Preparing Amines, and Their Reactions and Uses – Amines and Amino Acids (Senior 5 Chemistry) | Ganzaa

Amines and Amino Acids

Lessons

Senior 5 · Chemistry

Amines and Amino Acids

4 lessons

Lesson 2 of 4

Preparing Amines, and Their Reactions and Uses

25 min

One lone pair, many reactions

Hold a glass rod dipped in concentrated hydrochloric acid over an open bottle of aqueous methanamine and a white smoke forms, exactly as it does over ammonia. The smoke is a salt, methylammonium chloride. Behind the smoke, and behind almost every reaction of an amine, is the lone pair on nitrogen. A lone pair that accepts a proton makes the amine a base; a lone pair that attacks an electron-poor carbon makes it a nucleophile. This lesson shows how amines are made, how that lone pair reacts, and why dye factories and medicine cabinets are both full of amines.

Definition

Brønsted–Lowry base: a proton ( $\mathrm{H^{+}}$ ) acceptor. Nucleophile: a species that donates a lone pair to an electron-deficient carbon atom, forming a new covalent bond.

Acylation: replacing an H atom on N (or O) by an acyl group, $\mathrm{RCO-}$ . An amine is acylated by an acyl chloride or acid anhydride to give an N-substituted amide, $\mathrm{RCONHR'}$ .

Diazonium ion: an ion containing the group $\mathrm{-N^{+}{\equiv}N}$ , e.g. benzenediazonium, $\mathrm{C_6H_5N_2^{+}}$ . Azo compound: a compound with the group $\mathrm{-N{=}N-}$ joining two aryl rings; many are coloured dyes.

Preparing amines

Figure 9.2: Five laboratory routes to amines and their conditions.

1. From haloalkanes and ammonia. The lone pair of $\mathrm{NH_3}$ attacks the $\mathrm{C^{\delta+}}$ of the C–X bond and the halide leaves (nucleophilic substitution); a second ammonia molecule removes $\mathrm{H^{+}}$ :

$\mathrm{CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br}$

The product is itself a nucleophile, often a better one than ammonia, so it attacks more haloalkane, giving the 2° and 3° amines and finally a quaternary ammonium salt, $\mathrm{(CH_3CH_2)_4N^{+}Br^{-}}$ . A large excess of ammonia makes it more likely that each haloalkane molecule meets ammonia rather than product, so the primary amine dominates; the mixture still has to be separated.

2. Reducing nitriles. $\mathrm{RC{\equiv}N + 4[H] \rightarrow RCH_2NH_2}$ , with lithium tetrahydridoaluminate ( $\mathrm{LiAlH_4}$ ) in dry ether, or hydrogen over a nickel catalyst. This route gives only the primary amine. Because the nitrile is made from a haloalkane and potassium cyanide, the route also lengthens the carbon chain by one atom.

3. Reducing amides. $\mathrm{RCONH_2 + 4[H] \rightarrow RCH_2NH_2 + H_2O}$ with $\mathrm{LiAlH_4}$ : the C=O becomes $\mathrm{CH_2}$ and the carbon count stays the same.

4. Hofmann degradation. Warming a primary amide with bromine and aqueous sodium hydroxide removes the carbonyl carbon as carbonate:

$\mathrm{RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$

This is the way to shorten a chain by one carbon: ethanamide gives methanamine.

5. Reducing nitrobenzene. Tin and concentrated hydrochloric acid, heated under reflux, reduce the nitro group; the product leaves the flask as the salt $\mathrm{C_6H_5NH_3^{+}Cl^{-}}$ , and sodium hydroxide then frees the amine, which is separated by steam distillation:

$\mathrm{C_6H_5NO_2 + 6[H] \rightarrow C_6H_5NH_2 + 2H_2O}$

Worked example

Problem. 12.3 g of nitrobenzene is reduced, and 6.98 g of phenylamine is collected. Calculate the percentage yield. ( $M$ : $\mathrm{C_6H_5NO_2}$ = 123.11, $\mathrm{C_6H_5NH_2}$ = 93.13 g mol⁻¹)

Solution. One mole of nitrobenzene gives one mole of phenylamine.

$n(\mathrm{C_6H_5NO_2}) = \frac{12.3}{123.11} = 0.09991\,\mathrm{mol} \qquad m_{\text{theoretical}} = 0.09991 \times 93.13 = 9.30\,\mathrm{g}$

$\text{yield} = \frac{6.98}{9.30} \times 100 = 75.0\,\%$

Answer: 75.0 %. Losses come from incomplete reduction and from the steam distillation and drying.

Amines as bases

In water an amine accepts a proton, just as ammonia does, and the solution turns red litmus blue:

$\mathrm{CH_3NH_2(aq) + H_2O(l) \rightleftharpoons CH_3NH_3^{+}(aq) + OH^{-}(aq)} \qquad K_b = \frac{[\mathrm{CH_3NH_3^{+}}][\mathrm{OH^{-}}]}{[\mathrm{CH_3NH_2}]}$

Base	$\mathrm{p}K_b$ at 25 °C	$K_{b}$ / $\mathrm{mol\,dm^{-3}}$
N-methylmethanamine, $\mathrm{(CH_3)_2NH}$	3.27	$5.4 \times 10^{-4}$
methanamine, $\mathrm{CH_3NH_2}$	3.34	$4.6 \times 10^{-4}$
N,N-dimethylmethanamine, $\mathrm{(CH_3)_3N}$	4.20	$6.3 \times 10^{-5}$
ammonia, $\mathrm{NH_3}$	4.75	$1.8 \times 10^{-5}$
phenylamine, $\mathrm{C_6H_5NH_2}$	9.37	$4.3 \times 10^{-10}$

Table 9.2: Base strengths, from $\mathrm{p}K_b = 14.00 - \mathrm{p}K_a$ of each conjugate acid (CRC Handbook values). A smaller $\mathrm{p}K_b$ means a stronger base.

Alkyl amines are stronger bases than ammonia. Alkyl groups push electron density towards nitrogen (a positive inductive effect), so the lone pair is more available, and the positive charge of the ammonium ion is spread and stabilised.
The tertiary amine is weaker than expected in water. $\mathrm{(CH_3)_3NH^{+}}$ has only one N–H to hydrogen-bond with water, so it is less stabilised by solvation, and the three groups crowd the nitrogen. In the gas phase, where there is no water, the order is simply 3° > 2° > 1° > $\mathrm{NH_3}$ .
Phenylamine is a very weak base. Its lone pair overlaps with the π-system of the ring and is delocalised into it, so it is far less available to accept a proton.

Salts. Amines neutralise acids: $\mathrm{C_6H_5NH_2 + HCl \rightarrow C_6H_5NH_3^{+}Cl^{-}}$ . The salt is ionic, odourless, involatile and soluble in water, which is why phenylamine, only slightly soluble in water, dissolves in dilute hydrochloric acid, and why many amine medicines are sold as "hydrochlorides". A strong base gets the amine back: $\mathrm{C_6H_5NH_3^{+} + OH^{-} \rightarrow C_6H_5NH_2 + H_2O}$ .

Worked example

Problem. Calculate the pH at 25 °C of 0.100 mol dm⁻³ aqueous methanamine ( $K_b = 4.57 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ ).

Solution. Let $x = [\mathrm{OH^{-}}] = [\mathrm{CH_3NH_3^{+}}]$ . With the usual approximation $0.100 - x \approx 0.100$ :

$x = \sqrt{K_b c} = \sqrt{4.57 \times 10^{-4} \times 0.100} = 6.76 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$

$\mathrm{pOH} = -\log_{10}(6.76 \times 10^{-3}) = 2.17 \qquad \mathrm{pH} = 14.00 - 2.17 = 11.83$

Check: $x$ is 6.8 % of 0.100, a little above the usual 5 % limit. Solving the quadratic $x^{2} + K_{b} x - K_{b} c = 0$ gives $x = 6.54 \times 10^{-3}$ and pH 11.82, so the approximation changes the answer by only 0.01.

Answer: pH = 11.82.

Amines as nucleophiles

With haloalkanes (as in route 1): a 1° amine gives a 2° amine, then a 3° amine, then a quaternary ammonium salt. Quaternary ammonium salts with one long chain, such as benzalkonium chloride, are cationic surfactants used in disinfectants and fabric softeners.

With acyl chlorides and acid anhydrides, a vigorous reaction at room temperature gives an N-substituted amide:

$\mathrm{CH_3COCl + 2CH_3NH_2 \rightarrow CH_3CONHCH_3 + CH_3NH_3^{+}Cl^{-}}$

The second amine molecule takes up the HCl. Paracetamol is made this way, by acylating 4-aminophenol with ethanoic anhydride.

With nitrous acid: a test that tells the classes apart

Nitrous acid, $\mathrm{HNO_2}$ , is unstable, so it is made in the tube from sodium nitrite and dilute hydrochloric acid, kept cold.

Primary aliphatic amine: steady bubbles of nitrogen, and an alcohol forms: $\mathrm{CH_3CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2OH + N_2 + H_2O}$ .
Phenylamine at 0–5 °C: no gas; a stable solution of benzenediazonium chloride forms: $\mathrm{C_6H_5NH_2 + HNO_2 + HCl \rightarrow C_6H_5N_2^{+}Cl^{-} + 2H_2O}$ . Added to phenol in sodium hydroxide, it couples to give a yellow-orange azo dye, 4-hydroxyazobenzene. Above about 10 °C the diazonium ion decomposes to phenol and nitrogen.
Secondary amine: a yellow oil, an N-nitrosamine, separates; no gas.
Tertiary aliphatic amine: it simply dissolves as its salt; no gas, no oil.

Ring substitution of phenylamine. The lone pair that makes phenylamine a weak base makes its ring very electron-rich. With bromine water it decolourises the bromine at once and gives a white precipitate of 2,4,6-tribromophenylamine, with no catalyst:

$\mathrm{C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr}$

Uses

Amine	Use
phenylamine	azo dyes and indicators (methyl orange); the polyurethane raw material MDI; rubber chemicals
hexane-1,6-diamine	nylon-6,6, with hexanedioic acid
4-aminophenol	paracetamol
quaternary ammonium salts	disinfectants, fabric softeners, hair conditioners
2-aminoethanol	removing $\mathrm{CO_2}$ and $\mathrm{H_2S}$ from natural gas ("amine scrubbing")

Warning

Safety. Dry diazonium salts can explode: keep them in cold solution and use them at once. N-Nitrosamines are carcinogens, so the nitrous acid test on secondary amines is done only on a small scale, in a fume cupboard. Tin and concentrated hydrochloric acid give off hydrogen and acid fumes.

First aid for concentrated hydrochloric acid (CDC/ATSDR): flush skin or eyes with plenty of water for at least 15 minutes, removing contaminated clothing; if swallowed, do not induce vomiting; get medical help.

Tip

The nitrous acid test in one line: 1° aliphatic = bubbles, 1° aromatic (cold) = diazonium, then a coloured dye with phenol, 2° = yellow oil, 3° = nothing visible.

Alt Academy: Preparation of Amines (A Level). Watch for why excess ammonia is used with a haloalkane, and how nitriles and nitrobenzene are reduced.

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Practice

Write equations for the preparation of (a) propan-1-amine from 1-bromopropane; (b) propan-1-amine from propanenitrile; (c) ethanamine from propanamide.
Arrange in order of increasing base strength in water, and explain: ammonia, phenylamine, ethanamine.
Calculate the pH at 25 °C of 0.0500 mol dm⁻³ phenylamine ( $K_b = 4.3 \times 10^{-10}\,\mathrm{mol\,dm^{-3}}$ ).
Give the name and structure of the organic product of ethanoyl chloride with ethanamine.
Describe a single test, with its results, that distinguishes ethanamine, N-methylethanamine and phenylamine.
Starting from bromomethane, plan a two-step synthesis of ethanamine, giving reagents and conditions.
Explain why excess ammonia does not completely prevent the formation of secondary and tertiary amines when a haloalkane is heated with ammonia.

Show answersHide answers

(a) $\mathrm{CH_3CH_2CH_2Br + 2NH_3 \rightarrow CH_3CH_2CH_2NH_2 + NH_4Br}$ ; (b) $\mathrm{CH_3CH_2CN + 4[H] \rightarrow CH_3CH_2CH_2NH_2}$ ; (c) $\mathrm{CH_3CH_2CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2NH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$ .
phenylamine < ammonia < ethanamine. The ethyl group donates electron density to N; in phenylamine the lone pair is delocalised into the ring.
$[\mathrm{OH^{-}}] = \sqrt{4.3 \times 10^{-10} \times 0.0500} = 4.6 \times 10^{-6}$ ; pOH = 5.34; pH = 8.66.
N-ethylethanamide, $\mathrm{CH_3CONHCH_2CH_3}$ .
Add cold $\mathrm{NaNO_2}$ + HCl (below 5 °C). Ethanamine: bubbles of $\mathrm{N_2}$ . N-methylethanamine: yellow oil, no gas. Phenylamine: no gas; the solution added to alkaline phenol gives an orange-yellow dye.
Step 1: KCN in ethanol, heat under reflux: $\mathrm{CH_3Br + CN^{-} \rightarrow CH_3CN + Br^{-}}$ . Step 2: reduce with $\mathrm{LiAlH_4}$ in dry ether (or $\mathrm{H_2}$ /Ni): $\mathrm{CH_3CN + 4[H] \rightarrow CH_3CH_2NH_2}$ .
The primary amine formed is also a nucleophile, with a more available lone pair than ammonia, so even when ammonia is in excess some haloalkane molecules react with amine molecules. Excess ammonia lowers this fraction but cannot reduce it to zero.

Sources and credits

$\mathrm{p}K_a$ values of the conjugate acids: Dissociation Constants of Organic Bases in Aqueous Solution, CRC Handbook of Chemistry and Physics, CRC Press.
First aid: Medical Management Guidelines for Hydrogen Chloride, Agency for Toxic Substances and Disease Registry, US Centers for Disease Control and Prevention.
Image: Methyl-orange-sample.jpg by Ben Mills, via Wikimedia Commons, public domain.
Video: Preparation of Amines, Alt Academy, embedded from YouTube.
The text, Figure 9.2, the tables, worked examples and Practice items were written and drawn by Ganzaa.

1One lone pair, many reactions

2Preparing amines

3Amines as bases

4Amines as nucleophiles

5With nitrous acid: a test that tells the classes apart

6Uses