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`
C
`NaNH_(2), I-CH_(2)-CH_(2)-CH_(2)-Br`
D
`NaNH_(2), I-CH_(2)-CH_(2)-CH_(2)-I`
Text Solution
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The correct Answer is:
c
Acid -base reaction follower by S `therefore` I is a better leavin group than `overset(N^(2))Br`. (c) `therefore` is favourable
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