Question

Organic compounds containing carbonyl group as the functional group are aldehydes, ketones and carboxylic acids. The carbon-oxygen double bond is polarised due to higher electronegativity of oxygen as compared to carbon. Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. They react with HCN, Sodium bisulphite, Grignard reagent and a number of ammonia derivatives like Hydroxylamine, Hydrazine, Semicarbazide, etc. Aldehydes and ketones undergo various reactions due to the acidic nature of \(\alpha\)-hydrogen. Aldehydes and ketones having at least one \(\alpha\)-hydrogen atom undergo a reaction in the presence of dilute alkali as a catalyst to form \(\beta\)-hydroxy aldehydes or \(\beta\)-hydroxy ketones. On the other hand, aldehydes which do not possess an \(\alpha\)-hydrogen undergo self-oxidation and reduction in the presence of concentrated alkali.

(i) Arrange the following compounds in increasing order of reactivity towards nucleophilic addition reactions :

Butanone, Propanone, Propanal, Ethanal

Hint:

Aldehydes are more reactive than ketones. Larger alkyl groups lower reactivity by steric hindrance and the +I effect.

Answer 1

Step 1: Recall the factors:
In nucleophilic addition the nucleophile attacks the carbonyl carbon. Two things slow it down: bulky groups around that carbon (steric effect) and electron-donating alkyl groups (electronic effect).
Alkyl groups push electrons towards the carbon and reduce its positive charge.

Step 2: Compare the four compounds:
Ethanal \(\mathrm{CH_3CHO}\) has one methyl group and one hydrogen. It is the least hindered and has the most positive carbonyl carbon.
Propanal \(\mathrm{CH_3CH_2CHO}\) has an ethyl group in place of methyl, so it is a little less reactive than ethanal.
Propanone \(\mathrm{CH_3COCH_3}\) is a ketone with two methyl groups. Two alkyl groups donate electrons and block the approach.
Butanone \(\mathrm{CH_3COC_2H_5}\) has a methyl and an ethyl group, so it is the most crowded and the least reactive.

Step 3: Write the order:
Reactivity falls as steric hindrance and the +I effect rise.

Final Answer:
Increasing order of reactivity: \[ \boxed{\text{Butanone} < \text{Propanone} < \text{Propanal} < \text{Ethanal}} \]