MCQ
To carry out above conversion, $(A)$ and $(B)$ respectively, are
  • A
    $NaNH_2 ,Cl - CH_2 - CH_2 - CH_2 -Br$
  • B
    $NaNH_2 , F-CH_2 -CH_2 -CH_2 -Br$
  • $NaNH_2 , I - CH_2 -CH_2 -CH_2 -Br$
  • D
    $NaNH_2 , I-CH_2 -CH_2 -CH_2 -I$

Answer

Correct option: C.
$NaNH_2 , I - CH_2 -CH_2 -CH_2 -Br$
c
$(c)$ Acid-base reaction follower by $S_{N^2}$
$\therefore$ $I$ is better leaving group than $Br$.
$\therefore$ $(c)$ is favourable.

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