Note also that carbonyl compounds without alpha hydrogens do not react with bromine at all. It is the enol, not the carbonyl compound, which is reactive toward bromine. This is because the enol is so reactive toward bromine that it never has a chance to reverse step 2, i. So bromine reacts much more rapidly than does hydronium ion with the intermediate enol.
Once formed, the enol always goes on to brominated product. The rate of enol formation is exactly equal to the rate of formation of the brominated product. If we double the concentration of bromine, the rate remains exactly the same or if we cut the concentration in half, the rate is not diminished at all. This is because the formation of the enol is rate-determining. The rate is not only independent of the concentration of the halogen but also of the nature of the halogen. Even though bromine is more reactive than chlorine, in general, in electrophilic additions, chlorination and bromination both occur at exactly the same rate.
As we have noted, when hydroxide ion is present, both the enol and enolate are typically present in equilibrium with the carbonyl component. We have also seen that under these basic conditions the enolate is the predominant form.
However, the enolate is also the more reactive form toward electrophiles. For both of these reasons the enolate is the reactive species of interest in basic solutions. Please keep in mind also that the carbonyl component, which is present in great excess over both the enol and enolate, is essentially unreactive toward bromine and many other electrophiles excluding, of course, hydronium ion.
The enolate reacts much more rapidly with bromine step 2 than with water reverse of step 1. So step 1 is never reversed, and the enolate once formed always goes on and goes on rapidly to the product. So the enolate has much more carbanion character than the enol, and is much more reactive towards electrophiles.
We have seen that the enol form of a carbonyl compound, though only a minor constituent in the equilibrium mixture, is vitally important in the reaction with electrophiles. The same is true of the enolate in basic solutions. We have also seen that the enolate is a more potent nucleophile than the enol. Since the carbonyl group of the carbonyl form is strongly electrophilic and reacts with a variety of nucleophiles, it would be reasonable to expect the strongly nucleophilic enolate to be able to add to the carbonyl group of the carbonyl component which is the major component in the equilibrium.
Mechanism of the Aldol Addition Reaction. As with most carbonyl additions, this reaction is reversible, so it is not rate determining. It is specifically called an al dol addition because it works better with aldehydes than ketones recall that the carbonyl pi bond of ketones is thermodynamically more stable than that of aldehydes.
Therefore the simplest aldehyde, methanal formaldehyde cannot undergo the aldol reaction. The simplest case: Aldol Addition of Ethanal Acetaldehyde. These are: 1 the enolate role and 2 the carbonyl role. In viewing a given aldol product, you can see which portion of the product arose from the enolate and which from the carbonyl component. Consider the product of the aldol reaction of propanal.
It is there because only the alpha hydrogen can be removed to form the enolate, leaving any other more remote carbons of the aldehyde as a branch. But in this reaction, a single carbonyl compound is used as the source of both the enolate and the carbonyl components.
Consequently many structures which might be desirable are not accessible because they would require two different components. Consider the hydroxyaldehyde given below:. Consequently, the crossed aldol is not a generally feasible reaction. In many cases it is possible to perform crossed aldol reactions between one of these non-enolizable carbonyl compounds and a general aldehyde. In the examples below, benzaldehyde, pivaldehyde, and methanal are the non-enolizable carbonyl compounds.
That is, we can get the crossed aldol shown, but we can also get the normal aldol of the enolizable component. Thus, when the enolate is formed, it always has a greater opportunity to react with the non-enolizable carbonyl component. A condensation reaction is one in which water or another small molecule, such as methanol, is formed in a reaction between two organic molecules. Another way to figure out which direction the equilibrium goes is to think about which one is the weaker acid.
So we have these two acids here. Let me go ahead and change colors. We have a pKa of 17 and a pKa of The lower the pKa the more acidic something is, so ethanol is more acidic than R aldehyde and the equilibrium favors formation of the weaker acid, and so since the aldehyde is the weaker acid the equilibrium favors formation of this weaker acid.
So that's another way to think about which direction for the equilibrium. What if you wanted to completely make your enolate anion? So one thing you could do is add a base like a hydride, so here we have once again acetaldehyde here.
So we have acetaldehyde, and this time our base is the hydride anion, so we could get that from something like sodium hydride, Na plus H minus, or potassium hydride, K plus H minus. And so once again we identify our alpha carbon, which is this one, with three alpha protons. So we can just go ahead and draw one of them in there. And we could show our hydride anion functioning as a base, taking this proton, leaving these electrons behind on that carbon. So let's go ahead and show the enolate anion that results.
So we have our carbonyl here, and then we have our electrons on this carbon, giving it a negative one formal charge, so electrons in magenta move out on to this carbon, forming the carbanion.
And just to save time I won't draw any oxyanion, but that's the one that's actually a greater contributor to the resonance hybrid here. So we have our carbanion and then we would also form hydrogen gas, so if this hydride anion picks up a proton we would form H2.
And so let's show those electrons, so the electrons in red here on our hydride anion pick up this proton and forming this bond and so we get hydrogen gas, which would bubble out of solution, and since this is going to bubble out of solution we're going to drive the reaction to completion, so we're going to push the reaction to completion and so we're going to get our enolate anion so we're pretty much going to get complete formation of our enolate anion here.
So in the next video we're going to talk about another base called LDA which can also give you complete formation of your enolate anion. A crossed-aldol reaction simply implies that two different components were used, one to act as the nucleophile electron donor and one as the carbonyl electron acceptor. Write a complete mechanism for the crossed aldol reaction between acetaldehyde and benzaldehyde.
Indicate which you selected to be the donor and which the acceptor, and also indicate how you might arrange the experiment so that you avoid or minimize the extent of normal aldol reaction.
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