Question

Anisole $\xrightarrow[\text{(1mole)}]{HI} \text{P} + \text{Q}$ X (Organic compound (1mole) containing oxygen)

P $\xrightarrow{Br_2/H_2O} \text{R} \xrightarrow[ \Delta ]{Zn dust} \text{S}$

If the overall yield of the reaction (From X $\rightarrow$ S) is 80%. Find the mass (in grams) of S formed.

[Atomic masses: C - 12 u, H - 1 u, O - 16 u, Br - 80 u]

• A. 75.2 g
• B. 94 g
• C. 60.16 g
• D. 82.5 g

Answer 1

Answer: (A)

75.2 g

Answer 2

1. Identify X and S: X is Anisole ($C_6H_5OCH_3$). Reaction with HI yields Phenol ($C_6H_5OH$) and Methyl Iodide ($CH_3I$). The subsequent reaction of P with $Br_2/H_2O$ indicates P is Phenol. Phenol reacts with $Br_2/H_2O$ to form 2,4,6-tribromophenol (R). Dehalogenation of R with Zn dust yields Phenol (S). Thus, X = Anisole, S = Phenol.
2. Theoretical Yield: 1 mole of Anisole theoretically produces 1 mole of Phenol (S) through the sequence: Anisole $\rightarrow$ Phenol (P) $\rightarrow$ 2,4,6-tribromophenol (R) $\rightarrow$ Phenol (S).
3. Molar Mass of S: Molar mass of Phenol ($C_6H_6O$) = $6 \times 12 + 6 \times 1 + 16 = 94$ g/mol.
4. Theoretical Mass of S: From 1 mole of Anisole, theoretical mass of S = 1 mole $\times$ 94 g/mol = 94 g.
5. Actual Mass of S: Using the overall yield of 80%, the actual mass of S formed = 94 g $\times$ 0.80 = 75.2 g.