Reactions of the α-Hydrogen: the Aldol Reaction

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take ethanal, an aldehyde with two carbon atoms, and add dilute sodium hydroxide. Out comes a molecule with four carbon atoms! Two molecules have joined through a new carbon to carbon bond. But methanal and benzaldehyde do not join like this. So what does ethanal have that those two do not? Hold on to your guess, we will check it in a few minutes. This idea comes up often in JEE, so let's begin.

0:30Why the alpha hydrogen is acidic

So, let us start with one special hydrogen. In a carbonyl group, C double bond O, the equals sign is the double bond. The carbon next to it is the alpha carbon, written with the Greek letter alpha, and its hydrogens are the alpha hydrogens. Okay? These alpha hydrogens are weakly acidic, with a p K a, the small letter A written low, of about seventeen for aldehydes like ethanal, up to about twenty for ketones. Yet ethane's hydrogens are not acidic at all. So what does the carbonyl group do here? It does two things. First, the carbonyl group withdraws electrons. Second, and this is the big one, when a base removes the alpha hydrogen, the conjugate base, the enolate ion, is stabilised by resonance. Clear? On the left, minus C H two, single bond, C H double bond O, okay? The small minus up top is a negative charge on carbon, and the small two written low means two hydrogens. The double headed arrow means resonance. On the right, C H two double bond C H, single bond, O minus, with the charge now on oxygen. Ethane gets no such help. With me?

1:44The aldol reaction

Our puzzle is in the top row of the figure: two C H three CHO, that is ethanal, with dilute sodium hydroxide over the arrow. The two half arrows, pointing both ways, mean the reaction can go back. The product is C H three, C H carrying an O H, C H two, CHO: three hydroxybutanal. So, what does ethanal have that methanal and benzaldehyde lack? Yes, exactly, an alpha hydrogen! Methanal has no carbon next to its carbonyl group, and in benzaldehyde that carbon is a ring carbon with no hydrogen. So rule one: we need at least one alpha hydrogen. Got it? In dilute alkali, the base pulls off an alpha hydrogen to make the enolate, and the enolate of one molecule adds to the carbonyl carbon of another. Right? So a new carbon to carbon bond joins the alpha carbon of one molecule to the carbonyl carbon of the other. In the figure, the C H carrying the O H was the carbonyl carbon. Okay? The O H sits on the beta carbon, the next one after alpha, written with the Greek letter beta. So the product is a beta hydroxy aldehyde, an aldol. From a ketone, we get a beta hydroxy ketone, a ketol. Clear? Now, when we heat the aldol, it loses water, minus H two O, in aldol condensation. In the top row, that gives C H three, C H double bond C H, CHO: but two enal. With me? The double bond sits between the alpha and beta carbons, so this is an alpha beta unsaturated carbonyl compound, here an aldehyde. The bottom row, with concentrated KOH, is the Cannizzaro reaction, for aldehydes with no alpha hydrogen, our next concept. Okay?

3:35Ketones too: propanone

Ketones do this too. Two propanone molecules, C H three COCH three, with barium hydroxide over the arrow, B a, O H in brackets, two, give C H three in brackets with a small two, C with O H in brackets, then C H two COCH three. That is four hydroxy four methylpentan two one: the O H is on the beta carbon of a ketone, so it is a ketol. Right? On heating, water leaves, and the double bond forms between the alpha and beta carbons: C H three in brackets with a small two, C double bond C H, COCH three. That is four methylpent three en two one, or mesityl oxide. Got it?

4:23Cross aldol and intramolecular aldol

Now mix two different compounds, both with alpha hydrogens. Each makes its own aldol, so two products, right? No, careful, that is a classic trap. Each compound can make an enolate, and each can be the acceptor, so we get four products, a messy mixture. Clear? How would you get just one main product? Use a partner with no alpha hydrogen, like benzaldehyde or methanal! It cannot form an enolate, so it can only be the acceptor, and one product predominates. Okay? Benzaldehyde, C six H five CHO, where C six H five is phenyl, with acetophenone, gives one three diphenylprop two en one one, or benzalacetophenone. Acetophenone supplies the enolate. Right? And benzaldehyde with ethanal gives cinnamaldehyde. Got it? Last, the intramolecular aldol, inside one molecule. In a dicarbonyl compound, with two carbonyl groups, the enolate can attack the other carbonyl group and close a ring, when a five or six membered ring can form. With me?

5:33Example 1: aldol and condensation product of propanal

Time for Example one. Name the aldol and the condensation product from propanal. Pause the video and try it first. Done? Okay, let's check. Propanal is C H three C H two CHO, okay? The CHO carbon is carbon one, so the alpha carbon is carbon two. Right? It is tempting to join the chains end to end, six carbons in a line, and write three hydroxyhexanal. Is that right? No, and that is the trap. The new bond starts from the alpha carbon, so carbon two of one molecule bonds to carbon one, the carbonyl carbon, of the other. Clear? So the aldol is C H three C H two, C H with O H in brackets, C H with C H three in brackets, and CHO at the end. Five carbons in the chain, O H on carbon three, methyl on carbon two: three hydroxy two methylpentanal. Okay? Heat it, and water leaves from carbons two and three. We get C H three C H two C H, double bond, C with C H three in brackets, then the CHO group: two methylpent two enal. Easy, isn't it?

6:53Example 2: four aldols from ethanal and propanal

Now Example two, a harder one. Here a mixture of ethanal and propanal is treated with N a O H, dilute sodium hydroxide. Name the four aldols that can be formed. Pause the video and try it first. Done? Okay, let's check. Both aldehydes have alpha hydrogens, so each can give the enolate or be the acceptor: two times two is four aldols. Okay? The two self reactions we already know. Ethanal gives three hydroxybutanal, and propanal gives three hydroxy two methylpentanal, our Example one. Right? Now the crossed ones. Let the enolate of ethanal attack propanal. What do you get? The C H two end of ethanal's enolate bonds to the carbonyl carbon of propanal: C H three C H two, C H with O H in brackets, C H two, and the CHO group. Five carbons, with the O H on carbon three: three hydroxypentanal. Got it? Last, the enolate of propanal attacks ethanal: C H three, C H with O H in brackets, C H with C H three in brackets, and the CHO group. That is three hydroxy two methylbutanal. Four aldols, just as the rule said. With me?

8:17Quick revision: alpha hydrogen, aldol, cross aldol

So, quick revision. The aldol reaction needs an alpha hydrogen, and dilute alkali gives a beta hydroxy aldehyde or ketone. The new bond joins the alpha carbon to the carbonyl carbon, and heating removes water. And a cross aldol gives four products, unless one partner has no alpha hydrogen. Now go and try the practice problems on this page. All the best!