EduNinja

34.1 Primary and secondary amines

Primary and secondary amines can be formed either by adding an alkyl group to nitrogen through nucleophilic substitution or by reducing a C=O/CN nitrogen compound. Choose the route by tracking both the carbon skeleton and the number of carbon groups bonded to N.

Starting material	Reagents and conditions	Amine product	Structural accounting
halogenoalkane, RX	excess NH3 in ethanol, heat under pressure	primary amine, RNH2	halogen is replaced by NH2; excess NH3 limits further alkylation
halogenoalkane, RX + primary amine, R'NH2	ethanol, heat in a sealed tube/under pressure	secondary amine, RR'NH	the alkyl group R is added to nitrogen already bearing R'
primary or N-substituted amide	LiAlH4 (commonly in dry ether), then work-up	primary or secondary amine	amide C=O becomes CH2; groups already on N are retained
nitrile, RCN	LiAlH4, or H2/Ni	primary amine, RCH2NH2	the nitrile carbon becomes the CH2 carbon bonded to NH2

$\ce{RX + 2NH3 -> RNH2 + NH4X}$

$\ce{RCONH2 ->[LiAlH4] RCH2NH2}\qquad\ce{RCN ->[LiAlH4\ or\ H2/Ni] RCH2NH2}$

Do not lose the carbonyl or nitrile carbon when counting the product chain: both become the carbon directly attached to N. Ammonia substitution can continue to secondary/tertiary products unless excess ammonia favours the primary amine.

At room temperature, ammonia or an amine attacks an acyl chloride and forms an amide by condensation, with HCl eliminated. The groups already bonded to the attacking nitrogen determine the substitution level of the amide product.

Nitrogen nucleophile	Amide formed from RCOCl	Amide class
NH3	RCONH2	primary (unsubstituted) amide
primary amine, R'NH2	RCONHR'	secondary (N-substituted) amide
secondary amine, R'2NH	RCONR'2	tertiary (N,N-disubstituted) amide

The first ammonia/amine molecule forms the C-N bond. A second equivalent can neutralise the HCl: for ammonia, RCOCl + 2NH3 -> RCONH2 + NH4Cl. With an amine, the corresponding alkylammonium chloride salt forms.

Ethanoyl chloride plus methylamine gives N-methylethanamide. The methyl group remains bonded to N, while the CH3CO- acyl fragment supplies the ethanamide carbonyl skeleton.

Do not call the product an amine: nitrogen is bonded directly to a carbonyl carbon, so the functional group is an amide. Ammonia, primary amines and secondary amines give different amide classes.

An amine is a Bronsted-Lowry base: its nitrogen lone pair accepts a proton from water, producing an alkylammonium ion and hydroxide.

The equilibrium is partial, so aqueous amines are weak bases. The position depends on how available the lone pair is and on electron-donating or withdrawing groups.

Ethylamine + H2O <=> ethylammonium ion + OH-. The solution is alkaline even though most ethylamine molecules remain unprotonated.

A weak base is not a base that cannot react; it is one whose protonation equilibrium is incomplete.

34.2 Phenylamine and azo compounds

Phenylamine is prepared from benzene through nitrobenzene. The order matters: install -NO2 by nitration, reduce it in acid, then add aqueous alkali to release the free amine from its phenylammonium salt.

Stage	Conversion	Reagents and conditions	Purpose
1 nitration	benzene -> nitrobenzene	concentrated HNO3 and concentrated H2SO4, 25-60 °C	electrophilic substitution installs -NO2
2 reduction	nitrobenzene -> phenylammonium chloride in the acidic mixture	hot Sn and concentrated HCl, heat/reflux	reduce -NO2 to the amine oxidation level; acid protonates the amine
3 alkaline work-up	phenylammonium chloride -> phenylamine	NaOH(aq)	remove H+ and liberate C6H5NH2

$\ce{C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O}$

$\ce{C6H5NH3+Cl- + NaOH -> C6H5NH2 + NaCl + H2O}$

Do not stop at the acidic reduction mixture: it contains protonated phenylamine. NaOH(aq) is an essential final step, not the reducing agent.

Phenylamine has two distinct reaction patterns here. Its electron-rich ring undergoes rapid 2,4,6-substitution with bromine water, while its -NH2 group can be converted into a diazonium salt only under cold conditions.

Reaction	Reagents and conditions	Organic product	Key consequence
bromination	Br2(aq), room temperature	2,4,6-tribromophenylamine	bromine water is decolourised and a white precipitate forms
diazotisation	HNO2, or NaNO2 plus dilute acid, below 10 °C	benzenediazonium salt	the unstable diazonium ion is preserved by keeping it cold
hydrolysis of diazonium salt	H2O, then warm	phenol	N2 is released as the diazonium group is replaced by -OH

The nitrogen lone pair donates electron density into the benzene ring. This activates the ring and directs electrophilic substitution to the 2, 4 and 6 positions, so no halogen carrier is needed for bromine water at room temperature.

$\ce{C6H5N2+ + H2O ->[warm] C6H5OH + N2 + H+}$

Keep the temperature stages separate: below 10 °C forms and preserves the diazonium salt; warming it with water deliberately decomposes it to phenol. Do not warm during diazotisation.

$\text{aqueous basicity: ethylamine} > \text{ammonia} > \text{phenylamine}$

All three accept H+ through the nitrogen lone pair. Their relative basicity depends on how available that lone pair is for forming a dative covalent bond to a proton.

Base	Effect on the nitrogen lone pair	Relative result
ethylamine	the ethyl group donates electron density by the positive inductive effect	lone pair is more available than in NH3; strongest of the three
ammonia	no ethyl group donates electron density and no benzene ring delocalises the lone pair	intermediate
phenylamine	the lone pair overlaps with and is delocalised into the benzene pi system	less available to accept H+; weakest of the three

Phenylamine is still a base; delocalisation makes it less basic than ammonia rather than preventing protonation altogether. Compare lone-pair availability, not the number of hydrogen atoms on nitrogen.

An azo compound contains the azo group -N=N-. In the products here, this link joins two aromatic carbon groups, giving the general pattern Ar-N=N-Ar'. Azo compounds are often used as dyes.

Step	Material and condition	Chemical role or outcome
1	benzenediazonium chloride, kept below 10 °C	supplies the diazonium electrophile without allowing it to hydrolyse
2	phenol dissolved in NaOH(aq)	forms an alkaline, electron-rich phenoxide coupling component
3	add the cold diazonium solution to the alkaline phenol	electrophilic substitution couples the two aromatic rings through -N=N- and forms an azo compound

To identify the azo group in any structure, locate N=N with a carbon-containing group bonded on each side. Do not confuse the neutral -N=N- link in an azo product with the charged -N2+ group in a diazonium ion.

The same route can make other azo dyes: change the substituted aromatic diazonium salt or the activated aromatic coupling component while retaining diazotisation followed by coupling. Different substituents can change the dye produced.

Coupling is not the warm-water reaction from the previous objective. The diazonium salt is kept cold and reacted with phenol in NaOH(aq); warming it with water instead gives phenol and N2.

34.3 Amides

At room temperature, ammonia reacts with an acyl chloride to form an unsubstituted amide, while a primary amine forms an N-substituted amide. In each condensation, the nitrogen-containing nucleophile replaces Cl at the acyl carbon and HCl is eliminated.

Nitrogen reactant	Acyl chloride	Amide product	Acid captured by excess reactant
ammonia, NH3	RCOCl	RCONH2	NH4Cl
primary amine, R'NH2	RCOCl	RCONHR'	R'NH3Cl

$\ce{RCOCl + 2NH3 -> RCONH2 + NH4Cl}$

$\ce{RCOCl + 2R'NH2 -> RCONHR' + R'NH3Cl}$

Only one ammonia or amine molecule supplies the nitrogen in the amide; the second equivalent captures HCl. Do not include secondary amines in this specific 34.3.1 recall objective, which names ammonia and primary amines.

Hydrolysis breaks the acyl C-N bond and separates the carbonyl and nitrogen fragments. LiAlH4 reduction keeps the C-N skeleton together but changes the amide carbonyl carbon into CH2, forming an amine.

Amide reaction	Reagents and conditions	Carbonyl-side product	Nitrogen-side product
acid hydrolysis	aqueous acid, heat/reflux	carboxylic acid, RCOOH	NH4+ from RCONH2, or R'NH3+ from RCONHR'
alkaline hydrolysis	aqueous alkali, heat/reflux	carboxylate, RCOO-	NH3 from RCONH2, or R'NH2 from RCONHR'
reduction	LiAlH4 in dry ether, then work-up	C=O becomes CH2; no separate carbonyl fragment	RCONH2 -> RCH2NH2; RCONHR' -> RCH2NHR'

$\ce{RCONH2 + H2O + H+ -> RCOOH + NH4+}$

$\ce{RCONH2 + OH- -> RCOO- + NH3}$

In LiAlH4 reduction, retain the former carbonyl carbon: ethanamide, CH3CONH2, forms ethylamine, CH3CH2NH2. Reduction does not remove that carbon or split the C-N bond.

Do not write identical hydrolysis products for acid and alkali. Acid protonates ammonia/amine; alkali deprotonates the carboxylic acid to a carboxylate. Reduction is a separate reaction and gives an amine without hydrolytic cleavage.

An amine nitrogen lone pair is relatively localised and available to accept H+. In an amide, the nitrogen lone pair overlaps with the adjacent C=O pi system and is delocalised across the O-C-N unit.

This delocalisation gives the C-N bond partial double-bond character and stabilises the unprotonated amide. Using the nitrogen lone pair to bond to H+ would remove it from that conjugated system, so it is much less available than an amine lone pair.

$\text{basicity: amine} \gg \text{amide}$

The amide nitrogen still has a lone pair, but presence is not the same as availability. Do not explain the difference only by saying that both compounds contain nitrogen; compare what happens to the lone pair next to C=O.

34.4 Amino acids

An amino acid is amphoteric because its -NH2 group can accept H+ and its -COOH group can donate H+. Proton transfer gives a zwitterion containing both -NH3+ and -COO- on the same molecule.

pH condition	Predominant general form	Net charge	Acid-base change from zwitterion
pH below pI	+H3N-CH(R)-COOH	positive	-COO- accepts H+
pH at pI	+H3N-CH(R)-COO-	zero overall	zwitterionic form predominates
pH above pI	H2N-CH(R)-COO-	negative	-NH3+ loses H+

The isoelectric point, pI, is the pH at which the amino acid has zero average net charge and therefore shows no net movement in an electric field. Internal positive and negative charges can still be present.

If the side chain contains another acidic or basic group, include its protonation state when finding the total charge; do not infer net charge only from the alpha-amino and alpha-carboxyl groups.

A zwitterion is not uncharged at every atom: its separated charges sum to zero. The pI is a particular pH, not a claim that the amino acid is always neutral in solution.

The -COOH group of one amino acid reacts with the -NH2 group of another in condensation. OH from -COOH and H from -NH2 leave as H2O, and the new covalent amide link is the peptide bond -CO-NH-.

Product	Amino-acid residues	Peptide bonds	H2O molecules eliminated on formation
dipeptide	2	1	1
tripeptide	3	2	2

A peptide retains a free amino end (N-terminus) and a free carboxyl end (C-terminus), so a dipeptide can condense with a third amino acid to form a tripeptide.

Sequence matters: glycine followed by alanine (Gly-Ala) and alanine followed by glycine (Ala-Gly) contain the same residues but have different orders and are different dipeptides. Draw each -CO-NH- link between consecutive residues.

Do not join two -NH2 groups or two -COOH groups, and do not count residues as peptide bonds: a chain of n residues contains n-1 peptide bonds.

Electrophoresis separates charged amino acids and dipeptides in an electric field. A positive species moves toward the negative electrode, a negative species moves toward the positive electrode, and a zero-net-charge species has no net migration from the origin.

Compare buffer pH with the species' pI	Likely net charge	Direction
pH < pI	positive	toward negative electrode (cathode)
pH = pI	zero overall	remains at/near origin
pH > pI	negative	toward positive electrode (anode)

For every amino acid or dipeptide in the mixture: 1) identify every ionisable -NH3+/-COO- group, including side chains and free peptide termini; 2) determine its net charge at the stated pH; 3) assign the opposite electrode; 4) only then compare distance travelled. Greater charge tends to increase migration, while larger species tend to move more slowly.

A dipeptide's internal peptide-bond nitrogen and carbonyl are not counted as a free amino and carboxyl pair. Start with its free N-terminus, free C-terminus and any ionisable side chains when predicting charge.

Do not send every amino acid to the same side or assume every dipeptide is neutral. At a fixed buffer pH, different pI values and ionisable side chains can give different net charges; distance also cannot be interpreted from charge alone when sizes differ.

34. Nitrogen compounds

Exam analysis

Most tested topics

34.1 Primary and secondary amines

Choose an amine preparation route from the required carbon skeleton

Ammonia and amines form progressively N-substituted amides

Amines are weak bases because the nitrogen lone pair accepts H+

34.2 Phenylamine and azo compounds

Prepare phenylamine by nitration, reduction and alkaline work-up

Separate phenylamine bromination from cold diazotisation and warm hydrolysis

Ethylamine is more basic than ammonia, while phenylamine is weaker

Couple benzenediazonium chloride with alkaline phenol to make an azo dye

34.3 Amides

Ammonia and primary amines form amides with acyl chlorides at room temperature

Amides undergo hydrolysis or C=O reduction with different structural outcomes

Amides are much weaker bases because the nitrogen lone pair is delocalised

34.4 Amino acids

Use pH and the isoelectric point to choose an amino acid's ionic form

Condensation joins amino acids through amide peptide bonds

Predict each electrophoresis band from net charge at the buffer pH