The product of the reaction CH_3-C(CH_3)_2-O- | vedclass

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Question

(C) The reaction of an ether with $HI$ follows the $S_N1$ mechanism when one of the alkyl groups is tertiary,as the tertiary carbocation is highly sta

Vedclass · Accepted Answer

$CH_3-OH + (CH_3)_3C-I$

Vedclass · Answer

(C) The reaction of an ether with $HI$ follows the $S_N1$ mechanism when one of the alkyl groups is tertiary,as the tertiary carbocation is highly stable. $1$. Protonation of the ether oxygen occurs: $(CH_3)_3C-O(H^+)-CH_3$. $2$. The bond between the tertiary carbon and oxygen breaks to form a stable tertiary carbocation $(CH_3)_3C^+$ and methanol $CH_3OH$. $3$. The iodide ion $I^-$ attacks the tertiary carbocation to form tert-butyl iodide $(CH_3)_3C-I$. Thus,the products are $CH_3OH$ and $(CH_3)_3C-I$.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A) \ CH_3CH_2CH_2-O-CH_3 + HI \rightarrow$	$(i) \ (CH_3)_3C-I + CH_3CH_2OH$
$(B) \ C_6H_5OCH_2CH_3 + HBr \rightarrow$	$(ii) \ CH_3CH_2CH_2OH + CH_3I$
$(C) \ (CH_3)_3C-OCH_2CH_3 + HI \rightarrow$	$(iii) \ CH_3CH_2I + C_6H_5OH$
$(D) \ C_6H_5OCH_2CH_3 + \text{conc. } H_2SO_4 + \text{conc. } HNO_3 \rightarrow$	$(iv) \ p-NO_2-C_6H_4-OCH_2CH_3$

The product of the reaction $CH_3-C(CH_3)_2-O-CH_3 \xrightarrow{cold \ HI}$ is:

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