Some Important Organic Compounds Containing Nitrogen

ChemistryAminesFor JEE aspirants

Organic compounds containing nitrogen include nitro compounds, alkyl nitrites, cyanides, isocyanides, amines and amino acids. Their chemistry turns on which atom is bonded to carbon: nitrogen or oxygen in $- N O_{2}$ and $- ONO$ , carbon or nitrogen in $- CN$ and $- NC$ . This page covers the structure, preparation and key reactions of these organic compounds containing nitrogen, including nitration, reduction of nitrobenzene and amino acids, as tested in JEE Main and NEET.

Key Formulas - Quick Reference

Nitrite ion: $R - X + AgN O_{2} R - N O_{2} + AgX$ (nitroalkane); $R - X + KN O_{2} R - ONO + KX$ (alkyl nitrite)
Cyanide ion: $R - X + KCN R - CN + KX$ (nitrile); $R - X + AgCN R - NC + AgX$ (isocyanide)
Nitronium ion: $HN O_{3} + 2 H_{2} S O_{4} ⇌ N O_{2}^{+} + H_{3} O^{+} + 2HS O_{4}^{-}$
Nitrobenzene: acid gives $C_{6} H_{5} N H_{2}$ , neutral (Zn/ $N H_{4} Cl$ ) gives $C_{6} H_{5} NHOH$ , Zn/NaOH gives $C_{6} H_{5} NHNH C_{6} H_{5}$
Zinin reduction: $(N H_{4})_{2} S$ reduces only one $N O_{2}$ of m-dinitrobenzene
Nitrous acid: 1° nitroalkane gives nitrolic acid (red in NaOH), 2° gives pseudonitrole (blue), 3° no reaction
Hydrolysis: $RCN + 2 H_{2} O H^{+} RCOOH + N H_{3}$ ; $RNC + 2 H_{2} O H^{+} RN H_{2} + HCOOH$
Reduction: $RCN$ gives $RC H_{2} N H_{2}$ (1° amine); $RNC$ gives $RNHC H_{3}$ (2° amine)
Amino acid zwitterion: $H_{3} N^{+} - CHR - CO O^{-}$ ; peptide bond $- CO - NH -$
Strecker: $RCHO$ + $N H_{3}$ + $HCN$ gives $RCH (N H_{2}) CN$ , hydrolysed to $RCH (N H_{2}) COOH$

1. Nitrogen-Containing Functional Groups

Nitrogen can be joined to carbon in many ways. The table lists the groups met in this unit; compare the atom that is bonded to the carbon chain in each pair of isomeric groups.

Group	Structure	Example	Name of example
Cyanide (nitrile)	$- C \equiv N$	$C H_{3} CN$	ethanenitrile (acetonitrile)
Isocyanide (carbylamine)	$- N^{+} \equiv C^{-}$	$C H_{3} NC$	methyl isocyanide
Nitro	$- N O_{2}$	$C H_{3} N O_{2}$	nitromethane
Nitrite (ester)	$- O - N = O$	$C H_{3} C H_{2} ONO$	ethyl nitrite
Nitroso	$- N = O$	$C_{6} H_{5} NO$	nitrosobenzene
Primary amine	$- N H_{2}$	$C H_{3} N H_{2}$	methanamine
Secondary amine	$> NH$	$(C H_{3})_{2} NH$	N-methylmethanamine
Tertiary amine	$> N -$	$(C H_{3})_{3} N$	N,N-dimethylmethanamine
Imine	$> C = N -$	$C H_{3} CH = NH$	ethanimine
Diazonium	$- N^{+} \equiv N$	$C_{6} H_{5} N_{2}^{+} C l^{-}$	benzenediazonium chloride
Azo	$- N = N -$	$C_{6} H_{5} N = N C_{6} H_{5}$	azobenzene
Amide	$- CON H_{2}$	$C H_{3} CON H_{2}$	ethanamide

Ambident nucleophile: an ion that can form a bond through two different atoms. The nitrite ion

N O_{2}^{-}

attacks through N or O, and the cyanide ion

C N^{-}

through C or N. That is why one alkyl halide can give two isomeric products.

Figure 1: The same alkyl halide gives isomeric products: ionic

KN O_{2}

KCN

react through O/C, covalent

AgN O_{2}

AgCN

react through N.

Exam Trick

Silver sends in nitrogen. With silver salts the metal holds on to O (in $AgN O_{2}$ ) or C (in $AgCN$ ) by a covalent bond, so the free nitrogen lone pair attacks: $AgN O_{2}$ gives $R - N O_{2}$ and $AgCN$ gives $R - NC$ . Ionic $KN O_{2}$ and $KCN$ react through the atom that carries more negative charge: O gives the nitrite $R - ONO$ and C gives the cyanide $R - CN$ .

2. Nitro Compounds

2.1 Classification

Nitroalkanes are alkanes in which a hydrogen atom is replaced by $- N O_{2}$ . Like alcohols, they are classed by the carbon that carries the group: primary $RC H_{2} N O_{2}$ , secondary $R_{2} CHN O_{2}$ and tertiary $R_{3} CN O_{2}$ . Aromatic nitro compounds (nitroarenes) such as nitrobenzene $C_{6} H_{5} N O_{2}$ carry the group on a ring carbon.

Because $N O_{2}^{-}$ is ambident, nitroalkanes $R - N O_{2}$ and alkyl nitrites $R - O - N = O$ with the same formula are functional isomers: nitroethane and ethyl nitrite are both $C_{2} H_{5} N O_{2}$ .

2.2 Structure of the Nitro Group

Nitrogen in the nitro group is $sp^{2}$ hybridised. Its three $sp^{2}$ orbitals bond to carbon and to the two oxygen atoms, so the group is planar. The unhybridised p orbital of nitrogen overlaps sideways with p orbitals of both oxygen atoms, giving a $π$ system spread over O-N-O. The group is therefore a resonance hybrid of two equivalent structures: both N-O bonds have the same length (about $122 pm$ ) and each oxygen carries half a negative charge.

Figure 2: Resonance makes both N-O bonds identical. Cl pulls electrons away from the ring like

N O_{2}

does, so their dipoles cancel partly;

C H_{3}

pushes electrons in, so its dipole adds.

The nitro group strongly withdraws electrons (both $- I$ and $- R$ ). Nitrobenzene has a large dipole moment (about $4.2 D$ ), and in para-disubstituted benzenes the group dipoles add or cancel as vectors (Figure 2).

2.3 Preparation of Nitro Compounds

(i) From alkyl halides. Alkyl halides heated with silver nitrite in ethanol give nitroalkanes, with a little alkyl nitrite. The method works best for 1° halides; 2° and 3° halides give mostly alkenes by $β$ -elimination.

C_{2} H_{5} I + AgN O_{2} ethanol, heat C_{2} H_{5} N O_{2} + AgI

Sodium or potassium nitrite is ionic, so in ethanol it gives the alkyl nitrite as the main product. In polar aprotic solvents such as DMF or DMSO, however, even $NaN O_{2}$ or $KN O_{2}$ gives a good yield (about 60%) of the nitroalkane.

C_{2} H_{5} I + KN O_{2} ethanol C_{2} H_{5} - O - N = O + KI

(ii) Nitration of arenes. Benzene reacts with the nitrating mixture (conc. $HN O_{3}$ + conc. $H_{2} S O_{4}$ ) to give nitrobenzene. Ethylbenzene gives a mixture of o- and p-nitroethylbenzene, because the ethyl group directs ortho and para.

C_{6} H_{6} + HN O_{3} conc. H_{2} S O_{4} 323-333 K C_{6} H_{5} N O_{2} + H_{2} O

The electrophile is the nitronium ion. Sulphuric acid, the stronger acid, protonates nitric acid, which loses water:

HN O_{3} + 2 H_{2} S O_{4} ⇌ N O_{2}^{+} + H_{3} O^{+} + 2HS O_{4}^{-}

Figure 3:

H_{2} S O_{4}

turns

HN O_{3}

into the electrophile

N O_{2}^{+}

; the slow step breaks the ring's aromaticity, and losing

H^{+}

restores it.

Nitronium salts such as $N O_{2}^{+} Cl O_{4}^{-}$ and $N O_{2}^{+} B F_{4}^{-}$ , and acetyl nitrate $C H_{3} COON O_{2}$ , are other nitrating agents.

(iii) Vapour-phase nitration of alkanes. Alkanes react with nitric acid vapour at about 675 K. At this temperature C-C bonds also break, so a mixture forms and the yield of any one product is low. Propane gives 2-nitropropane, 1-nitropropane, nitroethane and nitromethane.

C H_{4} + HN O_{3} 675 K C H_{3} N O_{2} + H_{2} O

(iv) Oxidation of tert-alkylamines. 3° nitroalkanes are made by oxidising a primary amine whose nitrogen sits on a tertiary carbon, such as tert-butylamine:

(C H_{3})_{3} C - N H_{2} + 3 [O] KMn O_{4} (C H_{3})_{3} C - N O_{2} + H_{2} O

2-Methyl-2-nitropropane is obtained in about 83% yield.

2.4 Physical Properties

Nitroalkanes are colourless liquids with a pleasant smell; aromatic nitro compounds such as nitrobenzene are pale yellow with a characteristic odour (bitter almonds).
They are highly polar, so strong dipole-dipole attractions give them much higher boiling points than isomeric alkyl nitrites (nitroethane 387 K, ethyl nitrite 290 K).
They are only sparingly soluble in water.
Most nitroalkanes distil without decomposition, whereas alkyl nitrites are unstable and may explode on heating.
Nitrobenzene and o-nitrophenol are steam volatile and are purified by steam distillation.

2.5 Chemical Properties of Nitro Compounds

(i) Reduction of nitroalkanes. The product depends on the reagent:

C H_{3} C H_{2} N O_{2} + 3 H_{2} Ni or Pt C H_{3} C H_{2} N H_{2} + 2 H_{2} O

C H_{3} C H_{2} N O_{2} + 6 [H] Sn/HCl or Fe/HCl C H_{3} C H_{2} N H_{2} + 2 H_{2} O

C H_{3} C H_{2} N O_{2} + 4 [H] Zn/ N H_{4} Cl (neutral) C H_{3} C H_{2} NHOH + H_{2} O

Lithium aluminium hydride also reduces nitroalkanes to 1° amines.

Mulliken-Barker test for the nitro group. Reduce the compound with zinc dust and ammonium chloride, then warm the product with Tollens' reagent. The hydroxylamine formed is oxidised to a nitroso compound and reduces the reagent to a silver mirror.

C H_{3} C H_{2} NHOH + 2 [Ag (N H_{3})_{2}]^{+} + 2O H^{-} C H_{3} C H_{2} N = O + 2Ag + 4N H_{3} + 2 H_{2} O

(ii) Reduction of nitrobenzene. Aromatic nitro compounds give a family of products, and the medium decides which one forms (Figure 4).

C_{6} H_{5} N O_{2} + 6 [H] Sn/HCl or Fe/HCl C_{6} H_{5} N H_{2} + 2 H_{2} O

C_{6} H_{5} N O_{2} + 4 [H] Zn/ N H_{4} Cl C_{6} H_{5} NHOH + H_{2} O

In alkaline media two molecules are reduced together (bimolecular reduction):

2 C_{6} H_{5} N O_{2} + 6 [H] N a_{3} As O_{3} /NaOH azoxybenzene C_{6} H_{5} N (O) = N C_{6} H_{5} + 3 H_{2} O

2 C_{6} H_{5} N O_{2} + 8 [H] Zn/NaOH, C H_{3} OH azobenzene C_{6} H_{5} N = N C_{6} H_{5} + 4 H_{2} O

2 C_{6} H_{5} N O_{2} + 10 [H] Zn/NaOH hydrazobenzene C_{6} H_{5} NHNH C_{6} H_{5} + 4 H_{2} O

Unlike nitroalkanes, nitrobenzene gives azobenzene, not aniline, with $LiAl H_{4}$ . Electrolytic reduction in weakly acidic solution gives aniline, while in strongly acidic solution the phenylhydroxylamine formed rearranges to p-aminophenol.

Figure 4: The medium decides the product: acid gives aniline, neutral gives the hydroxylamine, and alkali joins two rings through nitrogen (azoxy, azo, hydrazo).

Exam Trick

Acid: amine. Neutral: N-OH. Alkali: two rings. Sn/HCl or Fe/HCl stops at aniline, Zn/ $N H_{4} Cl$ stops at phenylhydroxylamine, and every alkaline reagent joins two rings through nitrogen. The more hydrogen added, the further along azoxy (6[H]) to azo (8[H]) to hydrazo (10[H]) the product lies.

(iii) Selective (Zinin) reduction. Mild reagents such as ammonium sulphide, $N H_{4} HS$ or $NaHS$ reduce only one of two nitro groups:

C_{6} H_{4} (N O_{2})_{2} + 3 (N H_{4})_{2} S N O_{2} C_{6} H_{4} N H_{2} + 6N H_{3} + 2 H_{2} O + 3S

Here m-dinitrobenzene is reduced to m-nitroaniline; the second nitro group survives.

(iv) Action of nitrous acid. Only nitroalkanes with an $α$ -hydrogen react:

RC H_{2} N O_{2} + HN O_{2} 1°: nitrolic acid RC (N O_{2}) = NOH + H_{2} O

R_{2} CHN O_{2} + HN O_{2} 2°: pseudonitrole R_{2} C (N O_{2}) NO + H_{2} O

A nitrolic acid dissolves in NaOH to give a red solution; a pseudonitrole gives a blue colour; a 3° nitroalkane, with no $α$ -hydrogen, does not react.

(v) Hydrolysis with boiling acid. 1° nitroalkanes give a carboxylic acid and hydroxylamine; 2° nitroalkanes give a ketone and nitrous oxide; 3° nitroalkanes do not react.

C H_{3} C H_{2} N O_{2} + H_{2} O + HCl heat C H_{3} COOH + N H_{2} OH \cdot HCl

2 (C H_{3})_{2} CHN O_{2} boiling HCl 2 (C H_{3})_{2} CO + N_{2} O + H_{2} O

(vi) Tautomerism and acidity. A nitroalkane with an $α$ -hydrogen is neutral to litmus but dissolves in strong alkali. It is a pseudoacid: the $α$ -hydrogen moves to oxygen to give the aci-nitro form (a nitronic acid), whose salt is stabilised by resonance.

C H_{3} C H_{2} N O_{2} nitro form, aci form ⇌ C H_{3} CH = N (O) OH

C H_{3} C H_{2} N O_{2} + NaOH sodium nitronate [C H_{3} CH = N O_{2}]^{-} N a^{+} + H_{2} O

3° nitroalkanes have no $α$ -hydrogen, so they show no tautomerism and do not dissolve in NaOH.

Property	1° $RC H_{2} N O_{2}$	2° $R_{2} CHN O_{2}$	3° $R_{3} CN O_{2}$
$α$ -hydrogens	2	1	0
With $HN O_{2}$ , then NaOH	nitrolic acid: red	pseudonitrole: blue	no reaction: colourless
Boiling HCl	$RCOOH$ + $N H_{2} OH$	ketone + $N_{2} O$	no reaction
Dissolves in NaOH	yes (nitronate)	yes (nitronate)	no
Condenses with aldehydes	yes	yes	no

Exam Trick

Red, blue, colourless = 1°, 2°, 3°. Every test on nitroalkanes (nitrous acid, alkali, acid hydrolysis, condensation) needs an $α$ -hydrogen. Count the H atoms on the carbon carrying $N O_{2}$ and you can predict all of them.

JEE Advanced

Nitromethane has $p K_{a} \approx 10.2$ , close to phenol, because its conjugate base $C H_{2} = N O_{2}^{-}$ spreads the negative charge onto two oxygen atoms. If a nitronate salt is added to cold strong acid it gives a carbonyl compound and $N_{2} O$ (the Nef reaction); the hydrolysis of 2° nitroalkanes to ketones runs through the same intermediate. The same stabilised carbanion adds to aldehydes in the Henry (nitroaldol) reaction below.

(vii) Condensation with aldehydes (Henry or nitroaldol reaction). In alkali, a 1° or 2° nitroalkane loses its $α$ -hydrogen and the carbanion adds to the carbonyl carbon of an aldehyde. The $β$ -nitro alcohol formed loses water to give a nitroalkene.

C H_{3} N O_{2} + C_{6} H_{5} CHO dil. NaOH C_{6} H_{5} CH (OH) C H_{2} N O_{2}

C_{6} H_{5} CH (OH) C H_{2} N O_{2} heat β-nitrostyrene C_{6} H_{5} CH = CHN O_{2} + H_{2} O

C H_{3} N O_{2} + C H_{3} CHO O H^{-} 1-nitroprop-1-ene C H_{3} CH = CHN O_{2} + H_{2} O

(viii) Ring substitution in nitrobenzene. $- N O_{2}$ withdraws electrons from the ring, so nitrobenzene is deactivated and substitution goes to the meta position under harsher conditions:

C_{6} H_{5} N O_{2} + HN O_{3} fuming HN O_{3} /conc. H_{2} S O_{4} 363 K, m-dinitrobenzene C_{6} H_{4} (N O_{2})_{2} + H_{2} O

C_{6} H_{5} N O_{2} + H_{2} S O_{4} fuming H_{2} S O_{4}, heat m-nitrobenzenesulphonic acid N O_{2} C_{6} H_{4} S O_{3} H + H_{2} O

C_{6} H_{5} N O_{2} + C l_{2} FeC l_{3} m-chloronitrobenzene Cl C_{6} H_{4} N O_{2} + HCl

Nitrobenzene does not undergo Friedel-Crafts alkylation or acylation at all, which is why it can be used as a solvent for these reactions.

2.6 Distinguishing Nitroalkanes from Alkyl Nitrites

Test	Nitroethane $C_{2} H_{5} N O_{2}$	Ethyl nitrite $C_{2} H_{5} ONO$
Atom bonded to carbon	N (C-N bond)	O (C-O bond)
Boiling point	387 K (polar, higher)	290 K (lower)
Reduction with Sn/HCl	ethylamine $C_{2} H_{5} N H_{2}$	ethanol $C_{2} H_{5} OH$ + $N H_{3}$
Aqueous NaOH	dissolves as a salt, not hydrolysed	hydrolysed to $C_{2} H_{5} OH$ + $NaN O_{2}$
On heating	distils unchanged	unstable, may explode

C_{2} H_{5} ONO + 6 [H] Sn/HCl C_{2} H_{5} OH + N H_{3} + H_{2} O

C H_{3} - O - N = O + NaOH C H_{3} OH + NaN O_{2}

Figure 5: Reduction decides it: the atom bonded to carbon stays bonded to carbon, so a nitroalkane gives an amine and an alkyl nitrite gives an alcohol.

3. Cyanides and Isocyanides

3.1 Structure and Bonding

Alkyl cyanides $R - C \equiv N$ and alkyl isocyanides $R - N \equiv C$ are both derived from hydrogen cyanide and have the same molecular formula. In alkali cyanides the $C N^{-}$ ion is free, so it bonds through carbon, where the charge is concentrated. Silver cyanide is covalent ( $Ag - C \equiv N$ ), so only the nitrogen lone pair is free to attack, and isocyanides are the main product (Figure 1). In an isocyanide nitrogen has used its lone pair and carries a formal positive charge, while carbon carries the lone pair: $R - N^{+} \equiv C^{-}$ .

3.2 Preparation of Alkyl Cyanides

(i) Dehydration of amides with $P_{2} O_{5}$ , $SOC l_{2}$ or $A l_{2} O_{3}$ at 773 K; amides of high molar mass lose water on heating alone:

RCON H_{2} P_{2} O_{5}, heat RCN + H_{2} O

C_{6} H_{5} CON H_{2} + SOC l_{2} heat C_{6} H_{5} CN + S O_{2} + 2HCl

C H_{3} (C H_{2})_{6} CON H_{2} strong heat C H_{3} (C H_{2})_{6} CN + H_{2} O

(ii) From alkyl halides with alcoholic KCN or NaCN. This is satisfactory for 1° and 2° halides; 3° halides give alkenes by elimination.

C H_{3} C H_{2} Cl + KCN aq. ethanol, heat C H_{3} C H_{2} CN + KCl

(iii) Grignard reagent and cyanogen chloride, the best route to 3° alkyl cyanides:

(C H_{3})_{3} CMgCl + ClCN dry ether (C H_{3})_{3} CCN + MgC l_{2}

(iv) From diazonium salts (Sandmeyer reaction), for aryl cyanides; see Diazonium Salt:

C_{6} H_{5} N_{2}^{+} C l^{-} + CuCN KCN, heat C_{6} H_{5} CN + N_{2} + CuCl

3.3 Properties and Reactions of Alkyl Cyanides

Lower members are liquids and dissolve in water, because water hydrogen-bonds to the nitrogen lone pair; solubility falls as molar mass rises.
They cannot form hydrogen bonds with each other, so they are much more volatile than the corresponding carboxylic acids.

(i) Hydrolysis. Acid hydrolysis gives the amide and then the acid; alkaline hydrolysis gives the salt and ammonia. Alkaline $H_{2} O_{2}$ stops at the amide.

RCN + 2 H_{2} O H^{+}, heat RCOOH + N H_{3}

RCN + NaOH + H_{2} O heat RCOONa + N H_{3}

RCN + H_{2} O alkaline H_{2} O_{2} RCON H_{2}

(ii) Reduction. Stephen reduction gives an aldehyde; sodium and ethanol (Mendius reduction) or $LiAl H_{4}$ give a 1° amine with the same number of carbon atoms.

C H_{3} CN SnC l_{2} /HCl C H_{3} CH = NH \cdot HCl H_{3} O^{+} C H_{3} CHO

RCN + 4 [H] Na/ C_{2} H_{5} OH or LiAl H_{4} RC H_{2} N H_{2}

(C H_{3})_{2} CHCN + 4 [H] LiAl H_{4} (C H_{3})_{2} CHC H_{2} N H_{2}

(iii) With Grignard reagents, followed by hydrolysis, nitriles give ketones:

C H_{3} CN + C_{2} H_{5} MgBr dry ether C H_{3} C (= NMgBr) C_{2} H_{5}

C H_{3} C (= NMgBr) C_{2} H_{5} + 2 H_{2} O butan-2-one C H_{3} CO C_{2} H_{5} + N H_{3} + Mg (OH) Br

(iv) With alcohols. A nitrile heated with an alcohol and conc. $H_{2} S O_{4}$ or HCl gives an ester:

C H_{3} CN + C_{2} H_{5} OH + H_{2} O conc. H_{2} S O_{4} C H_{3} COO C_{2} H_{5} + N H_{3}

3.4 Preparation of Alkyl Isocyanides

(i) From alkyl iodides and silver cyanide in aqueous ethanol:

C_{2} H_{5} I + AgCN aq. ethanol, heat ethyl isocyanide C_{2} H_{5} NC + AgI

(ii) Carbylamine reaction. A primary amine, aliphatic or aromatic, heated with chloroform and alcoholic KOH gives the foul-smelling isocyanide. The reactive intermediate is dichlorocarbene $CC l_{2}$ , formed by $α$ -elimination. The full mechanism, and the use of this reaction as a test, are covered in Analysis of Amines.

RN H_{2} + CHC l_{3} + 3KOH alc., heat RNC + 3KCl + 3 H_{2} O

C_{6} H_{5} N H_{2} + CHC l_{3} + 3KOH alc., heat C_{6} H_{5} NC + 3KCl + 3 H_{2} O

CHC l_{3} + KOH dichlorocarbene CC l_{2} + KCl + H_{2} O

3.5 Properties and Reactions of Alkyl Isocyanides

Isocyanides are poisonous liquids with a very unpleasant smell, and they boil lower than the isomeric cyanides.
They are sparingly soluble in water: the nitrogen has no lone pair free for hydrogen bonding.

(i) Hydrolysis occurs only with acid; alkalis do not attack isocyanides.

C H_{3} NC + 2 H_{2} O dil. HCl C H_{3} N H_{2} + HCOOH

(ii) Reduction gives a 2° amine carrying an N-methyl group:

C H_{3} NC + 4 [H] H_{2} /Pt or Na/ C_{2} H_{5} OH dimethylamine C H_{3} NHC H_{3}

(iii) Isomerisation. On prolonged heating isocyanides rearrange to the more stable cyanides:

C H_{3} C H_{2} NC heat C H_{3} C H_{2} CN

(iv) Addition to carbon. The carbon of an isocyanide adds halogens, sulphur and oxygen:

C H_{3} NC + C l_{2} C H_{3} NCC l_{2}

C H_{3} NC + S methyl isothiocyanate C H_{3} NCS

C H_{3} NC + HgO methyl isocyanate C H_{3} NCO + Hg

C H_{3} NC + O_{3} C H_{3} NCO + O_{2}

Figure 6: The atom bonded to R stays bonded to R: a nitrile keeps its C-R bond (acid, 1° amine), an isocyanide keeps its N-R bond (amine,

N-CH_{3}

JEE Advanced

The terminal carbon of an isocyanide has a lone pair and only two bonds in the $R - N^{+} \equiv C^{-}$ form, so it behaves partly like a carbene. That is why it adds $C l_{2}$ , S and O to the same carbon (an $α$ -addition) and why isocyanides are strong ligands for metals. Nitriles are more stable, so isocyanides isomerise to nitriles on heating, never the reverse.

4. Amino Acids and Proteins

4.1 Structure and the Zwitterion

Amino acids are carboxylic acids that also carry an amino group. In $α$ -amino acids, the ones obtained from proteins, $N H_{2}$ is on the carbon next to $COOH$ : $RCH (N H_{2}) COOH$ . About 20 amino acids occur in proteins; about 10 of them are essential, because the body cannot make them and they must come from food.

Glycine $H_{2} NC H_{2} COOH$ (R = H) is the simplest and the only one that is not optically active.
Alanine $C H_{3} CH (N H_{2}) COOH$ and all other $α$ -amino acids have a chiral $α$ -carbon; natural amino acids have the L-configuration.
Amino acids contain an acidic and a basic group, so the proton moves from $COOH$ to $N H_{2}$ to give an internal salt, the zwitterion (dipolar ion) $H_{3} N^{+} - CHR - CO O^{-}$ .
Being ionic, they are crystalline solids with high melting points, soluble in water but insoluble in organic solvents. They are amphoteric: they react with both acids and bases.

Figure 7: Amino acids are amphoteric: they gain

H^{+}

in acid and lose it in base, and they link into peptides through the

- CO - NH -

amide bond.

At the isoelectric point (about pH 6.0 for glycine) the zwitterion has no net charge and does not move in an electric field.

4.2 Peptides and Proteins

The $COOH$ of one amino acid condenses with the $N H_{2}$ of another, losing water, to form an amide link $- CO - NH -$ called the peptide bond. Two amino acids give a dipeptide, three a tripeptide, and many a polypeptide. A polypeptide with more than about a hundred residues and a molecular mass above 10,000 u is called a protein; egg albumin, for example, has a molecular mass of about 45,000 u.

Biuret test for proteins. Warm the sample with dilute NaOH and a few drops of dilute

CuS O_{4}

solution. A violet colour shows peptide bonds, and so proteins.

4.3 Preparation of Amino Acids

(i) Amination of $α$ -halo acids with excess ammonia:

ClC H_{2} COOH + 2N H_{3} glycine H_{2} NC H_{2} COOH + N H_{4} Cl

(ii) Gabriel phthalimide synthesis. Potassium phthalimide displaces chloride from an $α$ -halo ester; acid hydrolysis then frees the amino acid:

C_{6} H_{4} (CO)_{2} NK + ClC H_{2} COO C_{2} H_{5} C_{6} H_{4} (CO)_{2} NC H_{2} COO C_{2} H_{5} + KCl

C_{6} H_{4} (CO)_{2} NC H_{2} COO C_{2} H_{5} + 3 H_{2} O HCl, heat C_{6} H_{4} (COOH)_{2} + H_{2} NC H_{2} COOH + C_{2} H_{5} OH

(iii) Knoop synthesis. An $α$ -keto acid such as pyruvic acid forms an imine with ammonia, which is reduced:

C H_{3} COCOOH + N H_{3} C H_{3} C (= NH) COOH + H_{2} O

C H_{3} C (= NH) COOH + H_{2} Pd (or Na/ C_{2} H_{5} OH) alanine C H_{3} CH (N H_{2}) COOH

(iv) Strecker synthesis. An aldehyde, hydrogen cyanide and ammonia (or $N H_{4} CN$ ) give an $α$ -aminonitrile, which is hydrolysed:

RCHO + HCN cyanohydrin RCH (OH) CN

RCH (OH) CN + N H_{3} aminonitrile RCH (N H_{2}) CN + H_{2} O

RCH (N H_{2}) CN + 2 H_{2} O + HCl heat RCH (N H_{2}) COOH + N H_{4} Cl

4.4 Reactions of Amino Acids

Amino acids show the reactions of both groups:

Reagent	Product from $RCH (N H_{2}) COOH$	Group that reacts
$C_{2} H_{5} OH$ , dry HCl	ester hydrochloride $RCH (N H_{3}^{+} C l^{-}) COO C_{2} H_{5}$ ; free ester with base	$- COOH$
HCl	hydrochloride $RCH (N H_{3}^{+} C l^{-}) COOH$	$- N H_{2}$
NaOH	sodium salt $RCH (N H_{2}) COONa$	$- COOH$
$Ba (OH)_{2}$ , heat	amine $RC H_{2} N H_{2}$ + $C O_{2}$ (decarboxylation)	$- COOH$
$C H_{3} Br$	N-methyl amino acid $RCH (NHC H_{3}) COOH$	$- N H_{2}$
$C H_{3} COCl$	N-acetyl amino acid $RCH (NHCOC H_{3}) COOH$	$- N H_{2}$
$PC l_{5}$	acid chloride hydrochloride $RCH (N H_{3}^{+} C l^{-}) COCl$	$- COOH$
$LiAl H_{4}$	amino alcohol $RCH (N H_{2}) C H_{2} OH$	$- COOH$
$HN O_{2}$	hydroxy acid $RCH (OH) COOH$ + $N_{2}$	$- N H_{2}$

RCH (N H_{2}) COOH + HN O_{2} RCH (OH) COOH + N_{2} + H_{2} O

4.5 Action of Heat

What happens on heating depends on how far apart the two groups are:

$α$ -Amino acids: two molecules lose two molecules of water to form a cyclic diamide, a 2,5-diketopiperazine (glycine gives glycine anhydride).
$β$ -Amino acids: lose ammonia to give an $α, β$ -unsaturated acid.
$γ$ - and $δ$ -amino acids: lose water within one molecule to give a cyclic amide (a lactam); 4-aminobutanoic acid gives pyrrolidin-2-one.

2 H_{2} NC H_{2} COOH heat diketopiperazine C_{4} H_{6} N_{2} O_{2} + 2 H_{2} O

H_{2} NC H_{2} C H_{2} COOH heat acrylic acid C H_{2} = CHCOOH + N H_{3}

H_{2} NC H_{2} C H_{2} C H_{2} COOH heat a lactam C_{4} H_{7} NO + H_{2} O

Figure 8: Heat joins two

α

-amino acids into a ring, eliminates

N H_{3}

from a

β

-amino acid, and closes a

γ

- or

δ

-amino acid into a lactam.

Figure 9: One page on one screen: the atom that carries the carbon (N or O, C or N) decides every product on this page.

5. Solved Examples

Solved Example 1

What is an ambident nucleophile? Give two examples.

Solution:

A nucleophile that can attack through two different atoms is called ambident. Examples: the cyanide ion $C N^{-}$ , which bonds through C (nitriles) or N (isocyanides), and the nitrite ion $N O_{2}^{-}$ , which bonds through N (nitroalkanes) or O (alkyl nitrites). See Figure 1.

Solved Example 2

p-Chloronitrobenzene has a smaller dipole moment (about 2.4 D) than p-nitrotoluene. Explain.

Solution:

Dipole moments of groups on a benzene ring add as vectors (Figure 2).

In p-chloronitrobenzene both Cl and $N O_{2}$ pull electrons out of the ring. Their dipoles point in opposite directions and partly cancel: about $4.2 - 1.7 = 2.5 D$ .
In p-nitrotoluene, $C H_{3}$ pushes electrons into the ring while $N O_{2}$ pulls them out. Both dipoles point the same way and add: about $4.2 + 0.4 = 4.6 D$ .

Solved Example 3

Nitrobenzene is reduced with zinc dust and ammonium chloride to give (A). Identify (A). Will (A) reduce Tollens' reagent?

Solution:

(A) is N-phenylhydroxylamine, $C_{6} H_{5} NHOH$ . Yes, it reduces Tollens' reagent to a silver mirror and is itself oxidised to nitrosobenzene. This is the Mulliken-Barker test for the nitro group.

C_{6} H_{5} N O_{2} Zn/ N H_{4} Cl C_{6} H_{5} NHOH Tollens’ reagent C_{6} H_{5} NO + Ag ↓

Solved Example 4

Nitrobenzene, but not benzene, is used as the solvent for Friedel-Crafts alkylation of bromobenzene. Why?

Solution:

Benzene is more reactive towards electrophilic substitution than bromobenzene, so in benzene the solvent itself would be alkylated. The strongly deactivating $N O_{2}$ group stops nitrobenzene from undergoing Friedel-Crafts reactions at all, so it dissolves the reactants without taking part.

Solved Example 5

An organic compound (A),

C_{2} H_{5} N O_{2}

, gives on hydrolysis an alcohol (B) of molecular mass 46 and an acid. On reduction it gives the same alcohol (B) and ammonia. What is (A)?

Solution:

$C_{2} H_{5} N O_{2}$ could be nitroethane $C_{2} H_{5} N O_{2}$ or ethyl nitrite $C_{2} H_{5} - O - N = O$ . A compound that is hydrolysed to an alcohol, and reduced to an alcohol plus $N H_{3}$ , has a C-O bond, not a C-N bond. The alcohol of molar mass 46 is ethanol. So (A) is ethyl nitrite.

C_{2} H_{5} ONO + H_{2} O H^{+} C_{2} H_{5} OH + HN O_{2}

C_{2} H_{5} ONO + 6 [H] Sn/HCl C_{2} H_{5} OH + N H_{3} + H_{2} O

Solved Example 6

Convert benzene into m-nitroaniline.

Solution:

Nitrate twice (the first $N O_{2}$ directs meta), then reduce one nitro group selectively (Zinin reduction):

C_{6} H_{6} HN O_{3} / H_{2} S O_{4} C_{6} H_{5} N O_{2} fuming HN O_{3} C_{6} H_{4} (N O_{2})_{2} (N H_{4})_{2} S N O_{2} C_{6} H_{4} N H_{2}

Step 1 runs at 323-333 K; step 2 needs fuming nitric acid with conc. $H_{2} S O_{4}$ at 363 K and gives m-dinitrobenzene; step 3 gives m-nitroaniline.

Solved Example 7

Alkyl cyanides are soluble in water but alkyl isocyanides are not. Explain.

Solution:

In $R - C \equiv N$ the nitrogen keeps its lone pair, so water molecules form hydrogen bonds to it and the lower cyanides dissolve. In $R - N^{+} \equiv C^{-}$ nitrogen has used its lone pair in bonding, so no such hydrogen bond forms and isocyanides are only sparingly soluble.

Solved Example 8

An acid chloride

RCOCl

is treated with sodium azide to give X, which on heating gives Y. X and Y are
(A)

RCON

and

RCN

(B)

RCN

and

RNC

(C)

RCO N_{3}

and

RCN

(D)

RCO N_{3}

and

RNCO

Solution:

Answer: (D). Azide ion replaces chloride to give the acyl azide X. On heating it loses nitrogen and the R group migrates from carbon to nitrogen, giving the isocyanate Y. This is the Curtius rearrangement, which is used to make amines (see Preparation of Amines).

RCOCl + Na N_{3} RCO N_{3} + NaCl

RCO N_{3} heat RNCO + N_{2}

Solved Example 9

Why does glycine exist as a zwitterion but anthranilic acid (2-aminobenzoic acid) does not?

Solution:

In glycine the $N H_{2}$ group is on an $sp^{3}$ carbon, so it is basic enough to take the proton released by $COOH$ , giving $H_{3} N^{+} C H_{2} CO O^{-}$ . In anthranilic acid the nitrogen lone pair is delocalised into the ring, and the electron-withdrawing $COOH$ group lowers its electron density further. This aryl nitrogen is too weakly basic to accept the proton, so no zwitterion forms.

Solved Example 10

The product NOT obtained when nitromethane reacts with chlorine and sodium hydroxide is
(A)

ClC H_{2} N O_{2}

(B)

C l_{2} CHN O_{2}

(C)

C l_{3} CN O_{2}

(D)

C H_{3} N H_{2}

Solution:

Answer: (D). In alkali nitromethane forms its nitronate ion, and the acidic $α$ -hydrogens are replaced by chlorine one by one to give (A), (B) and finally chloropicrin (C). Nothing in this mixture can reduce the nitro group, so methylamine cannot form.

C H_{3} N O_{2} + 3C l_{2} + 3NaOH chloropicrin CC l_{3} N O_{2} + 3NaCl + 3 H_{2} O

Solved Example 11

Hydrazobenzene can be obtained by reducing nitrobenzene with
(A) Sn + HCl
(B) Zn +

N H_{4} Cl

(C)

N a_{3} As O_{3}

+ NaOH
(D) Zn + NaOH

Solution:

Answer: (D). Zinc and alkali give the fully reduced dimer, hydrazobenzene (Figure 4). Sn/HCl gives aniline, Zn/ $N H_{4} Cl$ gives phenylhydroxylamine and sodium arsenite gives azoxybenzene.

2 C_{6} H_{5} N O_{2} + 10 [H] Zn/NaOH C_{6} H_{5} NHNH C_{6} H_{5} + 4 H_{2} O

Solved Example 12

When benzamide is heated with thionyl chloride, the main product is
(A)

C_{6} H_{5} CN

(B)

C_{6} H_{5} COCl

(C)

C_{6} H_{5} N H_{2}

(D)

C_{6} H_{5} CSN H_{2}

Solution:

Answer: (A). With an amide, $SOC l_{2}$ acts as a dehydrating agent and removes $H_{2} O$ from $- CON H_{2}$ to give the nitrile.

C_{6} H_{5} CON H_{2} + SOC l_{2} heat C_{6} H_{5} CN + S O_{2} + 2HCl

Solved Example 13

Which of the following reacts with ozone to form an isocyanate?
(A)

C H_{3} CN

(B)

C_{6} H_{5} CN

(C)

C H_{3} NC

(D)

C_{6} H_{5} N H_{2}

Solution:

Answer: (C). The carbene-like carbon of an isocyanide takes up an oxygen atom. Nitriles do not react this way.

C H_{3} NC + O_{3} C H_{3} NCO + O_{2}

Solved Example 14

C H_{3} COCOOH

is treated with (i)

N H_{3}

and then (ii)

H_{2}

/Pd to give X. What is X?
(A)

C H_{3} CON H_{2}

(B)

C H_{3} COCON H_{2}

(C)

C H_{3} C H_{2} CON H_{2}

(D)

C H_{3} CH (N H_{2}) COOH

Solution:

Answer: (D). This is the Knoop synthesis. Pyruvic acid forms an imine with ammonia at the keto group, and hydrogenation of the C=N bond gives alanine.

C H_{3} COCOOH N H_{3}, - H_{2} O C H_{3} C (= NH) COOH H_{2} /Pd C H_{3} CH (N H_{2}) COOH

Solved Example 15

An optically active compound (A),

C_{3} H_{7} O_{2} N

, forms a hydrochloride but dissolves in water to give a neutral solution. On heating with soda lime (A) gives (B),

C_{2} H_{7} N

. Both (A) and (B) react with

NaN O_{2}

and HCl: (A) gives (C),

C_{3} H_{6} O_{3}

, which on heating gives (D),

C_{6} H_{8} O_{4}

; (B) gives (E),

C_{2} H_{6} O

. Identify (A) to (E) and suggest a synthesis of (A).

Solution:

(A) has 1 degree of unsaturation (a C=O), contains N and two O atoms, forms a salt with HCl and yet gives a neutral solution, so it holds both $N H_{2}$ and $COOH$ : an amino acid. Losing $C O_{2}$ gives the saturated amine (B), and (B) gives ethanol with nitrous acid, so (B) is ethylamine. (A) is therefore alanine, which is chiral.

C H_{3} CH (N H_{2}) COOH soda lime, heat (A) to (B) C H_{3} C H_{2} N H_{2} + C O_{2}

C H_{3} C H_{2} N H_{2} + HN O_{2} (E) ethanol C H_{3} C H_{2} OH + N_{2} + H_{2} O

C H_{3} CH (N H_{2}) COOH + HN O_{2} (C) lactic acid C H_{3} CH (OH) COOH + N_{2} + H_{2} O

2C H_{3} CH (OH) COOH heat (D) lactide, a cyclic diester C_{6} H_{8} O_{4} + 2 H_{2} O

Synthesis of (A):

C H_{3} C H_{2} COOH B r_{2} /red P C H_{3} CHBrCOOH excess N H_{3} C H_{3} CH (N H_{2}) COOH

Solved Example 16

Three isomeric nitroalkanes

C_{4} H_{9} N O_{2}

, X, Y and Z, are treated with nitrous acid and then NaOH. X gives a red solution, Y a blue colour and Z no change. Suggest one structure for each.

Solution:

The colour shows how many $α$ -hydrogens each isomer has.

X, red (nitrolic acid), is a 1° nitroalkane: 1-nitrobutane $C H_{3} C H_{2} C H_{2} C H_{2} N O_{2}$ (2-methyl-1-nitropropane also fits).
Y, blue (pseudonitrole), is 2°: 2-nitrobutane $C H_{3} C H_{2} CH (N O_{2}) C H_{3}$ .
Z, no reaction, is 3°: 2-methyl-2-nitropropane $(C H_{3})_{3} CN O_{2}$ .

Solved Example 17

Bromoethane is treated separately with (a) KCN, then the product with (i)

C H_{3} MgBr

(ii)

H_{3} O^{+}

; and (b) AgCN, then the product with dilute HCl. Name every product.

Solution:

(a) KCN gives propanenitrile $C H_{3} C H_{2} CN$ . The Grignard reagent adds to C≡N, and hydrolysis gives the ketone butan-2-one $C H_{3} C H_{2} COC H_{3}$ .

(b) AgCN gives ethyl isocyanide $C H_{3} C H_{2} NC$ . Acid hydrolysis gives ethylamine $C H_{3} C H_{2} N H_{2}$ and formic acid $HCOOH$ .

C H_{3} C H_{2} NC + 2 H_{2} O dil. HCl C H_{3} C H_{2} N H_{2} + HCOOH

Practice Questions

Predict the product: $C H_{3} N O_{2}$ + $C_{6} H_{5} CHO$ with alcoholic KOH.Answer: $β$ -nitrostyrene $C_{6} H_{5} CH = CHN O_{2}$ (Henry reaction, then loss of water)
Benzene is treated with $B r_{2}$ /Fe to give A, and A with $HN O_{3}$ / $H_{2} S O_{4}$ to give B. Identify A and B.Answer: A = bromobenzene; B = 1-bromo-4-nitrobenzene (major) with some 1-bromo-2-nitrobenzene
$C H_{3} C H_{2} Cl$ reacts with $KN O_{2}$ to give A, which is reduced ( $H_{2}$ , $H^{+}$ ) to B. Identify A and B.Answer: A = ethyl nitrite $C H_{3} C H_{2} ONO$ (ionic $KN O_{2}$ attacks through O); B = ethanol, with $N H_{3}$ and water
Name three important nitro compounds used as explosives.Answer: TNT (2,4,6-trinitrotoluene), picric acid (2,4,6-trinitrophenol) and TNB (1,3,5-trinitrobenzene); nitroglycerine is a nitrate ester, not a nitro compound
How are nitriles and isonitriles related to each other?Answer: they are isomers: R is bonded to C in $R - C \equiv N$ and to N in $R - N \equiv C$ ; isonitriles rearrange to nitriles on heating
In the scheme RBr (W) to RCN (Y) to RCHO, and RBr (X) to RNC (Z) to $RNHC H_{3}$ , name reagents W, X, Y and Z.Answer: W = alcoholic KCN; Y = $SnC l_{2}$ /HCl then $H_{3} O^{+}$ (Stephen reduction); X = AgCN; Z = $H_{2}$ /Pt or Na/ $C_{2} H_{5} OH$
Write a reaction that forms methyl isocyanate (MIC). What is formed when it is hydrolysed?Answer: $C H_{3} N H_{2}$ + $COC l_{2}$ gives $C H_{3} NCO$ + 2HCl (or $C H_{3} NC$ + HgO); hydrolysis gives $C H_{3} N H_{2}$ + $C O_{2}$
2-Nitropropane is treated with NaOH and $B r_{2}$ to give A, and A with $(C H_{3})_{2} Zn$ to give B. Identify A and B. Could $C H_{3} MgI$ replace $(C H_{3})_{2} Zn$ ?Answer: A = 2-bromo-2-nitropropane $(C H_{3})_{2} CBrN O_{2}$ ; B = 2-methyl-2-nitropropane $(C H_{3})_{3} CN O_{2}$ ; no, a Grignard reagent attacks the nitro group, while dimethylzinc only replaces Br
Hydroxylamine $N H_{2} OH$ and acetone oxime $(C H_{3})_{2} C = NOH$ both have an O-H group. Which is more acidic, and why?Answer: acetone oxime: its anion $(C H_{3})_{2} C = N - O^{-}$ is stabilised by resonance (charge also on carbon), while the anion of hydroxylamine is not
Which common $α$ -amino acid is optically inactive?Answer: glycine, $H_{2} NC H_{2} COOH$ , whose $α$ -carbon carries two H atoms

Common Mistakes to Avoid

Watch out

Swapping the silver and potassium salts: $AgCN$ gives isocyanides and $KCN$ gives cyanides; $AgN O_{2}$ gives nitroalkanes and $KN O_{2}$ gives alkyl nitrites.
Expecting a 3° nitroalkane to react with nitrous acid, dissolve in NaOH or condense with aldehydes. All of these need an $α$ -hydrogen.
Writing aniline as the product of every reduction of nitrobenzene. Only acid or catalytic reduction gives aniline; neutral and alkaline media give other products.
Giving the same reduction product for a nitroalkane and its isomeric nitrite. The nitroalkane gives an amine; the nitrite gives an alcohol and ammonia.
Proposing a Friedel-Crafts reaction on nitrobenzene. Its ring is too deactivated, which is why it serves as a solvent.
Mixing up the reduction products: $RCN$ gives the 1° amine $RC H_{2} N H_{2}$ , while $RNC$ gives the 2° amine $RNHC H_{3}$ .
Hydrolysing an isocyanide with alkali. Isocyanides are hydrolysed only by acid.
Calling glycine optically active, or drawing solid amino acids as neutral $H_{2} N - CHR - COOH$ . Glycine is achiral, and amino acids exist as zwitterions.
Calling nitroglycerine a nitro compound. It contains $C - O - N O_{2}$ groups, so it is a nitrate ester.

Frequently Asked Questions

What is an ambident nucleophile?

An ambident nucleophile is an ion that can bond through two different atoms. The cyanide ion bonds through carbon to give nitriles or through nitrogen to give isocyanides, and the nitrite ion bonds through nitrogen to give nitroalkanes or through oxygen to give alkyl nitrites. The metal salt used decides which product dominates.

How can you distinguish a nitroalkane from an alkyl nitrite?

Reduce both with tin and hydrochloric acid: a nitroalkane gives a primary amine, while an alkyl nitrite gives an alcohol and ammonia. Aqueous NaOH hydrolyses an alkyl nitrite to an alcohol and sodium nitrite but only dissolves a nitroalkane as its salt. Nitroalkanes also boil at much higher temperatures.

Why are primary and secondary nitroalkanes acidic?

The nitro group pulls electron density from the alpha carbon, and removing an alpha hydrogen gives a nitronate ion whose negative charge spreads onto both oxygen atoms by resonance. Nitromethane has a pKa near 10. Tertiary nitroalkanes have no alpha hydrogen, so they are not acidic and do not dissolve in alkali.

How do you distinguish primary, secondary and tertiary nitroalkanes?

Treat them with nitrous acid and then sodium hydroxide. A primary nitroalkane forms a nitrolic acid that gives a red solution, a secondary one forms a pseudonitrole with a blue colour, and a tertiary one does not react, because it has no alpha hydrogen. Remember the order red, blue, colourless.

Why does silver cyanide give isocyanides but potassium cyanide gives cyanides?

Potassium cyanide is ionic, so the free cyanide ion attacks the alkyl halide through carbon and gives a nitrile. In silver cyanide the carbon is held by a covalent bond to silver, leaving only the nitrogen lone pair free to attack, so an isocyanide forms. Silver nitrite behaves the same way and gives nitroalkanes.

What is a zwitterion in amino acids?

A zwitterion is a dipolar ion that carries a positive and a negative charge in the same molecule. In an amino acid the carboxyl group gives its proton to the amino group, giving H3N+ and COO- groups. This ionic form explains the high melting points, water solubility and amphoteric behaviour of amino acids.

Which questions on nitrogen compounds are common in NEET?

NEET mostly asks about amino acids and peptides: the zwitterion, the peptide bond, essential amino acids and the biuret test. It also asks for the product of reducing nitrobenzene with Sn and HCl, and for the carbylamine reaction of primary amines with chloroform and alcoholic KOH, which gives a foul-smelling isocyanide.

What does JEE Main ask about nitro compounds, cyanides and isocyanides?

JEE Main likes reagent-based product questions: AgCN versus KCN and AgNO2 versus KNO2, the reduction products of nitrobenzene in acidic, neutral and alkaline media, Stephen and Mendius reductions of nitriles, and the different hydrolysis and reduction products of cyanides and isocyanides. Nitration mechanism questions also appear.

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