The major product [B] in the following reacti | vedclass

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(D) $1$. The reaction of the ether $CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$ with $HI$ proceeds via an $S_N2$ mechanism.$2$. The iodide ion $(I^-)$ attack

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$CH_3-CH=C(CH_3)_2$

Vedclass · Answer

(D) $1$. The reaction of the ether $CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$ with $HI$ proceeds via an $S_N2$ mechanism.$2$. The iodide ion $(I^-)$ attacks the less sterically hindered carbon atom,which is the ethyl group $(CH_2-CH_3)$,resulting in the formation of $CH_3-CH_2-I$ and the alcohol $[A]$,which is $2-methylbutan-1-ol$ $(CH_3-CH_2-CH(CH_3)-CH_2OH)$.$3$. The dehydration of $2-methylbutan-1-ol$ with concentrated $H_2SO_4$ at high temperature involves the formation of a carbocation intermediate followed by rearrangement to a more stable carbocation.$4$. The primary carbocation $CH_3-CH_2-CH(CH_3)-CH_2^+$ rearranges to the more stable tertiary carbocation $CH_3-CH_2-C^+(CH_3)_2$ via a hydride shift.$5$. Elimination of a proton from this tertiary carbocation yields the most stable alkene,$2-methylbut-2-ene$ $(CH_3-CH=C(CH_3)_2)$,as the major product $[B]$.

The major product $[B]$ in the following reaction is:
$CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$ $\xrightarrow[Heat]{HI} [A] (alcohol)$ $\xrightarrow[\Delta]{H_2SO_4} [B]$

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The major product $[B]$ in the following reaction is:$CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$ $\xrightarrow[Heat]{HI} [A] (alcohol)$ $\xrightarrow[\Delta]{H_2SO_4} [B]$