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Polarization in acrolein is :
A
$\overset{+\delta}{C}H_2=\overset{-\delta}{C}H-CHO$
B
$\overset{-\delta}{C}H_2=CH-\overset{+\delta}{C}HO$
C
$\overset{-\delta}{C}H_2=\overset{+\delta}{C}H-CHO$
D
$\overset{+\delta}{C}H_2=CH-CH\overset{-\delta}{O}$
Detailed Solution
Acrolein, $CH_2=CH-CHO$, is an $\alpha,\beta$-unsaturated aldehyde. The C=C double bond is in conjugation with the C=O group.
The electronegative oxygen withdraws the $\pi$ electrons through the conjugated system (–R effect):
$CH_2=CH-CH=O \leftrightarrow \overset{+}{C}H_2-CH=CH-O^-$
So the terminal $CH_2$ (the $\beta$-carbon) gets a partial positive charge and the oxygen gets a partial negative charge.
So the polarisation is $\overset{+\delta}{C}H_2=CH-CH\overset{-\delta}{O}$. (This is why nucleophiles can attack the $\beta$-carbon in conjugate addition.)
The electronegative oxygen withdraws the $\pi$ electrons through the conjugated system (–R effect):
$CH_2=CH-CH=O \leftrightarrow \overset{+}{C}H_2-CH=CH-O^-$
So the terminal $CH_2$ (the $\beta$-carbon) gets a partial positive charge and the oxygen gets a partial negative charge.
So the polarisation is $\overset{+\delta}{C}H_2=CH-CH\overset{-\delta}{O}$. (This is why nucleophiles can attack the $\beta$-carbon in conjugate addition.)