Question

The $S_{N}1$ reactivity of the following halides will be in the order
(i) $(C{H}_3{)}_{3}CBr$
(ii) $(C_6{H}_5{)}_{2}CHBr$
(iii) $(C_6{H}_5{)}_{2}C(C{H}_3)Br$
(iv) $(C{H}_3{)}_{2}CHBr$
(v) $C_2{H}_{5}Br$

• A. $(v)>(iv)>(i)>(ii)>(iii)$
• B. $(ii)>(i)>(iii)>(v)>(iv)$
• C. $(i)>(iii)>(v)>(ii)>(iv)$
• D. $(v)>(i)>(ii)>(iv)>(iii)$
• E. $(iii)>(ii)>(i)>(iv)>(v)$

Answer 1

Answer: (E)

$S_{N}1$ (Unimolecular nucleophilic substitution reactions)
$Rate\propto (substrate)$
Rate of determining step in the formation of carbocation depends on the stability of carbocation formed. The stability of carbocations follow the order-
${(C_6{H}_5)}_2{C}^{+}(C{H}_3)>{(C_6{H}_5)}_2\stackrel{+}{C}H>$
${(C{H}_3)}_2\stackrel{+}{C}H>C{H}_{3}C{H}_2^{+}$
$\because$ Order of $S_{N}1$ reactivity is ${(C_6{H}_5)}_{2}C(C{H}_3)Br>{(C_6{H}_5)}_{2}CHBr>$
${(C{H}_3)}_{3}CBr>{(C{H}_3)}_2\stackrel{+}{C}HBr>C_2{H}_{5}Br$ ie,
$iii>ii>i>iv>v$