What are A and B in the following reaction?CH | vedclass

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(C) The reaction of an ether with $HI$ proceeds via an $S_N1$ mechanism when one of the alkyl groups is tertiary,as it can form a stable carbocation.I

Vedclass · Accepted Answer

$A = CH_3CH_2CH_2OH, B = I-C(CH_3)_2-CH_2CH_3$

Vedclass · Answer

(C) The reaction of an ether with $HI$ proceeds via an $S_N1$ mechanism when one of the alkyl groups is tertiary,as it can form a stable carbocation.In the given ether,$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3$,the oxygen atom is attached to a primary alkyl group ($n$-propyl) and a tertiary alkyl group ($2$-methylbutan$-2-$yl).When $HI$ is added,the oxygen gets protonated. The bond between the oxygen and the tertiary carbon breaks to form a stable tertiary carbocation,which then reacts with the iodide ion $(I^-)$ to form an alkyl iodide,and the primary group forms an alcohol.Therefore,the products are $CH_3-CH_2-CH_2-OH$ (propan$-1-$ol) and $I-C(CH_3)_2-CH_2-CH_3$ ($2$-iodo$-2-$methylbutane).Comparing this with the given options,option $C$ represents the correct products.

What are $A$ and $B$ in the following reaction?
$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3 \xrightarrow{HI} A + B$

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What are $A$ and $B$ in the following reaction?$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3 \xrightarrow{HI} A + B$