Part 2: Nucleophilic addition, hydroxynitriles and stereochemistry

Track every electron pair and carbon atom through hydride and cyanide addition, then explain when a new chiral centre forms.

Build the hydride mechanism in two stages

Represent NaBH₄ as supplying H⁻ for the mechanism required here. H⁻ is a nucleophile because it donates an electron pair to the δ+ carbonyl carbon. Draw one full-headed curly arrow from H⁻ to that carbon and another from the C=O π bond to oxygen. Both movements are needed: carbon cannot keep its double bond and acquire a fifth bond.

The intermediate is an alkoxide, with a new C–H bond and a full negative charge on O. It is no longer a carbonyl with δ− oxygen. Show the carbon’s four bonds and all the original substituents.

Next an oxygen lone pair takes a proton from water, with an arrow from the O–H bond of that water to its oxygen. The alcohol and OH⁻ form. If the question supplies a separate acidic work-up, use its stated proton source. Do not let an unexplained H atom appear on oxygen.

Cyanide forms a new carbon–carbon bond

A carbonyl compound reacts with HCN in the presence of cyanide ions to form a hydroxynitrile. The nucleophile is CN⁻, attacking through its carbon atom. A curly arrow begins at the carbon lone pair of CN⁻ and ends at the carbonyl carbon; the C=O π pair moves onto oxygen.

The intermediate contains C–CN and O⁻ attached to the original carbonyl carbon. Protonation gives OH. The carbon–nitrogen triple bond is retained throughout: CN is attached through carbon, not through nitrogen. Use the proton source supplied, such as HCN or H⁺, in the second step; proton transfer from HCN regenerates CN⁻.

HCN/cyanide chemistry involves severe toxicity and belongs in appropriately controlled specialist laboratory conditions. For exam synthesis, identify the specified reagents and mechanism; this lesson is not an instruction to prepare these reagents outside that setting.

Worked example: propanal gains one carbon

Start with CH₃CH₂CHO: the carbonyl carbon is attached to H and an ethyl group. Addition of HCN gives CH₃CH₂CH(OH)CN. Count four carbons, including the nitrile carbon; forgetting that carbon is a common error when planning an extended chain.

For nitrile nomenclature, the nitrile carbon is carbon 1. This product is 2-hydroxybutanenitrile. Its OH-bearing carbon has four different groups: H, OH, CN and CH₂CH₃, so it is chiral.

Propanone instead gives (CH₃)₂C(OH)CN. It also gains one carbon, but the OH-bearing carbon has two identical CH₃ groups and is not a chiral centre. Forming an alcohol group does not automatically produce optical isomerism.

Why a planar starting group can give two enantiomers

The carbonyl carbon is trigonal planar. In an achiral environment a nucleophile can approach either face. If the product carbon acquires four different groups, the two approaches give mirror-image arrangements and a racemic mixture is expected under the simple model.

Explain the sequence: planar C=O → attack from either face → tetrahedral product → check four different groups. Simply writing “two products because it is flat” omits the structural condition for chirality. Existing chiral centres or a chiral reaction environment can alter this simple prediction; apply supplied information.

When drawing the mechanism, check atom conservation, full charges, arrow origins and valency before considering stereochemistry. A correct final formula cannot repair an arrow showing electrons moving from an electrophile towards a nucleophile.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources