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How to convert benzene to N,N-dimethylbenzamide?
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Larry King
How to convert benzene to N,N-dimethylbenzamide?
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
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
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).
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
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
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).
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.More
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!More
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
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
More
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