Propane molecule on chlorination under photoc | vedclass

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(D) The chlorination of propane $(CH_3-CH_2-CH_3)$ gives two dichloro products: $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$ and $1,3-dichloropropane$ $

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$3$

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(D) The chlorination of propane $(CH_3-CH_2-CH_3)$ gives two dichloro products: $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$ and $1,3-dichloropropane$ $(ClCH_2-CH_2-CH_2Cl)$.Among these,$1,2-dichloropropane$ is optically active because the carbon at position $2$ is chiral (bonded to $-H, -CH_3, -Cl, -CH_2Cl$). Thus,$x$ is $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$.When $x$ $(CH_3-CHCl-CH_2Cl)$ is further chlorinated,we look for structural isomers of trichloropropane formed by replacing one $H$ atom with a $Cl$ atom:$1.$ Replacing $H$ at $C-1$: $CH_3-CHCl-CHCl_2$ $(1,1,2-trichloropropane)$$2.$ Replacing $H$ at $C-2$: $CH_3-CCl_2-CH_2Cl$ $(1,2,2-trichloropropane)$$3.$ Replacing $H$ at $C-3$: $ClCH_2-CHCl-CH_2Cl$ $(1,2,3-trichloropropane)$Thus,there are $3$ structural isomers of trichloropropane obtained from $x$.

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