6.2.2 (a) Reactions of Amino Acids

Syllabus

(a) the general formula for an α-amino acid as RCH(NH₂)COOH and the following reactions of Amino Acids:

(i) reaction of the Carboxylic Acid group with alkalis and in the formation of Esters (see also 6.1.3 c)

(ii) reaction of the Amine group with acids

What does this mean?

Amino Acid structure

A must learn the general formula of amino acids, the examiner will not mind if you draw a displayed formula when asking for it.

Strictly speaking this is an alpha amino acid - if asked why it is because it contains an amino group and an Acid group seperarated by only 1 Carbon atom.

It wouldn't matter what was on the R group - even if there were other acid groups or base groups.

There are many naturally occurring alpha amino acids but you are not required to know any of their structures.

You should be able to name the simpler ones given the structure.

Glycine is really 2-amino ethanoic acid because the C atom in the acid group is automatically C1 (because we name the molecule as an acid). This puts the Amino group on C2.

All Alpha amino acids will be 2 amino something.

Alanine will be 2-amino propanoic acid.

It's unlikely you'll be asked to name anything more complicated than Valine.

A chain of four C atoms makes it a 2-amino butanoic acid.

But there's a methyl group on C3.

So, it should be 2-amino-3-methyl butanoic acid.

Remembering to list amino before methyl (alphabetically)

You'll see lots of past paper questions about Zwitterions and Isoelectric points.

You can ignore these.

They're no longer on the syllabus.

Reactions of Amino Acids

a) Neutralising the Amino Group(s)

The amino group can act as a base just as Ammonia and Amines can.

NH₃ + HCl --> NH₄⁺ + Cl^-

CH₃NH₂ + HCl --> CH₃NH₃⁺ + Cl^-

So, you can assume Glycine (or any other amino acid) will behave in the same way.

The examiner's favourite thing to do is include an amino acid with more than one amine group.

Image result for 2 3-diaminopropionic acid

In this case the question always asks you to draw the structure in "excess HCl" - other words, they will penalise you if you don't protonate both NH₂ groups

They don't usually mind if you draw the Cl^- ion as the two ions would be dissociated in water anyway.

b) Neutralising the Acid Group(s)

The Acid group can act as an acid just as any other carboxylic acid can.

CH₃COOH + NaOH--> CH₃COO^- + H₂O

So, you can assume Glycine (or any other amino acid) will behave in the same way.

Of course, it's still likely that an examiner will give you an amino acid with more than one acid group and ask you to draw the structure in excess NaOH.

c) Esterifying the Acid(s)

You'll remember that carboxylic acids can be esterified by reaction with alcohols in the presence of an Acid catalyst.

The Carboxylic acid group in an amino acid will behave in the same way.

So, in theory:

But an examiner is likely to penalise you one of the two available marks for writing your equation as above.

Why?

The acid is a catalyst for the esterification, but it remains acidic when the reaction is over.

It doesn't "know" it's there only as a catalyst so it will also react with the alkaline Amino group and needs to be shown as a reactant.