The one giving the maximum number of isomeric | vedclass

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(C) Dehydrohalogenation involves the removal of $H$ and $X$ from adjacent carbons to form an alkene. We analyze each option:$A$) $1-$Bromo$-2-$methylb

Vedclass · Accepted Answer

$2-$Bromopentane

Vedclass · Answer

(C) Dehydrohalogenation involves the removal of $H$ and $X$ from adjacent carbons to form an alkene. We analyze each option:$A$) $1-$Bromo$-2-$methylbutane $(CH_3-CH_2-CH(CH_3)-CH_2Br)$: Only one $\beta$-hydrogen is available,yielding $2-$methylbut$-1-$ene ($1$ product).$B$) $2-$Bromopropane $(CH_3-CH(Br)-CH_3)$: Only one type of $\beta$-hydrogen is available,yielding propene ($1$ product).$C$) $2-$Bromopentane $(CH_3-CH_2-CH_2-CH(Br)-CH_3)$: Two types of $\beta$-hydrogens are available. Removal from $C_1$ gives pent$-1-$ene. Removal from $C_3$ gives pent$-2-$ene,which exists as $cis$ and $trans$ isomers. Total = $3$ products.$D$) $2-$Bromo$-3,3-$dimethylpentane: Only one $\beta$-hydrogen is available at $C_3$ (if possible) or $C_1$,but given the structure,it yields only $1$ alkene product.Thus,$2-$Bromopentane gives the maximum number of isomeric alkenes.

The one giving the maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement):

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