Question

Consider the following sequence of reactions: 
i) \( \text{Mg}, \text{dry ether} \)  
ii) \( \text{CO}_2, \text{H}_2\text{O}^+ \) 
iii) \( \text{NH}_3, \Delta \) 
11.25 mg of chlorobenzene will produce \( x \times 10^{-1} \) mg of product B. Given molar mass of C, H, O, N, and Cl as 12, 1, 16, 14, and 35.5 g mol\(^{-1}\), respectively, the value of \( x \) is:

Answer 1

Correct Answer: 121

The objective is to identify product B resulting from the specified reaction sequence and subsequently compute the mass of B generated from 11.25 mg of chlorobenzene.

1. Reaction Identification:

i) Reaction with \( \text{Mg} \) in dry ether: Formation of a Grignard reagent from chlorobenzene. \( \text{C}_6\text{H}_5\text{Cl} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{Mg}\text{Cl} \) (Phenylmagnesium chloride)

ii) Reaction with \( \text{CO}_2 \) followed by \( \text{H}_2\text{O}^+ \): Carboxylation of the Grignard reagent. \( \text{C}_6\text{H}_5\text{Mg}\text{Cl} + \text{CO}_2 \xrightarrow{\text{H}_2\text{O}^+} \text{C}_6\text{H}_5\text{COOH} + \text{Mg(OH)Cl} \) (Benzoic acid, product A)

iii) Reaction with \( \text{NH}_3 \) under heat (\( \Delta \)): Reaction of benzoic acid with ammonia to form benzamide (product B). \( \text{C}_6\text{H}_5\text{COOH} + \text{NH}_3 \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{CONH}_2 + \text{H}_2\text{O} \) (Benzamide, product B)

2. Molecular Weight Determination:

Chlorobenzene (\(\text{C}_6\text{H}_5\text{Cl}\)): (6 * 12) + (5 * 1) + 35.5 = 112.5 g/mol

Benzamide (\(\text{C}_6\text{H}_5\text{CONH}_2\)): (6 * 12) + (5 * 1) + 12 + 16 + 14 + 2 = 121 g/mol

3. Molar Calculation for Chlorobenzene:

Mass of chlorobenzene = 11.25 mg = 0.01125 g

Moles of chlorobenzene = 0.01125 g / 112.5 g/mol = 0.0001 mol

4. Molar Determination for Benzamide:

Stoichiometry dictates a 1:1 molar ratio between chlorobenzene and benzamide production. Thus, moles of benzamide = 0.0001 mol.

5. Mass Calculation for Benzamide:

Mass of benzamide = moles of benzamide * molar mass of benzamide

Mass of benzamide = 0.0001 mol * 121 g/mol = 0.0121 g = 12.1 mg

6. Result Formatting:

The mass of benzamide (B) is \(12.1 \text{ mg} = x \times 10^{-1} \text{ mg} \)

Solving for x: \( 12.1 = x \times 10^{-1} \Rightarrow x = 121 \)

Final Answer:
The value of \( x \) is 121.