The number of possible monohalogen derivative | vedclass

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(C) The molecular formula $C_4H_9X$ corresponds to the alkyl group derived from butane $(C_4H_{10})$. There are two isomeric alkanes with the formula

Vedclass · Accepted Answer

$4$

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(C) The molecular formula $C_4H_9X$ corresponds to the alkyl group derived from butane $(C_4H_{10})$. There are two isomeric alkanes with the formula $C_4H_{10}$: $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ and isobutane $(CH_3-CH(CH_3)-CH_3)$. For $n$-butane,there are two non-equivalent sets of hydrogen atoms,leading to two isomers: $1$-halobutane and $2$-halobutane. For isobutane,there are two non-equivalent sets of hydrogen atoms,leading to two isomers: $1$-halo-$2$-methylpropane and $2$-halo-$2$-methylpropane. Thus,the total number of possible monohalogen derivatives is $2 + 2 = 4$.

Column-$I$ (Structure)	Column-$II$ ($IUPAC$ Name) and Column-$III$ (Common Name)
$A$. $CH_3CHCl_2$	$i$. $1,1$-Dichloroethane and $r$. Ethylidene chloride
$B$. $CH_2Cl-CH_2Cl$	$ii$. $1,2$-Dichloroethane and $s$. Ethylene dichloride
$C$. $CH_2=CHCl$	$iv$. Chloroethene and $q$. Vinyl chloride
$D$. $CH_2=CHCH_2Cl$	$iii$. $3$-Chloropropene and $p$. Allyl chloride

The number of possible monohalogen derivatives for the alkyl halide having molecular formula $C_4H_9X$ is

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