OCR A Level Chemistry A: Amino acids, amides and chirality (H432)

This guide teaches section 6.2.2, Amino acids, amides and chirality, from the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Module 6, Organic chemistry and analysis, places it under 6.2, Nitrogen compounds, polymers and synthesis. Each of the four lettered outcomes, 6.2.2(a) to (d), gets an explanation and a worked example. The course is the full A level. The assessment overview puts Module 6 content on two papers, Unified chemistry (03) as well as Synthesis and analytical techniques (02). The specification lists no PAG for this section.

Other H432 sections are listed on the OCR chemistry hub. Record progress with the printable checklist, and spot weak earlier topics with the free diagnostics.

What 6.2.2 asks of you

Outcome	Skill required	Additional guidance and links in the specification
6.2.2(a)	Give the general formula of an α-amino acid, RCH(NH₂)COOH; write reactions of (i) the COOH group with alkalis and to form esters, and (ii) the NH₂ group with acids	Cross-reference to esterification, 6.1.3(c)
6.2.2(b)	Recognise and draw the structures of primary and secondary amides	Cross-references to 6.1.3(f) and 6.2.3(a)–(b)
6.2.2(c)	Explain optical isomerism as a kind of stereoisomerism: non-superimposable mirror images about a chiral centre	M4.2, M4.3; draw 3-D diagrams to illustrate stereoisomerism; HSW1, HSW8; see also 4.1.3(c)–(d)
6.2.2(d)	Identify the chiral centres in a molecule of any organic compound	M4.2, M4.3

M4.2 is drawing and reading 2-D pictures of 3-D shapes; M4.3 is symmetry, which you test whenever you ask whether a molecule matches its mirror image.

6.2.2(a): the α-amino acid and its two functional groups

An α-amino acid has the general formula RCH(NH₂)COOH. The amine group and the carboxylic acid group are both bonded to the same carbon, the α-carbon, which also carries one hydrogen and the side chain R.

        H
        |
   H₂N—C—COOH
        |
        R

R is H in glycine, CH₂CH₂SCH₃ in methionine and CH₂CH₂CH₂CH₃ in 2-aminohexanoic acid. With an acidic group and a basic group, the molecule reacts with both alkalis and acids. The equations here use the un-ionised form RCH(NH₂)COOH, as the specification does.

Worked example 1: building formulae from R

Task: give the structural and molecular formulae of methionine.

1. Put R in place of the letter: CH₃SCH₂CH₂CH(NH₂)COOH.
2. Count the atoms: C 5, H 11, N 1, O 2, S 1.
3. Molecular formula: C₅H₁₁NO₂S, so M = 5(12.0) + 11(1.0) + 14.0 + 2(16.0) + 32.1 = 149.1 g mol⁻¹.

(a)(i) The carboxylic acid group with alkalis

The COOH group is a weak acid. With an alkali such as NaOH(aq) it forms a carboxylate salt and water; the NH₂ group is unchanged.

RCH(NH₂)COOH + NaOH → RCH(NH₂)COO⁻Na⁺ + H₂O

Worked example 2: an amino acid and sodium hydroxide

Task: 2.62 g of 2-aminohexanoic acid, CH₃(CH₂)₃CH(NH₂)COOH, reacts completely with NaOH(aq). Write the equation, then find the mass of NaOH used and the mass of salt that forms.

1. Equation: CH₃(CH₂)₃CH(NH₂)COOH + NaOH → CH₃(CH₂)₃CH(NH₂)COO⁻Na⁺ + H₂O
2. M(C₆H₁₃NO₂) = 131.0 g mol⁻¹, so n = 2.62 ÷ 131.0 = 0.0200 mol.
3. The ratio is 1 : 1, so n(NaOH) = 0.0200 mol and mass = 0.0200 × 40.0 = 0.800 g.
4. M(C₆H₁₂NO₂Na) = 153.0 g mol⁻¹, so mass of sodium 2-aminohexanoate = 0.0200 × 153.0 = 3.06 g.

(a)(i) The carboxylic acid group forming esters

This is the esterification of 6.1.3(c): heat the amino acid with an alcohol and a little concentrated sulfuric acid as catalyst. The reaction is reversible.

RCH(NH₂)COOH + R′OH ⇌ RCH(NH₂)COOR′ + H₂O

Name the alcohol part first: methanol and 2-aminohexanoic acid give methyl 2-aminohexanoate. The acid catalyst can also protonate the NH₂ group, so in the mixture the ester may be present as its ammonium salt; when asked for the ester, show COOR′ with NH₂ unchanged.

Worked example 3: ester yield

Task: 5.24 g of 2-aminohexanoic acid is heated with excess methanol and concentrated H₂SO₄. After work-up, 4.06 g of methyl 2-aminohexanoate is collected. Find the percentage yield.

1. Equation: CH₃(CH₂)₃CH(NH₂)COOH + CH₃OH ⇌ CH₃(CH₂)₃CH(NH₂)COOCH₃ + H₂O
2. n(acid) = 5.24 ÷ 131.0 = 0.0400 mol.
3. M(C₇H₁₅NO₂) = 145.0 g mol⁻¹, so n(ester) = 4.06 ÷ 145.0 = 0.0280 mol.
4. Yield = 0.0280 ÷ 0.0400 × 100 = 70.0%. The reversible reaction is one reason it falls short of 100%.

(a)(ii) The amine group with acids

The nitrogen lone pair accepts a proton, as in the amines of 6.2.1. With dilute HCl the NH₂ group becomes NH₃⁺ and a chloride salt forms; COOH is unchanged.

RCH(NH₂)COOH + HCl → RCH(NH₃⁺)COOH Cl⁻

Worked example 4: methionine and hydrochloric acid

Task: write the equation for methionine with dilute HCl, and find the mass of salt from 2.982 g of methionine.

1. CH₃SCH₂CH₂CH(NH₂)COOH + HCl → CH₃SCH₂CH₂CH(NH₃⁺)COOH Cl⁻
2. n(methionine) = 2.982 ÷ 149.1 = 0.0200 mol.
3. M(salt) = 149.1 + 36.5 = 185.6 g mol⁻¹, so mass = 0.0200 × 185.6 = 3.71 g.

A good check: alkali changes only the COOH end; acid only the NH₂ end.

6.2.2(b): primary and secondary amides

An amide contains the group C=O bonded directly to nitrogen, written –CONH– in condensed formulae. The class depends on how many carbon atoms are bonded to that nitrogen.

Type	General structure	Carbons bonded to N	Example
Primary amide	RCONH₂	one (the C=O carbon)	3-methylpentanamide, CH₃CH₂CH(CH₃)CH₂CONH₂
Secondary amide	RCONHR′	two	N-methylhexanamide, CH₃(CH₂)₄CONHCH₃

The “N-” in a secondary amide’s name shows which group sits on the nitrogen. Both kinds are made from acyl chlorides (6.1.3(f)): ammonia gives a primary amide, a primary amine a secondary one. The links in the polyamides of 6.2.3 are secondary amide groups.

Worked example 5: classifying and making amides

Task: classify each compound, then write an equation that makes B from an acyl chloride.

• A CH₃C₆H₄CONH₂
• B CH₃CH₂CH(CH₃)CONHCH₂CH₃
• C CH₃CH₂CH₂CONHC₆H₅
• D CH₃CH₂CH₂NHCH₃

1. A (4-methylbenzamide): N is bonded to one carbon, the C=O carbon. Primary amide.
2. B (N-ethyl-2-methylbutanamide): N is bonded to the C=O carbon and to an ethyl carbon. Secondary amide.
3. C: N is bonded to the C=O carbon and to a benzene ring carbon. Secondary amide (N-phenylbutanamide).
4. D: there is no C=O next to the nitrogen. It is a secondary amine, not an amide.
5. Making B: CH₃CH₂CH(CH₃)COCl + 2CH₃CH₂NH₂ → CH₃CH₂CH(CH₃)CONHCH₂CH₃ + CH₃CH₂NH₃⁺Cl⁻. The second mole of ethylamine reacts with the HCl formed.

6.2.2(c): optical isomerism

Two compounds are stereoisomers when their structural formulae match but the 3-D positions of their atoms do not. After E/Z and cis–trans isomerism (4.1.3), optical isomerism is the second kind in the course.

A chiral centre is a carbon atom carrying four different atoms or groups. Such a carbon has two possible 3-D arrangements. They are mirror images of each other, and however you rotate them the two never line up group for group: they are non-superimposable. These two forms are optical isomers, also called enantiomers.

Your hands are the everyday model: mirror images that cannot be overlaid. If two groups on the carbon are the same, the mirror image can be rotated to match the original, so there is no optical isomerism.

Drawing 3-D diagrams

Use the standard convention around the tetrahedral carbon:

• two neighbouring bonds as plain lines, in the plane of the page;
• one solid wedge, pointing towards you;
• one hashed (dashed) wedge, pointing away from you, with both wedges on the side opposite the plain pair.

Draw one isomer, a dotted mirror line, then the reflection: wedges stay wedges and hashed bonds stay hashed; only left and right swap.

Worked example 6: optical isomers of 2-aminohexanoic acid

Task: show the two optical isomers of 2-aminohexanoic acid.

1. The α-carbon carries H, NH₂, COOH and C₄H₉. Four different groups, so it is a chiral centre.
2. Sketch, where | and — are bonds in the page, ◀ or ▶ is a solid wedge towards you and ⋮ is a hashed bond away from you:

          H          :          H
          |          :          |
  C₄H₉ —  C ◀ NH₂    :    H₂N ▶ C — C₄H₉
          ⋮          :          ⋮
        COOH         :        COOH

3. The right-hand structure is the mirror image. Swapping any two groups also gives the other isomer.

Glycine is the exception: R is H, so its α-carbon carries two hydrogens and it has no optical isomers. Every α-amino acid with R other than H has a chiral α-carbon.

HSW1 and HSW8 point to using the tetrahedral model to explain the isomers and communicating 3-D shape clearly. Optical isomers of complexes with bidentate ligands appear in 5.3.1(f).

6.2.2(d): finding chiral centres in any molecule

Check each carbon in turn. Rule out any carbon that:

• is part of a C=C or C=O (it has only three groups);
• carries two or more hydrogens (CH₂ or CH₃);
• carries two identical groups, such as two ethyl groups.

For the rest, compare whole groups, not just the first atom: methyl and ethyl differ. In a ring, follow both ring paths away from the carbon; if they differ, they are two different groups. Mark each chiral centre with an asterisk (*).

Worked example 7: counting chiral centres

Molecule	Structure	Chiral centres	Reason
2-methylhexan-1-ol	CH₃(CH₂)₃C*H(CH₃)CH₂OH	1	C2 has H, CH₃, C₄H₉ and CH₂OH
3-methylpentane	CH₃CH₂CH(CH₃)CH₂CH₃	0	C3 carries two ethyl groups
1-phenylpropan-1-ol	C₆H₅C*H(OH)CH₂CH₃	1	H, OH, C₂H₅ and C₆H₅
4-methylheptan-2-one	CH₃COCH₂C*H(CH₃)CH₂CH₂CH₃	1	C4 has H, CH₃, C₃H₇ and CH₂COCH₃; the C=O carbon is not chiral
3-chlorobutan-2-ol	CH₃C*H(OH)C*HClCH₃	2	C2 and C3 each carry four different groups
3-methylcyclopentan-1-ol	ring, OH on C1, CH₃ on C3	2	From C1, one route round the ring meets the CH(CH₃) carbon after one CH₂, the other after two; the same unevenness makes C3 chiral

Compare 4-methylcyclohexan-1-ol: there the two routes round the ring from C1 are identical, so it has no chiral centre.

Common errors

• Writing the general formula as RCH(NH₃)COOH or putting R on nitrogen. It is RCH(NH₂)COOH.
• Letting NaOH react with the NH₂ group, or HCl with the COOH group.
• Calling CH₃CH₂NHCH₃ an amide. No C=O on the nitrogen means it is an amine.
• Drawing all four bonds as plain lines, so the mirror images look identical.
• Marking a CH₂, a C=O carbon or a carbon with two identical groups as a chiral centre.

Where to go next

The Amino acids, amides and chirality revision notes condense this page, and the practice questions test it. Related pages: the Amines study guide, Carboxylic acids and esters and Alkenes for E/Z isomerism.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.2.2 Amino acids, amides and chirality, learning outcomes (a)(i)–(ii), (b), (c) and (d). Published by OCR.