Suggest possible reaction mechanisms for the reaction of acetone and semicarbazide?

Answer 1

Reaction: CH3C(=O)CH3 + NH2NHC(=O)NH2 --> (CH3)2C=NNHC(=O)NH2

1) N from NH2 acts as the nucleophile, attacks the carbonyl group and changes into a tetrahedral intermediate. Positive charge on NH2, negative charge on oxygen.

(CH3)2(O-C)--(NH2)+NHC(=O)NH2

2) oxygen takes the hydrogen from nitrogen through rearrangement because it is more electronegative.

(CH3)2(HOC)--(NH)NHC(=O)NH2

3) Acid protonates the alcohol once more to make it a better leaving group.

(CH3)2(H2O+C)--(NH)NHC(=O)NH2

4) Nitrogen from NH2 donates its electrons to generate a double bond, bumping off the water molecule and leaving a positive charge on nitrogen.

(CH3)2C=NH+NHC(=O)NH2

5) Water pulls off the hydrogen that was still attached to the nitrogen, thus generating the condensation product between acetone and semicarbazide.

(CH3)2C=NNHC(=O)NH2

Answer 2

One possible reaction mechanism involves the initial formation of an imine intermediate between acetone and semicarbazide, followed by a nucleophilic attack of the amine group on the carbonyl carbon to form a semicarbazone. An alternative mechanism involves the condensation of acetone with semicarbazide to form an enamine intermediate, which subsequently reacts with semicarbazide in a nucleophilic addition to form the semicarbazone product.