Can someone explain (and preferably draw) the intermediates of this reaction?
I understand that it is a Claisen Reaction, and likely a condensation due to the heat, but I am not sure why there isn't an ester in the final product.
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1$\begingroup$ The Claisen reaction is a type of acyl substitution ... as opposed to acyl addition. $\endgroup$ – Dissenter Apr 9 '15 at 19:36
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$\begingroup$ Thanks for that, although that doesn't quite fill in the blanks for me. Could you draw the intermediates? $\endgroup$ – 2567655222 Apr 9 '15 at 20:07
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2$\begingroup$ The Claisen is step 1. Step 2 might appear just to be a workup, but several things actually happen when the product from step 1 is heated in acidic aqueous conditions. Start by drawing the Claisen product (as you say, it should be an ester), and then think about what could happen when reacted with water. If you include those thoughts in your question, someone might take you the rest of the way. $\endgroup$ – jerepierre Apr 9 '15 at 20:14
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$\begingroup$ @UofTStudent - I meant the Claisen condensation is a type of acyl substitution ... and try thinking of the differences between substitution and addition ... that will help. $\endgroup$ – Dissenter Apr 10 '15 at 2:53
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Step 1 is a Claisen ester condensation with loss of ethoxide from the ester acting as the carbonyl part. The product is the $\beta$-ketoester (2).
Upon acidification and heating, the ester is cleaved, the resulting $\beta$-ketoacid decarboxylates to yield the symmetrical ketone.
answered Apr 9 '15 at 22:09
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Can someone explain (and preferably draw) the intermediates of this reaction?
Ethoxide - a base comparable in strength to hydroxide ion - abstracts a proton from the alpha position of the ester. This forms an enolate ion. Note that while protons alpha to a carbonyl center are mildly acidic - they are still not very acidic. Definitely not acidic enough for ethoxide to completely deprotonate. So we have both enolate anion and untouched, still-protonated ester.
Enolate anions are nucleophilic at both their oxygens and carbons. However, we mostly consider the reactions of enolates at their negatively charged carbons rather than their oxygens as we are interested in forming C-C bonds (as opposed to O-C bonds). Also, forming O-C bonds isn't particularly favorable versus C-C bonds - you'll have to study a bit of hard-soft acid-base theory to understand that.
Given that we have enolate anions in the system and untouched esters, which have electrophilic carbonyl carbons, we can make carbon-carbon bonds. The enolate anions can attack the carbonyl carbons, forming a tetrahedral intermediate. This intermediate collapses and ejects ethoxide ion.
answered Apr 10 '15 at 2:59
