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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?

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  • $\begingroup$ Your answer looks correct to me. The "official" answer looks to be in error $\endgroup$ – Waylander Jun 4 at 7:08
  • $\begingroup$ @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. $\endgroup$ – Safdar Faisal Jun 4 at 7:20
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    $\begingroup$ BTW I’ll get some hint (or solution) tomorrow. I’ll be sure to share it here. $\endgroup$ – WhySee Jun 4 at 9:06
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    $\begingroup$ 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. $\endgroup$ – James Gaidis Jun 4 at 13:34
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    $\begingroup$ So are we proposing that P2 is triphenylacetic acid and it undeergoes an acid catalysed decarboxylation? $\endgroup$ – Waylander Jun 4 at 15:08
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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.

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    $\begingroup$ Wow. This is actually the correct explanation. I just confirmed today. Thanks. $\endgroup$ – WhySee Jun 5 at 3:39

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