Question

${(C{H_3})_3}C - O - C{H_3}$ on reaction with $HI$ gives ${(C{H_3})_3}C - I$ and $C{H_3}OH$ as the main product and not ${(C{H_3})_3}C - OH$ and $C{H_3} - I$.

Answer 1

Hint: The given is an example of reaction of ether with $HI$ . One degree alcohol and alkyl iodide are formed. Here,the alkyl group is tertiary , the tertiary halide will form because the departure of the leaving group will create a tertiary carbocation and reaction will follow the $S{N_1}$ mechanism .

Complete step-by-step answer:
Any chemical reaction follows a few basic steps. These are :
Protonation: Addition of ${H^ + }$.
${(C{H_3})_3}C - O - C{H_3} + HI\xrightarrow{{{H^ + }}}{(C{H_3})_3}COC{H_3}^ + $
Formation of the carbocation : tertiary carbocation will be formed because it is most stable.
${(C{H_3})_3}COC{H_3}^ + \xrightarrow{{}}C{H_3}OH + {(C{H_3})_3}{C^ + }$
Nucleophilic attack : ${I^ - }$ is a good nucleophile and a tertiary carbocation is a great electrophile. Both will add with each other and form tertiary iodide.
${(C{H_3})_3}{C^ + } + {I^ - } \to {(C{H_3})_3}CI$

These three are the basic steps in any chemical reaction. When ether and $HI$ react with each other , primary alkyl iodide and primary alcohol are formed . The reaction proceeds through the $S{N_2}$ mechanism.

${I^ - }$ is a good nucleophile . It displaces $ - OH$ molecule by $S{N_2}$ mechanism. If the primary and secondary alkyl group are present , the lower alkyl group forms the halide followed by $S{N_2}$. This is the general case of reaction. But when the alkyl halide is tertiary , the reaction proceeds through the $S{N_1}$ mechanism and hence the product is different from expected.

Note: The above reaction has proceeded through nucleophilic substitution of first order. In the\[S{N_1}\] mechanism , the rate of reaction is fastest for tertiary carbocation. Hence , In step 2, the formation of tertiary carbocation is favourable and forms the basis of the final product.