MCQMediumJEE 2023Reaction Mechanisms (Substitution, Addition, Elimination)

JEE Chemistry 2023 Question with Solution

Choose the halogen which is most reactive towards SN1S_N1 reaction in the given compounds (A, B, C, & D):

Four organic compounds labeled A, B, C and D with halogen positions marked as (a) and (b), asking which halogen is most reactive toward SN1 reaction in each structure.
  • A

    A-Br(a); B-I(a); C-Br(b); D-Br(a)

  • B

    A-Br(a); B-I(a); C-Br(a); D-Br(a)

  • C

    A-Br(b); B-I(b); C-Br(b); D-Br(b)

  • D

    A-Br(a); B-Br(a); C-Br(a); D-I(a)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: We must choose the halogen most reactive towards the SN1S_N1 reaction in compounds A, B, C, and D.

Find: The correct option listing the most reactive halogen position in each compound.

For an SN1S_N1 reaction, the rate depends on the stability of the carbocation formed after the leaving group departs. Better leaving groups and more stable carbocations favor the reaction.

The extracted explanation states:

  • Compound A forms a benzyl carbocation, which is highly stable due to resonance.
  • Compound B forms a primary carbocation, but reacts due to iodine's leaving group strength.
  • Compound C forms a tertiary carbocation, which is very stable and reactive.
  • Compound D forms a primary carbocation, less stable but reacts due to bromine's moderate leaving group strength.

Thus, the most reactive halogens are:

  • A: Br(a)
  • B: I(a)
  • C: Br(b)
  • D: Br(a)

Therefore, the correct option is A.

Carbocation Stability Basis

Given: Reactivity towards SN1S_N1 depends mainly on carbocation stability and leaving group ability.

Find: Which labeled halogen leaves most readily in each structure.

Use the principle:

More stable carbocationfaster SN1\text{More stable carbocation} \Rightarrow \text{faster } S_N1

Also, among halides, iodide is generally a better leaving group than bromide.

Applying the extracted solution statements:

  1. A gives a benzylic carbocation, stabilized by resonance, so Br(a) is preferred.
  2. B is influenced by the stronger leaving ability of iodine, so I(a) is preferred.
  3. C gives a tertiary carbocation, so the halogen at Br(b) is most reactive.
  4. D reacts through the position indicated as Br(a) according to the extracted explanation.

Hence the sequence matches: A-Br(a); B-I(a); C-Br(b); D-Br(a)

Therefore, the correct option is A.

Common mistakes

  • Assuming only leaving group ability decides SN1S_N1 reactivity is incorrect because carbocation stability is the primary factor. Compare the carbocation formed after halogen departure before ranking the sites.

  • Ignoring resonance stabilization in benzylic systems leads to a wrong choice for compound A. A benzylic carbocation is much more stable than an ordinary primary carbocation.

  • Treating all primary halides as equally unreactive in SN1S_N1 is incorrect here. A better leaving group such as iodine can still make one position more reactive than another.

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