The $S_{N}1$ reactivity of following halides will be in the order (i) $\left( CH_{3} \right)_{3}CBr$ (ii)$\left( C_{6}H_{5} \right)_{2}CHBr\$ $(iii)\left( C_{6}H_{5} \right)_{2}C\left( CH_{3} \right)Br$ (iv) $\left( CH_{3} \right)_{2}CHBr$ $(v)\ C_{2}H_{5}Br$

Question

Dr. Abubacker · Accepted Answer

(d)
$\mathbf{S}_{\mathbf{N}}\mathbf{1}$(Unimolecular nucleophilic substitution reactions)
Rate$\propto$ (substrate)
Rate determining step in the formation of carbocation depends on the stability of carbocation formed. The stability of carbocations follow the order
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$\because$ Order of $S_{N}1$ reacticity is
$\left( C_{6}H_{5} \right)_{2}C\left( CH_{3} \right)Br > \left( C_{6}H_{5} \right)_{2}CHBr > \left( CH_{3} \right)_{3}CBr > \left( CH_{3} \right)_{2}CHBr > C_{2}H_{5}Br$
$i.e.,$ iii>ii>i>iv>v

Q. The $S_{N}1$ reactivity of following halides will be in the order (i) $\left( CH_{3} \right)_{3}CBr$ (ii)$\left( C_{6}H_{5} \right)_{2}CHBr\$ $(iii)\left( C_{6}H_{5} \right)_{2}C\left( CH_{3} \right)Br$ (iv) $\left( CH_{3} \right)_{2}CHBr$ $(v)\ C_{2}H_{5}Br$

Explanation

Reviewed by Dr. Abubacker

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