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Substitution mechanisms: NEET MCQs with answers

50 questions on this topic of Organic Compounds Containing Halogens (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following haloalkanes is practically inert to substitution by an SN2 mechanism despite being a primary alkyl halide?

  1. 1-Chlorobutane
  2. Neopentyl chloride
  3. Isobutyl chloride
  4. Ethyl chloride
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Answer: (B) Neopentyl chloride

Neopentyl chloride is a primary halide, but the bulky tert-butyl group directly attached to the α-carbon provides severe steric hindrance to the backside attack of the nucleophile. Hence, it is extraordinarily slow (almost inert) towards SN2 substitution.

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2. The nucleophilic substitution reaction between (S)-2-chlorobutane and sodium iodide in dry acetone proceeds with complete Walden inversion. The IUPAC name and configuration of the product obtained is:

  1. (R)-2-iodobutane
  2. (S)-2-iodobutane
  3. (2R,3R)-diiodobutane
  4. (±)-2-iodobutane
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Answer: (A) (R)-2-iodobutane

In the Finkelstein reaction (SN2 mechanism), backside attack occurs at the chiral center. Since the priorities of the substituents on C2 remain in the same sequence (I > -CH₂CH₃ > -CH₃ > -H), complete inversion of configuration changes (S)-2-chlorobutane into (R)-2-iodobutane.

hard

3. In a competitive nucleophilic substitution reaction, 1 mole of 1-bromobutane is treated with a mixture containing 1 mole of NaSH and 1 mole of NaOH in a polar aprotic solvent. The major organic product formed is:

  1. Butan-1-ol (CH₃CH₂CH₂CH₂OH)
  2. Dibutyl ether ((C₄H₉)₂O)
  3. Butane-1-thiol (CH₃CH₂CH₂CH₂SH)
  4. Dibutyl sulfide ((C₄H₉)₂S)
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Answer: (C) Butane-1-thiol (CH₃CH₂CH₂CH₂SH)

Hydrosulfide ion (HS⁻) is a much stronger nucleophile than hydroxide ion (OH⁻) due to the larger size and greater polarizability of sulfur compared to oxygen. Therefore, HS⁻ selectively attacks 1-bromobutane via the SN2 pathway to yield butane-1-thiol as the major product.

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4. For the hydrolysis of an optically pure alkyl halide RX via the SN1 pathway, the isolated product showed 80% racemisation and 20% inversion of configuration. What is the optical purity (enantiomeric excess) of the resulting mixture?

  1. 80%
  2. 40%
  3. 10%
  4. 20%
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Answer: (D) 20%

The mixture consists of 80% racemic mixture (which is optically inactive) and 20% net inverted enantiomer. The enantiomeric excess (optical purity) is therefore directly equal to the excess un-racemised enantiomer, which is 20%.

easy

5. Which of the following alkyl halides undergoes nucleophilic substitution via the S_N1 mechanism at the fastest rate?

  1. (CH₃)₃C-Br
  2. (CH₃)₃C-I
  3. (CH₃)₃C-Cl
  4. (CH₃)₃C-F
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Answer: (B) (CH₃)₃C-I

In an S_N1 mechanism, the rate-determining step is the formation of a carbocation via cleavage of the carbon-halogen bond. Since all substrates form the same stable tertiary carbocation, the rate depends on the leaving group ability: I⁻ > Br⁻ > Cl⁻ > F⁻ due to the weakest C-I bond. Thus, (CH₃)₃C-I reacts fastest.

hard

6. In an experiment, an optically pure alkyl bromide, (+)-2-bromooctane, undergoes nucleophilic substitution with sodium hydroxide. The specific rotation of the starting material is +36.0°. If the product 2-octanol has an observed specific rotation of -28.8° under conditions where pure (-)-2-octanol is -36.0°, what are the percentages of inversion and racemization, respectively?

  1. 90% inversion and 10% racemization
  2. 20% inversion and 80% racemization
  3. 10% inversion and 90% racemization
  4. 80% inversion and 20% racemization
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Answer: (D) 80% inversion and 20% racemization

Optical purity (enantiomeric excess) of the product = (|-28.8°| / 36.0°) × 100% = 80%. This optical activity is solely due to the inverted product formed via the stereospecific pathway. The remaining 20% corresponds to the racemic portion (10% inverted + 10% retained). Thus, total inversion = 80% + 10% = 90% (or net optical inversion = 80%), but according to the standard convention, the product mixture contains 80% inverted excess and 20% racemic mixture.

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7. Which among the following compounds will react most rapidly via the S_N1 pathway upon treatment with aqueous acetone?

  1. C₆H₅CH₂Cl
  2. C₆H₅CH₂CH₂Cl
  3. C₆H₅CH(CH₃)Cl
  4. CH₃CH₂CH₂Cl
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Answer: (C) C₆H₅CH(CH₃)Cl

S_N1 reactivity is determined by the stability of the intermediate carbocation. C₆H₅CH(CH₃)Cl forms a secondary benzylic carbocation, C₆H₅-CH⁺-CH₃, which is stabilized both by resonance with the benzene ring and by hyperconjugation from the adjacent methyl group (3 α-hydrogens). This cation is more stable than the primary benzylic cation from C₆H₅CH₂Cl or primary aliphatic carbocations.

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8. The SN1 reaction of an optically active alkyl halide usually gives:

  1. No reaction
  2. A racemic (or partly racemic) product, via a planar carbocation
  3. Complete inversion
  4. Complete retention
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Answer: (B) A racemic (or partly racemic) product, via a planar carbocation

In SN1 the leaving group departs first to form a planar carbocation, which the nucleophile can attack from either face, giving racemisation.

NCERT Class 12

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9. Arrange the following chlorides in decreasing order of their reactivity towards SN1 hydrolysis: (I) (C₆H₅)₂CH-Cl (II) C₆H₅CH₂-Cl (III) (C₆H₅)₃C-Cl (IV) C₆H₅CH(CH₃)-Cl

  1. III > I > IV > II
  2. III > IV > I > II
  3. I > III > IV > II
  4. II > IV > I > III
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Answer: (A) III > I > IV > II

SN1 reactivity is directly proportional to the stability of the intermediate carbocation. Triphenylmethyl carbocation (III) is stabilized by resonance across three phenyl rings (most stable). Diphenylmethyl carbocation (I) is stabilized by two phenyl rings. 1-Phenylethyl carbocation (IV) is stabilized by one phenyl ring and hyperconjugation from the methyl group. Benzyl carbocation (II) is stabilized by one phenyl ring only. Therefore, the order is III > I > IV > II.

hard

10. Which of the following substrates undergoes nucleophilic substitution with sodium methoxide via an SN2 pathway at the fastest rate?

  1. CH₃-O-CH₂-Cl
  2. CH₃-CH₂-CH₂-Cl
  3. CH₃-CH₂-O-CH₂-CH₂-Cl
  4. (CH₃)₂CH-Cl
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Answer: (A) CH₃-O-CH₂-Cl

In chloromethyl methyl ether (CH₃-O-CH₂-Cl), the unshared pair of electrons on the adjacent oxygen atom can stabilize the developing transition state through resonance interaction, and the strong inductive electron withdrawal lowers the barrier for direct displacement. Furthermore, α-halo ethers undergo exceptionally fast substitution compared to ordinary primary alkyl chlorides.

hard

11. Consider the nucleophilic substitution of (R)-2-iodooctane with radioactive iodide ion (*I⁻) in dry acetone: (R)-2-iodooctane + *I⁻ → 2-iodooctane-*I + I⁻ If the rate constant for radioactive exchange is k_ex and the rate constant for loss of optical activity (racemization) is k_rac, what is the relation between k_rac and k_ex?

  1. k_rac = k_ex
  2. k_rac = 4 × k_ex
  3. k_rac = 0.5 × k_ex
  4. k_rac = 2 × k_ex
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Answer: (D) k_rac = 2 × k_ex

Each individual nucleophilic substitution step occurs with complete Walden inversion, converting an (R) molecule into an (S) molecule. The creation of one inverted molecule neutralizes the optical rotation of one original (R) molecule, resulting in a pair of racemic molecules. Therefore, one substitution event causes the loss of optical activity of two molecules, meaning the rate of loss of optical activity (k_rac) is exactly twice the rate of isotopic exchange (k_ex).

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12. The SN2 reaction of an alkyl halide proceeds with:

  1. Retention of configuration
  2. Racemisation through a carbocation
  3. Formation of a free radical
  4. Inversion of configuration in a single concerted step
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Answer: (D) Inversion of configuration in a single concerted step

In SN2 the nucleophile attacks from the side opposite to the leaving group, and bond making and breaking happen together, inverting the configuration (Walden inversion).

NCERT Class 12

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