OCR A Level Chemistry A: Carbon–carbon bond formation (H432)

This study guide teaches section 6.2.4, Carbon–carbon bond formation, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. You will find it in Module 6, Organic chemistry and analysis. Its four lettered outcomes, 6.2.4(a) to (d), each get an explanation and a worked example here. Because H432 is the full A level, this content carries no AS or A2 label. The specification assigns Module 6 to Synthesis and analytical techniques (02); Unified chemistry (03) draws on all six modules, this one included. This section has no PAG link.

The H432 course hub leads to every other section; to log which 6.2.4 outcomes you have secured, use the outcome-by-outcome checklist.

Outcomes in 6.2.4

Outcome	What you must be able to do	Cross-references in the specification
6.2.4(a)	Explain why forming a C–C bond is used in synthesis to make a carbon chain longer	See also 6.1.1 d and 6.1.2 b
6.2.4(b)(i)	Form C–C≡N from a haloalkane with CN⁻ and ethanol, including the nucleophilic substitution mechanism	See also 4.2.2 c
6.2.4(b)(ii)	Form C–C≡N from a carbonyl compound with HCN, including the nucleophilic addition mechanism	See also 6.1.2 b–c
6.2.4(c)(i)	Reduce the nitriles from (b), e.g. with H₂/Ni, to amines	—
6.2.4(c)(ii)	Hydrolyse the nitriles from (b) with acid to carboxylic acids	—
6.2.4(d)	Form a C–C bond to a benzene ring by alkylation (haloalkane) or acylation (acyl chloride) with a halogen carrier: the Friedel–Crafts reaction	See also 6.1.1 d

The additional guidance column is blank for this section.

6.2.4(a): Why make C–C bonds?

Oxidising an alcohol, hydrolysing a haloalkane or reducing a carbonyl compound all keep the same carbon skeleton. If a target has more carbon atoms than your starting material, you need a step that joins carbon to carbon.

This section gives you three such steps:

• a haloalkane plus cyanide ions adds one carbon (as –C≡N) in place of the halogen
• an aldehyde or ketone plus HCN adds one carbon (as –C≡N) to the carbonyl carbon
• benzene plus a haloalkane or acyl chloride, with a halogen carrier, joins a whole carbon chain to the ring

The nitrile can then be reduced to a primary amine (–CH₂NH₂) or hydrolysed to a carboxylic acid (–COOH). Either way the nitrile carbon stays in the chain.

Counting rule. Count the carbons in the target and the starting material. One extra carbon at a chain end usually means a nitrile step; a chain joined directly to a benzene ring usually means a Friedel–Crafts step.

6.2.4(b)(i): Haloalkanes with cyanide ions

Reagent and conditions: sodium or potassium cyanide dissolved in ethanol, heated under reflux.

General equation: R–X + CN⁻ → R–CN + X⁻ (X = Cl, Br or I)

The cyanide ion is a nucleophile (an electron pair donor); the lone pair on its carbon atom forms the new C–C bond.

Naming nitriles. The carbon of the –C≡N group counts as carbon 1 of the chain. So a nitrile always has one more carbon in its stem name than the haloalkane it came from. Bromoethane gives propanenitrile, CH₃CH₂CN.

The nucleophilic substitution mechanism

The pattern is the one you learned for hydrolysis of a primary haloalkane with hydroxide ions (4.2.2 c), with CN⁻ as the nucleophile.

1. Show the C–X bond as polar: Cδ+ and Xδ−, because the halogen is more electronegative.
2. Start a curly arrow at the carbon lone pair of :CN⁻ and end it on the δ+ carbon.
3. Draw a curly arrow from the middle of the C–X bond to the halogen atom. The bond breaks heterolytically.
4. Show the products: the nitrile and a halide ion, X⁻.

Worked example 1: a nitrile and its yield

8.25 g of 1-bromo-3-methylpentane, CH₃CH₂CH(CH₃)CH₂CH₂Br, is heated under reflux with excess KCN in ethanol. 4.12 g of nitrile is collected. Name the product and calculate the percentage yield. Use Mᵣ values C₆H₁₃Br = 164.9 and C₇H₁₃N = 111.0.

1. Product: CH₃CH₂CH(CH₃)CH₂CH₂CN. Number from the nitrile carbon: C1 is CN, C2 and C3 are CH₂, C4 carries the methyl. Name: 4-methylhexanenitrile.
2. Equation: CH₃CH₂CH(CH₃)CH₂CH₂Br + KCN → CH₃CH₂CH(CH₃)CH₂CH₂CN + KBr
3. Amount of haloalkane = 8.25 ÷ 164.9 = 0.0500 mol. The ratio is 1 : 1.
4. Theoretical mass of nitrile = 0.0500 × 111.0 = 5.55 g
5. Percentage yield = 4.12 ÷ 5.55 × 100 = 74.2%

The methyl branch moves from C3 in the haloalkane to C4 in the nitrile, because the chain gained a carbon at the end.

6.2.4(b)(ii): Carbonyl compounds with HCN

Reagent: HCN, supplied in practice as NaCN(aq) with H⁺(aq). The product is a hydroxynitrile: the former carbonyl carbon now carries both –OH and –C≡N.

General equation: R₂C=O + HCN → R₂C(OH)CN

This works for aldehydes and ketones; again the chain gains one carbon.

The nucleophilic addition mechanism

1. Show the C=O bond as polar: Cδ+ and Oδ−.
2. Start an arrow at the cyanide’s carbon lone pair; end it on the carbonyl carbon.
3. Curly arrow from the C=O double bond onto the oxygen. This gives an intermediate with O⁻.
4. The O⁻ is then protonated. The specification allows either H₂O (arrow from an O⁻ lone pair to an H of water, with an arrow from the O–H bond onto the oxygen, releasing OH⁻) or H⁺ (one arrow from an O⁻ lone pair to H⁺).

Worked example 2: addition to a branched ketone

3-Methylpentan-2-one, CH₃COCH(CH₃)CH₂CH₃, reacts with NaCN(aq)/H⁺(aq). Give the product and describe the mechanism.

1. The C=O carbon is C2 of the ketone. CN adds to it and H adds to the oxygen.
2. Product: CH₃C(OH)(CN)CH(CH₃)CH₂CH₃, molecular formula C₇H₁₃NO.
3. Name it from the nitrile carbon: C1 is CN, C2 carries OH and a methyl, C3 carries a methyl, C4 is CH₂ and C5 is the end CH₃. Name: 2-hydroxy-2,3-dimethylpentanenitrile.
4. Mechanism: :CN⁻ carbon lone pair attacks the δ+ carbonyl carbon; the π pair moves onto O; the intermediate is CH₃C(O⁻)(CN)CH(CH₃)CH₂CH₃; the O⁻ takes H⁺ from water or from H⁺(aq).

6.2.4(c): Taking the nitrile further

(c)(i) Reduction to amines

Reagent and conditions: hydrogen gas with a nickel catalyst.

R–C≡N + 2H₂ → R–CH₂NH₂

Two H₂ molecules add across the triple bond. The nitrile carbon becomes the CH₂ next to NH₂, and the amine is numbered from that carbon.

(c)(ii) Acid hydrolysis to carboxylic acids

Reagent and conditions: heat under reflux with a dilute acid, such as dilute hydrochloric acid.

R–C≡N + 2H₂O + H⁺ → R–COOH + NH₄⁺

With hydrochloric acid, write it as R–CN + 2H₂O + HCl → R–COOH + NH₄Cl. The nitrogen leaves as ammonium ions, and the nitrile carbon becomes the –COOH carbon.

Worked example 3: two products from one nitrile

Take the 4-methylhexanenitrile from worked example 1.

(a) Reduction. 3.33 g of the nitrile (Mᵣ 111.0) is reduced completely with H₂/Ni. Give the product and the volume of hydrogen used at room temperature and pressure, where 1 mol of gas occupies 24.0 dm³.

1. CH₃CH₂CH(CH₃)CH₂CH₂CN + 2H₂ → CH₃CH₂CH(CH₃)CH₂CH₂CH₂NH₂
2. Name: 4-methylhexan-1-amine, C₇H₁₇N.
3. Amount of nitrile = 3.33 ÷ 111.0 = 0.0300 mol
4. Amount of H₂ = 2 × 0.0300 = 0.0600 mol
5. Volume = 0.0600 × 24.0 = 1.44 dm³

(b) Hydrolysis. Give the equation for hydrolysis with dilute hydrochloric acid.

CH₃CH₂CH(CH₃)CH₂CH₂CN + 2H₂O + HCl → CH₃CH₂CH(CH₃)CH₂CH₂COOH + NH₄Cl

The product is 4-methylhexanoic acid, C₇H₁₄O₂: a five-carbon chain has become a six-carbon chain.

6.2.4(d): Friedel–Crafts reactions

A benzene ring is too stable to react with a haloalkane or acyl chloride alone. A halogen carrier, such as AlCl₃, FeCl₃ or FeBr₃, generates a strong electrophile. Keep the conditions dry, because aluminium chloride reacts with water.

Alkylation uses a haloalkane and adds an alkyl group:

C₆H₆ + R–Cl → C₆H₅–R + HCl

Electrophile formation: R–Cl + AlCl₃ → R⁺ + AlCl₄⁻

Acylation uses an acyl chloride and adds an acyl group, giving an aromatic ketone:

C₆H₆ + R–COCl → C₆H₅–COR + HCl

Electrophile formation: R–COCl + AlCl₃ → RCO⁺ + AlCl₄⁻

The carrier is regenerated: H⁺ + AlCl₄⁻ → AlCl₃ + HCl. The electrophile attacks the ring as in the halogenation mechanism from 6.1.1: π electrons bond to the electrophile, then the ring loses H⁺. Outcome 6.1.1(g) expects you to predict mechanisms like this for unfamiliar electrophiles.

Worked example 4: alkylation and acylation

(a) Benzene reacts with chlorocyclopentane and AlCl₃. Write the equation and name the product.

C₆H₆ + C₅H₉Cl → C₆H₅–C₅H₉ + HCl. Product: cyclopentylbenzene, C₁₁H₁₄. The electrophile is the cyclopentyl cation, C₅H₉⁺.

(b) 5.46 g of benzene reacts with excess 2-ethylbutanoyl chloride, (CH₃CH₂)₂CHCOCl, and AlCl₃. 8.87 g of ketone is obtained. Find the percentage yield. Mᵣ: benzene 78.0, C₁₂H₁₆O 176.0.

1. C₆H₆ + (CH₃CH₂)₂CHCOCl → C₆H₅COCH(CH₂CH₃)₂ + HCl
2. Product name: 2-ethyl-1-phenylbutan-1-one.
3. Amount of benzene = 5.46 ÷ 78.0 = 0.0700 mol
4. Theoretical mass = 0.0700 × 176.0 = 12.32 g
5. Percentage yield = 8.87 ÷ 12.32 × 100 = 72.0%

Putting it together: a route that adds a carbon

Worked example 5: planning and overall yield

Plan a route from pentan-2-ol to 2-hydroxy-2-methylpentanoic acid, CH₃CH₂CH₂C(OH)(CH₃)COOH. The steps have yields of 85%, 70% and 62%.

1. Pentan-2-ol has five carbons and the target six, so one C–C bond is needed.
2. The target has OH and COOH on the same carbon, which points to a hydroxynitrile made from a ketone.
3. Step 1: oxidise pentan-2-ol to pentan-2-one with acidified dichromate under reflux.
4. Step 2: add HCN (NaCN(aq)/H⁺(aq)) to give 2-hydroxy-2-methylpentanenitrile.
5. Step 3: heat under reflux with dilute HCl to hydrolyse the nitrile to the acid.
6. Overall yield = 0.85 × 0.70 × 0.62 × 100 = 36.9%

To make 5.00 g of the acid (Mᵣ 132.0): 5.00 ÷ 132.0 = 0.03788 mol, so 0.03788 ÷ 0.3689 = 0.1027 mol of pentan-2-ol (Mᵣ 88.0), which is 9.04 g.

Common errors

• Starting the cyanide curly arrow at nitrogen or at the minus sign. It starts at the lone pair on carbon.
• Using aqueous KCN for the haloalkane reaction. The specification pairs CN⁻ with ethanol.
• Naming the nitrile with the haloalkane’s carbon count. The nitrile carbon is carbon 1, so the stem gains one carbon.
• Writing RCN + H₂ → RCH₂NH₂. Two H₂ are needed for the triple bond.
• Leaving out NH₄⁺ or NH₄Cl from the hydrolysis equation, or showing NH₃ as the product under acid conditions.
• Writing Friedel–Crafts equations without the HCl product, or forgetting to regenerate the AlCl₃.
• Calling the acylation product an aldehyde. It is a ketone.

Next steps

• Condense this page: Carbon–carbon bond formation revision notes
• Test yourself: Carbon–carbon bond formation practice questions
• Related: OCR carbonyl compounds study guide
• Check the whole course: all free 10-minute diagnostics

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.2.4 Carbon–carbon bond formation, learning outcomes (a) to (d). Published by OCR.