Module 6: Organic Chemistry and Analysis · Year 13
6.2.2 Amino Acids, Amides and Chirality
Use the –NH₂ and –COOH groups independently: predict their reactions, recognise amides, and then check for chiral centres and enantiomers.
What you need to know
Open a line for a quick recap. If it feels obvious, move straight to the linked practice.
6.2.2(a) Draw the general structure of an α-amino acid and identify its acid and base functional groups. Quick revision
An α-amino acid has the general formula RCH(NH₂)COOH. The carbon carrying –NH₂ is directly next to the carboxyl carbon, which is what the α label tells you.
The –COOH group is acidic and the –NH₂ group is basic. Keep those two jobs clear because the reaction conditions decide which end of the molecule changes.
6.2.2(a)(i) Draw the product formed when an amino acid reacts with alkali. Quick revision
Aqueous alkali reacts with the carboxylic acid group. The –COOH becomes a carboxylate salt, while the amino group remains –NH₂.
For glycine with NaOH: H₂NCH₂COOH + NaOH → H₂NCH₂COO⁻ Na⁺ + H₂O. Check that you have removed the acidic H from –COOH, not added a proton to nitrogen.
6.2.2(a)(i) Draw the ester formed when an amino acid reacts with an alcohol under acidic conditions. Quick revision
At the carboxyl group, esterification replaces the acid –OH by –OR from the alcohol. With methanol, the –COOH end becomes –COOCH₃.
Because the reaction mixture is explicitly acidic, the basic –NH₂ group is protonated. Draw the major organic species as RCH(NH₃⁺)COOCH₃ for methanol, with the appropriate counter-ion if the question asks for the full salt.
6.2.2(a)(ii) Draw the salt formed when an amino acid reacts with acid. Quick revision
In acid, the amine group accepts H⁺ and becomes –NH₃⁺. The carboxylic acid group stays as –COOH.
For glycine in HCl: H₂NCH₂COOH + HCl → [H₃NCH₂COOH]⁺ Cl⁻. I’d check the conditions before drawing charges: acidic solution should not leave the carboxyl group as –COO⁻.
6.2.2(b) Recognise and draw primary amides. Quick revision
A primary amide contains the group –CONH₂. The nitrogen is bonded to the carbonyl carbon and to two H atoms: RCONH₂.
For example, ethanamide is CH₃CONH₂. Do not classify it from the carbon chain as though it were an amine; the carbonyl directly attached to N is what makes it an amide.
6.2.2(b) Recognise and draw secondary amides. Quick revision
A secondary amide has one H and one carbon-containing group attached to the amide nitrogen: RCONHR′. The –CONH– link is the feature to spot.
CH₃CONHCH₃ is a simple example. Find the C=O directly beside N before you classify it; that stops a secondary amide being mistaken for a secondary amine.
6.2.2(c) Explain optical isomerism and draw a pair of enantiomers. Quick revision
Optical isomers are non-superimposable mirror images about a chiral centre. The two molecules have the same structural formula and connectivity; only their three-dimensional arrangement differs.
To draw a pair, use wedge-and-dash bonds around the chiral carbon and swap the 3-D arrangement to make the mirror image. A carbon attached to four different groups is the usual chiral centre you will meet here.
6.2.2(d) Identify chiral centres in unfamiliar organic molecules. Quick revision
Look for a tetrahedral carbon bonded to four different groups. Check the whole attached groups, not just the first atom: two groups can both start with carbon and still be different.
Work through every suitable carbon because an unfamiliar structure can contain more than one chiral centre. I’d mark the centres first; you do not need to redraw the whole molecule just to decide whether each carbon is chiral.