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The reduction of a regular amide with $\ce{LiAlH4}$ yields an amine. However, with a Weinreb amide, the product is an aldehyde. How can this be justified? I can't find the mechanism for the Weinreb amide reduction - is it known?

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up vote 4 down vote accepted

A dimethylamide anion is a highly active nucleophile, which makes it a bad leaving group. Remember that the two alkyl groups are electron-donating, thus destabilizing the negative charge on N. (blue mechanism favored)

enter image description here

However, in the case of a Weinreb amide, the oxygen atom adjacent to the nitrogen atom decreases the electron density on nitrogen, so that the attack from the nitrogen atom becomes unfavorable. Moreover, the methoxylamide anion becomes a much better leaving group, due to the oxygen atom stabilizing the negative charge. (red mechanism followed)

enter image description here

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Where did you get such nice pictures? – Mithoron Jan 8 at 21:08
    
You'll have to ask my chemistry teacher for them. – cfcief Jan 8 at 21:34
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If the methoxylamide anion leaves in the course of the reaction, then the aldehyde would be formed in the presence of LiAlH4. The aldehyde is much more reactive than an amide, so it would be reduced to an alcohol. This is precisely what the Weinreb amide tries to avoid. – jerepierre Jan 8 at 21:40
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Good point. Due to the good affinity of oxygen with metals (for instance, Al), the oxygen atom next to the nitrogen may actually form a O - - - Al - - - -O complex, which stabilizes the intermediate state. It is after the hydrolysis that the aldehyde is formed, thus preventing the over reduction to an alcohol. – cfcief Jan 8 at 22:09

Reaction of acyl carbonyl compounds with strong nucleophiles (hydrides, organometallics) is a very common and useful reaction, although it has some limitations. One of the simplest cases is the reaction of an ester with a strong nucleophile. As one would expect, the strong nucleophile attacks the electrophilic carbonyl carbon, giving a tetrahedral intermediate. In most cases, the tetrahedral intermediate breaks down readily to give a new carbonyl compound. In this case, that carbonyl intermediate will be a ketone (from a Grignard, e.g.) or an aldehyde (from LiAlH4). Both of those functional groups are more reactive than the ester starting material, so the next equivalent of nucleophile will react faster with the aldehyde/ketone, eventually giving a primary (from LiAlH4) or tertiary (from a Grignard) alcohol product.

enter image description here

The situation at the tetrahedral intermediate changes slightly when an amide is the electrophile. As cfcief points out, the less electronegative nitrogen will kick out the more electronegative oxygen (stabilized by aluminum) to give an iminium intermediate. Again, this intermediate is more reactive than the starting material, so a second nucleophilc attack takes place.

enter image description here

The key feature of the Weinreb amide is that after the nucleophilic attack at the carbonyl, the tetrahedral intermediate is stabilized as a five-membered cyclic chelate. Without decomposition of the tetrahedral intermediate, the only active electrophile is more of the Weinreb amide starting material. On workup, the strong nucleophile is quenched, the metal washed out, and the tetrahedral intermediate is allowed to decompose to the aldehyde/ketone product.

enter image description here

Although the N-methoxy-N-methylamide was specifically designed by Weinreb to behave this way (ref), it wasn't until more than 20 years later that mechanistic evidence was observed (ref).

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