Compound A on reaction with chlorine in the p | vedclass

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(A) The reaction sequence is as follows:$1$. Compound $A$ is $n$-butane $(CH_{3}CH_{2}CH_{2}CH_{3})$.$2$. Reaction with $Cl_{2}$ in the presence of $U

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Butane

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(A) The reaction sequence is as follows:$1$. Compound $A$ is $n$-butane $(CH_{3}CH_{2}CH_{2}CH_{3})$.$2$. Reaction with $Cl_{2}$ in the presence of $U.V.$ light (free radical substitution) gives $2$-chlorobutane $(B)$ as the major product: $CH_{3}CH_{2}CH_{2}CH_{3} + Cl_{2} \xrightarrow{h
u} CH_{3}CH_{2}CHClCH_{3} (B) + HCl$.$3$. $2$-chlorobutane $(B)$ reacts with $NaNO_{2}$ in $DMF$ (a polar aprotic solvent) via an $S_{N}2$ mechanism to form $2$-nitrobutane: $CH_{3}CH_{2}CHClCH_{3} + NaNO_{2} \xrightarrow{DMF} CH_{3}CH_{2}CH(NO_{2})CH_{3} + NaCl$.Thus,compound $A$ is butane.

Compound $A$ on reaction with chlorine in the presence of $U.V.$ light gives $B$; which when reacted with $NaNO_{2}$ in the solvent $N, N$-dimethylformamide $(DMF)$,gives $2$-nitrobutane. The compound $A$ is:

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