Which of the following alkyl halides gives th | vedclass

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(B) complex mixture of alkenes in an $E_2$ reaction occurs when there are multiple non-equivalent $\beta$-hydrogens available for abstraction,leading

Vedclass · Accepted Answer

$CH_3-CH_2-CH_2-CH(Br)-CH_3$

Vedclass · Answer

(B) complex mixture of alkenes in an $E_2$ reaction occurs when there are multiple non-equivalent $\beta$-hydrogens available for abstraction,leading to different structural isomers and stereoisomers (cis/trans).$A$. $CH_3-CH_2-CH_2-CH_2-Br$: Only one type of $\beta$-hydrogen,gives $1$-butene.$B$. $CH_3-CH_2-CH_2-CH(Br)-CH_3$: Two types of $\beta$-hydrogens (at $C_1$ and $C_3$). Abstraction at $C_1$ gives $1$-pentene. Abstraction at $C_3$ gives $2$-pentene (as a mixture of $cis$ and $trans$ isomers).$C$. $CH_3-CH_2-CH(Br)-CH_2-CH_3$: Two types of $\beta$-hydrogens (at $C_2$ and $C_4$),but the molecule is symmetric. It gives $2$-pentene (mixture of $cis$ and $trans$ isomers).$D$. $CH_3-C(Br)(CH_3)-CH_2-CH_3$: Two types of $\beta$-hydrogens (at $C_1$ and $C_3$). Abstraction at $C_1$ gives $2$-methyl-$1$-butene. Abstraction at $C_3$ gives $2$-methyl-$2$-butene.Comparing $B$ and $D$,option $B$ produces $1$-pentene,$cis$-$2$-pentene,and $trans$-$2$-pentene ($3$ products). Option $D$ produces $2$-methyl-$1$-butene and $2$-methyl-$2$-butene ($2$ products). Thus,option $B$ yields the most complex mixture.

Which of the following alkyl halides gives the most complex mixture of alkenes in an $E_2$ reaction?

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