Question 13

Step 1 ($X \rightarrow P$): 16 moles of $X$ undergo Wurtz-type coupling followed by acid hydrolysis of the acetal groups. Two moles of $X$ ($C_9H_9BrO_2$) react to form one mole of the dialdehyde $P$ ($[1,1'-biphenyl]-2,2'-dicarbaldehyde$, $C_{14}H_{10}O_2$).

• Molar mass of $P = (14 \times 12) + (10 \times 1) + (2 \times 16) = 168 + 10 + 32 = 210$ g/mol (matches given data).
• Moles of $P = \frac{16}{2} \times 1.00 = 8$ moles.

Step 2 ($P \rightarrow Q$): Intramolecular Cannizzaro reaction of $P$ gives $Q$ ($2'-(hydroxymethyl)-[1,1'-biphenyl]-2-carboxylic acid$).

• Moles of $Q = 8 \times 0.50 = 4$ moles.

Step 3 ($Q \rightarrow R$): Soda-lime decarboxylation of $Q$ removes the $-COOH$ group to form $R$ ($[1,1'-biphenyl]-2-ylmethanol$, $C_{13}H_{12}O$).

• Moles of $R$ produced $= 4 \times 0.50 = 2$ moles.

Step 4 ($R \rightarrow T$): Conversion of alcohol $R$ to bromide $T$ ($2-(bromomethyl)-1,1'-biphenyl$, $C_{13}H_{11}Br$).

• Moles of $T = 2 \times 0.50 = 1$ mole.
• Note: From the 2 moles of $R$ produced in Step 3, 1 mole was consumed to make $T$, leaving 1 mole of $R$ unreacted.

Step 5 ($T \rightarrow S$): Williamson ether synthesis between $T$ (1 mole) and the remaining $R$ (1 mole) using $NaH$. $S$ is bis([1,1'-biphenyl]-2-ylmethyl) ether ($C_{26}H_{22}O$).

• Moles of $S = 1 \times 0.50 = 0.5$ moles.
• Molar mass of $S = (26 \times 12) + (22 \times 1) + 16 = 312 + 22 + 16 = 350$ g/mol.

Final Calculation:

• Amount of $S = 0.5 \text{ moles} \times 350 \text{ g/mol} = 175$ grams.