Chapter 13 Amines Solutions
Question - 11 : - Give one chemical test to distinguish between thefollowing pairs of compounds.
Answer - 11 : -
(i) Methylamineand dimethylamine
(ii) Secondaryand tertiary amines
(iii) Ethylamine and aniline
(iv) Anilineand benzylamine
(v) Anilineand N-methylaniline.
Answer
(i) Methylamineand dimethylamine can be distinguished by the carbylamine test.
Carbylamine test: Aliphatic and aromatic primary amines onheating with chloroform and ethanolic potassium hydroxide form foul-smellingisocyanides or carbylamines. Methylamine (being an aliphatic primary amine)gives a positive carbylamine test, but dimethylamine does not.
(ii) Secondaryand tertiary amines can be distinguished by allowing them to react withHinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl).
Secondary amines react with Hinsberg’s reagent to form aproduct that is insoluble in an alkali. For example, N, N−diethylamine reactswith Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which isinsoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’sreagent.
(iii) Ethylamineand aniline can be distinguished using the azo-dye test. A dye is obtained whenaromatic amines react with HNO2 (NaNO2 +dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphaticamines give a brisk effervescence due (to the evolution of N2 gas)under similar conditions.
(iv) Anilineand benzylamine can be distinguished by their reactions with the help ofnitrous acid, which is prepared in situ from a mineral acid and sodium nitrite.Benzylamine reacts with nitrous acid to form unstable diazonium salt, which inturn gives alcohol with the evolution of nitrogen gas.
On the other hand, aniline reacts with HNO2 ata low temperature to form stable diazonium salt. Thus, nitrogen gas is notevolved.
(v) Anilineand N-methylaniline can be distinguished using the Carbylamine test. Primaryamines, on heating with chloroform and ethanolic potassium hydroxide, formfoul-smelling isocyanides or carbylamines. Aniline, being an aromatic primaryamine, gives positive carbylamine test. However, N-methylaniline, being asecondary amine does not.
Account for the following:
Answer - 12 : -
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamineis soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride toprecipitate hydrated ferric oxide.
(iv) Althoughamino group is o, p− directing in aromaticelectrophilic substitution reactions, aniline on nitration gives a substantialamount of m-nitroaniline.
(v) Anilinedoes not undergo Friedel-Crafts reaction.
(vi) Diazoniumsalts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred forsynthesising primary amines.
Answer
(i) pKb of aniline is more than that of methylamine:
Aniline undergoes resonance and as a result, the electronson the N-atom are delocalized over the benzene ring. Therefore, the electronson the N-atom are less available to donate.
On the other hand, in case of methylamine(due to the +I effect of methyl group), the electron density on the N-atom isincreased. As a result, aniline is less basic than methylamine. Thus, pKb ofaniline is more than that of methylamine.
(ii) Ethylamineis soluble in water whereas aniline is not:
Ethylamine when added to water forms intermolecularH−bonds with water. Hence, it is soluble in water
But aniline does not undergo H−bonding withwater to a very large extent due to the presence of a large hydrophobic −C6H5 group.Hence, aniline is insoluble in water.
(iii) Methylaminein water reacts with ferric chloride to precipitate hydrated ferric oxide:
Due to the +I effect of −CH3 group,methylamine is more basic than water. Therefore, in water, methylamine producesOH− ions by accepting H+ ionsfrom water.
Ferric chloride (FeCl3)dissociates in water to form Fe3+ and Cl− ions.
Then, OH− ion reacts withFe3+ ion to form a precipitate of hydrated ferric oxide.
(iv) Althoughamino group is o,p− directing in aromatic electrophilicsubstitution reactions, aniline on nitration gives a substantial amountof m-nitroaniline:
Nitration is carried out in an acidic medium. In anacidic medium, aniline is protonated to give anilinium ion (which is meta-directing).
For this reason, aniline on nitration gives a substantialamount of m-nitroaniline.
(v) Anilinedoes not undergo Friedel-Crafts reaction:
A Friedel-Crafts reaction is carried out inthe presence of AlCl3. But AlCl3 is acidic innature, while aniline is a strong base. Thus, aniline reacts with AlCl3 toform a salt (as shown in the following equation).
Due to the positive charge on the N-atom, electrophilicsubstitution in the benzene ring is deactivated. Hence, aniline does notundergo the Friedel-Crafts reaction.
(vi) Diazoniumsalts of aromatic amines are more stable than those of aliphatic amines:
The diazonium ion undergoes resonance as shown below:
This resonance accounts for the stability of the diazoniumion. Hence, diazonium salts of aromatic amines are more stable than those ofaliphatic amines.
(vii) Gabrielphthalimide synthesis is preferred for synthesising primary amines:
Gabriel phthalimide synthesis results in the formation of1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferredfor synthesizing primary amines.
Question - 13 : - Arrange the following:
Answer - 13 : -
(i) Indecreasing order of the pKbvalues:
C2H5NH2, C6H5NHCH3, (C2H5)2NHand C6H5NH2
(ii) Inincreasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2NHand CH3NH2
(iii) In increasing order of basic strength:
(a) Aniline, p-nitroanilineand p-toluidine
(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2.
(iv) Indecreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH,(C2H5)3N and NH3
(v) Inincreasing order of boiling point:
C2H5OH,(CH3)2NH, C2H5NH2
(vi) Inincreasing order of solubility in water:
C6H5NH2,(C2H5)2NH,C2H5NH2
Answer
(i) In C2H5NH2,only one −C2H5 group is present while in (C2H5)2NH,two −C2H5 groups are present. Thus, the +I effect is more in(C2H5)2NH than in C2H5NH2.Therefore, the electron density over the N-atom is more in (C2H5)2NHthan in C2H5NH2. Hence, (C2H5)2NHis more basic than C2H5NH2.
Also, both C6H5NHCH3 andC6H5NH2 are less basic than (C2H5)2NHand C2H5NH2 due to the delocalization of the lone pair in theformer two. Further, among C6H5NHCH3 andC6H5NH2, the former will be more basic due to the +T effect of−CH3 group. Hence, the order of increasing basicity ofthe given compounds is as follows:
C6H5NH2 6H5NHCH3 < C2H5NH2 <(C2H5)2NH
We know that the higher the basic strength,the lower is the pKb values.
C6H5NH2 >C6H5NHCH3 > C2H5NH2 >(C2H5)2NH
(ii) C6H5N(CH3)2 ismore basic than C6H5NH2 due to the presence of the +I effect of two −CH3 groupsin C6H5N(CH3)2. Further, CH3NH2 containsone −CH3 group while (C2H5)2NHcontains two −C2H5 groups. Thus, (C2H5)2 NHis more basic than C2H5NH2.
Now, C6H5N(CH3)2 isless basic than CH3NH2 because of the−R effect of −C6H5 group.
Hence, the increasing order of the basic strengths of thegiven compounds is as follows:
C6H5NH2 6H5N(CH3)2 < CH3NH2 < (C2H5)2NH
(iii) (a)
In p-toluidine, the presence ofelectron-donating −CH3 group increases the electron density on the N-atom.
Thus, p-toluidine is more basicthan aniline.
On the other hand, the presence of electron-withdrawing
−NO2 group decreasesthe electron density over the N−atom in p-nitroaniline. Thus, p-nitroanilineis less basic than aniline.
Hence, the increasing order of the basic strengths of thegiven compounds is as follows:
p-Nitroaniline< Aniline < p-Toluidine
(b) C6H5NHCH3 ismore basic than C6H5NH2 due to the presence of electron-donating −CH3 groupin C6H5NHCH3.
Again, in C6H5NHCH3, −C6H5 groupis directly attached to the N-atom. However, it is not so in C6H5CH2NH2.Thus, in C6H5NHCH3, the −R effect of −C6H5 groupdecreases the electron density over the N-atom. Therefore, C6H5CH2NH2 ismore basic than C6H5NHCH3.
Hence, the increasing order of the basic strengths of thegiven compounds is as follows:
C6H5NH2 6H5NHCH3 < C6H5CH2NH2.
(iv) Inthe gas phase, there is no solvation effect. As a result, the basic strengthmainly depends upon the +I effect. The higher the +I effect, the stronger isthe base. Also, the greater the number of alkyl groups, the higher is the +Ieffect. Therefore, the given compounds can be arranged in the decreasing orderof their basic strengths in the gas phase as follows:
(C2H5)3N> (C2H5)2NH > C2H5NH2 >NH3
(v) Theboiling points of compounds depend on the extent of H-bonding present in thatcompound. The more extensive the H-bonding in the compound, the higher is theboiling point. (CH3)2NH contains only one H−atom whereas C2H5NH2 containstwo H-atoms. Then, C2H5NH2 undergoes more extensive H-bonding than (CH3)2NH.Hence, the boiling point of C2H5NH2 ishigher than that of (CH3)2NH.
Further, O is more electronegative than N.Thus, C2H5OH forms stronger H−bonds than C2H5NH2. Asa result, the boiling point of C2H5OHis higher than that of C2H5NH2 and (CH3)2NH.
Now, the given compounds can be arranged in the increasingorder of their boiling points as follows:
(CH3)2NH< C2H5NH2 < C2H5OH
(vi) Themore extensive the H−bonding, the higher is the solubility. C2H5NH2 containstwo H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2 undergoesmore extensive H−bonding than (C2H5)2NH.Hence, the solubility in water of C2H5NH2 ismore than that of (C2H5)2NH.
Further, the solubility of amines decreaseswith increase in the molecular mass. This is because the molecular mass ofamines increases with an increase in the size of the hydrophobic part. The molecularmass of C6H5NH2 is greater than that of C2H5NH2 and(C2H5)2NH.
Hence, the increasing order of their solubility in wateris as follows:
C6H5NH2 <(C2H5)2NH < C2H5NH2
How will you convert:
Answer - 14 : -
(i) Ethanoicacid into methanamine
(ii) Hexanenitrileinto 1-aminopentane
(iii) Methanol to ethanoic acid
(iv) Ethanamineinto methanamine
(v) Ethanoicacid into propanoic acid
(vi) Methanamineinto ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoicacid into ethanoic acid
Answer
(i)
(iv)
(v)
(vi)
(vii)
(viii)
Question - 15 : - Describe a method for the identification of primary,secondary and tertiary amines. Also write chemical equations of the reactionsinvolved.
Answer - 15 : -
Primary, secondary and tertiary amines canbe identified and distinguished by Hinsberg’s test. In this test, the aminesare allowed to react with Hinsberg’s reagent, benzenesulphonyl chloride (C6H5SO2Cl).The three types of amines react differently with Hinsberg’s reagent. Therefore,they can be easily identified using Hinsberg’s reagent.
Primary amines react with benzenesulphonyl chloride toform N-alkylbenzenesulphonyl amide which is soluble in alkali.Due to the presence of a strong electron-withdrawingsulphonyl group in the sulphonamide, the H−atom attached to nitrogen can beeasily released as proton. So, it is acidic and dissolves in alkali.
Secondary amines react with Hinsberg’s reagent to give asulphonamide which is insoluble in alkali.
There is no H−atom attached to the N-atom in thesulphonamide. Therefore, it is not acidic and insoluble in alkali.
On the other hand, tertiary amines do not react withHinsberg’s reagent at all.
Write short notes on the following:
Answer - 16 : -
(i) Carbylamine reaction (ii) Diazotisation
(iii) Hofmann’s bromamide reaction (iv) Coupling reaction
(v) Ammonolysis (vi) Acetylation
(vii) Gabriel phthalimide synthesis.
Answer
(i) Carbylaminereaction
Carbylamine reaction is used as a test for theidentification of primary amines. When aliphatic and aromatic primary aminesare heated with chloroform and ethanolic potassium hydroxide, carbylamines (orisocyanides) are formed. These carbylamines have very unpleasant odours.Secondary and tertiary amines do not respond to this test.
For example,
(ii) Diazotisation
Aromatic primary amines react with nitrousacid (prepared in situ from NaNO2 and a mineralacid such as HCl) at low temperatures (273-278 K) to form diazonium salts. Thisconversion of aromatic primary amines into diazonium salts is known asdiazotization.
For example, on treatment with NaNO2 andHCl at 273−278 K, aniline produces benzenediazonium chloride, with NaCl and H2O asby-products.
(iii) Hoffmannbromamide reaction
When an amide is treated with bromine in an aqueous orethanolic solution of sodium hydroxide, a primary amine with one carbon atomless than the original amide is produced. This degradation reaction is known asHoffmann bromamide reaction. This reaction involves the migration of an alkylor aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.
For example,
(iv) Couplingreaction
The reaction of joining two aromatic rings through the−N=N−bond is known as coupling reaction. Arenediazonium salts such as benzenediazonium salts react with phenol or aromatic amines to form coloured azocompounds.
It can be observed that, the para-positions of phenol andaniline are coupled with the diazonium salt. This reaction proceeds throughelectrophilic substitution.
(v) Ammonolysis
When an alkyl or benzyl halide is allowed to reactwith an ethanolic solution of ammonia, it undergoes nucleophilic substitutionreaction in which the halogen atom is replaced by an amino (−NH2)group. This process of cleavage of the carbon-halogen bond is known asammonolysis.
When this substituted ammonium salt is treated with astrong base such as sodium hydroxide, amine is obtained.
Though primary amine is produced as the major product,this process produces a mixture of primary, secondary and tertiary amines, andalso a quaternary ammonium salt as shown.
(vi) Acetylation
Acetylation (or ethanoylation) is the process ofintroducing an acetyl group into a molecule.
Aliphatic and aromatic primary and secondaryamines undergo acetylation reaction by nucleophilic substitution when treatedwith acid chlorides, anhydrides or esters. This reaction involves the replacementof the hydrogen atom of −NH2 or > NH groupby the acetyl group, which in turn leads to the production of amides. To shiftthe equilibrium to the right hand side, the HCl formed during the reaction isremoved as soon as it is formed. This reaction is carried out in the presenceof a base (such as pyridine) which is stronger than the amine.
When amines react with benzoyl chloride, the reaction isalso known as benzoylation.
For example,
(vii) Gabrielphthalimide synthesis
Gabriel phthalimide synthesis is a very useful method forthe preparation of aliphatic primary amines. It involves the treatment ofphthalimide with ethanolic potassium hydroxide to form potassium salt ofphthalimide. This salt is further heated with alkyl halide, followed by alkalinehydrolysis to yield the corresponding primary amine.
Question - 17 : - Accomplish the following conversions:
Answer - 17 : -
(i) Nitrobenzeneto benzoic acid
(ii) Benzeneto m-bromophenol
(iii) Benzoic acid to aniline
(iv) Anilineto 2,4,6-tribromofluorobenzene
(v) Benzylchloride to 2-phenylethanamine
(vi) Chlorobenzeneto p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii) Benzamide to toluene
(ix) Anilineto benzyl alcohol.
Answer
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Give the structures of A, B and C in the followingreactions:
Answer - 18 : - (i) 
(ii) 
(iii)
(iv) 
(v) 
(vi) 
Answer
(i) 
Question - 19 : - An aromatic compound ‘A’ on treatment with aqueous ammoniaand heating forms compound
Answer - 19 : - ‘B’ which on heating with Br_2 and KOH forms a compound ‘C’ of molecular formula C_6H_7N. Write the structures and IUPAC names of compounds A, B and C.
Answer
It is given that compound ‘C’ having themolecular formula, C6H7N is formed by heating compound ‘B’ with Br2 andKOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound ‘B’is an amide and compound ‘C’ is an amine. The only amine having the molecularformula, C6H7N is aniline, (C6H5NH2).
Therefore, compound ‘B’ (from which ’C’ isformed) must be benzamide, (C6H5CONH2).
Further, benzamide is formed by heating compound ‘A’ withaqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.
The given reactions can be explained with the help of thefollowing equations:
Complete the following reactions:
Answer - 20 : - (i) 
(ii) 
(iii)
(iv)
(v)
(vi) 
(vii)
Answer
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
