Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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 # 756

For each amine below, show all Hofmann elmination products.

If more than one product is formed, predict which one will be the major product.

Solution

The Hoffman elimination first turns an amine into a better leaving group by methylating it with excess methly iodide, and then eliminates the methylated amine using base (such as ^-OH fomed from Ag₂O in H₂O). The products are alkenes.

This reaction is similar to an E2 elimination in that a beta proton is ripped off, and the alkene forms between alpha and beta. One notable difference from the E2 is that the less substituted product is favored, so Hoffman products are usually anti-Zaitsev.

a) There are three different types of beta protons, so three possible products can form. Product A is the least substituted alkene, so it's the major product.

b) There is only one possible product. The Hoffman elimination can't be done on aryl hydrogens (benzene rings).