The following reaction is known as the benzoin condensation. T...

Question

The following reaction is known as the benzoin condensation. The reaction does not take place if sodium hydroxide is used instead of sodium cyanide. Propose a mechanism for the reaction and explain why the reaction does not occur if hydroxide ion is the base.

Benzoin condensation reaction diagram showing reactants, sodium cyanide, and product benzoin.

Accepted Answer

All right. Hi, everyone. So this question is asking us in the benzoin condensation reaction. What is the role of the cyanide ion? Why does the reaction fail if the hydroxide ion is used instead? And here two equivalents of benz aldehyde react with methanol and sodium cyanide to yield the following product. We have four answer choices. Each proposing a different role of the cyanide ion. Now to understand the role of the cyanide ion, let's go ahead and actually review the mechanism of this reaction, right. So here's Benz aldehyde and recall that the first step of the benzoin condensation is going to be the reaction of cyanide with benz aldehyde specifically, right, the coupling of two equivalents of benz aldehyde requires the removal of the aldehyde hydrogen. First and foremost, right. So here my cyanide ion is going to go ahead and attack my carbonyl carbon of benz aldehyde and that forces the pi bond or pi electrons of my carbon to shift upwards towards oxygen. And then my next step, right, I end up with the following intermediate. Now, here in this intermediate, my carbonyl oxygen now has a negative charge which needs to be pro So from here, I can recall the fact that I have methanol in my system, right, because here my negatively charged carbonyl oxygen is going to take a proton from methanol and it's going to become a hydroxy group instead. So when I do that, I get a hydroxy group instead of it negatively charged oxygen. No recall that once methanol has been dep coordinated here, it's going to yield its conjugate base which is methanol. And also recall that the principle of this reaction requires the removal of the aldehyde hydrogen, which was still there. Right. So Methanol is going to go ahead and remove that all the Hyde hydrogen. And that in turn is going to create a pie bond between the carbonyl carbon and the carbon from cyanide, which is in turn, going to push a pair of pi electrons from cyanide towards nitrogen. So that is going to result in the following. Here's my hydroxy group and now my nitrogen is going to have a negative charge. Now, from here, right, I can go ahead and introduce my second molecule of benz aldehyde because now the pie bond between the initial carbonyl carbon and the carbon from cyanide is what's going to go ahead and attack the carbonyl carbon of my second molecule of Benzal hide. And in the meantime, right before I proceed with the next step, my nitrogen from cyanide and my carbon from cyanide is going to go ahead and form a triple bond once again because the lone pair on nitrogen is going to go ahead and move between carbon and nitrogen once again. So in my next step, right, I've connected my two molecules of benz aldehyde. The second one is now going to have a negatively charged carbonyl oxygen, whereas my first one is going to have both a hydroxy group and a cyano group attached or a nitrile. Now, once again, right, my negatively charged carbonyl oxygen can be pronated using methanol that's in solution. So that's going to take place. Here are my two benzene rings, right? And now I'm going to have a second hydroxy group. I'm going to have a second hydroxy group. But let's go ahead and recall what the end product is supposed to look like, right? The end product, if I scroll up a little bit more, the end product contains both a car bottle and a hydroxy group. So if I go back to where we're at in the reaction mechanism, we have that nitrile and hydroxy group where the carbonnel is supposed to be at the end. So I am running out of space honestly. But here, what's going to happen is that my hydroxy group, my first one I should say is going to go ahead and be derotated by the molecule of methanol that formed in the prior step because that is going to push the electrons between oxygen and hydrogen to form the carbonnel. And that in turn is going to go ahead and displace my night trial and that is going to yield the product that we see at the end of the screen or end of the reaction. Now, the main role that cyanide plays in this reaction is that it behaves as an um right, because it's responsible for creating the polarity inversion that we see with the carbonyl carbon because here, right, normally that carno carbon has a positive charge. But if it had a positive charge, there was no way it could have attacked the carbonic carbon of the second molecule of Benzal hide the way that it did in the mechanism we we went through together, right? So therefore, if I scroll up once again, therefore, the answer to this question is going to be option c because the cyanide ion creates an um belong to the carbonyl carbon allowing the reaction to occur. If a hydroxide ion is used, the polarity of the carbonic carbon will not be inverted and the reaction will not continue. Because if we didn't reverse the polarity of the carbon carbon, then it would have remained positively charge and therefore it would have never attacked the second molecule of Benzal. So I know there was a lot, right? But if you stuck around to the end, thank you so much for watching. And I hope you found this helpful

The following reaction is known as the benzoin condensation. The reaction does not take place if sodium hydroxide is used instead of sodium cyanide. Propose a mechanism for the reaction and explain why the reaction does not occur if hydroxide ion is the base.

Key Concepts

Benzoin Condensation

Role of Base in Organic Reactions

Nucleophilicity vs. Basicity

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