Topic 19 · page 13 of 13

Routes and exam technique

Put the reactions together into synthesis routes, learn what you see at each step, and learn the exact wording that reagent, condition and observation questions are marked on.

15 min read19.2(iv)19.2(v)19.4(i)19.4(ii)19.519.619.7

Read first: Phenylamine and azo dyes, Peptide bonds

Every reagent and condition in one table 19.4(i)19.4(ii)19.519.619.7

To make	From	Reagent	Conditions
primary aliphatic amine	halogenoalkane	excess concentrated NH₃ in ethanol	heat in a sealed tube
primary aliphatic amine	nitrile	LiAlH₄	dry ether
nitrile (one more C)	halogenoalkane	KCN in ethanol	heat under reflux
phenylamine	nitrobenzene	Sn and concentrated HCl; then NaOH(aq)	heat under reflux
benzenediazonium ion	phenylamine	NaNO₂ and HCl (HNO₂)	below 10 °C
azo dye	diazonium ion	phenol in NaOH(aq)	cold
amide	acyl chloride	concentrated NH₃	room temperature
N-substituted amide	acyl chloride	primary amine	room temperature

A halogenoalkane gives an amine with the same number of carbons. Going through the nitrile adds one carbon, because the cyanide group brings its own carbon atom.

Butylamine at the centre. In: a halogenoalkane with excess ammonia in ethanol, or a nitrile with LiAlH4 in dry ether. Out: a salt with hydrochloric acid, a secondary amine with a halogenoalkane, an N-substituted amide with ethanoyl chloride, and a deep blue complex with copper(II) ions. — Aliphatic routes: two ways into butylamine, four ways out.

Multi-step synthesis 19.2(iv)19.519.619.7

Write each step on its own line: reagent, condition, product. Name every intermediate.

Worked exampleBenzene to N-phenylethanamide

Outline a three-step synthesis of N-phenylethanamide from benzene.

Nitration: concentrated nitric acid and concentrated sulfuric acid, about 50 °C, giving nitrobenzene.
Reduction: tin and concentrated hydrochloric acid, heat under reflux, then add excess sodium hydroxide solution, giving phenylamine.
Acylation: ethanoyl chloride at room temperature, giving N-phenylethanamide, CH₃CONHC₆H₅, and HCl.

Worked example1-bromobutane to N-butylethanamide

Heat 1-bromobutane in a sealed tube with a large excess of concentrated ammonia in ethanol, giving butylamine.
React butylamine with ethanoyl chloride at room temperature, giving N-butylethanamide and HCl.

Worked exampleBenzene to the azo dye

Nitration, as above, gives nitrobenzene.
Sn and concentrated HCl under reflux, then NaOH(aq), gives phenylamine.
NaNO₂ and HCl below 10 °C gives the benzenediazonium ion.
Add the cold diazonium ion to phenol in NaOH(aq): a yellow-orange precipitate of 4-hydroxyazobenzene forms.

The aromatic route: nitrobenzene is reduced by tin and concentrated hydrochloric acid under reflux, then sodium hydroxide, to give phenylamine. Phenylamine with sodium nitrite and hydrochloric acid below 10 degrees Celsius gives the benzenediazonium ion. The diazonium ion couples with phenol in sodium hydroxide, cold, to give the yellow-orange azo dye 4-hydroxyazobenzene. — The aromatic route in one picture.

What you see 19.2(v)19.619.7

Test or reaction	Observation
butylamine solution	alkaline: pH about 11 to 12
phenylamine and water	oily, only slightly soluble; very slightly alkaline
phenylamine and hydrochloric acid	dissolves: soluble salt forms
butylamine, a little, with copper(II) ions	pale blue precipitate
butylamine, excess, with copper(II) ions	precipitate dissolves: deep blue solution
ethanoyl chloride and ammonia or an amine	vigorous reaction, misty fumes or white smoke
diazonium ion and phenol in NaOH(aq)	yellow-orange precipitate
ninhydrin on an amino acid	purple spot
chiral amino acid in a polarimeter	the plane of polarisation is rotated

Exam technique

Command word	What to write
state	the answer only: a reagent, a formula or a colour
explain	a reason, with the structure → lone pair → proton chain for base strength
describe	steps in order, with reagents, conditions and observations
outline a synthesis	a numbered route with reagent, condition and product at each step
draw	check that the formula type is the one asked for: structural, displayed or skeletal

Common mix-ups, side by side:

Wrong	Right
amine and amide used as the same word	amine: –NH₂ on a carbon chain; amide: –CO–NH₂
Cl₂ as the by-product of acylation	HCl
tin as a catalyst	tin as the reducing agent
dilute HCl for the reduction	concentrated HCl
diazotisation at room temperature	below 10 °C
N-butylethanamide drawn as butyl ethanoate	an amide, with the butyl group on the nitrogen
repeat unit with the side group in the backbone	two chain carbons in brackets, side group hanging off

Check yourself

Give the reagent and conditions for each step in turning 1-bromopropane into butylamine.
Outline the synthesis of N-methylethanamide from methylamine.
Give the three steps, with reagents and conditions, that convert nitrobenzene into an azo dye with phenol.
A student writes the product of ethanoyl chloride with ammonia as ethanoic acid. What is wrong?
State what you would see when excess butylamine is added to copper(II) sulfate solution, a little at a time.
Explain why LiAlH₄ must be used in dry ether.

Answers cover these until you have tried the questions

Step 1: heat under reflux with potassium cyanide in ethanol, giving butanenitrile. Step 2: reduce with LiAlH₄ in dry ether, giving butylamine.
React methylamine with ethanoyl chloride at room temperature: $\ce{CH3COCl + CH3NH2 -> CH3CONHCH3 + HCl}$ . The product is N-methylethanamide.
Step 1: tin and concentrated HCl under reflux, then NaOH(aq), giving phenylamine. Step 2: NaNO₂ and HCl below 10 °C, giving the benzenediazonium ion. Step 3: add the cold solution to phenol in NaOH(aq), giving the yellow-orange azo dye.
Ethanoic acid is the wrong product. The acyl chloride is substituted by NH₂, so the product is the amide, ethanamide, CH₃CONH₂, with HCl (or NH₄Cl with excess ammonia).
A pale blue precipitate forms first, then dissolves in excess to give a deep blue solution.
LiAlH₄ reacts with water, so the ether must be dry or the reagent is destroyed.