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As part of my homework, I had to plan the synthesis of labetol, a blood pressure drug, from its component parts. One of the steps involved was the following $\mathrm{S_N2}$ reaction:

enter image description here

The solutions manual just shows the two being added together, but my mind harkened back to previous reactions I had learned in which triethylamine and pyridine were used when $\ce{HCl}$ was being eliminated. Would that be a reasonable thing to do in this instance, or would these amines compete with 4-phenylbutan-2-amine?

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    $\begingroup$ I'm more worried about multiple alkylations of the nitrogen. $\endgroup$ – bon Apr 26 '16 at 7:28
  • $\begingroup$ If the solution manual shows these two added together, then I wouldn't worry about multiple alkylations in this problem (IRL is a different matter.) That chloride is going to be extra electrophilic because of the adjacent carbonyl. I'd say this reaction would probably be fine without any base, so triethylamine or pyridine should be fine to do the final deprotonation. $\endgroup$ – SendersReagent Apr 26 '16 at 12:36
  • $\begingroup$ I would actually be curious to see what the "component parts" are for your retrosynthesis... $\endgroup$ – Zhe Feb 27 '17 at 16:32
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I disagree with IT Tsoi's answer. Triethylamine absolutely will compete, and form ammonium salts. Those salts may react with 4-phenylbutan-2-amine to lead to product, but more likely they will precipitate. And be general gunk in the reaction flask. Pyridine will probably be less of an issue.

In general direct alkylation of amines is a pretty ugly reaction with overalkylation products likely. A better reaction in this vein would be to make the acetamide and alkylate that with the chloroacetophenone, and then deacetylate, or similar strategy.

If you must alyklate the amine, use Hunig's base (N,N-diisopropylethylamine) which is too sterically hindered to alyklate. Also, add the electrophile to excess nucleophile, if the reaction is faster this will decrease polylaklyation probabilistically, and the carbonyl alpha to the amine may have enough electron withdrawing character to slow down the increase in nucleophiliciity that you get from a more substituted amine.

Further mucking up this route is the phenol with two carbonyls on the ring, that is probably quite acidic as well, and may be deprotonated with the base scavengers. Use a phosgene protecting group to cover up both the phenol and the aramide, and this problem will be side-stepped.

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  • $\begingroup$ I don't understand how the ammonium salts will react with the amine. Triethylamine will not react with the chloride. I don't think overalkylation is an issue in general with haloacetates. Also we shouldn't worry about phenol (and the amide) deprotonation since triethylamine or the amine reagent used are not basic enough $\endgroup$ – K_P Apr 27 '16 at 0:10
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Triethylamine and Pyridine will not compete with the primary amine - their main purpose would simply be to capture the acid. Neither of them have a proton they can readily lose, and compared to the primary amine they are sterically hindered so their rate of attack (if ever) will be slower too.

It is a reasonable thing to do (adding base) because acidification will lead to the protonation of the amine rendering it unable to attack so your reaction won't run to completion or be slower than it should be.

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    $\begingroup$ Afaik, pyridine is actually sterically less hindered. Otherwise the rest you say holds true. $\endgroup$ – Jan Apr 26 '16 at 13:33

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