Question

(i) Arrange the following compounds in increasing order of their acidic strengths :

\( \mathrm{CH_3CH(Br)CH_2COOH} \), \( \mathrm{CH_3-CH(CH_3)-COOH} \), \( \mathrm{CH_3CH_2CH(Br)COOH} \)

(ii) Why is \( \mathrm{CH_3CHO} \) more reactive than acetone towards reaction with HCN ?

(iii) Complete the equation :

\( \mathrm{CH_3{-}CH{=}O + NH_2-NH_2 \xrightarrow{H^+}} \)

Hint:

Compare the inductive effect and its distance from COOH for part (i). Aldehydes have less +I effect and less steric hindrance than ketones. Hydrazine gives a hydrazone with loss of water.

Answer 1

Step 1: Part (i), the idea:
A stronger acid gives a more stable carboxylate ion after it loses H+.
An electron-withdrawing group (-I effect) spreads the negative charge, so the acid becomes stronger.
An electron-releasing group (+I effect) pushes charge onto the carboxylate, so the acid becomes weaker.
The effect fades as the group moves away from the COOH group.

Step 2: Part (i), compare the three acids:
\( \mathrm{CH_3CH(Br)CH_2COOH} \) is 3-bromobutanoic acid. Br sits on the beta carbon, so its -I effect is only moderate.
\( \mathrm{(CH_3)_2CHCOOH} \) is 2-methylpropanoic acid. The two methyl groups push electrons (+I), so this is the weakest acid.
\( \mathrm{CH_3CH_2CH(Br)COOH} \) is 2-bromobutanoic acid. Br sits on the alpha carbon, right next to COOH, so the -I effect is strongest.
\[ \mathrm{(CH_3)_2CHCOOH} < \mathrm{CH_3CH(Br)CH_2COOH} < \mathrm{CH_3CH_2CH(Br)COOH} \]

Step 3: Part (ii), why acetaldehyde reacts faster with HCN:
HCN adds to the C=O group by nucleophilic attack of \( \mathrm{CN^-} \) on the carbonyl carbon.
Acetaldehyde has one methyl group and one H on the carbonyl carbon. Acetone has two methyl groups.
Two +I groups lower the positive charge on the carbonyl carbon of acetone, so it attracts \( \mathrm{CN^-} \) less.
Two methyl groups also crowd the carbon, so the attacking ion meets more steric hindrance in acetone.
The tetrahedral product of acetaldehyde is also less crowded and more stable. So acetaldehyde is more reactive.

Step 4: Part (iii), complete the equation:
An aldehyde reacts with hydrazine in weak acid by addition, then loss of water. This forms a hydrazone.
\[ \mathrm{CH_3CHO + H_2N-NH_2 \xrightarrow{H^+} CH_3CH{=}N-NH_2 + H_2O} \]
The product is acetaldehyde hydrazone.

Final Answer:
(i) 2-methylpropanoic acid < 3-bromobutanoic acid < 2-bromobutanoic acid. (ii) Acetaldehyde has less +I effect and less steric hindrance. (iii) \( \mathrm{CH_3CH{=}N-NH_2} \) and water form.