Consider the reactions :(i) CH32CH−CH2Br⟶C2H5OH CH32CH−CH2OC2H5+HBr (ii) CH32CH−CH2Br⟶C2H5O− CH32CH−CH2OC2H5+Br−The mechanisms of reactions (i) and (ii) are respectively :

First reaction is SN1 reaction because, C2H5OH, is a polar protic solvent and a weak nucleophile. Moreover, in SN1 reaction, rearrangement takes place to form a more stable carbocation.Second reaction is SN2 reaction because, C2H5O−is a strong nucleophile and the reaction proceeds with the formation of transition state.

Consider the reactions :(i) CH32CH−CH2Br⟶C2H5OH CH32CH−CH2OC2H5+HBr (ii) CH32CH−CH2Br⟶C2H5O− CH32CH−CH2OC2H5+Br−The mechanisms of reactions (i) and (ii) are respectively :

First reaction is SN1 reaction because, C2H5OH, is a polar protic solvent and a weak nucleophile. Moreover, in SN1 reaction, rearrangement takes place to form a more stable carbocation.Second reaction is SN2 reaction because, C2H5O−is a strong nucleophile and the reaction proceeds with the formation of transition state.

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Consider the reactions :
(i) ${({CH}_{3})}_{2} CH - {CH}_{2} Br \overset{C_{2} H_{5} OH}{⟶}$ ${({CH}_{3})}_{2} CH - {CH}_{2} {OC}_{2} H_{5} + HBr$
(ii) ${({CH}_{3})}_{2} CH - {CH}_{2} Br \overset{C_{2} H_{5} O^{-}}{⟶}$ ${({CH}_{3})}_{2} CH - {CH}_{2} {OC}_{2} H_{5} + {Br}^{-}$
The mechanisms of reactions (i) and (ii) are respectively :

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$S_{N} 1 and S_{N} 1$

$S_{N} 2 and S_{N} 2$

$S_{N} 1 and S_{N} 2$

$S_{N} 2 and S_{N} 1$

answer is A.

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Detailed Solution

First reaction is S_N1 reaction because, $C_{2} H_{5} OH$ , is a polar protic solvent and a weak nucleophile. Moreover, in S_N1 reaction, rearrangement takes place to form a more stable carbocation.
Second reaction is S_N2 reaction because, $C_{2} H_{5} O^{-}$ is a strong nucleophile and the reaction proceeds with the formation of transition state.