Nitriles and hydroxynitriles

There are four carbon atoms including the nitrile carbon, so the stem is butane and the name is butanenitrile. It is made from 1-bromopropane and KCN, which is why it has one more carbon than the halogenoalkane. Propanenitrile counts only the three carbons in the propyl chain, the classic mistake. Propyl cyanide nitrile repeats the functional group and is not a systematic name. Butylamine is CH₃CH₂CH₂CH₂NH₂, the product of reducing butanenitrile.

CH₃CH₂Br + KCN → CH₃CH₂CN + KBr. The CN⁻ ion is the nucleophile in a substitution, and heating under reflux in ethanol gives a reasonable rate. Ethanol is the solvent because in water, OH⁻ ions would also substitute, giving ethanol as a by-product, and the reaction is slow at room temperature. HCN with a trace of KCN is the reagent for addition to aldehydes and ketones, not substitution of halogenoalkanes. Excess ammonia gives ethylamine.

The C=O bond is polar, so the carbon is δ+. The lone pair on the carbon of CN⁻ attacks it, the π bond electrons move onto oxygen, and an O⁻ intermediate forms. In step 2 this takes H⁺ from HCN, giving CH₃CH(OH)CN and regenerating CN⁻, so KCN is a catalyst. The first attack is by the nucleophile, not by H⁺, so this is not electrophilic addition; HCN is a weak acid and gives few H⁺ ions, which is why KCN is added. Nothing is replaced: the H stays on the carbon and the C=O becomes C-O, so it is addition, not substitution. A nucleophile attacks the δ+ carbon, never the δ− oxygen.

CH₃CH₂CN + 2H₂O + HCl → CH₃CH₂COOH + NH₄Cl. The nitrile carbon becomes the carboxyl carbon, so the three-carbon chain is kept, and the nitrogen leaves as NH₄⁺, which forms ammonium chloride with the acid. Propylamine is the product of reduction (H₂/Ni or LiAlH₄), not hydrolysis. Dilute HCl is not oxidised to chlorine. The C-C bond to the CN group is not broken, so no carbon is lost and HCN is not released.

CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂. Four hydrogen atoms add across C≡N, turning it into CH₂NH₂, so the amine keeps all three carbons. The same reduction can be done with H₂ and a nickel catalyst. Ethylamine has lost a carbon atom, which reduction does not do. Propanoic acid comes from hydrolysis with dilute acid. Propan-1-ol has no nitrogen; reduction does not remove the N atom.

Step 1 is nucleophilic substitution: CH₃CH₂CH₂Br + KCN → CH₃CH₂CH₂CN + KBr, giving butanenitrile (four carbons, counting the nitrile C). Step 2 is acid hydrolysis under reflux: the C≡N carbon becomes COOH, giving butanoic acid. Propanenitrile would be CH₃CH₂CN, which has only three carbons. LiAlH₄ reduces a nitrile to an amine, butylamine, not an acid. NaOH(aq) then dichromate gives propanoic acid via propan-1-ol: it keeps three carbons, so it cannot make butanoic acid.

n(ethanal) = 2.20 / 44.0 = 0.0500 mol, so the theoretical mass = 0.0500 × 71.0 = 3.55 g. Yield = 2.84 / 3.55 × 100 = 80.0%. 129% divides the mass of product by the mass of ethanal, ignoring the different M_r values; a yield over 100% is impossible. 62.0% divides the mass of ethanal by the theoretical mass of product. 77.5% divides the mass of ethanal by the actual mass of product.

Butanone gives CH₃C(OH)(CN)CH₂CH₃: the central carbon carries OH, CN, CH₃ and CH₃CH₂, four different groups, so it is chiral. Because the C=O is planar and attacked equally from both faces, a racemic mixture forms, which shows no optical activity overall. Propanone gives (CH₃)₂C(OH)CN, with two identical CH₃ groups. Methanal gives HOCH₂CN, with two H atoms on that carbon. Pentan-3-one gives a carbon with two identical ethyl groups. None of these three has a chiral centre.