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Question:

Give reasons for the following: (i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide. (ii) (±)2-Butanol is optically inactive. (iii) C−X bond length in halobenzene is smaller than C−X bond length in CH3−X.

Solution:

  1. Because in ethyl iodide, iodide (I) is acting as a best leaving group among all the halide ions so the rate of SN2 reaction is directly proportional to leaving group ability.
  2. (±) 2-butanol is a racemic mixture which is optically inactive due to external compensation.
  3. Due to resonance in halobenzene it has less bond length value as compared to CH3−X.