n-Butane on monochlorination under photochemi | vedclass

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Question

(C) $n$-Butane $(CH_3-CH_2-CH_2-CH_3)$ on monochlorination gives $2$-chlorobutane $(CH_3-CHCl-CH_2-CH_3)$, which is the optically active compound "$P$

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(C) $n$-Butane $(CH_3-CH_2-CH_2-CH_3)$ on monochlorination gives $2$-chlorobutane $(CH_3-CHCl-CH_2-CH_3)$, which is the optically active compound "$P$".Further chlorination of $2$-chlorobutane involves the substitution of a hydrogen atom by a chlorine atom at different positions.The possible dichloro products are:$1,1$-dichlorobutane $(CH_3-CH_2-CH_2-CHCl_2)$$1,2$-dichlorobutane $(CH_3-CH_2-CHCl-CH_2Cl)$$1,3$-dichlorobutane $(CH_3-CHCl-CH_2-CH_2Cl)$$2,2$-dichlorobutane $(CH_3-CCl_2-CH_2-CH_3)$$2,3$-dichlorobutane $(CH_3-CHCl-CHCl-CH_3)$Ignoring stereoisomers, there are $5$ distinct constitutional isomers.

$n$-Butane on monochlorination under photochemical conditions gives an optically active compound "$P$". "$P$" on further chlorination gives dichloro compounds. The number of dichloro compounds obtained (ignore stereoisomers) is:

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