Question

Polarisation of electrons in acrolein may be written as

• A. $^{\delta+}_{CH_2 =CH -}\, ^{ \, \, \, \, \, \, \, \, \, \, \, \delta-}_{CH=O}$
• B. $^{\delta+}_{CH_2 =}\, ^{ \delta+}_{CH-CH=O}$
• C. $^{\delta+}_{CH_2 =CH -}\, ^{ \, \, \, \, \, \, \, \, \, \, \, \delta+}_{CH=O}$
• D. $^{\delta-}_{CH_2 =CH -}\, ^{\delta+}_{CH=O}$

Answer 1

The Correct Option is A

To determine the polarisation of electrons in acrolein, let's analyze the molecular structure and electron distribution in this compound, which is also known as propenal with the formula CH_2=CH-CHO. Polarisation in organic molecules often arises due to the differences in electronegativity between atoms.

1. The double bond between the two carbon atoms (CH_2=CH) involves shared electrons. However, due to resonance, the electron density is slightly shifted toward the CH group.
2. The carbonyl group (C=O) is highly polarised because oxygen is more electronegative than carbon. This results in a partial positive charge on the carbon atom and a partial negative charge on the oxygen atom.
3. Considering the entire structure of acrolein, the electron density in the molecule can be represented with partial charges (δ+ and δ-) :

The correct polarisation of acrolein is:

$^{\delta+}_{CH_2 =CH -}\, ^{ \, \, \, \, \, \, \, \, \, \, \, \delta-}_{CH=O}$

Now, let us justify why this is the correct option:

• The CH_2=CH segment shows a slight positive charge due to the conjugation and the slight electron-withdrawing effect of the carbonyl group.
• The CH=O segment shows a resonance effect, where the carbonyl group has a polarised double bond, with the carbon having a partial positive charge (δ+) and the oxygen having a partial negative charge (δ-).

Thus, option $^{\delta+}_{CH_2 =CH -}\, ^{ \, \, \, \, \, \, \, \, \, \, \, \delta-}_{CH=O}$ is the correct representation of polarisation in acrolein due to the resonance and electronic structure discussed.