Part 5: Halogenoalkanes: substitution and elimination

Classify a halogenoalkane, choose a reagent for each product and follow lone pairs through the required hydroxide and ammonia mechanisms.

Classify the carbon bearing the halogen

In a halogenoalkane a halogen is bonded to a saturated carbon. A primary halogenoalkane has one carbon group attached to the carbon bearing X; a secondary has two; a tertiary has three. Count carbon neighbours of that carbon, not the total number of carbons or halogen atoms in the molecule.

CH₃CH₂CH₂Br is primary, CH₃CHBrCH₃ is secondary and (CH₃)₃CBr is tertiary. CH₃CH(CH₃)CH₂Br is still primary even though the next carbon is branched. Halomethanes are a separate simple case with no carbon neighbour, so do not try to call them tertiary because they have three hydrogen atoms.

The halogen is more electronegative than carbon, making Cδ+–Xδ−. This leaves the carbon susceptible to a nucleophile: an electron-pair donor. OH⁻, CN⁻, H₂O and NH₃ have available lone pairs. A nucleophile need not have a negative charge; ammonia and water are neutral.

One starting material can give five different products

Conditions are part of the answer. Aqueous hydroxide favours substitution to an alcohol; hot ethanolic hydroxide favours elimination to an alkene. These pathways can compete, so the named condition describes the favoured transformation rather than a promise of a single quantitative product.

The cyanide carbon joins the organic chain: bromoethane has two carbons, but propanenitrile has three. By contrast hydroxide and ammonia substitution retain the carbon count. Potassium cyanide is highly toxic and requires specialist controlled practical arrangements; learn this as a reagent transformation, not as an unsupervised experiment. A sealed ammonia reaction needs purpose-designed pressure equipment, not a stoppered heated test tube.

Primary substitution: bond making and breaking together

For bromoethane reacting with OH⁻, draw a lone pair on oxygen and a curly arrow from that pair to the carbon bonded to Br. Draw another arrow from the C–Br bond to Br. The C–O bond forms as C–Br breaks, giving ethanol and Br⁻. Both arrows belong in the same step for the required primary-halogenoalkane mechanism; do not insert a stable primary carbocation.

Charge check: OH⁻ contributes −1 overall initially; Br⁻ carries −1 finally, while the organic product is neutral. Oxygen begins with one O–H bond and gains one O–C bond. The electron pair in C–Br goes to Br, not to carbon. Incorrect arrow direction would reverse the physical account of electron movement.

Ammonia gives a charged intermediate before the amine

First, the nitrogen lone pair in NH₃ attacks the carbon bearing Br while the C–Br pair moves to Br. The intermediate is CH₃CH₂NH₃⁺, paired with Br⁻. Nitrogen now makes four bonds and is positively charged. It is not the neutral primary amine yet.

Second, another NH₃ molecule accepts a proton from the intermediate. Draw a lone-pair arrow from that ammonia nitrogen to an H on –NH₃⁺, and an arrow from the original N–H bond back to the original nitrogen. Products are CH₃CH₂NH₂ and NH₄⁺; together with Br⁻ the inorganic salt is NH₄Br. Two ammonia molecules are used in the overall equation.

The amine product still has a nitrogen lone pair and can react with more halogenoalkane. A large excess of ammonia increases the chance that the halogenoalkane encounters NH₃ and reduces further alkylation. It does not mean the amine loses its nucleophilic character.

Hydroxide can act as a base instead

In elimination, hydroxide accepts an H⁺ from a carbon adjacent to the carbon bearing X. The electrons left by C–H form C=C while X leaves with the C–X pair. Overall the organic molecule loses H and X, producing an alkene; the base becomes water. A β hydrogen on a neighbouring carbon is required.

Worked example: 2-bromobutane with hot ethanolic KOH can eliminate from either neighbouring carbon. Loss of H from C1 gives but-1-ene; loss from C3 gives but-2-ene. But-2-ene itself has E and Z stereoisomers. Keep four carbons in every product; changing solvent does not remove a carbon atom.

The required mechanisms in 6.36 are hydroxide and ammonia substitution on primary halogenoalkanes. The electron-flow description of elimination helps explain the products, but the central AS obligation is recognising OH⁻ as a base, the ethanolic conditions and the resulting C=C.

Plan the route by checking both group and carbon count

To convert 1-bromopropane to propan-1-ol, select aqueous KOH and reflux. To obtain butanenitrile from the same starting material, use KCN in ethanol: the extra carbon comes from CN⁻. To obtain propene, choose hot ethanolic KOH. A reagent name without the solvent leaves the intended hydroxide pathway ambiguous.

Worked yield: 10.9 g bromoethane, M = 109.0 g mol⁻¹, reacts with excess KCN. Starting amount = 10.9/109.0 = 0.100 mol. Propanenitrile, C₃H₅N, has M = 55.0 g mol⁻¹ and forms 1:1, so theoretical mass = 5.50 g. An isolated mass of 4.40 g gives 4.40/5.50 × 100 = 80.0%. This is yield, not atom economy: KBr is also formed.

Pearson’s June 2023 AS Q4(d) separated reagents from solvent conditions and accepted specific causes of low yield, including side reactions, incomplete conversion and handling losses. Explain a named loss such as volatile product escaping during transfer; “human error” does not identify what happened or its effect.

Quick checks

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

Sources