Question

what are the products of the reaction of propanoic acid with methylamine?
select the correct answer below:
○ n, n - dimethylpropanamide and water
○ n - methylpropanamide and water
○ propanamide and water
○ n - propylmethanamide and water

Explanation:

Step1: Understand the reaction mechanism

Carboxylic acid (propanoic acid) reacts with amine (methylamine). The general reaction is \(R - COOH+R' - NH_2
ightarrow R - CONHR'+H_2O\).

Step2: Analyze the substituents

For propanoic acid (\(CH_3CH_2COOH\)) and methylamine (\(CH_3NH_2\)), the acid part (\(CH_3CH_2CO -\)) and the amine part (\(-NHCH_3\)) combine.
The amide formed is \(CH_3CH_2CONHCH_3\) (N - methylpropanamide) and water (\(H_2O\)) is also produced as a by - product.

• \(N,N\) - dimethylpropanamide would require two methyl groups on the nitrogen from a secondary amine with two methyl groups, but methylamine (\(CH_3NH_2\)) has only one methyl group.
• Propanamide (\(CH_3CH_2CONH_2\)) would be formed if the amine was \(NH_3\).
• \(N\) - propylmethanamide (\(HCONHCH_2CH_2CH_3\)) has an incorrect carbon - chain assignment (the acid part is from propanoic acid, not methanoic acid for the amide part naming in this wrong option).

Answer:

N - methylpropanamide and water