating methyl group and reacts readily.

Teacher's Note:
1. Rings with -M deactivating groups do not undergo Friedel-Crafts reactions.
2. Amines poison the AlCl₃ catalyst.

Answer: (c) 3-bromocyclohexene

Explanation:
1. NBS gives a low, steady concentration of Br₂, which prevents ionic addition to the double bond.
2. A bromine radical removes an allylic hydrogen to give the resonance-stabilised allyl radical, which takes Br from Br₂: 3-bromocyclohexene.

Teacher's Note:
1. NBS is used for allylic and benzylic bromination.
2. The double bond stays intact.

Answer: (a) trans-2-chloro-1-methylcyclopentan-1-ol

Explanation:
1. Cl⁺ adds to form a bridged chloronium ion; water attacks from the opposite face at the more substituted carbon (the tertiary carbon, which bears more positive charge).
2. So OH ends up on the tertiary carbon and Cl on the adjacent carbon, trans to each other (anti addition).

Teacher's Note:
1. Halohydrin: OH goes to the more substituted carbon, X to the less substituted carbon.
2. Anti addition means the two groups are trans.

Answer: (d) but-2-ene

Explanation:
1. Anti-Markovnikov and Markovnikov products are different only for unsymmetrical alkenes.
2. But-2-ene is symmetrical (CH₃CH=CHCH₃), so both modes of addition give the same product, 2-bromobutane.

Teacher's Note:
1. Symmetrical alkenes show no regiochemistry.
2. The peroxide effect needs an unsymmetrical alkene.

Answer: (b) 4-(1-bromopropyl)phenol

Explanation:
1. HBr adds to the C=C by Markovnikov addition so that Br ends up at the benzylic carbon: C₆H₄(OCH₃)-CHBr-CH₂CH₃.
2. With excess HBr and heat the aryl methyl ether is also cleaved (the methyl group is removed as CH₃Br), giving the phenol: 4-(1-bromopropyl)phenol.

Teacher's Note:
1. Aryl alkyl ethers are cleaved by hot concentrated HBr or HI to a phenol and an alkyl halide.
2. The aryl C-O bond is not cleaved.

Answer: (c) 1-bromobutane

Explanation:
1. In the presence of peroxide, HBr adds by a free-radical chain mechanism: Br· adds to the terminal carbon so as to give the more stable secondary radical.
2. This radical takes H from HBr to give CH₃CH₂CH₂CH₂Br (anti-Markovnikov).

Teacher's Note:
1. Peroxide effect: bromine goes to the carbon with more hydrogens.
2. Only HBr shows this effect.

Answer: (b) propan-1-ol

Explanation:
1. Boron adds to the less hindered, terminal carbon (anti-Markovnikov) and oxidation replaces B by OH with retention.
2. The product is CH₃CH₂CH₂OH.

Teacher's Note:
1. Hydroboration-oxidation gives the anti-Markovnikov alcohol.
2. No rearrangement occurs.

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

Explanation:
1. Toluene is activated by the methyl group, benzene is the reference, chlorobenzene is mildly deactivated (-I > +M) and nitrobenzene is strongly deactivated (-I and -M).

Teacher's Note:
1. Activating > H > halogen > strongly deactivating groups.
2. Nitro slows the reaction by a factor of about a million.

Answer: (d) C₆H₅-CO-CH₂CH₃ (propiophenone)

Explanation:
1. Hg²⁺ hydration adds OH to the carbon that can better bear positive charge, the benzylic carbon, giving the enol C₆H₅C(OH)=CHCH₃.
2. The enol tautomerises to the ketone C₆H₅COCH₂CH₃.

Teacher's Note:
1. Enols are not isolated; they tautomerise to carbonyls.
2. OH goes to the carbon stabilising the positive charge.

Answer: (c) kinetic control for (1) and thermodynamic control for (2)

Explanation:
1. At low temperature the faster-forming 1,2-adduct (from attack at the carbon next to the cation centre) predominates: kinetic control.
2. At higher temperature the addition is reversible and the more stable, more substituted 1,4-adduct predominates: thermodynamic control.

Teacher's Note:
1. Low temperature favours the kinetic product.
2. The 1,4-adduct has the more substituted (more stable) double bond.

Answer: (c) 3-bromocyclohexene

Explanation:
1. In light at a low concentration of Br₂ (and high temperature) allylic hydrogen abstraction by Br· gives a resonance-stabilised allyl radical, which gives the allylic bromide.
2. Addition of Br₂ to give vicinal dibromides is the ionic reaction in the dark in CCl₄.

Teacher's Note:
1. Light means radical substitution at the allylic position.
2. The double bond is retained.

Answer: (a) neopentane (2,2-dimethylpropane)

Explanation:
1. C₅H₁₂ has molecular mass 72. Of its three isomers the branched ones are isopentane and neopentane.
2. Neopentane has twelve equivalent hydrogens, so it gives only one monohalide, (CH₃)₃CCH₂X.

Teacher's Note:
1. Mass 72 means C₅H₁₂.
2. One kind of hydrogen gives one monosubstituted product.

Answer: (c) Toluene

Explanation:
1. All alkyl groups activate the ring, but the activation decreases (CH₃ > C₂H₅ > i-C₃H₇ > t-C₄H₉) as the number of C-H bonds for hyperconjugation falls and as bulk hinders the incoming SO₃.
2. Toluene has the greatest hyperconjugation and least steric hindrance.

Teacher's Note:
1. Hyperconjugation increases with the number of alpha C-H bonds.
2. Bulky groups also block ortho attack.

Answer: (c) -Cl

Explanation:
1. -OR, -NH₂ and -NHR donate their lone pair (+M) and strongly activate the ring.
2. -Cl withdraws electrons by -I more strongly than it donates by +M, so it deactivates the ring (while still directing ortho and para).

Teacher's Note:
1. Halogens are the exception among o/p directors.
2. Look for lone-pair donors versus electronegative atoms.

Answer: (b) P is 1,1-dichlorobutan-2-one

Explanation:
1. Excess HOCl adds twice and the enol gives the alpha,alpha-dihaloketone: P = CH₃CH₂COCHCl₂ (1,1-dichlorobutan-2-one).
2. Excess HCl gives the gem-dihalide Q = CH₃CH₂CCl₂CH₃ (2,2-dichlorobutane), so Q is not 1,2-dichlorobutane and P (a ketone) and Q (a dihalide) are not isomers.

Teacher's Note:
1. HOX adds twice to a terminal alkyne to give a dihaloketone.
2. HX adds twice to give a gem-dihalide at the internal carbon.

Answer: (d) ortho to the NHCOCH₃ group (2-bromo-4-methylacetanilide)

Explanation:
1. Both groups are activating and ortho-para directing, and they are para to each other, so the incoming Br goes ortho to one of them.
2. NHCOCH₃ is the stronger activator, so substitution occurs ortho to NHCOCH₃.

Teacher's Note:
1. When both groups activate, the stronger activator decides.
2. Amide nitrogen is a stronger donor than the methyl group.

Answer: (c) 2-bromo-2-methylbutane

Explanation:
1. Br· removes the tertiary hydrogen to give the tertiary radical, which reacts with Br₂ to give 2-bromo-2-methylbutane.

Teacher's Note:
1. Bromine is highly selective for tertiary C-H.
2. The same selectivity applies to isobutane.

Answer: (c) CH₂=CH-C(Br)=CH₂ (2-bromobuta-1,3-diene)

Explanation:
1. Protonation at the terminal alkyne carbon gives the vinyl cation CH₂=CH-C⁺=CH₂, which is stabilised by conjugation with the vinyl group (allylic character).
2. Bromide then adds to this cation, giving 2-bromobuta-1,3-diene (the same type of reaction used to make chloroprene with HCl).

Teacher's Note:
1. Protonation of an enyne occurs at the alkyne to give a conjugated cation.
2. Chloroprene is made by HCl addition to vinylacetylene.

Answer: (b) CH₃-C≡CH + 2 HBr

Explanation:
1. Propyne adds two molecules of HBr by Markovnikov addition, with both bromines ending on C-2: CH₃CBr₂CH₃.
2. 1-Bromopropene gives 1,1-dibromopropane (the bromine on C-1 stabilises the cation), acetylene gives 1,1-dibromoethane, and propene gives 2-bromopropane.

Teacher's Note:
1. A gem-dihalide at C-2 comes from a terminal alkyne with two HX.
2. A halogen on the double bond directs the second H to the other carbon.

Answer: (a) 3-bromo-3-methylcyclohex-1-ene

Explanation:
1. Bromine radicals abstract the allylic hydrogen that gives the most stable radical: the allylic hydrogen at C-3 (tertiary and allylic) gives a tertiary allylic radical.
2. Bromine then bonds to this carbon giving 3-bromo-3-methylcyclohexene (the allylic shift product is a minor alternative).

Teacher's Note:
1. Allylic and tertiary position gives the most stable radical.
2. Light favours substitution over addition.

Answer: (c) position isomers

Explanation:
1. Ozonolysis fragments fit 2-methylbut-1-ene (CH₂=O + butanone), 2-methylbut-2-ene (acetone + CH₃CHO) and 3-methylbut-1-ene (CH₂=O + 2-methylpropanal).
2. All three have the same carbon skeleton (2-methylbutane after hydrogenation) and differ only in the position of the double bond: position isomers.

Teacher's Note:
1. Same skeleton, different position of C=C: position isomers.
2. Hydrogenation to the same alkane shows the same skeleton.

Answer: (d) (1-bromopropyl)benzene, C₆H₅CHBrCH₂CH₃

Explanation:
1. H⁺ adds to the carbon next to CH₃ so that the positive charge sits on the benzylic carbon, which is stabilised by the ring.
2. Bromide then attacks the benzylic carbon, giving C₆H₅CHBrCH₂CH₃.

Teacher's Note:
1. Markovnikov addition gives the benzylic cation.
2. Ring bromination does not occur without a Lewis acid.

Answer: (b) C₆H₅COCHBr₂ (alpha,alpha-dibromoketone)

Explanation:
1. HOBr adds across the triple bond (Br⁺ to the terminal carbon, OH to the benzylic carbon) to give an enol, which tautomerises to the alpha-bromoketone C₆H₅COCH₂Br.
2. With excess HOBr a second addition gives the enol of the dibromo compound, which becomes C₆H₅COCHBr₂, the major product.

Teacher's Note:
1. Two molecules of HOX add to a terminal alkyne to give an alpha,alpha-dihaloketone.
2. The intermediate enol tautomerises to the carbonyl form.