How many of the following reagents react with propanone to give a pro…

The correct answer is Option (4) → 5

Ammonia and its derivatives, H₂N–Z, add to the carbonyl group, and the intermediate then loses water to give a >C=N–Z product. Other nucleophiles add to the C=O group but bring in no nitrogen double bond.

    \( \displaystyle \ce{(CH3)2C=O + H2N-Z -> (CH3)2C=N-Z + H2O} \)

>C=N– product:

• Hydroxylamine — propanone oxime, \( \ce{(CH3)2C=N-OH} \)

• Semicarbazide — semicarbazone, \( \ce{(CH3)2C=N-NHCONH2} \)

• Methylamine — substituted imine (Schiff’s base), \( \ce{(CH3)2C=N-CH3} \)

• Phenylhydrazine — phenylhydrazone, \( \ce{(CH3)2C=N-NHC6H5} \)

• 2,4-Dinitrophenylhydrazine — 2,4-dinitrophenylhydrazone

No >C=N– product:

• Hydrogen cyanide — gives the cyanohydrin \( \ce{(CH3)2C(OH)CN} \); its nitrogen is in a C≡N triple bond.

• Ethylene glycol (with dry HCl) — gives the cyclic ethylene glycol ketal; no nitrogen.

• Methylmagnesium bromide — gives 2-methylpropan-2-ol after hydrolysis; no nitrogen.

>C=N– products: hydroxylamine, semicarbazide, methylamine, phenylhydrazine and 2,4-dinitrophenylhydrazine = 5.

NCERT (2026-27 edition):
8.4 Chemical Reactions – “Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins.” … “Ketones react with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals.”

“(e) Addition of ammonia and its derivatives: Nucleophiles, such as ammonia and its derivatives H₂N-Z add to the carbonyl group of aldehydes and ketones. The reaction is reversible and catalysed by acid. The equilibrium favours the product formation due to rapid dehydration of the intermediate to form >C=N-Z. Z = Alkyl, aryl, OH, NH₂, C₆H₅NH, NHCONH₂, etc.”

Table 8.2: Some N-Substituted Derivatives of Aldehydes and Ketones (>C=N-Z)