The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is. [figure] Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
Answer
175
Solution
Step 1 (Fittig coupling, then acetal hydrolysis): two aryl bromides join; H3O+ frees both CHO groups.
[figure] [(ii) H3O+](i) Na, dry ether [figure] (P)
P = C14H10O2, M = 210. 16 mol X gives 8 mol P (100 %).
Step 2 (intramolecular Cannizzaro): one CHO is reduced, the other oxidised.
P [(ii) H3O+](i) NaOH, [figure] (Q), 50 %: 4 mol
Step 3 (decarboxylation with soda lime): COOH is replaced by H.
Q [(ii) ](i) NaOH, CaO [figure] (R), 50 %: 2 mol
Step 4: OH is replaced by Br.
R PBr3 [figure] (T), 50 %: 1 mol
Step 5 (Williamson synthesis): NaH makes the alkoxide of R, which displaces Br⁻ from T.
T + R NaH [figure] (S), 50 %: 0.5 mol
S = C26H22O, M = 312 + 22 + 16 = 350
Mass of S = 0.5 × 350 = 175 g
Answer: 175