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In Clayden et al. on p. 696 the following nitroaldol reaction is listed:

Nitroaldol reaction

Later in the same chapter they explain the Cannizzaro reaction. Why is the yield of this nitroaldol reaction particularly good, while a Cannizzaro reaction is also competing for the benzaldehyde? Is the nitroaldol reaction that much faster?

I looked in some literature and they mostly seem to use nitrogen bases (cyclohexylamine, $\ce{Et3N}$, $\ce{n-BuNH2}$, methylamine, etc.) for these kind of reactions (nitroaldol reaction with aromatic electrophile). The reported yields are also typically lower. See for example the "Literature method" section of the synthesis of 1-phenyl-2-nitropropene on Erowid (which seems to be some kind of precursor to amphetamines, so there is plentiful information about it in comparison to the 1-phenyl-2-nitroethene synthesis listed in Clayden. Can the yields be so different due to the steric hindrance of the additional methyl group?). What is the benefit of these nitrogen bases over $\ce{NaOH}$? Clayden et al. only mentions them with the comment that $\ce{NaOH}$ is not really necessary since nitrogen bases are sufficiently strong for nitroaldol reaction, because the typical pKa value of short chain nitro alkyl compounds is ~10.

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Is the nitroaldol reaction that much faster?

My guess would be that the nitroaldol is faster than the Cannizzaro. The Cannizzaro reaction involves

  • attack by hydroxide on the carbonyl to reversibly form the tetrahedral intermediate
  • alignment of the tetrahedral intermediate with a second aldehyde molecule (entropically costly)
  • hydride transfer in a redox reaction (relatively high energy, even complexed hydride is not particularly stable) to form molecules (alcohol and carboxylic acid) that are not that different in stability than the starting material

The nitroaldol involves

  • generation of the nitroalkyl carbanion
  • attack by the carbanion on the carbonyl to form a tetrahedral intermediate
  • capture of a proton by the negatively charged oxygen to produce a neutral molecule
  • elimination of water to produce a more stable (more conjugated) molecule than we started with (e.g. energetically downhill)

The nitroaldol doesn't require any special molecular alignment after the tetrahedral intermediate is formed and is energetically downhill.

Can the yields be so different due to the steric hindrance of the additional methyl group?

Further alkylation at the nitroalkyl carbanion center will not help the reaction in a steric sense just as you suggest, but also a secondary carbanion will be less stable (higher energy, more difficult to form) than a primary carbanion.

What is the benefit of these nitrogen bases over $\ce{NaOH}$?

They are somewhat milder reagents than hydroxide and therefore less likely to produce side reactions.

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  • $\begingroup$ One small addition to Ron's excellent answer. Sodium hydroxide has limited solubility in organic solvents, while the amine bases will be very soluble in most organic solvents. $\endgroup$ – jerepierre Dec 24 '14 at 3:33
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Although it is some time ago already, seeing this question active today made me decide to include the answer given by Prof. Clayden himself over e-mail to me.

The reaction you ask about appears in the classic text on practical organic chemistry by Vogel: in fact this preparation is online at http://www.erowid.org/archive/rhodium/chemistry/nitrostyrene.vogel5.html

It’s not clear to me why amines would be preferable bases in general, although you are right that they avoid the Cannizzaro reaction if the nitro compound is hindered. They could also be preferable if the alcohol were the desired product, rather than the eliminated alkene.

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