8
$\begingroup$

What are the products of reaction of 4-nitrobenzaldehyde with 3-nitrobenzaldehyde in the presence of $\ce{NaOH}$ when they show Cannizzaro reaction?

To my understanding, 4-nitrobenzaldehyde has more electrophilic carbonyl because the nitro group is an electron withdrawing group and at para position it exerts a very strong mesomeric effect, so it should form the carboxylic acid. To my surprise, I was told that it's the other way around.

Can someone explain me why this is true?

$\endgroup$
7
$\begingroup$

The first thing to point out is that the products of the Cannizzaro reaction do not necessarily correspond to the most stable products, i.e. the reaction is not necessarily thermodynamically controlled. Instead, one should consider the transition state—a six-membered ring as taught generally in organic chemistry—and note that after the reaction the compound which formed the hydrate will end up as the carboxylic acid and the one which did not form a hydrate is reduced to the alcohol. The transition state is shown in figure 1 below.

Transition state for the Cannizzaro reaction
Figure 1: transition state of the Cannizzaro reaction

Thus, the underlying question is which compound is more likely to form an aldehyde hydrate. You argued with the nitro group of 4-nitrobenzaldehyde exhibiting a strong $-M$ (mesomeric) effect. However, this mesomeric effect cannot, in fact, cross the $\ce{C-C(=O)}$ bond and thus cannot influence the carbonyl group as much as the simple $-M$ label implies.

mesomeric structures of 4-nitrobenzaldehyde
Figure 2: some mesomeric structures of 4-nitrobenzaldehyde, demonstrating that the mesomeric effect cannot ‘cross the $\ce{C-C}$ bond’.

Now it is known that aldehydes with a more electron deficient carbon are more likely to form hydrates, hence why we are doing a mesomeric/inductive discussion in the first place. But since we actually do not have any $-M$ arguments that can help our case, we need to turn to $-I$ arguments. When considering inductive effects, distance matters much more so 3-nitrobenzaldehyde experiences a stronger $-I$ effect than 4-nitrobenzaldehyde. Therefore, this compound is hydrated more favourably and form 3-nitrobenzoic acid while the reduced product should be 4-nitrobenzyl alcohol.

$\endgroup$
3
$\begingroup$

@Jan has provided a well-reasoned explanation for the OP's question regarding the Cannizzaro reaction. However, the transition state provided in Figure 1 is not in accord with the experimental results. In 1956 Hauser, et al.1 demonstrated that benzaldehyde-d1 (1-d1) in the presence of aqueous KOH efficiently affords benzyl alcohol-d2 (3-d2). On the other hand, benzaldehyde-1-d0 in the presence of sodium deuteroxide in deuterated hydroxylic solvent failed to incorporate deuterium into the resultant benzyl alcohol (3-d0).

I have presumed that the mono-alkoxy hydrate is operable and that the sodium counterion chelates with the carbonyl oxygen. [Equations 3 & 4 are from the Hauser paper, loc. cit..]

[Note: @Jan has subsequently corrected Figure 1.]



enter image description here

1) C. R. Hauser, P. J. Hamrick, Jr. and A. T. Stewart, J. Org. Chem., 1956, 21, 260.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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