6.2.1 (b) Preparation of Amines

Syllabus

(b) the preparation of:

(i) aliphatic Amines by substitution of Haloalkanes with excess Ethanolic Ammonia and Amines

(ii) aromatic Amines by reduction of Nitroarenes using Tin and concentrated Hydrochloric Acid.

{Including formation of primary Amines from Ammonia and secondary/tertiary Amines from Amines. }

{See also reduction of Nitriles (see 6.2.4 c)}

What does this mean?

You should probably recall that Aliphatic refers to organic substances that are not Aromatic, and that Aromatic means "containing a Benzene Ring"

Image result for amine as base

So, all the above compounds (except Ammonia) are Amines; the ones on the left being aliphatic and the ones on the right aromatic

Preparing Aliphatic Amines.

Nucleophilic substitution of Haloalkanes with ethanolic Ammonia makes primary Amines.

As you can see from the mechanism above, excess Ammonia is needed partly because each molecule of haloalkane reacts with two moles of Ammonia.

This would make the overall equation: RCH2Br + 2 NH3 --> RCH2NH2 + NH4Br

But mostly we use excess Ammonia because without it the primary amine we are trying to make would also react with the haloalkane , forming a secondary amine we don't want.

RCH2NH2 + RCH2Br --> (RCH2)2NH + HBr

Ethanolic Ammonia (sometimes called Alcoholic Ammonia, which has rather sad connotations) is also required because using aqueous ammonia introduces water as a competing nucleophile and we would substitute an -OH group rather than an -NH2 group, making alcohols instead of amines (or probably as well as amines).

H2O + RCH2Br --> RCH2OH + HBr

Preparing Aromatic Amines

The examiner only really wants you to know how to make Phenylamine (C6H5NH2) from Nitrobenzene (C6H5NO2)

This reaction involves both removing Oxygen atoms and adding Hydrogen atoms - both definitions of Reduction.

Most A level reductions can be done with NaBH4, but this requires a stronger reducing agent: Tin and Conc HCl, followed by Sodium Hydroxide.

No details other than this are required except a redox equation using [H].

Image result for nitrobenzene to phenylamine

Note that 6 [H] are needed, 2 to add to the Nitrogen atom and 4 to make the 2 moles of water.

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