SolutionsAddition of ammonia, amines and derivs. to aldehydes and ketones
1. Imines, enamines, hydrazones and oximes
Give the products that each ketone forms with each of the amines or amine derivatives in its box.
- Cycloheptanone: with 2-butanamine (primary amine) it gives an imine, N-(2-butyl)cycloheptanimine; with (2-butyl)hydrazine, a hydrazone. The stereogenic carbon of the chain does not take part and keeps its configuration.
- Cyclohexanone: with 4-nitroaniline (primary amine) it gives the imine N-cyclohexylidene-4-nitroaniline; with 4-methoxy-N-methylaniline (secondary amine) it gives an enamine, N-(cyclohex-1-en-1-yl)-4-methoxy-N-methylaniline.
- Cyclopentanone: with hydroxylamine it gives the oxime (cyclopentanone oxime); with pyrrolidine (secondary amine) it gives the enamine 1-(cyclopent-1-en-1-yl)pyrrolidine.
Primary amines and their derivatives (NH2–R, NH2–NHR, NH2–OH) give compounds with a C=N double bond; secondary amines cannot lose a second H from nitrogen and give enamines.
2. The Wolff-Kishner reduction
When acetophenone is treated with hydrazine a hydrazone is obtained. If the hydrazone is treated with hydroxide, nitrogen is evolved and ethylbenzene forms. Give a mechanistic explanation of these facts.
- Hydrazine adds to the carbonyl and, by losing water, gives the hydrazone PhC(CH3)=N–NH2.
- Hydroxide removes a proton from the NH2; the delocalized anion is protonated on carbon to give a diazene, PhCH(CH3)–N=NH.
- A second deprotonation of the N–H and loss of the very stable N2 (gas evolution) give a benzylic carbanion.
- The carbanion is protonated by water: ethylbenzene.
The C=O has been reduced to CH2: this is the Wolff-Kishner reduction, complementary to the Clemmensen reduction (Zn(Hg)/HCl), which is carried out in acid.