Amines
Class 12 Chemistry Notes • NEB Syllabus 2083
Syllabus, 20 solved multiple choice questions, 7 short and 3 long answer questions with structures drawn for every benzene-ring compound, handwritten and typed PDF notes and frequently asked questions on aliphatic amines, Hoffmann’s method and aniline.
1. Syllabus
Unit 16: Amines
These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Practise the multiple choice questions first, then work through the solved short and long answer questions, and finish with the PDF notes for quick revision.
2. Multiple Choice Questions
- a) 2
- b) 3
- c) 4
- d) 5
Correct Answer: c) 4
Reason: \(C_3H_9N\) is a saturated amine, and four structures are possible: two primary amines (propan-1-amine, \(CH_3CH_2CH_2NH_2\), and propan-2-amine, \((CH_3)_2CHNH_2\)), one secondary amine (N-methylethanamine, \(CH_3NHCH_2CH_3\)) and one tertiary amine (N,N-dimethylmethanamine, \((CH_3)_3N\)). So there are four isomers.
| Type | IUPAC name | Structure |
|---|---|---|
| Primary | Propan-1-amine | \(CH_3CH_2CH_2NH_2\) |
| Primary | Propan-2-amine | \((CH_3)_2CHNH_2\) |
| Secondary | N-Methylethanamine | \(CH_3NHCH_2CH_3\) |
| Tertiary | N,N-Dimethylmethanamine | \((CH_3)_3N\) |
- a) \(CH_3CH_2CH_2NH_2\)
- b) \(CH_3CH_2NHCH_3\)
- c) \((CH_3)_3N\)
- d) \((CH_3)_4N^+Cl^-\)
Correct Answer: c) \((CH_3)_3N\)
Reason: A tertiary amine has three alkyl groups bonded to nitrogen and no N-H bond. In \((CH_3)_3N\) all three hydrogens of ammonia are replaced by methyl groups. Option a is a primary amine (one alkyl group), b is a secondary amine (two alkyl groups) and d is a quaternary ammonium salt (four alkyl groups on \(N^+\)).
- a) Nitrous acid
- b) Nitrolic acid
- c) Diethyl oxalic acid
- d) Diethyl oxalate
Correct Answer: d) Diethyl oxalate
Reason: In Hofmann’s method the mixture is heated with diethyl oxalate, \((COOC_2H_5)_2\). A primary amine gives a solid dialkyl oxamide, a secondary amine gives an oily liquid dialkyl oxamic ester and a tertiary amine does not react. The three products are then separated by filtration and distillation. Nitrous acid only distinguishes the three types of amines, it does not separate them.
- a) Ethanamine
- b) N,N-diethylethanamine
- c) N-ethylethanamine
- d) N,N,N,N-tetraethylammonium chloride
Correct Answer: d) N,N,N,N-tetraethylammonium chloride
Reason: Ammonia is alkylated again and again by chloroethane. Ethanamine, diethylamine and triethylamine are formed one after the other, and with excess alkyl halide the tertiary amine takes one more ethyl group and becomes the quaternary ammonium salt, tetraethylammonium chloride, \((C_2H_5)_4N^+Cl^-\). The primary amine is the main product only when ammonia is in excess.
- a) Ethyl isocyanide
- b) N-ethylethanamine
- c) Ethanamine
- d) N,N-diethylethanamine
Correct Answer: c) Ethanamine
Reason: Tin and hydrochloric acid reduce the nitro group to the amino group: \(CH_3CH_2NO_2+6[H]\xrightarrow{Sn/HCl}CH_3CH_2NH_2+2H_2O\). Nitroethane therefore gives the primary amine ethanamine, with the same number of carbon atoms.
- a) \(NH_3\)
- b) \(CH_3NH_2\)
- c) \((CH_3)_2NH\)
- d) \((CH_3)_3N\)
Correct Answer: c) \((CH_3)_2NH\)
Reason: The basic character depends on how easily the lone pair on nitrogen takes a proton. Alkyl groups push electrons to nitrogen (\(+I\) effect), so every amine is more basic than ammonia. In aqueous solution the cation formed is also stabilised by hydration, which is best for primary, then secondary ions, and the bulky tertiary amine suffers steric hindrance. The balance of these three effects makes dimethylamine the strongest base of this set in water (\(pK_b\): \(NH_3\) 4.75, \(CH_3NH_2\) 3.36, \((CH_3)_2NH\) 3.27, \((CH_3)_3N\) 4.22). In the gas phase, with no hydration, the order is \((CH_3)_3N>(CH_3)_2NH>CH_3NH_2>NH_3\), and trimethylamine would be the strongest.
- a) Propanoic acid
- b) Propanal
- c) Propanol
- d) Propene
Correct Answer: c) Propanol
Reason: A primary aliphatic amine reacts with nitrous acid (\(NaNO_2+HCl\), \(0-5^\circ C\)) and gives the alcohol with the same carbon chain, with evolution of nitrogen gas: \(CH_3CH_2CH_2NH_2+HNO_2\rightarrow CH_3CH_2CH_2OH+N_2\uparrow+H_2O\). Propan-1-ol is the main product (a little propan-2-ol and propene are formed as by-products).
- a) \(NH_3\)
- b) \(CH_3NH_2\)
- c) \((CH_3)_2NH\)
- d) \((CH_3)_3N\)
Correct Answer: b) \(CH_3NH_2\)
Reason: Only primary amines (aliphatic and aromatic) give the carbylamine reaction. They are heated with chloroform and alcoholic \(KOH\) and give an alkyl isocyanide with a very bad smell: \(CH_3NH_2+CHCl_3+3KOH\xrightarrow{\Delta}CH_3NC+3KCl+3H_2O\). Ammonia, secondary amines and tertiary amines do not give it.
- a) Alkanes
- b) Alkenes
- c) Amides
- d) Amines
Correct Answer: d) Amines
Reason: Reduction of an alkyl cyanide (nitrile) with \(LiAlH_4\) (or \(H_2/Ni\), or \(Na/C_2H_5OH\)) converts the \(-C\equiv N\) group into \(-CH_2NH_2\), and a primary amine with one more carbon than the alkyl halide is formed: \(R-C\equiv N+4[H]\rightarrow R-CH_2NH_2\). For example, \(CH_3CN\) gives \(CH_3CH_2NH_2\).
- a) \(1^\circ>2^\circ>3^\circ>NH_3\)
- b) \(1^\circ>3^\circ>2^\circ>NH_3\)
- c) \(3^\circ>2^\circ>1^\circ>NH_3\)
- d) \(2^\circ>1^\circ>3^\circ>NH_3\)
Correct Answer: d) \(2^\circ>1^\circ>3^\circ>NH_3\)
Reason: In water three factors work together: the \(+I\) effect of the alkyl groups (increases basicity), the hydration of the ammonium ion formed (best for \(RNH_3^+\), poorer for \(R_3NH^+\)), and the steric hindrance of the alkyl groups (largest in the tertiary amine). For the methylamines the result is \((CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3\), that is \(2^\circ>1^\circ>3^\circ>NH_3\) (\(pK_b\) = 3.27, 3.36, 4.22 and 4.75). Compare this with the gas phase, where the order is \(3^\circ>2^\circ>1^\circ>NH_3\) because there is no hydration.
- a) P-nitroethane, Q-ethanamine, R-ethylisocyanide, S-ethylethanamine
- b) P-nitromethane, Q-ethanamine, R-ethylisocyanide, S-ethylethanamine
- c) P-nitroethane, Q-ethanamine, R-ethylisocyanide, S-ethylethanamine
- d) P-methane, Q-nitromethane, R-methanamine, S-methylisocyanide
Correct Answer: d) P-methane, Q-nitromethane, R-methanamine, S-methylisocyanide
Reason: Work backwards from the last clue. A secondary amine with two carbon atoms is dimethylamine, \((CH_3)_2NH\), and it is formed by the reduction of methyl isocyanide, \(CH_3NC\) (\(R-NC+4[H]\rightarrow R-NHCH_3\)). So S is methyl isocyanide. It is made by the carbylamine reaction of the primary amine methanamine (R), which is obtained from nitromethane (Q) by reduction with \(Sn/HCl\). Nitromethane is made by nitration of methane (P). Options a, b and c start with a two-carbon chain and name an ethyl compound, so their S would be an ethyl isocyanide, not a compound that reduces to a two-carbon secondary amine.
- a) Ortho
- b) Para
- c) Meta
- d) Ortho and para
Correct Answer: d) Ortho and para
Reason: The lone pair of nitrogen of the \(-NH_2\) group is delocalised into the ring (\(+M\) effect). The resonating structures show negative charge on the carbons at the ortho and para positions, so these are the electron-rich positions and the electrophile attacks them. The \(-NH_2\) group is a strong activating, ortho-para directing group (so aniline gives 2,4,6-tribromoaniline with bromine water).
Figure: Resonating structures of aniline (negative charge at the ortho and para carbons)
| Reactant(s) | Conditions | Product |
|---|---|---|
| \(CH_3NH_2\) | alc. \(KOH/CHCl_3\) | M |
| N | \(HNO_2+HCl\) at freezing temperature | \(C_6H_5N_2^+Cl^-+H_2O\) (O) |
- a) M – Methanol, N – Sn/HCl, O – Nitrobenzene
- b) M – Ethylcyanide, N – Zn/NH4Cl, O – Aniline
- c) M – Methyl isocyanide, N – Aniline, O – Benzene diazonium chloride
- d) M – Methanal, N – Zn/NH4Cl, O – Aniline
Correct Answer: c) M – Methyl isocyanide, N – Aniline, O – Benzene diazonium chloride
Reason: (1) Methanamine, a primary amine, heated with chloroform and alcoholic \(KOH\) gives methyl isocyanide (carbylamine reaction), so M is methyl isocyanide. (2) The product O is benzene diazonium chloride, which is formed by diazotisation of a primary aromatic amine with \(HNO_2+HCl\) at \(273-278\,K\), so N is aniline.
Figure: Diazotisation of aniline
- a) o-aminodiazobenzene
- b) Azobenzene
- c) p-aminodiazobenzene
- d) Hydrazobenzene
Correct Answer: c) p-aminodiazobenzene
Reason: Aniline couples with benzene diazonium chloride in a weakly acidic medium (\(H^+\), \(273-278\,K\)) and the diazonium ion attacks the para position of the activated ring. The product is \(p\)-aminoazobenzene (4-aminoazobenzene), a yellow azo dye, and hydrochloric acid: \(C_6H_5N_2^+Cl^-+C_6H_5NH_2\rightarrow C_6H_5-N=N-C_6H_4-NH_2+HCl\).
Figure: Coupling of benzene diazonium chloride with aniline
- a) Ethanal
- b) Ethanamide
- c) Ethanoic acid
- d) Ethanamine
Correct Answer: d) Ethanamine
Reason: The alcohol of mass \(46\) amu is ethanol, \(C_2H_5OH\) (\(2\times12+6\times1+16=46\)). Nitrous acid changes a primary aliphatic amine into an alcohol, so M is ethanamine: \(C_2H_5NH_2+HNO_2\rightarrow C_2H_5OH+N_2\uparrow+H_2O\). Ethanal, ethanamide and ethanoic acid do not give ethanol with nitrous acid (ethanamide gives ethanoic acid).
- a) N-methyl methanamine
- b) Ethanamide
- c) Methyl cyanide
- d) Ethanoic acid
Correct Answer: a) N-methyl methanamine
Reason: The reduction of an isocyanide (\(R-N^+\equiv C^-\)) with \(H_2/Ni\) or \(LiAlH_4\) gives a secondary amine with a methyl group on the nitrogen: \(R-NC+4[H]\rightarrow R-NH-CH_3\). So methyl isocyanide gives \((CH_3)_2NH\), N-methylmethanamine (dimethylamine). Reduction of an alkyl cyanide would give a primary amine instead.
- a) Amino group is meta-directing group
- b) Amino group is reduced
- c) Amino group is oxidized
- d) Amino group is basic in nature
Correct Answer: c) Amino group is oxidized
Reason: Nitric acid is a strong oxidising agent, and the \(-NH_2\) group of aniline is easily oxidised, so direct nitration gives tarry oxidation products and a poor yield. Also in the strongly acidic nitrating mixture aniline changes to the anilinium ion (\(-NH_3^+\)), which is meta directing, and a lot of the meta product is formed. The amino group is therefore first protected by acetylation (acetanilide), the ring is nitrated at the para position, and the protecting group is removed by hydrolysis.
Figure: Protection of the amino group (acetylation of aniline)
Figure: Nitration of acetanilide and removal of the protecting group
- a) Nitrous acid
- b) Diethyl oxalate
- c) Fehling’s reagent
- d) Tollen’s reagent
Correct Answer: b) Diethyl oxalate
Reason: The three compounds are a primary, a secondary and a tertiary amine. Diethyl oxalate reacts with the primary amine to give a solid N,N’-dimethyloxamide, with the secondary amine to give an oily liquid ester (ethyl N,N-dimethyloxamate) and does not react with the tertiary amine. Fehling’s and Tollens’ reagents are for aldehydes, and nitrous acid only distinguishes the three types (it destroys the primary amine).
- a) To make \(-NH_2\) group active
- b) To make \(-NH_2\) group less reactive
- c) To make benzene ring less active
- d) Nitration is addition of acyl group
Correct Answer: b) To make \(-NH_2\) group less reactive
Reason: When the \(-NH_2\) group is changed to \(-NHCOCH_3\) in acetanilide, the lone pair of nitrogen is partly drawn towards the carbonyl group, so the amino group becomes less reactive (less electron releasing) and cannot be oxidised by nitric acid or protonated in the acid mixture. It is still ortho-para directing, so the nitro group goes mainly to the para position. After nitration the acetyl group is removed by hydrolysis.
Figure: Protection of the amino group
- a) N-methyl methanamine
- b) Ethanamide
- c) Methyl cyanide
- d) Ethanoic acid
Correct Answer: a) N-methyl methanamine
Reason: This question is the same as Question 16 in the textbook. Methyl isocyanide, \(CH_3-N^+\equiv C^-\), takes four hydrogen atoms on reduction, the isocyanide carbon becomes a methyl group, and the secondary amine N-methylmethanamine, \((CH_3)_2NH\), is formed: \(CH_3NC+4[H]\xrightarrow{H_2/Ni}(CH_3)_2NH\).
3. Numerical and Subjective Questions
Short Answer Questions
V. Nitrobenzene reduced with \(Sn/HCl\) gives aniline, \(C_6H_5NH_2\). W. A primary aromatic amine with \(NaNO_2\) and dilute \(HCl\) at \(273-278\,K\) (freezing temperature) gives the diazonium salt, so W is benzene diazonium chloride. X. On hydrolysis (warming with water) the diazonium group is replaced by \(-OH\), so X is phenol. Y. Phenol gives a violet colour with neutral \(FeCl_3\) because of the complex ion, so Y is hexaphenoxoferrate(III) ion, \([Fe(OC_6H_5)_6]^{3-}\). Z. The diazonium salt (W) couples with phenol (X) in a weakly alkaline medium at the para position, so Z is the azo dye p-hydroxyazobenzene (4-hydroxyazobenzene).
Figure: Reduction of nitrobenzene to aniline
Figure: Diazotisation of aniline
Figure: Hydrolysis of benzene diazonium chloride
Figure: Ferric chloride test of phenol
Figure: Coupling of benzene diazonium chloride with phenol
| Compound | Name | Formula |
|---|---|---|
| V | Aniline | \(C_6H_5NH_2\) |
| W | Benzene diazonium chloride | \(C_6H_5N_2^+Cl^-\) |
| X | Phenol | \(C_6H_5OH\) |
| Y | Hexaphenoxoferrate(III) ion (violet) | \([Fe(OC_6H_5)_6]^{3-}\) |
| Z | 4-Hydroxyazobenzene | \(C_6H_5-N=N-C_6H_4-OH\) |
| Reactant(s) | Reagents / conditions | Products |
|---|---|---|
| P | \(Cu_2Cl_2/HCl\) | Q |
| Q | \(NH_3/Cu_2O\) | R |
| R | \(NaNO_2/dil.\ HCl\) | P |
| P | S | P-amino diazobenzene |
P. The diazonium group is replaced by chlorine with \(Cu_2Cl_2/HCl\) (Sandmeyer reaction), and P must be a diazonium salt. The row R with \(NaNO_2/dil.\ HCl\) gives P shows that P is made by the diazotisation of a primary aromatic amine. So P is benzene diazonium chloride. Q is chlorobenzene (Sandmeyer reaction). R. Chlorobenzene heated with ammonia and \(Cu_2O\) at about \(473\,K\) under high pressure gives aniline, so R is aniline, and aniline with \(NaNO_2/HCl\) gives P again. S. p-Aminoazobenzene (\(C_6H_5-N=N-C_6H_4-NH_2\)) is formed from the diazonium salt P and aniline, so S is aniline.
Figure: Sandmeyer reaction
Figure: Q to R
Figure: R to P
Figure: Coupling of benzene diazonium chloride with aniline
The reaction between P (benzene diazonium chloride) and S (aniline) is the coupling reaction (azo coupling). It is an electrophilic substitution: the diazonium ion is the electrophile and it attacks the para position of aniline.
| Compound | Name | Formula |
|---|---|---|
| P | Benzene diazonium chloride | \(C_6H_5N_2^+Cl^-\) |
| Q | Chlorobenzene | \(C_6H_5Cl\) |
| R | Aniline | \(C_6H_5NH_2\) |
| S | Aniline | \(C_6H_5NH_2\) |
Figure: Aniline to A to B
Figure: B to C to D
Figure: D to E
| Step | Reaction | Compound |
|---|---|---|
| 1 | Diazotisation of aniline at \(273-278\,K\) | A = benzene diazonium chloride, \(C_6H_5N_2^+Cl^-\) |
| 2 | Hydrolysis of the diazonium salt (warm water) | B = phenol, \(C_6H_5OH\) |
| 3 | Reduction of phenol with zinc dust | C = benzene, \(C_6H_6\) |
| 4 | Nitration of benzene | D = nitrobenzene, \(C_6H_5NO_2\) |
| 5 | Reduction of nitrobenzene | E = aniline, \(C_6H_5NH_2\) |
The sequence ends with the aniline we started with, so the cycle is aniline, phenol, benzene, nitrobenzene and aniline again.
The beaker contains a primary amine (ethanamine), a secondary amine (N-ethylethanamine, diethylamine) and a tertiary amine (N,N-diethylethanamine, triethylamine). They are separated by Hofmann’s method with diethyl oxalate.
Step 1. Add diethyl oxalate to the mixture and warm it. The primary amine gives a solid crystalline N,N’-diethyloxamide. The secondary amine gives an oily liquid, ethyl N,N-diethyloxamate (dialkyl oxamic ester). The tertiary amine has no N-H bond and does not react.
Step 2. Filter the mixture. The residue on the filter paper is the solid dialkyl oxamide. The filtrate contains the oily oxamic ester and the free tertiary amine.
Step 3. Heat the solid with aqueous \(KOH\) and distil. Potassium oxalate stays behind and the primary amine, ethanamine, distils over.
Step 4. Fractionally distil the filtrate. The tertiary amine (triethylamine, boiling point \(89^\circ C\)) distils first and is collected. The oily ester remains as the residue.
Step 5. Heat the residue with \(KOH\) and distil. The secondary amine, N-ethylethanamine, distils over.
| Amine | With diethyl oxalate | How it is recovered |
|---|---|---|
| Primary (ethanamine) | Solid dialkyl oxamide | Filter, heat with \(KOH\), distil |
| Secondary (N-ethylethanamine) | Oily liquid oxamic ester | Residue after distillation, heat with \(KOH\), distil |
| Tertiary (N,N-diethylethanamine) | No reaction | Fractional distillation of the filtrate |
a) What happens when benzamide is heated with aq. KOH in presence of bromine at about 70°C?
Figure: Hofmann bromamide reaction of benzamide
This is the Hofmann bromamide degradation. Benzamide is changed into aniline, a primary amine with one carbon atom less (the carbonyl carbon is lost as potassium carbonate):
b) Between ammonia and aniline, which one is more basic? Explain.
Figure: Resonating structures of aniline (negative charge at the ortho and para carbons)
Ammonia is more basic than aniline (\(pK_b\) 4.75 for \(NH_3\) and 9.38 for aniline). In ammonia the lone pair on nitrogen is fully available to accept a proton. In aniline the lone pair is delocalised into the benzene ring (resonance, see the structures above), so it is less available for protonation. The anilinium ion \(C_6H_5NH_3^+\) also has no such resonance stabilisation, so the equilibrium \(C_6H_5NH_2+H^+\rightleftharpoons C_6H_5NH_3^+\) lies on the left.
c) Convert ethanamine into methanamine.
Ethanamine has two carbon atoms and methanamine has one, so the carbon chain must be shortened by one carbon. This is done by Hofmann bromamide degradation of ethanamide, which is made from ethanamine through ethanol and ethanoic acid:
| Step | Reaction | Product |
|---|---|---|
| 1 | Ethanamine with \(HNO_2\) (\(NaNO_2+HCl\)) | Ethanol (\(N_2\) escapes) |
| 2 | Oxidation of ethanol | Ethanoic acid |
| 3 | Ethanoic acid with \(NH_3\) gives ammonium ethanoate, which loses water on heating | Ethanamide |
| 4 | Ethanamide with \(Br_2\) and \(KOH\) (about \(343\,K\)) | Methanamine |
a) Convert methanamine to ethanamine [2]
One carbon atom has to be added, which is a chain-lengthening (step-up) conversion. The amine group is first replaced by a halogen, the nitrile is formed with \(KCN\), and the nitrile is reduced to the amine, which has one more carbon:
| Step | Reaction | Product |
|---|---|---|
| 1 | Methanamine with \(HNO_2\) | Methanol |
| 2 | Methanol with \(PCl_5\) (or \(SOCl_2\)) | Chloromethane |
| 3 | Chloromethane with alcoholic \(KCN\) | Ethanenitrile (methyl cyanide) |
| 4 | Reduction with \(LiAlH_4\) (or \(Na/C_2H_5OH\)) | Ethanamine |
b) Increasing order of basicity [1]
The \(pK_b\) values are about \(9.38\) for aniline, \(3.36\) for methanamine and \(3.19\) for ethanamine, and the smaller \(pK_b\) means the stronger base.
c) Justification [2]
Aniline is the weakest because the lone pair of nitrogen is delocalised into the benzene ring by resonance, so it is not easily available for protonation. Methanamine and ethanamine are aliphatic amines. Their alkyl groups push electrons towards nitrogen (\(+I\) effect), so the lone pair is more available and the ammonium ion formed is stabilised. The ethyl group has a larger \(+I\) effect than the methyl group, so ethanamine is more basic than methanamine. So the order is aniline < methanamine < ethanamine.
C. An aliphatic primary amine of mass \(45\) amu is \(C_2H_7N\) (\(2\times12+7+14=45\)), which is ethanamine, \(CH_3CH_2NH_2\). B. C is formed from an amide by the Hofmann reaction (\(Br_2/aq.KOH\)), which removes one carbon, so B is propanamide, \(CH_3CH_2CONH_2\). A. The amide is made from ammonia and a carboxylic acid (via the ammonium salt, heated), so A is propanoic acid, \(CH_3CH_2COOH\) (propanoyl chloride, an anhydride or an ester would also give propanamide with \(NH_3\)). D. A primary amine with nitrous acid gives an alcohol, so D is ethanol. E. Ethanol has the \(CH_3CH(OH)-\) group, so it gives the iodoform test with \(NaOH/I_2\), and E is iodoform, \(CHI_3\) (yellow precipitate).
| Compound | Name | Formula |
|---|---|---|
| A | Propanoic acid | \(CH_3CH_2COOH\) |
| B | Propanamide | \(CH_3CH_2CONH_2\) |
| C | Ethanamine | \(CH_3CH_2NH_2\) |
| D | Ethanol | \(CH_3CH_2OH\) |
| E | Iodoform | \(CHI_3\) |
Long Answer Questions
A. An aliphatic gaseous compound ‘L’ is passed into red hot iron tube under pressure and produces a compound ‘M’. ‘M’ is under nitration and gives ‘N’. By the reduction of ‘N’, ‘O’ is obtained. Again for the nitration of ‘O’, the active site is protected and the product ‘P’ is obtained in which major product is p-nitroaniline. Identify the L, M, N, O and P from the information given.
L and M. The gaseous aliphatic compound that polymerises in a red-hot iron tube to an aromatic compound is ethyne (acetylene): three molecules join to give benzene (M). N. Nitration of benzene with \(conc.\ HNO_3+conc.\ H_2SO_4\) gives nitrobenzene. O. Reduction of nitrobenzene with \(Sn/HCl\) gives aniline. P. The amino group of aniline is protected by acetylation (acetanilide), the ring is nitrated, mainly at the para position, and the acetyl group is removed by hydrolysis. The final product P is p-nitroaniline (4-nitroaniline), with a little o-nitroaniline.
Figure: L (ethyne) to M (benzene) to N (nitrobenzene)
Figure: N to O and protection of the amino group
Figure: Nitration of acetanilide to give P, p-nitroaniline
| Compound | Name | Formula |
|---|---|---|
| L | Ethyne | \(HC\equiv CH\) |
| M | Benzene | \(C_6H_6\) |
| N | Nitrobenzene | \(C_6H_5NO_2\) |
| O | Aniline | \(C_6H_5NH_2\) |
| P | p-Nitroaniline (through p-nitroacetanilide) | \(p\text{-}NO_2C_6H_4NH_2\) |
B. Give an example of each of the following. i. Coupling reaction of aniline ii. Separation of primary amine only from the mixture. iii. Test of secondary amine.
i. Coupling reaction of aniline. Aniline couples with benzene diazonium chloride at the para position, in weakly acidic solution at \(273-278\,K\), and gives the yellow azo dye p-aminoazobenzene:
Figure: Coupling of benzene diazonium chloride with aniline
ii. Separation of the primary amine only from a mixture. The mixture is warmed with diethyl oxalate. Only the primary amine forms a solid dialkyl oxamide, which is filtered off, and heating it with \(KOH\) gives back the pure primary amine. For example, from a mixture of ethanamine and triethylamine:
Another method is Hinsberg’s test: benzenesulphonyl chloride, \(C_6H_5SO_2Cl\), gives with a primary amine a sulphonamide that has an acidic N-H and dissolves in aqueous \(KOH\), while the secondary amine derivative is insoluble and the tertiary amine does not react.
iii. Test of a secondary amine. A secondary amine reacts with nitrous acid (\(NaNO_2+HCl\), \(273-278\,K\)) and gives a yellow oily N-nitrosoamine, which also gives the green colour in Liebermann’s nitroso test (a primary amine gives nitrogen gas and an alcohol instead):
A. Write down the detail description of separation of the mixture of the following compounds. Methanamine (\(CH_3NH_2\)), N-methylethanamine (\(CH_3-NH-CH_2CH_3\)) and N,N-dimethylmethanamine (\((CH_3)_3N\)).
The mixture has a primary amine (methanamine), a secondary amine (N-methylethanamine) and a tertiary amine (trimethylamine). It is separated by Hofmann’s method, using diethyl oxalate.
1. Treatment with diethyl oxalate. The mixture is warmed with diethyl oxalate. The primary amine gives a solid, N,N’-dimethyloxamide. The secondary amine gives an oily liquid, ethyl N-ethyl-N-methyloxamate. The tertiary amine does not react.
2. Filtration. The solid dialkyl oxamide stays on the filter paper. The filtrate has the oily ester and trimethylamine.
3. Primary amine. The solid is heated with \(KOH\) and distilled, and methanamine distils over: \((CONHCH_3)_2+2KOH\rightarrow(COOK)_2+2CH_3NH_2\uparrow\).
4. Tertiary amine. The filtrate is fractionally distilled, and trimethylamine (boiling point \(3^\circ C\), the most volatile) is collected first.
5. Secondary amine. The oily residue is heated with \(KOH\) and distilled, and N-methylethanamine distils over:
| Amine | Product with diethyl oxalate | Recovered by |
|---|---|---|
| Methanamine (1°) | N,N’-Dimethyloxamide, solid | Filtration, then \(KOH\) and distillation |
| N-Methylethanamine (2°) | Ethyl N-ethyl-N-methyloxamate, oily liquid | Residue, then \(KOH\) and distillation |
| N,N-Dimethylmethanamine (3°) | No reaction | Fractional distillation of the filtrate |
B. What product would you obtain when nitrobenzene is treated with a. Zn/NH₄Cl/Δ b. NaOH/Na₃AsO₃ c. Conc. H₂SO₄ in presence of heat
Nitrobenzene gives different products with different reducing conditions, because the medium decides how far the \(-NO_2\) group is reduced.
a. \(Zn/NH_4Cl\), heat (neutral medium) gives N-phenylhydroxylamine, \(C_6H_5NHOH\):
Figure: Nitrobenzene with zinc and ammonium chloride
b. \(NaOH/Na_3AsO_3\) (alkaline medium, mild reducing agent) gives azoxybenzene, \(C_6H_5-N(O)=N-C_6H_5\):
Figure: Nitrobenzene with sodium arsenite in alkali
c. Conc. \(H_2SO_4\), heat sulphonates the ring. The \(-NO_2\) group is meta directing and strongly deactivating, so the product is 3-nitrobenzenesulphonic acid (m-nitrobenzenesulphonic acid); fuming sulphuric acid at about \(380\,K\) is used in practice:
Figure: Sulphonation of nitrobenzene
i) Diethyl oxalate with A and with B
A, \((CH_3)_2NH\), is a secondary amine and B, \(CH_3NH_2\), is a primary amine.
With A (dimethylamine): only one \(-OC_2H_5\) group of the ester is replaced, and an oily liquid, ethyl N,N-dimethyloxamate (dialkyl oxamic ester), is formed.
With B (methanamine): both \(-OC_2H_5\) groups are replaced, and a white crystalline solid, N,N’-dimethyloxamide (dialkyl oxamide), is formed.
ii) Obtaining B (methanamine) by Hofmann’s method
Hofmann bromamide degradation of ethanamide (acetamide) gives methanamine. The amide is heated with bromine and aqueous alkali, and the amine has one carbon atom less than the amide:
iii) Steps to obtain methanamine from ethanamine
Ethanamine has two carbon atoms and methanamine has one, so the carbon chain must be shortened by one carbon. This is done by Hofmann bromamide degradation of ethanamide, which is made from ethanamine through ethanol and ethanoic acid:
| Step | Reaction | Product |
|---|---|---|
| 1 | Ethanamine with \(HNO_2\) (\(NaNO_2+HCl\)) | Ethanol (\(N_2\) escapes) |
| 2 | Oxidation of ethanol | Ethanoic acid |
| 3 | Ethanoic acid with \(NH_3\) gives ammonium ethanoate, which loses water on heating | Ethanamide |
| 4 | Ethanamide with \(Br_2\) and \(KOH\) (about \(343\,K\)) | Methanamine |
iv) Three methods of preparation of aniline
1. Reduction of nitrobenzene with tin and concentrated hydrochloric acid:
Figure: Reduction of nitrobenzene to aniline
2. From phenol by heating with ammonia in the presence of anhydrous zinc chloride (nucleophilic substitution of \(-OH\)):
Figure: Preparation of aniline from phenol
3. Hofmann bromamide degradation of benzamide (benzamide with bromine and aqueous \(KOH\), about \(343\,K\)):
Figure: Hofmann bromamide reaction of benzamide
Another method is the ammonolysis of chlorobenzene with ammonia and \(Cu_2O\) at about \(473\,K\) under high pressure.
Figure: Ammonolysis of chlorobenzene
4. View PDF Notes
i. Handwritten Notes
ii. Typed PDF Notes
5. Frequently Asked Questions
What are amines? (Amines class 12 notes)
What topics are included in Unit 16 Amines of NEB Class 12 Chemistry?
How are amines named?
What is the classification of amines?
What are the isomers of C3H9N?
How can primary, secondary and tertiary amines be separated by Hoffman's method? (Separation of 1 2 3 amine by Hoffman's method)
How are primary amines prepared?
What is Hofmann bromamide degradation?
What are the physical properties of aliphatic amines?
Why are amines basic?
What is the order of basic strength of aliphatic amines in the gaseous state and in aqueous solution?
Why is aniline less basic than ammonia and aliphatic amines?
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How do 1, 2 and 3 amines react with nitrous acid?
What is the reaction of amines with acid chlorides and alkyl halides?
What are the uses of aliphatic amines?
What is aniline? (Aromatic amine)
How is aniline prepared?
What are the chemical reactions of aniline?
What is diazotisation and what is a coupling reaction?
Why is the amino group of aniline protected during nitration?
What is the ortho-para directing nature of aniline?
What products does nitrobenzene give with different reducing agents?
Where can I get Amines class 12 notes PDF for NEB? (Amines PDF)
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