JEE Chemistry Organic Chemistry Some Basic Principles And Techniques MCQs Set 15

Answer: (b) substitution

Explanation:
1. Reaction with aqueous KOH replaces the halogen by OH, which is substitution. Answer: (b)

Teacher's Note:
1. Dehydrohalogenation would give an alkene.

Answer: (d) C₆H₅CH₂Cl

Explanation:
1. Benzyl chloride forms a delocalisation-stabilised benzyl carbocation and also undergoes SN2 easily
2. Vinyl and aryl chlorides are inert because of partial double-bond character
3. C₆H₅CH₂Cl is the most reactive. Answer: (d)

Teacher's Note:
1. Allylic and benzylic halides are very reactive.

Answer: (c) (C₆H₅)₃C⁺

Explanation:
1. The triphenylmethyl cation is stabilised by electron delocalisation over three phenyl rings
2. It is more stable than the t-butyl, diphenylmethyl and benzyl cations. Answer: (c)

Teacher's Note:
1. More aryl groups mean more delocalisation stabilisation.

Answer: (c) a racemic product

Explanation:
1. The carbocation intermediate is planar
2. The nucleophile can attack from either side with equal probability, giving a racemic product. Answer: (c)

Teacher's Note:
1. SN2 gives inversion, and SN1 gives racemisation.

Answer: (d) nucleophilic substitution reaction

Explanation:
1. OH⁻ is a nucleophile that replaces the halide, so this is nucleophilic substitution. Answer: (d)

Teacher's Note:
1. OH- is the nucleophile and X- is the leaving group.

Answer: (a) ether

Explanation:
1. This is the Williamson synthesis
2. C₂H₅ONa + C₂H₅I → C₂H₅–O–C₂H₅ + NaI
3. The product is diethyl ether. Answer: (a)

Teacher's Note:
1. Williamson synthesis gives ethers by SN2.

Answer: (d) vinyl chloride

Explanation:
1. In vinyl chloride the C–Cl bond has partial double-bond character due to delocalisation of the chlorine lone pair with the double bond
2. So the chlorine is very unreactive. Answer: (d)

Teacher's Note:
1. Vinyl and aryl halides are inert to nucleophilic substitution.

Answer: (a) (a) < (b) < (c) < (d)

Explanation:
1. SN2 needs unhindered backside attack
2. Reactivity decreases with the number of alkyl substituents on the carbon
3. Increasing order: 3° < 2° < 1° < methyl. Answer: (a)

Teacher's Note:
1. Steric hindrance slows SN2.

Answer: (c) isobutyl alcohol

Explanation:
1. Lucas reagent reacts through carbocations
2. Isobutyl alcohol is primary and reacts very slowly at room temperature
3. Tertiary and benzylic alcohols react almost at once, and secondary ones in a few minutes. Answer: (c)

Teacher's Note:
1. Primary alcohols give no turbidity at room temperature with Lucas reagent.

Answer: (b) Williamson's continuous etherification process

Explanation:
1. Heating excess alcohol with conc. H₂SO₄ at 140 °C and adding more alcohol continuously forms ether
2. This is the Williamson continuous etherification process. Answer: (b)

Teacher's Note:
1. Williamson synthesis uses an alkoxide and an alkyl halide.

Answer: (c) a nucleophilic substitution reaction

Explanation:
1. The ethoxide ion is a nucleophile that displaces iodide
2. This is nucleophilic substitution. Answer: (c)

Teacher's Note:
1. Alkoxide ions are good nucleophiles and strong bases.

Answer: (c) SN2 displacement

Explanation:
1. An alkoxide ion attacks a primary alkyl halide from the back side in one step
2. This is an SN2 displacement. Answer: (c)

Teacher's Note:
1. Use primary alkyl halides to avoid elimination.

Answer: (a) CH₃X > 1° > 2° > 3°

Explanation:
1. The reaction is SN2
2. Yield decreases as the alkyl halide becomes more hindered, since elimination competes
3. Order: CH₃X > 1° > 2° > 3°. Answer: (a)

Teacher's Note:
1. Tertiary halides give alkenes instead of ethers.

Answer: (a) phenol and ethyl bromide

Explanation:
1. The ether oxygen is protonated
2. Br⁻ attacks the alkyl carbon (SN2) and not the aryl carbon
3. Products: phenol and ethyl bromide. Answer: (a)

Teacher's Note:
1. The aryl C-O bond is stronger because of delocalisation.

Answer: (b) tertiary alcohol by SN1

Explanation:
1. Tertiary alcohols give the most stable carbocations
2. So they react fastest through the SN1 mechanism. Answer: (b)

Teacher's Note:
1. Lucas test: 3° immediately, 2° in about 5 minutes, 1° does not react at room temperature.

Answer: (b) R–F < R–Cl < R–Br < R–I

Explanation:
1. The better the leaving group, the faster the elimination
2. The C–X bond strength decreases from F to I, so reactivity increases
3. Order: R–F < R–Cl < R–Br < R–I. Answer: (b)

Teacher's Note:
1. Iodide is the best leaving group among the halides.

Answer: (c) a carbocation

Explanation:
1. E1 is a two-step process in which the leaving group departs first to give a carbocation
2. A base then removes a β-hydrogen. Answer: (c)

Teacher's Note:
1. E1 and SN1 share the same first step.

Answer: (d) t-butyl bromide

Explanation:
1. Dehydrohalogenation is fastest for tertiary halides because they give the most stable (most substituted) alkene and carbocation
2. Order: 3° > 2° > 1°. Answer: (d)

Teacher's Note:
1. Zaitsev's rule favours more substituted alkenes.

Answer: (a) 1-phenylpropene

Explanation:
1. Alcoholic KOH causes dehydrohalogenation
2. The β-hydrogen is lost from the benzylic carbon, giving the alkene conjugated with the ring: PhCH=CH–CH₃ (1-phenylpropene)
3. This is the Zaitsev and conjugated product. Answer: (a)

Teacher's Note:
1. Conjugation with the ring makes the benzylic alkene the major product.

Answer: (c) s-butyl bromide

Explanation:
1. s-Butyl bromide, CH₃CH₂CHBrCH₃, has two different β-carbons
2. It gives both but-1-ene and but-2-ene
3. n-Propyl, isopropyl and t-butyl halides give only one alkene. Answer: (c)

Teacher's Note:
1. A different alkene forms for each non-equivalent β-hydrogen set.

Answer: (c) B > C > A

Explanation:
1. Dehydration goes through carbocations
2. B gives a tertiary carbocation, C a secondary one and A a primary one
3. Order: B > C > A. Answer: (c)

Teacher's Note:
1. Tertiary alcohols dehydrate most easily.

Answer: (c) 1-methylcyclohex-1-ene (double bond to the carbon bearing CH₃)

Explanation:
1. Protonation and loss of water give a primary carbocation
2. A hydride shift gives the tertiary 1-methylcyclohexyl cation
3. Loss of H⁺ then gives the most substituted alkene, 1-methylcyclohexene. Answer: (c)

Teacher's Note:
1. Rearrangement leads to the tertiary cation, and Zaitsev's rule gives the alkene.