Question

118 g of CH3CONH₂ undergoes Hoffmann bromamide degradation to produce CH3NH2. Find the sum of number of moles of Bromine and sodium hydroxide used for this reaction. [Nearest integer]

Answer 1

10

Answer 2

The Hoffmann bromamide degradation reaction converts an amide into a primary amine with one less carbon atom. The general balanced chemical equation for this reaction is:

$\text{RCONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{RNH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$

For $\text{CH}_3\text{CONH}_2$ (acetamide), R = $\text{CH}_3$. So the specific reaction is:

$\text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{CH}_3\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$

Step 1: Calculate the molar mass of $\text{CH}_3\text{CONH}_2$.

Molar mass of C = 12 g/mol Molar mass of H = 1 g/mol Molar mass of N = 14 g/mol Molar mass of O = 16 g/mol

Molar mass of $\text{CH}_3\text{CONH}_2$ = $(2 \times 12) + (7 \times 1) + 14 + 16 = 59 \text{ g/mol}$

Step 2: Calculate the number of moles of $\text{CH}_3\text{CONH}_2$.

Given mass of $\text{CH}_3\text{CONH}_2$ = 118 g Number of moles = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{118 \text{ g}}{59 \text{ g/mol}} = 2 \text{ moles}$

Step 3: Determine the moles of $\text{Br}_2$ and $\text{NaOH}$ used.

From the balanced chemical equation, 1 mole of $\text{CH}_3\text{CONH}_2$ reacts with 1 mole of $\text{Br}_2$ and 4 moles of $\text{NaOH}$.

Since we have 2 moles of $\text{CH}_3\text{CONH}_2$:

Moles of $\text{Br}_2$ used = $2 \text{ moles } \text{CH}_3\text{CONH}_2 \times \frac{1 \text{ mole } \text{Br}_2}{1 \text{ mole } \text{CH}_3\text{CONH}_2} = 2 \text{ moles}$

Moles of $\text{NaOH}$ used = $2 \text{ moles } \text{CH}_3\text{CONH}_2 \times \frac{4 \text{ moles } \text{NaOH}}{1 \text{ mole } \text{CH}_3\text{CONH}_2} = 8 \text{ moles}$

Step 4: Calculate the sum of moles of $\text{Br}_2$ and $\text{NaOH}$ used.

Sum = Moles of $\text{Br}_2$ + Moles of $\text{NaOH}$ Sum = $2 \text{ moles} + 8 \text{ moles} = 10 \text{ moles}$

The nearest integer to 10 is 10.