nucleophilic acyl substitution

When enolates attack, the arrow is drawn from the double bond to the electrophile, and NOT from the negatively charged oxygen! The enolate's carbonyl reforms, and a new bond is created at the alpha position.

But Aldehydes and ketones react differently than ester and acid chlorides.

a) Aldehydes and ketones are oxidation state II carbonyls (the carbonyl has carbon-heteroatom bonds- the double bonded oxygen counts twice).

They do not have a built-in leaving group, and so undergo nucleophilic acyl addition reactions. The carbonyl on the electrophile becomes an alcohol.

So the product from this reaction is a beta hydroxyl carbonyl. If this β-hydroxyl carbonyl is heated up, the hydroxyl will eliminate to from an alpha, beta unsaturated carbonyl.

The mechanism for a) is very similiar to that of an aldol condensation reaction.

b) Esters (and acid chlorides) are oxidation state III carbonyls (the carbonyl carbon has three carbon-heteroatom bonds). They have a built-in leaving group (^-OR in the case of an ester, and Cl^- for an acid chloride), and so undergo nucleophilic acyl substitution reactions. ("up, down, kick"). The carbonyl on the electrophile is reformed.

So the product from this reaction is a beta keto carbonyl.

The mechanism for b) is very similiar to that of a Claisen condensation reaction.

To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism. (see problem 725)

Water reacts with an ester to form a carboxylic acid. This is what happens here. 

If there is only trace amounts of water, it's possible that the water reacts with the ester to form a carboxylic acid, and then goes back to reform the ester. But, in this process, leaves an isotopically labeled oxygen in the carbonyl position.

These reactions are both nucleophilic acyl substitution under basic conditions. See problem 705 for the mechanism in detail.

Esters are prepared from a carboyxlic acid and an alcohol (Fischer esterification).

So the problem becomes, "how can you prepare this ester from an alochol and a carboxylic acid, each containing three carbons? (propene)"

The "pair" of carboxylic acid and alcohol to make this ester is propanoic acid and propanol.

From propene, to make propanol, just do an anti-Markovnikov H₂O addition using borane (BH₃ or B₂H₆) then hydrogen peroxide (H₂O₂).

To make propanoic acid, oxidize propanol to propanoic acid using Jones reagent (aqueous chromium).

You can also convert propanoic acid to its acid chloride using SOCl₂, and then add propanol.

Adding LiAlH₄ to an ester yields two alcohols. The carbonyl "side" becomes one alcohol, and the ester "side" becomes the other alcohol.

So there are two esters that would yield these two alcohols after reduction with LiAlH₄.

To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism.

In this mechanism, the nucleophile (methanol) becomes the -OCH₃ group in the ester.

At least 80% of second semester organic chemistry is two mechnanisms: nucleophilic acyl addition and nucleophilic acyl substitution. It's worth your time to become familiar with these mechanisms. See problems 705 (basic conditions), 706 (acidic conditions), 707, and 708.

This is a good mechanism to know (nucleophilic acyl substitution). If you don't understand it, see problems 706 and 708.

Acidic mechanisms only appear complicated because they contain several proton transfer steps.

Nucleophilic acyl substitution mechanisms have only three real steps- the "up, down, and kick."

First, the nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate (the "up").

Then the carbonyl reforms (the "down") and a leaving group leaves (the "kick").