Consider the following sequence of reactions:
11.25 mg of chlorobenzene will produce ___________× 10⁻¹ mg of product B.
(Consider the reactions result in complete conversion.)
[Given molar mass of C, H, O, N and Cl as 12,1,16,14 and 35.5 g mol⁻¹ respectively]

Solution

Chemistry ma’am · 4 steps

Track the carbon count: CO2 adds one carbon, Hofmann removes it, so B is aniline and moles stay 1 : 1.

1. Make A. Mg in dry ether gives the Grignard reagent C₆H₅MgCl. With CO₂ and then H₃O⁺ it becomes benzoic acid, C₆H₅COOH. Heating with NH₃ gives benzamide, C₆H₅CONH₂, which is A.
2. Make B. Br₂ with NaOH on an amide is the Hofmann bromamide degradation. It removes the C=O carbon and leaves an amine with one carbon fewer, so B is aniline, C₆H₅NH₂.
3. Count moles. Molar mass of C₆H₅Cl is 72 + 5 + 35.5 = 112.5 g mol⁻¹, so 11.25 mg is 0.1 mmol. Every step is 1 : 1, so 0.1 mmol of aniline forms.

   m_(C₆H₅NH₂) = 0.1 mmol × 93 g mol⁻¹ = 9.3 mg

4. So. 9.3 mg = 93 × 10⁻¹ mg, so the answer is 93.

The common slip: 121. Stopping at benzamide (121 g mol⁻¹) gives 12.1 mg, that is 121. The Hofmann step still has to remove a carbon to give aniline.

The idea to keep. In a reaction sequence, find the final product first, then use the 1 : 1 mole ratio. Remember: Hofmann bromamide always gives an amine with one carbon fewer than the amide.

Answer from the NTA final answer key. Solution by PrepWiser.