JEE Chemistry Hydrocarbons MCQs Set 31

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

Explanation:
1. The rate follows the stability of the carbocation formed: 2-methylpropene gives a tertiary cation (fastest).
2. But-2-ene and propene both give secondary cations, but the more substituted alkene (but-2-ene) has more alkyl groups to stabilise the transition state; ethene gives a primary cation (slowest).

Teacher's Note:
1. More alkyl substituents on C=C mean faster electrophilic addition.
2. Ethene is the least reactive of the simple alkenes.

Answer: (b) diphenylmethane, C₆H₅CH₂C₆H₅

Explanation:
1. Both C-Cl bonds of CH₂Cl₂ take part in Friedel-Crafts alkylation: the first gives C₆H₅CH₂Cl, which reacts again with benzene (benzylic chlorides are very reactive).
2. With excess benzene the final product is C₆H₅CH₂C₆H₅.

Teacher's Note:
1. Dihalomethanes alkylate two arene molecules.
2. Excess arene prevents further reaction at the new CH₂ group.

Answer: (b) 1-bromo-1-methylcyclohexane

Explanation:
1. Bromine is highly selective and abstracts the hydrogen that gives the most stable radical.
2. The tertiary hydrogen on the ring carbon bearing the methyl group gives a tertiary radical, so the tertiary bromide is the major product.

Teacher's Note:
1. Bromination selectivity: 3 degree >> 2 degree > 1 degree.
2. Chlorination is much less selective.

Answer: (d) 4

Explanation:
1. Ortho-para directing groups have a lone pair or donate electrons: -Cl (+M), -OH, -OC₂H₅ and -NHCOCH₃.
2. -NO₂, -CN and -CHO withdraw electrons by -M and -I effects and are meta directing.

Teacher's Note:
1. Atoms with lone pairs directly on the ring direct ortho and para.
2. A carbonyl, nitrile or nitro group on the ring directs meta.

Answer: (b) 4-chloro-2-nitrotoluene (nitro group ortho to CH₃)

Explanation:
1. The activating group (CH₃) controls the position when an activating and a deactivating group compete.
2. The incoming NO₂⁺ goes ortho to CH₃ (para is already occupied by Cl).

Teacher's Note:
1. Activating group wins over deactivating group.
2. Mononitration is the main reaction under controlled conditions.

Answer: (a) 2-phenylpropan-2-ol

Explanation:
1. Protonation of CH₂=C(CH₃)Ph gives the tertiary benzylic cation (CH₃)₂C⁺-Ph, which is captured by water to give (CH₃)₂C(OH)Ph.

Teacher's Note:
1. Markovnikov: OH on the carbon that forms the more stable cation.
2. Tertiary benzylic cation is very stable.

Answer: (b) structural (positional) isomers: 2-bromo-1-phenylpropane and 1-bromo-1-phenylpropane

Explanation:
1. With peroxide, Br· adds to the carbon next to CH₃ to give the benzylic radical, so A = C₆H₅CH₂-CHBr-CH₃ (2-bromo-1-phenylpropane).
2. Without peroxide, H⁺ adds to the carbon next to CH₃ to give the benzylic cation, so B = C₆H₅CHBr-CH₂CH₃ (1-bromo-1-phenylpropane).
3. The two have the bromine at different carbons, so they are positional isomers.

Teacher's Note:
1. Radical addition and ionic addition give opposite regiochemistry.
2. Compare the carbon bearing Br in each product.

Answer: (a) free radical substitution

Explanation:
1. NBS with peroxides brominates the allylic position by a radical chain mechanism, replacing an allylic hydrogen by Br (giving 3-bromopropene).

Teacher's Note:
1. NBS gives allylic substitution, not addition to C=C.
2. Peroxide or light is the radical initiator.

Answer: (d) 1-phenylpropan-1-ol

Explanation:
1. Protonation first gives the secondary cation C₆H₅CH₂-CH⁺-CH₃.
2. A hydride shift from the benzylic carbon gives the more stable benzylic cation C₆H₅CH⁺-CH₂CH₃, which is captured by water: 1-phenylpropan-1-ol.

Teacher's Note:
1. Rearrangement to a benzylic cation is favourable.
2. Direct Markovnikov product would be 1-phenylpropan-2-ol, but it is minor.

Answer: (a) the bridged bromonium ion from an alkyne (a three-membered ring with a C=C) is more strained and less stable than that from an alkene

Explanation:
1. The first step is formation of a cyclic bromonium ion. In the alkyne case the ring contains an sp²-like double bond (a cyclopropene-like cation), which is highly strained.
2. A less stable intermediate means a higher activation energy and a slower reaction; also sp carbons hold their pi electrons more tightly.

Teacher's Note:
1. Stability of the intermediate decides the rate.
2. Alkynes: pi electrons are held more tightly (more s character).

Answer: (c) Both propyne and 2-bromopropene

Explanation:
1. Propyne adds two molecules of HBr by Markovnikov addition: CH₃C≡CH to CH₃C(Br)=CH₂ to CH₃CBr₂CH₃.
2. 2-Bromopropene adds one HBr with H to the CH₂ carbon (the Br-stabilised cation), also giving 2,2-dibromopropane.
3. Propene gives only 2-bromopropane.

Teacher's Note:
1. Gem-dihalides arise from alkynes with two HX.
2. Halogen on the double bond directs the next H to the other carbon.

Answer: (d) (A), (B) and (C) only

Explanation:
1. (A) and (B) are intramolecular Friedel-Crafts alkylation and acylation, ring closure on the benzene ring: electrophilic substitution.
2. (C) Br₂ with Fe brominates the ring: electrophilic substitution.
3. (D) Br₂ in light brominates the benzylic C-H by a radical path: not electrophilic substitution.

Teacher's Note:
1. Lewis acid means ionic (ring) chemistry; light means radical (side chain) chemistry.
2. Ring closure onto the arene is electrophilic substitution.

Answer: (d) 4

Explanation:
1. Br· adds to C-2 to give the more stable tertiary radical at C-3, and the radical takes H from HBr: the product is 2-bromo-3-methylpentane, CH₃CHBrCH(CH₃)CH₂CH₃.
2. It has two stereocentres (C-2 and C-3) with different substituents, so 2² = 4 stereoisomers (RDKit enumeration).

Teacher's Note:
1. Anti-Markovnikov addition puts Br on C-2 here.
2. Two different stereocentres give four stereoisomers.

Answer: (d) CH₂=CH-Cl (vinyl chloride)

Explanation:
1. Allyl chloride forms the stable allyl cation, and the two acid chlorides form acylium ions, so all three can alkylate or acylate benzene.
2. Vinyl chloride would need a vinyl cation (very unstable); the C-Cl bond is strong because of conjugation, so no reaction occurs.

Teacher's Note:
1. Aryl and vinyl halides do not ionise in Friedel-Crafts conditions.
2. Acid chlorides always work (acylium ion).

Answer: (a) CF₃CH=CH₂ + HCl

Explanation:
1. The CF₃ group is strongly electron withdrawing, so a cation next to it is destabilised; H⁺ adds to the carbon bearing CF₃, giving CF₃CH₂CH₂Cl (anti-Markovnikov).
2. Vinyl chloride and methyl vinyl ether give the Markovnikov products because Cl and OCH₃ stabilise the adjacent cation by +M.

Teacher's Note:
1. Strong -I groups reverse the orientation.
2. Lone-pair donors (O, Cl) next to C=C give Markovnikov products.

Answer: (c) 1-phenylpropan-2-ol

Explanation:
1. Oxymercuration is a Markovnikov hydration without rearrangement, because a bridged mercurinium ion is formed.
2. OH therefore goes to C-2 (the more substituted carbon), giving C₆H₅CH₂CH(OH)CH₃.

Teacher's Note:
1. Oxymercuration-demercuration: Markovnikov hydration with no rearrangement.
2. Hydroboration gives the anti-Markovnikov alcohol instead.

Answer: (d) CH₃CH₂C≡CH (but-1-yne)

Explanation:
1. Hydroboration-oxidation giving an aldehyde and Hg²⁺ hydration giving a ketone shows X is a terminal alkyne.
2. C₄H₆ as a terminal alkyne is but-1-yne: Y = butanal and Z = butan-2-one (functional isomers C₄H₈O).

Teacher's Note:
1. Only terminal alkynes give an aldehyde on hydroboration.
2. But-2-yne is internal and gives only a ketone.

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

Explanation:
1. Bromination is highly selective for the weakest C-H bond, the tertiary C-H at C-2, which leads to the tertiary radical.
2. The product is 2-bromo-2-methylbutane.

Teacher's Note:
1. Selectivity of bromine radicals is very high.
2. Tertiary > secondary > primary.

Answer: (a) 3-bromobut-1-ene and 1-bromobut-2-ene

Explanation:
1. The allylic hydrogen on C-3 is removed to give the resonance-stabilised radical CH₂=CH-CH·-CH₃ ↔ ·CH₂-CH=CH-CH₃.
2. Bromine can bond to either end of the allyl radical, giving CH₂=CH-CHBr-CH₃ (3-bromobut-1-ene) and BrCH₂-CH=CH-CH₃ (1-bromobut-2-ene).

Teacher's Note:
1. Allyl radicals have two reactive ends.
2. Both products keep a C=C bond.

Answer: (c) CH₃-CH⁺-CH₂-Cl

Explanation:
1. The electrophile Cl⁺ adds first to the terminal carbon, giving the more stable secondary carbocation CH₃-CH⁺-CH₂Cl (a bridged chloronium ion in practice).
2. Water then attacks the secondary carbon to give CH₃CH(OH)CH₂Cl.

Teacher's Note:
1. Cl⁺ is the electrophile.
2. OH ends up on the more substituted carbon.

Answer: (a) deactivating and ortho-para directing

Explanation:
1. Chlorine withdraws electrons by -I (more strongly than it donates by +M), so the ring is deactivated.
2. The lone pair donates by +M into the ortho and para positions, so the orientation is ortho-para.

Teacher's Note:
1. Halogens: -I controls rate, +M controls orientation.
2. Do not confuse halogens with other ortho-para directors, which are activating.

Answer: (b) (B), (C) and (D) only

Explanation:
1. The reaction is a radical chain reaction, so the intermediate is a benzylic radical, not a carbocation: (A) is wrong.
2. The benzylic radical is planar and Br₂ attacks from either face, giving a racemic mixture: (B) is correct. The radical is stabilised by the ethyl's +I effect, hyperconjugation and resonance with the ring: (C) correct.
3. Br₂ at high temperature (or light) also brominates the benzylic position: (D) correct.

Teacher's Note:
1. Benzylic radicals are stabilised by resonance with the ring.
2. Radical intermediates are planar so racemic products form.

Answer: (d) an acetal (2-alkoxytetrahydropyran)

Explanation:
1. Protonation of the enol-ether double bond gives an oxocarbenium ion, stabilised by the ring oxygen.
2. The alcohol adds to this cation to give a carbon bearing two OR groups (the ring O and the new OCH₂R): an acetal.

Teacher's Note:
1. Two OR groups on one carbon means acetal.
2. Enol ethers add alcohols under acid catalysis.

Answer: (d) (B) and (D) only

Explanation:
1. Anti addition to the trans alkene gives the meso dibromide (M and N are the same meso compound), while anti addition to the cis alkene gives the pair of enantiomers (2R,3R) and (2S,3S) (O and P).
2. So O and P are enantiomers (C is false), and the meso compound is a diastereomer of each of them: (M, O) and (N, P) are diastereomer pairs (D true). Both reactions are anti additions (B true); (A) is false.

Teacher's Note:
1. Meso and chiral forms of the same constitution are diastereomers.
2. Enantiomers are non-superimposable mirror images.

Answer: (a) butan-2-one, CH₃CH₂COCH₃

Explanation:
1. Mercuric-ion catalysed hydration is Markovnikov: OH goes to C-2, giving the enol CH₃CH₂C(OH)=CH₂.
2. The enol tautomerises to the ketone, butan-2-one.

Teacher's Note:
1. Terminal alkynes give methyl ketones (ethyne gives acetaldehyde).
2. Enol to keto tautomerism is fast.

Answer: (a) (A) and (D) only

Explanation:
1. (A) is incorrect and (B) is correct: the intermediate vinyl cation or strained bromonium ion makes alkynes less reactive.
2. (C) is correct: alkynes add bromine.
3. (D) is incorrect: in the presence of peroxide the radical mechanism gives anti-Markovnikov addition.

Teacher's Note:
1. Peroxide reverses the orientation of HBr addition.
2. Alkenes are more nucleophilic than alkynes.

Answer: (a) III > I > II

Explanation:
1. Pyrrole is electron rich: the nitrogen lone pair is part of the aromatic sextet and raises the electron density on every carbon, so it is much more reactive than benzene.
2. In pyridine the nitrogen is more electronegative than carbon, withdraws electrons and (in acid) is protonated, so the ring is strongly deactivated.

Teacher's Note:
1. Five-membered heteroaromatics (pyrrole, furan, thiophene) are more reactive than benzene.
2. Pyridine behaves like nitrobenzene in EAS.

Answer: (c) benzyl bromide, C₆H₅CH₂Br

Explanation:
1. NBS carries out free-radical bromination at the benzylic position, where the radical is stabilised by the ring.
2. The ring is not brominated under these conditions (no Lewis acid).

Teacher's Note:
1. Ring bromination needs Br₂ and FeBr₃.
2. Side-chain bromination needs light, heat or NBS.

Answer: (d) Cl⁺ in the first step

Explanation:
1. In HOCl the chlorine is the positive end (Cl^δ+-OH^δ-), so Cl⁺ is the electrophile and attacks the pi bond first to give a chloronium ion or carbocation.
2. Water or hydroxide then attacks the more substituted carbon to complete the addition.

Teacher's Note:
1. Electrophile = the less electronegative atom of the reagent.
2. Halohydrin formation is a two-step process.

Answer: (c) (B), (C) and (D) only

Explanation:
1. (A) gives elimination (isobutene) with the alkoxide base; benzene is not alkylated.
2. In (B) and (D) the isobutyl cation rearranges by a hydride shift to the tertiary butyl cation, and in (C) protonation of isobutene gives the tert-butyl cation directly; all three then alkylate benzene to give tert-butylbenzene.

Teacher's Note:
1. Friedel-Crafts alkylation can proceed through rearranged carbocations.
2. Alkoxide bases promote elimination.

Answer: (c) a secondary or tertiary alcohol

Explanation:
1. By Markovnikov's rule the proton adds to give the more stable carbocation, which is secondary or tertiary for every alkene other than ethene.
2. Water adds to that cation to give a secondary or tertiary alcohol; a primary alcohol is formed only from ethene.

Teacher's Note:
1. Primary carbocations are too unstable to form here.
2. Ethene is the only alkene that gives a primary alcohol (ethanol).

Answer: (d) propane

Explanation:
1. Propane has two kinds of hydrogen (CH₃ and CH₂), so it gives two monobromo isomers: 1-bromopropane and 2-bromopropane.
2. Ethane gives only one; n-pentane gives three; 2-methylbutane gives four. Pentane isomers are also liquids, not gases.

Teacher's Note:
1. Count distinct hydrogens to count monosubstituted isomers.
2. Propane and butane are gases at room temperature.