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

To solve this problem, we need to determine the product B formed from the given reaction sequence and then calculate the mass of B produced from 11.25 mg of chlorobenzene.

1. Identify the Reactions:

i) \( \text{Mg}, \text{dry ether} \): This reaction forms 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) \( \text{CO}_2, \text{H}_2\text{O}^+ \): This is the 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) \( \text{NH}_3, \Delta \): This is the reaction of benzoic acid with ammonia under heat, forming 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. Determine the Molecular Weights:

Chlorobenzene (\(\text{C}_6\text{H}_5\text{Cl}\)): (6 * 12) + (5 * 1) + 35.5 = 72 + 5 + 35.5 = 112.5 g/mol

Benzamide (\(\text{C}_6\text{H}_5\text{CONH}_2\)): (6 * 12) + (5 * 1) + 12 + 16 + 14 + 2 = 72 + 5 + 12 + 16 + 14 + 2 = 121 g/mol

3. Calculate Moles of Chlorobenzene:

Mass of chlorobenzene = 11.25 mg = 0.01125 g

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

4. Determine Moles of Benzamide:

From the stoichiometry of the reactions, 1 mole of chlorobenzene produces 1 mole of benzamide. Therefore, moles of benzamide = 0.0001 mol.

5. Calculate Mass of 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. Express the answer in the required format:

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

Therefore, \( 12.1 = x \times 10^{-1} \Rightarrow x = 121 \)

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

Answer 2

Step 1: Write the given reaction sequence.
The sequence of reactions is as follows:
(i) \( \text{Mg}, \text{dry ether} \)
(ii) \( \text{CO}_2, \text{H}_2\text{O}^+ \)
(iii) \( \text{NH}_3, \Delta \)

Starting compound: Chlorobenzene (\( C_6H_5Cl \))

Step 2: Reaction pathway.
1. Formation of Grignard reagent:
\[ C_6H_5Cl + Mg \xrightarrow{\text{dry ether}} C_6H_5MgCl \]
2. Reaction with carbon dioxide and hydrolysis:
\[ C_6H_5MgCl + CO_2 \xrightarrow{\text{H}_2O^+} C_6H_5COOH \] This gives benzoic acid as product (A).
3. Reaction with ammonia and heat:
\[ C_6H_5COOH + NH_3 \xrightarrow{\Delta} C_6H_5CONH_2 \] This gives benzamide (B) as the final product.

Step 3: Molar mass calculations.
- For chlorobenzene (\( C_6H_5Cl \)):
\[ M = (6 \times 12) + (5 \times 1) + 35.5 = 72 + 5 + 35.5 = 112.5 \text{ g/mol} \]
- For benzamide (\( C_6H_5CONH_2 \)):
\[ M = (7 \times 12) + (7 \times 1) + (1 \times 16) + (1 \times 14) = 84 + 7 + 16 + 14 = 121 \text{ g/mol} \]

Step 4: Calculate product mass from 11.25 mg of chlorobenzene.
\[ \text{Mass of product} = \frac{11.25 \times 121}{112.5} = 12.1 \text{ mg} \]
Thus, \( 11.25 \, \text{mg of chlorobenzene} \) produces \( 12.1 \, \text{mg of product} \).

Step 5: Express in the required form.
\[ 12.1 = x \times 10^{-1} \, \text{mg} \Rightarrow x = 121 \]

Final Answer:
\[ \boxed{x = 121} \]