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(B) The degree of unsaturation (double bond equivalent) for $C_3H_5Cl$ is calculated as: $DU = C + 1 - (H + X - N)/2 = 3 + 1 - (5 + 1)/2 = 4 - 3 = 1$.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The degree of unsaturation (double bond equivalent) for $C_3H_5Cl$ is calculated as: $DU = C + 1 - (H + X - N)/2 = 3 + 1 - (5 + 1)/2 = 4 - 3 = 1$. This indicates the presence of either one double bond or one ring. The possible structural isomers are: $1$. $CH_2=CH-CH_2Cl$ ($3$-chloroprop$-1-$ene) $2$. $CH_3-CCl=CH_2$ ($2$-chloroprop$-1-$ene) $3$. $CH_3-CH=CHCl$ ($1$-chloroprop$-1-$ene) $4$. $Cl-CH_2-CH=CH_2$ (same as $1$,but structural isomers include cyclic forms) $5$. Chlorocyclopropane $6$. $1-$chlorocyclopropane (not possible) Actually,the structural isomers are: $(i)$ $CH_2=CH-CH_2Cl$ (ii) $CH_3-CCl=CH_2$ (iii) $CH_3-CH=CHCl$ (iv) Chlorocyclopropane $(v)$ $1-$chloroprop$-1-$ene (cis/trans are stereoisomers,not structural) Wait,the structural isomers are: $1$. $CH_2=CH-CH_2Cl$ $2$. $CH_3-CCl=CH_2$ $3$. $CH_3-CH=CHCl$ $4$. Chlorocyclopropane Thus,there are $4$ structural isomers.

The number of possible structural isomers for the molecular formula $C_3H_5Cl$ is ...............

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