Draw the resonance structures of the following compounds.
$(a)$ $CH_2=CH-Cl:$
$(b)$ $CH_2=CH-CH=CH_2$
$(c)$ $CH_2=CH-CHO$

(N/A) In $CH_2=CH-Cl:$,the lone pair on $Cl$ participates in resonance: $CH_2=CH-\ddot{Cl}: \leftrightarrow \overset{-}{C}H_2-CH=\overset{+}{Cl}:$
$(b)$ For $CH_2=CH-CH=CH_2$,the resonance structures are:
$CH_2=CH-CH=CH_2 \leftrightarrow \overset{+}{C}H_2-CH=CH-\overset{-}{C}H_2 \leftrightarrow \overset{-}{C}H_2-CH=CH-\overset{+}{C}H_2$
$(c)$ In $CH_2=CH-CHO$,the $\pi$-electrons of the $C=C$ bond shift towards the $C=O$ group:
$CH_2=CH-CH=O \leftrightarrow \overset{+}{C}H_2-CH=CH-O^-$