A) \[MeBr>M{{e}_{2}}CHBr>\] \[M{{e}_{3}}CBr>E{{t}_{3}}CBr\left( {{S}_{N}}2 \right)\]
B) \[PhC{{H}_{2}}Br>Ph\,CHBrMe>PhCBrM{{e}_{2}}>\]\[PhCBrMePh\left( {{S}_{N}}1 \right)\]
C) \[MeI>MeBr>MeCl>MeF({{S}_{N}}2)\]
D) All are correct
Correct Answer: B
Solution :
[b] The more is the stability of intermediate (carbonium ion), the more is the chance of \[{{S}_{N}}1\] mechanism The intermediates obtained will be \[Ph\overset{+}{\mathop{C}}\,{{H}_{2}}(i),Ph\overset{+}{\mathop{C}}\,H-Me(ii),Ph\overset{+}{\mathop{C}}\,-M{{e}_{2}}(iii),\]\[Ph\overset{+}{\mathop{C}}\,MePh\left( iv \right)\]. The stabilty is of the order iv > iii > ii > i.
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