reductive amination

Problem # 757

Propose a synthesis to accomplish each transformation. The only carbon sources allowed are alkenes and NaCN.

a) The starting alkene has 4 carbons, and the product has 5 carbons and a nitrogen. So you have to add 1C and 1N. This screams "add cyanide!"

Based on the location of the alcohol on the product, it looks like the way to go is to add NaCN to an epoxide. This puts the -CN and the -OH in the right positions.

b) The best way to prepare 2º and 3º amines is via reductive amination. Working backwards, the product can be prepared from a 4-carbon aldehyde (butyraldehyde) and a 3-carbon amine (propyl amine).

Performing hydroboration on the butene results in the anti-Markovnikov alcohol, which can be oxidized to the aldehyde using PCC.

To prepare propyl amine from an alkene, add HBr/peroxides to make the bromide (anti-Markovnikov), and then add NaN₃ follow by H₂/Pd.

Problem Details

MendelSet practice problem # 757 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): nitriles, reductive amination

Problem # 754

Show how each amine can be prepared from a carbonyl and an amine via reductive amination.

Solution

Reductive amination is a two step process. First, an imine or enamine is formed from a carbonyl and an imine.

Then, a the imine (or enamine) is reduced to an amine (using reducing agents such as H₂, NaBH₄, NaBH₃(CN), NaBH(OAc)₃, etc.). For practice on forming imine and enamines, see problem 712).