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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set practises section 6.2.2, Amino acids, amides and chirality, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Every Module 6 outcome in this section, 6.2.2(a) to (d), is tested, and the course is the full A level. Questions 9, 10 and 11 are extended problems; question 7 also uses acyl chlorides from 6.1.3(f).

The study guide and revision notes cover the content. Other topics are on the OCR chemistry hub; log progress on the printable checklist and test wider gaps with the free diagnostics.

Data for all questions:

Element	H	C	N	O	Na	S	Cl
A_r	1.0	12.0	14.0	16.0	23.0	32.1	35.5

Questions

1. Methionine is an α-amino acid in which the side chain R is –CH₂CH₂SCH₃.

(a) State the general formula of an α-amino acid. [1] (b) Give the structural formula and the molecular formula of methionine. [2]

2. Classify each compound as a primary amide, a secondary amide or neither.

A CH₃CH₂CH(CH₃)CONH₂ B C₆H₅CONHCH₃ C CH₃CH₂CH₂CONHCH(CH₃)₂ D CH₃COCH₂NH₂ [4]

3. 2-Amino-4-phenylbutanoic acid has the structure C₆H₅CH₂CH₂CH(NH₂)COOH.

(a) Write an equation for its reaction with dilute hydrochloric acid. [1] (b) Write an equation for its reaction with aqueous sodium hydroxide. [1] (c) Name the organic product of (b). [1]

4. Methionine, CH₃SCH₂CH₂CH(NH₂)COOH, is converted into its propyl ester.

(a) State the other reagent, the catalyst and the condition needed. [2] (b) Give the equation for this esterification. [1] (c) 4.473 g of methionine gives 3.44 g of the propyl ester, C₈H₁₇NO₂S. Calculate the percentage yield. [3]

5. 1-Phenylethanamine has the structure C₆H₅CH(NH₂)CH₃.

(a) Explain why 1-phenylethanamine shows optical isomerism. [2] (b) Draw 3-D diagrams of the two optical isomers. [2]

6. State the number of chiral centres in each molecule.

(a) 2,3-dimethylpentane, CH₃CH(CH₃)CH(CH₃)CH₂CH₃ (b) 2-hydroxy-3-methylbutanoic acid, (CH₃)₂CHCH(OH)COOH (c) pentane-2,4-diol, CH₃CH(OH)CH₂CH(OH)CH₃ (d) 3-bromopentane, CH₃CH₂CHBrCH₂CH₃ (e) 2-methylcyclohexan-1-one [5]

7. 2-Methylpentanoyl chloride, CH₃CH₂CH₂CH(CH₃)COCl, is used to make two amides.

(a) Write an equation for its reaction with excess ammonia, and state the type of amide formed. [2] (b) Write an equation for its reaction with propylamine, CH₃CH₂CH₂NH₂, using two moles of the amine, and state the type of amide formed. [2] (c) Identify, by its position in the chain, the chiral centre in the amide from (b). [1]

8. Three 3-D drawings of heptan-2-amine are described. Each has the chiral carbon central, two neighbouring in-plane bonds, and one wedge and one hashed bond on the opposite side.

• Drawing 1: H up and CH₃ left (in plane), NH₂ on a wedge to the right, C₅H₁₁ on a hashed bond pointing down.
• Drawing 2: H up and CH₃ left (in plane), C₅H₁₁ on a wedge to the right, NH₂ on a hashed bond pointing down.
• Drawing 3: H down and CH₃ right (in plane), NH₂ on a wedge to the left, C₅H₁₁ on a hashed bond pointing up.

(a) State the relationship between drawings 1 and 2, and explain your answer. [2] (b) State the relationship between drawings 1 and 3. [1]

9. 2.90 g of an α-amino acid, X, in which R is a saturated alkyl group, reacts completely with aqueous sodium hydroxide. The water is evaporated, leaving 3.34 g of the dry sodium salt.

(a) Write a general equation for the reaction, using RCH(NH₂)COOH. [1] (b) Use the mass increase to calculate the amount of X and its molar mass. [3] (c) Deduce the formula of R. [1] (d) Suggest a structure for R that gives X two chiral centres. [1]

10. Compound V is H₂NCH(CH₂C₆H₅)CONHCH(CH₃)COOCH₃.

(a) State whether the amide group in V is primary or secondary, and give a reason. [2] (b) State how many chiral centres V has, and identify each one. [2] (c) Give the organic product of V with dilute hydrochloric acid at room temperature. [1] (d) V is made from U, H₂NCH(CH₂C₆H₅)CONHCH(CH₃)COOH. State the reagent and conditions, and write an equation for the reaction. [2]

11. An α-amino acid, Z, contains 10.7% nitrogen by mass. Its side chain R is a saturated, non-cyclic alkyl group.

(a) Calculate the molar mass of Z. [1] (b) Deduce the molecular formula of R. [1] (c) Draw the structures of the four possible amino acids that fit this data. [2] (d) Identify the isomer with two chiral centres and mark both with asterisks. [2] (e) Write an equation for this isomer with aqueous sodium hydroxide. [1] (f) Explain why all four isomers show optical isomerism. [1]

Answers

1. (a) RCH(NH₂)COOH [1] (b) CH₃SCH₂CH₂CH(NH₂)COOH [1]; C₅H₁₁NO₂S [1] Examiner insight: NH₂ and COOH must sit on the same carbon; otherwise it is not an α-amino acid.

2. A primary amide [1]; B secondary amide [1]; C secondary amide [1]; D neither: a ketone and an amine, with no C=O on N [1] Examiner insight: For D, C=O and NH₂ in one molecule are not an amide unless bonded to each other.

3. (a) C₆H₅CH₂CH₂CH(NH₂)COOH + HCl → C₆H₅CH₂CH₂CH(NH₃⁺)COOH Cl⁻ [1] (b) C₆H₅CH₂CH₂CH(NH₂)COOH + NaOH → C₆H₅CH₂CH₂CH(NH₂)COO⁻Na⁺ + H₂O [1] (c) Sodium 2-amino-4-phenylbutanoate [1] Examiner insight: Each reagent changes one group only; changing both NH₂ and COOH with one mole of reagent loses the mark.

4. (a) Propan-1-ol with concentrated sulfuric acid as catalyst [1]; heat [1] (b) CH₃SCH₂CH₂CH(NH₂)COOH + CH₃CH₂CH₂OH ⇌ CH₃SCH₂CH₂CH(NH₂)COOCH₂CH₂CH₃ + H₂O [1] (c) n(methionine) = 4.473 ÷ 149.1 = 0.0300 mol [1]; M(ester) = 191.1 g mol⁻¹, so n(ester) = 3.44 ÷ 191.1 = 0.0180 mol [1]; yield = 0.0180 ÷ 0.0300 × 100 = 60.0% [1] Examiner insight: Show both mole values; a ratio of masses ignores the change in molar mass.

5. (a) The CH carbon is a chiral centre: it carries four different groups, H, CH₃, NH₂ and C₆H₅ [1]; so the molecule exists as two mirror images that are non-superimposable [1] (b) A 3-D tetrahedral drawing using in-plane, wedged and hashed bonds [1]; its reflection beside a mirror line, every group labelled [1]. For example: H up and CH₃ left in the plane, NH₂ wedge right, C₆H₅ hashed down; then H up, CH₃ right, NH₂ wedge left, C₆H₅ hashed down. Examiner insight: Four plain bonds cannot show the mirror relationship; use wedges and label every group.

6. (a) 1 (C3) [1] (b) 1 (C2) [1] (c) 2 (C2 and C4) [1] (d) 0: C3 carries two ethyl groups [1] (e) 1 (C2: H, CH₃, the C=O side of the ring and the CH₂ side of the ring) [1] Examiner insight: In (a), C2 holds two CH₃ groups, so it is not chiral; test every carbon for four different groups.

7. (a) CH₃CH₂CH₂CH(CH₃)COCl + 2NH₃ → CH₃CH₂CH₂CH(CH₃)CONH₂ + NH₄Cl [1]; primary amide [1] (b) CH₃CH₂CH₂CH(CH₃)COCl + 2CH₃CH₂CH₂NH₂ → CH₃CH₂CH₂CH(CH₃)CONHCH₂CH₂CH₃ + CH₃CH₂CH₂NH₃⁺Cl⁻ [1]; secondary amide [1] (c) C2: H, CH₃, C₃H₇ and CONHC₃H₇ [1] Examiner insight: With two moles of amine, the second must appear as the alkylammonium chloride, not as unused amine beside HCl.

8. (a) They are optical isomers, non-superimposable mirror images [1]; drawing 2 is drawing 1 with two groups (NH₂ and C₅H₁₁) exchanged, and one swap converts a chiral centre into its mirror-image form [1] (b) They are the same molecule: drawing 3 is drawing 1 turned through 180° in the plane of the page [1] Examiner insight: Try a rotation before deciding; say how one drawing turns into the other.

9. (a) RCH(NH₂)COOH + NaOH → RCH(NH₂)COO⁻Na⁺ + H₂O [1] (b) Na replaces H, so each mole gains 22.0 g; gain = 3.34 − 2.90 = 0.44 g [1]; n(X) = 0.44 ÷ 22.0 = 0.0200 mol [1]; M = 2.90 ÷ 0.0200 = 145 g mol⁻¹ [1] (c) CH(NH₂)COOH accounts for 74.0, so R = 145 − 74 = 71 = C₅H₁₁ [1] (d) For example –CH(CH₃)CH₂CH₂CH₃; also acceptable: –CH₂CH(CH₃)CH₂CH₃ or –CH(CH₃)CH(CH₃)₂ [1] Examiner insight: In (d), a straight pentyl chain gives only one chiral centre; R must contain a carbon with four different groups.

10. (a) Secondary [1]: the amide N is bonded to two carbons [1] (b) 2 [1]: the CH carrying NH₂ and CH₂C₆H₅, and the CH carrying CH₃ and COOCH₃ [1] (c) Cl⁻ ⁺H₃NCH(CH₂C₆H₅)CONHCH(CH₃)COOCH₃ [1] (d) Methanol with concentrated H₂SO₄ catalyst, heated [1]; U + CH₃OH ⇌ V + H₂O, with U and V written out in full [1] Examiner insight: The benzyl CH₂ and the ring carbons are not chiral; identify each chiral carbon by the groups on it.

11. (a) M = 14.0 ÷ 0.107 = 130.8, so 131 g mol⁻¹ [1] (b) R = 131 − 74 = 57, so C₄H₉ [1] (c) CH₃CH₂CH₂CH₂CH(NH₂)COOH and CH₃CH₂CH(CH₃)CH(NH₂)COOH [1]; (CH₃)₂CHCH₂CH(NH₂)COOH and (CH₃)₃CCH(NH₂)COOH [1] (d) CH₃CH₂C*H(CH₃)C*H(NH₂)COOH [1]; asterisks on the α-carbon and on the side-chain CH carrying CH₃ and C₂H₅ [1] (e) CH₃CH₂CH(CH₃)CH(NH₂)COOH + NaOH → CH₃CH₂CH(CH₃)CH(NH₂)COO⁻Na⁺ + H₂O [1] (f) Each α-carbon carries four different groups: H, NH₂, COOH and C₄H₉ [1] Examiner insight: Show the subtraction of 74 for CH(NH₂)COOH; a bare C₄H₉ with no working gives nothing to credit if it is wrong.

Where marks are usually lost

• Writing RCH(NH₃)COOH as the general formula (question 1).
• Calling compound D an amide (question 2).
• Showing HCl at the COOH group or NaOH at the NH₂ group (questions 3 and 9).
• Omitting “concentrated” for the H₂SO₄ catalyst (question 4).
• Finding percentage yield from masses, not moles (question 4).
• Omitting “non-superimposable” from the definition (question 5).
• Marking a carbon with two identical groups as chiral (questions 6 and 9).
• Missing a side-chain chiral centre (questions 9 to 11).

Next steps

• Recap with the Amino acids, amides and chirality revision notes
• Full explanations: Amino acids, amides and chirality study guide
• Related practice: Amines practice and Carboxylic acids and esters practice
• Find gaps across H432 with all free 10-minute diagnostics
• Learn with a tutor: Book a free trial class
• Every OCR chemistry topic: the H432 topic hub
• Tick off outcomes: H432 printable checklist

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.