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Case 1: In alcohols, the negative charge on 'Oxygen atom' increases due to '+I effect' of alkyl groups. As a result the polarity of C-O bond increases and bond breakage becomes easy. Case 2: In aldehydes(and acids), the '+I effect' of alkyl groups decreases the positive charge on the 'Carbon atom' thus making it difficult to break the 'C=O' bond In the first case the electron density is more towards 'O' and in the second case its more towards 'C' while the procedure remains similar. Please explain.

Case 1: In alcohols, the negative charge on 'Oxygen atom' increases due to '+I effect' of alkyl groups. As a result the polarity of C-O bond increases and bond breakage becomes easy.


Case 2: In aldehydes(and acids), the '+I effect' of alkyl groups decreases the positive charge on the 'Carbon atom' thus making it difficult to break the 'C=O' bond


In the first case the electron density is more towards 'O' and in the second case its more towards 'C' while the procedure remains similar. Please explain.

Grade:11

2 Answers

PRIYANKA TATA ASKIITIAN
78 Points
10 years ago

in the second case, u have the effect of resonance....aldehydes are resonance stabilised and have more canonical structures wich makes the carbon - oxygen bond stronger enough for bond breakage...so this is the main reason.

priyanka tata

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

Dear student,

The greater the electronegativity difference between atoms in a bond, the more polar the bond. Partial negative charges are found on the most electronegative atoms, the others are partially positive. In general, the presence of an oxygen is more polar than a nitrogen because oxygen is more electronegative than nitrogen.

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