Question

What is the number of moles of H atoms required to prepare one mole ethylamine from one mole acetamide?

• A. 2
• B. 4
• C. 3
• D. 1

Hint:

Amide-to-amine reduction replaces a C=O bond entirely with a $\text{-CH}_2\text{-}$ group. To turn an oxygen atom into water ($\text{H}_2\text{O}$) and add two hydrogens to carbon, you always need exactly $2 + 2 = 4$ nascent hydrogens!

Answer 1

The Correct Option is B

Step 1: Identify the change.
Acetamide $CH_3CONH_2$ is reduced to ethylamine $CH_3CH_2NH_2$. The C=O group must become $CH_2$.

Step 2: Count the hydrogens.
Two H atoms remove the oxygen as water, and two more H atoms attach to the carbon. That is four H atoms in all.

Step 3: Confirm with the equation.
\[ CH_3CONH_2 + 4[H] \rightarrow CH_3CH_2NH_2 + H_2O \]
\[ \boxed{4,\ \text{option 2}} \]