JEE Chemistry Hydrocarbons MCQs Set 40

Answer: (b) A mixture of benzene and Mg(OMe)Br

Explanation:
1. A Grignard reagent is a very strong base and takes the acidic hydrogen of the OH group of methanol.
2. C₆H₅MgBr + CH₃OH → C₆H₆ + Mg(OCH₃)Br.
3. Answer: (b)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) Ph–C(CH₃)₂OH

Explanation:
1. The reaction goes through a carbocation (SN1), so the alcohol that gives the most stable carbocation reacts fastest.
2. Ph–C(CH₃)₂OH gives a tertiary benzylic cation, stabilised by both alkyl groups and the phenyl ring.
3. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) Benzoic acid

Explanation:
1. Vigorous side-chain oxidation by KMnO₄ converts any alkyl group with a benzylic hydrogen to COOH.
2. The whole ethyl side chain is oxidised to –COOH, giving benzoic acid.
3. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) Toluene

Explanation:
1. Br₂ in CCl₄ is decolourised by C=C and C≡C bonds through addition.
2. But-2-ene, styrene (vinyl group) and phenylacetylene (ethynyl group) all add bromine.
3. Toluene has only the aromatic ring, which resists addition, so the colour stays.
4. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (d) All of these

Explanation:
1. A is acetophenone (Friedel-Crafts acylation).
2. The C=O group can be reduced to CH₂ by HI/red P, by Clemmensen reduction (Zn-Hg/HCl) or by Wolff-Kishner reduction (NH₂NH₂/OH⁻).
3. All three reagents give ethylbenzene.
4. Answer: (d)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) 2,3-Dimethylbut-2-ene

Explanation:
1. The stability of an alkene increases with the number of alkyl groups on the double-bond carbons (hyperconjugation).
2. Ethene has none, 2-methylpropene has two, but-2-ene has two, and 2,3-dimethylbut-2-ene has four.
3. It is the most stable.
4. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) CH₂=CH–CH₂–Br

Explanation:
1. The Wurtz reaction needs an alkyl (or allyl/benzyl) halide, which couples with sodium in dry ether.
2. Allyl bromide is such a halide. Aryl halides need the Wurtz-Fittig conditions with an alkyl halide, and the other two are not halides.
3. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (d) CH₂=CH–OH

Explanation:
1. Hydration of ethyne first gives the enol, vinyl alcohol.
2. Vinyl alcohol is unstable and tautomerises to acetaldehyde, so x is CH₂=CH–OH.
3. Answer: (d)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (d) Ammoniacal AgNO₃

Explanation:
1. Both decolourise bromine water or CCl₄ solution and Baeyer's reagent.
2. Only propyne has a terminal ≡C–H and gives a white precipitate with ammoniacal AgNO₃.
3. Answer: (d)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (a) NaOH

Explanation:
1. Acetylene is only weakly acidic (pK_a about 25), much weaker than water.
2. NaOH is too weak a base to remove its hydrogen, so no reaction takes place.
3. Na and the ammoniacal silver solution form acetylides, and HCl adds across the triple bond.
4. Answer: (a)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) Cl₂CHCHO

Explanation:
1. HOCl adds to the triple bond to give CHCl=CH(OH) and then, with a second HOCl, Cl₂CH–CH(OH)₂.
2. The gem-diol loses water to give dichloroacetaldehyde, Cl₂CHCHO.
3. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (a)

Explanation:
1. Oxymercuration-demercuration follows Markovnikov's rule without rearrangement.
2. OH adds to the more substituted carbon of the double bond (C-2 of the allyl chain).
3. The product is 1-phenylpropan-2-ol.
4. Answer: (a)

Teacher's Note:
1. Oxymercuration gives the Markovnikov alcohol; hydroboration gives the anti-Markovnikov alcohol.

Answer: (d) Benzylamine

Explanation:
1. In benzylamine the nitrogen is attached to an sp³ carbon, so its lone pair is fully available.
2. In the other three, the lone pair is delocalised into one or more benzene rings (and into NO₂ in p-nitroaniline).
3. Answer: (d)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) Cyclic halonium ion

Explanation:
1. The halogen molecule is polarised by the π-bond and forms a bridged three-membered cyclic halonium ion.
2. The halide ion then attacks from the opposite side, giving anti addition.
3. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (b) A ketone

Explanation:
1. Hydration of a terminal alkyne other than ethyne follows Markovnikov's rule, so OH goes to C-2.
2. The enol CH₃–C(OH)=CH₂ tautomerises to the ketone, acetone.
3. Answer: (b)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) Friedel-Crafts reaction

Explanation:
1. AlCl₃ generates the acylium ion CH₃CO⁺, which attacks benzene to give acetophenone.
2. This is Friedel-Crafts acylation.
3. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (c) Homologues

Explanation:
1. Soda lime removes CO₂ to give propane (A).
2. Red P + HI reduces the COOH group to CH₃, giving butane (B).
3. Propane and butane differ by a CH₂ unit, so they are homologues.
4. Answer: (c)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (a) Ionic followed by free radical mechanism

Explanation:
1. At the anode the carboxylate ion loses an electron (an ionic process) and then loses CO₂ to give an alkyl radical.
2. Two alkyl radicals combine to give the alkane.
3. Answer: (a)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (b) Isopentane

Explanation:
1. Alkanes with a tertiary hydrogen are oxidised by KMnO₄ to the tertiary alcohol.
2. Isopentane has a tertiary C–H and gives 2-methylbutan-2-ol.
3. Neopentane, propane and n-pentane have no tertiary C–H.
4. Answer: (b)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (b) HCHO

Explanation:
1. Controlled oxidation of methane with a Mo₂O₃ catalyst stops at the aldehyde stage.
2. The product is formaldehyde (HCHO).
3. Answer: (b)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).

Answer: (b) It gives an oily liquid on reaction with Cl₂

Explanation:
1. Ethene reacts with chlorine to form 1,2-dichloroethane, an oily liquid.
2. The name olefin means 'oil forming'.
3. Answer: (b)

Teacher's Note:
1. Identify the functional group and the reagent first, then apply the rule for that reaction (Markovnikov, anti-Markovnikov, stereochemistry of the reagent).