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.

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

Butanone, Propanone, Propanal, Ethanal

Hint:

Alkyl groups lower the positive charge on carbonyl carbon and add steric hindrance. So aldehydes beat ketones, and smaller alkyl groups give higher reactivity.

Answer 1

Step 1: Know what controls reactivity:
In nucleophilic addition, the nucleophile attacks the carbonyl carbon.
Alkyl groups push electrons towards that carbon (+I effect), which lowers its positive charge.
Bulky groups also block the approach of the nucleophile (steric hindrance).

Step 2: Aldehydes versus ketones:
Propanal and ethanal are aldehydes, with one alkyl group and one H.
Propanone and butanone are ketones, with two alkyl groups. So ketones have more +I effect and more crowding.
Hence both aldehydes are more reactive than both ketones.

Step 3: Order within each pair:
Ethanal has a methyl group, which is smaller than the ethyl group of propanal. So ethanal is more reactive than propanal.
Butanone has an ethyl group in place of one methyl of propanone. It has more +I effect and more crowding, so it is the least reactive.

Final Answer:
Increasing order of reactivity:

\[ \boxed{\text{Butanone} < \text{Propanone} < \text{Propanal} < \text{Ethanal}} \]