Clearly the amide will be unreactive given the poor quality of the leaving group and poor electrophilcity due to delocalization. However, does it react with organolithiums for example?


2 Answers 2


Amides will react with three equivalents n-butyllithium to give nitriles via geminal lithium oxyimide intermediates (J. Org. Chem., 1967, 32 (11)). n-butyllithium is a strong enough base to deprotonate the molecule twice at nitrogen and once alpha to the carbonyl group. Grignard reagents may well behave similarly.

I do not think this is a particularly useful transformation though because n-butyllithium is a rather indiscriminate reagent. There are a wide range of other methods available - see this paper if you're interested.

In the special case of a N,O-dimethylhydroxyamide, you can make ketones using either.


For a N,N-alkyl, a N,N-diaryl or a N-alkyl-N-aryl amide that what will happen is that the grignard will react attacking the carbonyl carbon. This will form R''C(NR2)(R')-OMgX from RMgX and R''-CONR2.

What we normally do is to treat the product with dilute acid, this will then convert the initial additon product into either a ketone or an aldehyde.

One common method of making aldehydes is to react Grignards and other similar reagents with DMF, then treating the intial product with aqueous acid.


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