CH_3-C(CH_3)_2-O-CH_3 xrightarrow{text{Cold } | vedclass

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Question

(C) The reaction of tert-butyl methyl ether with cold $HI$ proceeds via the $S_N1$ mechanism.Since the tert-butyl group can form a stable tertiary car

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The reaction of tert-butyl methyl ether with cold $HI$ proceeds via the $S_N1$ mechanism.Since the tert-butyl group can form a stable tertiary carbocation $(CH_3)_3C^+$,the cleavage occurs such that the iodide ion attacks the tertiary carbon.The reaction is: $CH_3-C(CH_3)_2-O-CH_3 + HI \xrightarrow{	ext{Cold}} (CH_3)_3C-I + CH_3-OH$.Thus,the products are tert-butyl iodide and methanol.

$CH_3-C(CH_3)_2-O-CH_3 \xrightarrow{\text{Cold } HI} \text{Product is}$

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