# Finding the weight of the final product in a reaction sequence

The question is:

What is the molecular weight of the final product? $$\ce{Ph3C-Cl ->[NaCN] P1 ->[H3O+] P2 ->[Conc H2SO4][CH3OH]P3}$$

My products were:

$$\mathrm P_1$$ = $$\ce{Ph3C-CN}$$ (by SN); $$\mathrm P_2$$ = $$\ce{Ph3C-COOH}$$ (by hydrolysis); $$\mathrm P_3$$ = $$\ce{Ph3C-COOCH3}$$ (By esterification)

If this was correct, the answer should have been 302. But the correct answer is 274. What did I do wrong?

• Your answer looks correct to me. The "official" answer looks to be in error – Waylander Jun 4 at 7:08
• @Waylander, However, is the first step feasible? I was unable to find literature referencing the reaction between trityl chloride and sodium cyanide. Any preparation of triphenyl acetonitrile involved copper cyanide. – Safdar Faisal Jun 4 at 7:20
• BTW I’ll get some hint (or solution) tomorrow. I’ll be sure to share it here. – WhySee Jun 4 at 9:06
• 274 corresponds to 3 phenyls + C + 31. 31 = 16 (O) + 15 (CH3). So it looks like if CN goes on, it comes off because Ph3C+ is so stable, but in conc H2SO4 - probably adjusted carefully - OCH3 can go on, while the H2O produced reduces the acidity of the H2SO4 somewhat. But I just played with the numbers. – James Gaidis Jun 4 at 13:34
• So are we proposing that P2 is triphenylacetic acid and it undeergoes an acid catalysed decarboxylation? – Waylander Jun 4 at 15:08

You are correct in your first two steps. Cyanide adds to trityl chloride to give triphenylacetonitrile which is then hydrolysed in aqueous acid to triphenylacetic acid. In $$\ce{H2SO4/MeOH}$$ the acid is protonated and loses water to give the acylium cation ($$\ce{Ph3C-CO+}$$). This cation loses $$\ce{CO}$$ to give the very stable trityl cation ($$\ce{Ph3C+}$$) which is eventually captured by $$\ce{MeOH}$$ to give trityl methyl ether.