Question

Draw / write the structure of the semicarbazone of acetone.

Hint:

Replace acetone's =O with =N−NH−CO−NH₂.

Answer 1

Concept: A semicarbazone forms when the $\mathrm{C{=}O}$ of a ketone joins with semicarbazide, $\mathrm{H_2N{-}NH{-}CO{-}NH_2}$, and a water molecule is thrown out.

Step 1: See what reacts
Acetone is $\mathrm{(CH_3)_2C{=}O}$. The part that reacts is its carbonyl oxygen. The $\mathrm{-NH_2}$ end of semicarbazide attacks this carbon.

Step 2: Swap oxygen for nitrogen
The carbonyl oxygen leaves as water, and in its place comes a $\mathrm{{=}N{-}NH{-}CO{-}NH_2}$ group. So the double bond is now between carbon and nitrogen instead of carbon and oxygen.

Step 3: Write the structure
Putting it together, the two methyl groups stay on the carbon, and that carbon is now double bonded to the nitrogen chain. The structure is $\mathrm{(CH_3)_2C{=}N{-}NH{-}CO{-}NH_2}$.

Answer: Acetone semicarbazone is $\mathrm{(CH_3)_2C{=}N{-}NH{-}CO{-}NH_2}$.