The reaction between ammonia and ethyl ethanoate produces

Analysis

The question asks for the products of the reaction between ammonia ($NH_3$) and ethyl ethanoate ($CH_3COOCH_2CH_3$). This is a classic example of an ammonolysis reaction, which is a type of nucleophilic acyl substitution.

Concept Explanation

When an ester reacts with ammonia, the reaction is called ammonolysis. In this process:

1. The ammonia molecule acts as a nucleophile and attacks the carbonyl carbon of the ester.
2. The ester bond (the C–O bond) breaks.
3. The alkoxy group (the part derived from the alcohol) is replaced by an amino group ($-NH_2$) from the ammonia.

The general equation is: Ester + Ammonia $\rightarrow$ Amide + Alcohol

Step-by-Step Solution

1. Identify the reactants: • Ethyl ethanoate: $CH_3COOCH_2CH_3$ • Ammonia: $NH_3$

2. Break down the ester: • The "ethanoate" part ($CH_3CO-$) will combine with the $-NH_2$ group from ammonia to form an amide. • The "ethyl" part ($-OCH_2CH_3$) will combine with the remaining hydrogen ($H$) from the ammonia to form an alcohol.

3. Write the reaction: $CH_3COOCH_2CH_3 + NH_3 \rightarrow CH_3CONH_2 + CH_3CH_2OH$

4. Identify the products: • $CH_3CONH_2$ is ethanamide. • $CH_3CH_2OH$ is ethanol.

Comparing this to your options, the products are ethanol and ethanamide.

Conclusion

The reaction between ammonia and ethyl ethanoate produces ethanol and ethanamide. This matches option D.

Exam Tip

Whenever you see a reaction between an ester and ammonia, remember the "split": the ester splits into an amide (containing the acid part of the ester) and an alcohol (containing the alkyl part of the ester). Don't get confused by the names—just look at the number of carbons in the acid part (ethanoate = 2 carbons $\rightarrow$ ethanamide) and the alcohol part (ethyl = 2 carbons $\rightarrow$ ethanol).