Read the passage carefully and answer the Questions.
Aldehydes, carboxylic acids and ketones are widespread in the animal and plant kingdoms. They play a pivotal role in various biochemical processes of life. Their presence in nature adds fragrance and flavor. These compounds are widely used in food products, pharmaceuticals, paints, resins and other important product industries. These compounds are prepared by various laboratory methods, which mainly include oxidation, formylation, acylation and reduction. Due to the polar nature of the carbonyl group in aldehydes and ketones, they can exhibit different reactions like nucleophilic addition. They do exhibit redox and various condensation reactions which lead to the formation of various important compounds. On the other hand, the carboxylic acids are mainly prepared by oxidation and hydrolysis of different compounds. The carboxylic acid consists of a carbonyl group and the hydroxy group (attached to the carbonyl carbon atom). This makes it possible for the carboxylic acid to participate in various chemical reactions which involve cleavage of the C-OH bond and the O-H bond along with the reactions involving the complete -COOH group.
The products formed when 2,2-Dimethylpropan-1-al is heated with concentrated KOH are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Potassium 2,2-Dimethylpropan-1-oate and 2,2-Dimethylpropan-1-ol
Aldehydes that have no α-hydrogen cannot form an enolate, so they do not undergo aldol condensation. On heating with concentrated alkali they undergo the Cannizzaro reaction, a self oxidation-reduction (disproportionation): one molecule is oxidised to the salt of the carboxylic acid and another is reduced to the alcohol.
2,2-Dimethylpropanal is \( \ce{(CH3)3C-CHO} \). The carbon next to the –CHO group carries three methyl groups and no hydrogen, so it gives the Cannizzaro reaction:
\( \displaystyle \ce{2(CH3)3C-CHO + KOH} \xrightarrow{\Delta} \ce{(CH3)3C-COOK + (CH3)3C-CH2OH} \)
The products are potassium 2,2-dimethylpropanoate (the oxidised half) and 2,2-dimethylpropan-1-ol (the reduced half). Both keep the five-carbon skeleton of the aldehyde.
Why the other options are incorrect
Option (2) 2,2-Dimethylprop-1-ene and 2,2-Dimethylpropan-1-oate — Incorrect. No alkene is formed; besides, a double bond at C-1 would need five bonds on C-2, which already carries three methyl groups.
Option (3) 2,2-Dimethylpropan-1-ol and Potassium 2-Methylbutan-1-oate — Incorrect. The alcohol is right, but the acid salt must have the same carbon skeleton as the aldehyde; the skeleton is not rearranged in this reaction.
Option (4) Potassium 2,2-Dimethylpropan-1-oate and 3-Methylbutan-2-one — Incorrect. The reduced half of a Cannizzaro reaction is always a primary alcohol, never a ketone.
NCERT (2026-27 edition):
Other Reactions
