2
$\begingroup$

enter image description here

This is my solution to convert benzene to N,N-dimethylbenzamide. If you think is the right way, even just saying the equations are correct would help me! Thanks in advance.

$\endgroup$
8
  • $\begingroup$ What is exactly that you're unsure of? $\endgroup$
    – Mithoron
    May 12, 2020 at 19:15
  • $\begingroup$ How many steps are you proposing doing this in? $\endgroup$
    – Waylander
    May 12, 2020 at 19:46
  • 3
    $\begingroup$ This can be done in one step, but it is pretty sophisicated chemistry and probably not what your teacher is thinking of. It is more easily done in 2 steps which I'm happy to tell you about, but you need to make more of an effort first. $\endgroup$
    – Waylander
    May 12, 2020 at 21:51
  • 1
    $\begingroup$ I would recommend changing the Friedel-Crafts alkylation into a Vilsmeier-Haack formylation. Because the alkyl groups are activating the ring more, this will lead to poly-substitution. If, on the other hand, you use a Vilsmeier-Haack formylation, followed by oxidation of the resulting formyl group, you should be able to avoid the issue of poly-alkylation (If you are curious, a Vilsmier-Haack formylation uses N,N-dimethylformamide and phosphoryl chloride to form a reactive complex that will formylate your benzene ring) $\endgroup$
    – Eli Jones
    May 13, 2020 at 17:37
  • 1
    $\begingroup$ @Eli Jones: The Vilsmeier-Haack formylation requires activated benzene rings (anilines, phenols, etc). Benzene probably doesn't have enough nucleophilicity. $\endgroup$
    – user55119
    May 13, 2020 at 23:09

1 Answer 1

3
$\begingroup$

The scheme you have written will satisfy your teacher in my opinion but it can be shortened. The suggestion of @Eli Jones of the V-H formylation to give benzaldehyde is a good one. However, it can be done in 2 steps using reasonably well-known chemistry:

Step 1 - Brominate with $\ce{FeBr3/Br2}$ to give bromobenzene

Step 2 - Either lithiate by Li-Halogen exchange using $\ce{n-BuLi}$, or form the Grignard (directly with Mg or by Knochel exchange with $\ce{i-PrMgBr}$) then react the resulting benzene organometallic with dimethylcarbamoyl chloride ($\ce{(Me)2NCOCl}$) to give the benzamide.

There are a couple of ways of doing this in one step from benzene by the use of biscarbamoyl diselenides/Lewis acids$\ce{^{[1]}}$ and by $\ce{(Me)3SiOTf}$ activated carbamoyl chloride$\ce{^{[2]}}$.

References

  1. Biscarbamoyl diselenides as new carbamoylating reagents. Lewis acid promoted carbamoylation of aromatic compounds by Shin-Ichi Fujiwara, Akiya Ogawa, Nobuaki Kambe, Ilhyong Ryu, Noboru Sonoda, Volume 29, Issue 47, 1988, Pages 6121-6124, DOI: https://doi.org/10.1016/S0040-4039(00)82282-0
  2. Jin-Wei Yuan, Qian Chen, Chuang Li, Jun-Liang Zhu, Liang-Ru Yang, Shou-Ren Zhang, Pu Mao, Yong-Mei Xiao and Ling-Bo Qu, Silver-catalyzed direct C–H oxidative carbamoylation of quinolines with oxamic acids, Organic & Biomolecular Chemistry, 10.1039/D0OB00358A, (2020).
$\endgroup$
3
  • $\begingroup$ Thank you for this answer! I have to perform an acylation in my research and I have the issue of regioisomer formation (which makes purification very difficult); however, I have not considered performing a lithiation followed by treatment with an acyl chloride. This would definitely solve the regioisomer problem. It is Interesting how it it hard to consider different synthetic routes when you are so set on using a particular reaction! $\endgroup$
    – Eli Jones
    May 13, 2020 at 21:41
  • 1
    $\begingroup$ Aryl lithiums and Grignards do not react well with acyl chlorides. You'll get a higher yield if you transmetallate to the arylmanganese with MnCl2 onlinelibrary.wiley.com/doi/abs/10.1002/adsc.201400654 $\endgroup$
    – Waylander
    May 13, 2020 at 21:51
  • 1
    $\begingroup$ If you have to use an acylation in your scheme, form acetophenone from benzene via an FC reaction. Then do a haloform reaction to get benzoic acid and continue as you planned. $\endgroup$
    – user55119
    May 14, 2020 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.