Cinemugam (Kollywood): Amines Class 12th Chemistry Part Ii CBSE Solution

Amines Class 12th Chemistry Part Ii CBSE Solution

Class 12^th Chemistry Part Ii CBSE Solution

Intext Questions Pg-384
Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii)…
(i) Write structures of different isomeric amines corresponding to the molecular formula,…
Intext Questions Pg-387
Benzene into aniline How will you convert
Benzene into N, N-dimethylaniline How will you convert
Cl-(CH2)4-Cl into hexan-1,6-diamine? How will you convert
Intext Questions Pg-396
C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH Arrange the following in increasing order…
C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 Arrange the following in increasing order of their…
CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. Arrange the following in increasing order…
Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl →…
Write reactions of the final alkylation product of aniline with an excess of methyl iodide…
Write chemical reaction of aniline with benzoyl chloride and write the name of the product…
Write structures of different isomers corresponding to the molecular formula, C3H9N. Write…
Intext Questions Pg-399
3-Methylaniline into 3-nitrotoluene. Convert
Aniline into 1,3,5 - tribromobenzene Convert
Exercises
(CH3)2CHNH2 Write IUPAC names of the following compounds and classify them into primary,…
CH3(CH2)2NH2 Write IUPAC names of the following compounds and classify them into primary,…
CH3NHCH(CH3)2 Write IUPAC names of the following compounds and classify them into primary,…
(CH3)3CNH2 Write IUPAC names of the following compounds and classify them into primary,…
C6H5NHCH3 Write IUPAC names of the following compounds and classify them into primary,…
(CH3CH2)2NCH3 Write IUPAC names of the following compounds and classify them into primary,…
m-BrC6H4NH2 Write IUPAC names of the following compounds and classify them into primary,…
Methylamine and dimethylamine Give one chemical test to distinguish between the following…
Secondary and tertiary amines Give one chemical test to distinguish between the following…
Ethylamine and aniline Give one chemical test to distinguish between the following pairs…
Aniline and benzylamine Give one chemical test to distinguish between the following pairs…
Aniline and N-methylaniline. Give one chemical test to distinguish between the following…
pKb of aniline is more than that of methylamine. Account for the following:…
Ethylamine is soluble in water whereas aniline is not. Account for the following:…
Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.…
Although amino group is o- and p- directing in aromatic electrophilic substitution…
Aniline does not undergo Friedel-Crafts reaction. Account for the following:…
Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Account…
Gabriel phthalimide synthesis is preferred for synthesising primary amines. Account for…
In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 Arrange…
In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 Arrange…
In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine (b)…
In decreasing order of basic strength in gas phase: C6H5NH2, (C2H5)2NH, (C2H5)3N and NH3…
In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 Arrange the following:…
In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2 Arrange the…
Ethanoic acid into methanamine How will you convert:
Hexanenitrile into 1-aminopentane How will you convert:
Methanol to ethanoic acid How will you convert:
Ethanamine into methanamine How will you convert:
Ethanoic acid into propanoic acid How will you convert:
Methanamine into ethanamine How will you convert:
Nitromethane into dimethylamine How will you convert:
Propanoic acid into ethanoic acid? How will you convert:
Describe the method for the identification of primary, secondary and tertiary amines. Also…
Carbylamine reaction Write short notes on
Diazotization Write short notes on
Hoffman Bromamide Reaction Write short notes on
Coupling Reaction Write short notes on
Ammonolysis Write short notes on
Acetylation Write short notes on
Gabriel phthalimide reaction Write short notes on
Nitrobenzene to benzoic acid Accomplish the following conversions:…
Benzene to m-bromophenol Accomplish the following conversions:
Benzoic acid to aniline Accomplish the following conversions:
Aniline to 2,4,6-tribromofluorobenzene Accomplish the following conversions:…
Benzyl chloride to 2-phenylethanamine Accomplish the following conversions:…
Chlorobenzene to p-chloroaniline Accomplish the following conversions:…
Aniline to p-bromoaniline Accomplish the following conversions:
Benzamide to toluene Accomplish the following conversions:
Aniline to benzyl alcohol. Accomplish the following conversions:
Give the structures of A, B and C in the following reactions:
Give the structures of A, B and C in the following reactions:
Give the structures of A, B and C in the following reactions:
Give the structures of A, B and C in the following reactions:
Give the structures of A, B and C in the following reactions:
Give the structures of A, B and C in the following reactions:
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’…
C6H5NH2 + CHCl3 + alc.KOH → Complete the following reactions:
C6H5N2Cl + H3PO2 + H2O → Complete the following reactions:
C6H5NH2 + H2SO4(conc.) → Complete the following reactions:
C6H5N2Cl + C2H5OH → Complete the following reactions:
C6H5NH2 + Br2(aq) → Complete the following reactions:
C6H5NH2 + (CH3CO)2O → Complete the following reactions:
C6H5N2Cl Complete the following reactions:
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?…
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.…
Give plausible explanation for each of the following: (i) Why are amines less acidic than…

Intext Questions Pg-384

Question 1.
Classify the following amines as primary, secondary or tertiary:

(i)

(ii)

(iii) (C₂H₅)₂CHNH₂

(iv) (C₂H₅)₂NH

Answer:
(i) Primary. Because the nitrogen atom is attached to the only 1 carbon atom.

Note: The degree of amines depends on the no of atoms attached to the nitrogen atom of amine (except hydrogen).

(ii) Tertiary. Because the Nitrogen atom is attached to the 3 carbon atoms.

(iii) Primary. Because the nitrogen atom is attached to the only 1 carbon atom.

(iv) Secondary. Because the nitrogen atom is attached to the only 1 carbon.

Question 2.
(i) Write structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.

(ii) Write IUPAC names of all the isomers.

(iii) What type of isomerism is exhibited by different pairs of amines?

Answer:
(i) & (ii) There are total 8 geometrical isomers of the given compound.

(iii) a) Pairs 1,2,6,7 Exhibit Position isomerism; means the change in position of the substituent.

b) Pairs 1,3 and 1,4 and 2,3 and 2,4 exhibit chain isomerism i.e. in this type of isomerism the different structures can be produced by changing the chain of the atoms.

c) Pairs 5,6 and 5,7 exhibit metamerism; i.e. different group on either side of the central atom.

d) All Primary amines exhibit functional isomers. All secondary amines share functional isomerism and same for tertiary. The functional isomerism means same functional group.

Intext Questions Pg-387

Question 1.
How will you convert

Benzene into aniline

Answer:
Benzene into aniline

When Benzene is treated with HNO₃/H₂SO₄ it forms nitrobenzene. When Nitrobenzene reduced with Sn/HCL it forms Aniline. Because Sn/HCl is a reducing Mixture.

Question 2.
How will you convert

Benzene into N, N-dimethylaniline

Answer:
Benzene into N, N-dimethylaniline

When Benzene is reacted with nitrating mixture it forms nitrobenzene. When it Reduced H₂/Pd in ethanol or Sn/HCl, it forms Aniline. When Aniline reacts 2 times with CH₃Cl It forms N, N-dimethylaniline.

Question 3.
How will you convert

Cl–(CH₂)₄–Cl into hexan-1,6-diamine?

Answer:
Cl–(CH₂)₄–Cl into hexan-1,6-diamine?

When 1,4-dichlorobutane reacts with NaCN it forms Di cyanide compound, After Hydrogenation it forms the Hexane 1,6-Diamine.

Intext Questions Pg-396

Question 1.
Arrange the following in increasing order of their basic strength:

C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH

Answer:
Alkyl group contribute inductive effect which increases the basic strength of compounds.

NH₃<C₂H₅NH₂<(C₂H₅)₂NH.

Then C₆H₅NH₂ is having –I effect that reduces strength. And C₆H₅CH₂NH₂ increases the basic strength but not as much as C₂H₅ group.

Hence final order will be C₆H₅NH₂<NH₃<C₆H₅CH₂NH₂<C₂H₅NH₂<(C₂H₅)₂NH.

Question 2.
Arrange the following in increasing order of their basic strength:

C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂

Answer:
By taking into consideration –R effect and steric hindrance of groups we can arrange them in the order C₆H₅NH₂< C₂H₅NH₂<(C2H5)₃N<(C₂H₅)2NH. Because (C₂H₅)₃N has a lot of steric hindrances that reduces the basic strength.

Question 3.
Arrange the following in increasing order of their basic strength:

CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂.

Answer:
In C₆H₅NH₂ , N is directly attached to the ring that causes delocalization of electrons of the benzene ring. Whereas in case of C₆H₅CH₂NH₂ it is not directly connected to benzene ring Hence it has more basic strength.

Due to –I effect of (CH₃)₃^– a group it has more basic strength than C₆H₅CH₂NH₂. Hence final order will be C₆H₅NH₂< C₆H₅CH₂NH₂<(CH₃)₃N< CH₃NH₂< (CH₃)2NH

Question 4.
Complete the following acid-base reactions and name the products:

(i) CH₃CH₂CH₂NH₂ + HCl → ?

(ii) (C₂H₅)₃N + HCl → ?

Answer:
(i) CH₃CH₂CH₂NH₂ + HCl → CH₃CH₂CH₂N⁺H₃Cl^-

The final product is (N-propyl ammonium chloride.)

(ii) (C₂H₅)₃N + HCl → (C₂H₅)₃N⁺HCl^-

The final product is (Tri ethyl ammonium chloride)

Question 5.
Write reactions of the final alkylation product of aniline with an excess of methyl iodide in the presence of sodium carbonate solution.

Answer:

On excessive alkylation with methyl iodide aniline gets converted into N,N,N-Trimethylanilinium iodide. After reacting it with sodium carbonate it get converted into N,N,N-Trimethylanilinium carbonate.

Question 6.
Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.

Answer:
When aniline is treated with benzoyl chloride in the presence of base it gets converted into N-Phenylbenzamide.

Question 7.
Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Answer:
The different isomers of the molecular formula: C₃H₉N are given in the table. However only 1° amines will liberate nitrogen gas on the treatment with h=the nitrous acid are given as follows:

Note: Only primary amines liberate nitrogen gas on treatment with nitrous acid.

Intext Questions Pg-399

Question 1.
Convert

3-Methylaniline into 3-nitrotoluene.

Answer:
When 3-methylaniline treated with NaNO₂ + HCl it gets converted into chlorine complex. When that complex reacted with HBF₄ It gets converted into Barium Fluoride complex. This complex reacts with NaNO₂ in presence of copper to give 3-Nitrotoluene.

Question 2.
Convert

Aniline into 1,3,5 – tribromobenzene

Answer:

When aniline reacts with Br₂ water it gets converted into 2,4,6 tribromobenzamine. When this further reacted with NaNO₂/HCl it forms Chloride complex. This complex forms 1,3,5 tribromobenzene after treating with H₃PO₂ in presence of water.

Exercises

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(CH₃)2CHNH₂

Answer:
1-Methylethanamine
The root name is based on the longest chain with the -NH₂ attached. The chain is numbered so as to give the amine unit the lowest possible number. The longest chain is ethane chain which is further suffixed with ‘amine’.

Question 2.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

CH₃(CH₂)2NH₂

Answer:
Propan-1-amine
The longest chain here is propane. The naming is such that amine unit should get a lowest possible number. Propane-1-amine can also be written as 1-propylamine.

Question 3.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

CH₃NHCH(CH₃)₂

Answer:
N−Methyl-2-methylethanamine
The chain is numbered so as to give the amine unit the lowest possible number. The other alkyl group is treated as a substituent, with N as the locant. The N locant is listed before numerical locants. The longest chain is ethane which has a Methyl substituent group.

Question 4.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(CH₃)3CNH₂

Answer:
2-Methylpropan-2-amine
Propane is the longest chain and is joined to the amine group at the 2^nd position. Another substituent methyl group is also at the same position.

Question 5.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

C₆H₅NHCH₃

Answer:
N−Methylbenzamine or N-methylaniline
Aniline is considered as the principle group. The other alkyl group is treated as a substituent, with N as the locant.

Question 6.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(CH₃CH₂)₂NCH₃

Answer:
N-Ethyl-N-methylethanamine
This is a tertiary group of amine. The root name is based on the longest chain with an amine having the lowest possible number. The alkyl groups are treated as substituents with N as the locant.

Question 7.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

m–BrC₆H₄NH₂

Answer:
3-Bromobenzenamine or 3-bromoaniline
Benzene is considered as the principle group. The substituent is bromine which is numbered 3.Conventional method of naming is followed.

Question 8.
Give one chemical test to distinguish between the following pairs of compounds.

Methylamine and dimethylamine

Answer:
The best test for distinguishing methyl amine and dimethylamine is the Carbylamines test.

Carbylamine Test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines.

In this case, Methylamine (which is an aliphatic primary amine) gives a positive carbylamine test while dimethylamine wont.

Reaction:

Question 9.
Give one chemical test to distinguish between the following pairs of compounds.

Secondary and tertiary amines

Answer:
Hinsberg’s reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl). can be used to distinguish secondary and tertiary amines.

Hinsberg Test: Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N,

N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent.

Reaction:

Question 10.
Give one chemical test to distinguish between the following pairs of compounds.

Ethylamine and aniline

Answer:
Azo-dye test can be used to distinguish ethylamine and aniline.

Azo-dye test: A dye is obtained when aromatic amines react with HNO₂ (NaNO₂ + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due to the evolution of N₂ gas under similar conditions.

Reaction:

Ethylamine;

Aniline;

CH₃CH₂ – NH₂ + HONO C₂H₅OH + N₂↑ + H₂O

Question 11.
Give one chemical test to distinguish between the following pairs of compounds.

Aniline and benzylamine

Answer:
Aniline and benzylamine can be distinguished with the help of nitrous acid.

Nitrous acid test: Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas, while aniline reacts with nitrous acid to form a stable diazonium salt without releasing nitrogen gas.

Reaction:

(Benzylamine)

(Aniline)

Question 12.
Give one chemical test to distinguish between the following pairs of compounds.

Aniline and N-methylaniline.

Answer:
Carbylamine test can be used to distinguish between Aniline and N-methylaniline.

Carbylamine Test: Primary amines, on heating with chloroform and ethanolic potassium hydroxide, form foul smelling isocyanides or carbylamines. Aniline, being an aromatic primary amine, gives positive carbylamine test. However, N-methylaniline, being a secondary amine does not.

Reaction:

Question 13.
Account for the following:

pK_b of aniline is more than that of methylamine.

Answer:
Pk_b value is the negative logarithm of the basicity constant (K_b) .i.e., pK_b = -logK_b

Evidently, smaller the value of pK_b , stronger is the base (strong tendency to donate electrons)

⇒ The structure of methyl amine is CH₃NH₂

⇒ The structure of aniline is:

Aniline (C₆H₅NH₂) shows resonance:

As a result of resonance, the lone pair of electrons on the nitrogen atom gets delocalized over benzene ring. As a result, electron density on the nitrogen decreases and thus is less easily available to donate electrons making it less basic. In contrast, in methyl amine (CH₃NH₂), delocalization of the lone pair of electrons on the nitrogen atom by resonance is not possible. Furthermore, CH₃ being an electron- releasing group, due to which +I effect of CH₃ increases the electron density on the N- atom and thus is more easily available to donate electrons making it more basic.

Therefore, aniline is weaker base than methylamine and hence its pK_b value is higher than that of methylamine.

Question 14.
Account for the following:

Ethylamine is soluble in water whereas aniline is not.

Answer:
_⇒ The structure of ethylamine is CH₃CH₂NH₂

⇒ The structure of aniline is:

Ethylamine form H-bonds with water. It dissolves in water due to intermolecular H-bonding as shown below:

Thus, ethylamine is soluble in water.

However, in aniline, due to the larger hydrophobic part, i.e., hydrocarbon part (C₆H₅ group), which tends to retard the formation of H-bonds. The extent of H-bonding decreases and hence aniline is insoluble in water.

Question 15.
Account for the following:

Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Answer:
⇒ The structure of methylamine is CH₃NH₂

Methylamine is more basic than water due to the presence of CH₃ (electron releasing group) and +I effect (pushing electrons). Being more basic, methylamine accepts a proton from water liberating OH^- ions.

Dissociation of ferric chloride in water to give Fe³⁺ and Cl^-

FeCl₃→ Fe³⁺ + 3Cl^-

Now, the liberated OH^- ions combine with Fe³⁺ ions present in H₂O to form brown precipitate of hydrated ferric oxide.

Question 16.
Account for the following:

Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

Answer:
The below diagram shows the position of ortho(o), para(p) and meta(m) derivatives of amino group:

Electrophilic addition reaction of amines: In addition to the reaction of the amino group (NH₂ group), aromatic amines also undergo typical electrophilic substitution reactions of the aromatic ring. In all these reactions, the NH₂ group strongly activates the aromatic ring through delocalization of the lone pair of electrons of the N-atom over the aromatic ring.

As a result, electron density increases more at ortho and para positions. Therefore NH₂ group directs the incoming group to ortho and para positions, i.e., NH₂ is an o-, p-directing group.

But it has been observed that on nitration of aniline gives a substantial amount of m-nitroanilne.

Explanation: Nitration is usually carried out in acidic medium in the presence of concentrated HNO₃ and concentrated H₂SO₄. As a result, most of the aniline is converted into anilinium ion(gets protonated) and since - ⁺NH₃ is m-directing group, therefore, an unexpected large amount of m-nitroaniline is obtained.

Nitration of anilne gives mainly p-nitroaniline

Nitration of anilinium ions give m-nitroanilne(due to protonation)

Hence, nitration of aniline gives a mixture of p-nitroaniline and m-nitroaniline in approx. 1:1 ratio.

Question 17.
Account for the following:

Aniline does not undergo Friedel-Crafts reaction.

Answer:
Friedel- Crafts reaction: When any benzene or its derivative is treated with alkyl halide (R-X, X=Cl) or acetyl chloride (CH₃-COCl) in the presence of annhydrous aluminium chloride (AlCl₃) to form alkyl or acetyl substituted benzene or its derivative, this reaction is called Friedel-Crafts reaction.

⇒ Friedel-Crafts alkylation: When any benzene or its derivative is treated with alkyl halides (R-X, X=Cl,Br) in the presence of annhydrous aluminium chloride (AlCl₃) to form alkyl substituted benzene or its derivatives, this reaction is called Friedel-Crafts alkylation. For example:

⇒ Friedel-Crafts acylation: When any benzene or its derivative is treated with acetyl chloride (R-COCl) in the presence of annhydrous aluminium chloride (AlCl₃) to form acetyl substituted benzene or its derivatives, this reaction is called Friedel-Crafts acylation. For example:

Aniline does not undergo Friedel-Crafts reaction.

Explanation: Aniline is a Lewis base (electron-pair acceptor) while AlCl₃ is Lewis acid (electron-pair donor). They combine with each to form a salt.

Due to presence of positive charge on nitrogen (N) atom in the salt, the group N⁺H₂AlCl₃^- acts as a strong electron withdrawing group (strong deactivating group). As a result, it reduces the electron density in the benzene ring and hence aniline does not undergo Friedel-Crafts (alkylation or acetylation) reaction

Question 18.
Account for the following:

Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Answer:
Diazonium salts: Diazonium salts have the general formula

(R/Ar—N₂⁺Cl^-) where R stands for the alkyl group and Ar stands for the aryl group. The structure is given below:

Formation of Diazonium salts:

Diazonium salt is obtained by treating aromatic amine(aniline) dissolved in dil. HCl with HNO₂ at 273-278K (0° -5° C)

Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Explanation:

Aromatic amine form arenediazonium salts, which are stable for a short time in solution at low temperature(273-278K).The diazonium salts of aromatic amines are more stable due to the dispersal of the positive charge on the benzene ring (resonance) as shown below:

The aliphatic amines, on the other hand, form highly unstable alkane diazonium salt (R—N₂⁺Cl^-). They rapidly decompose even at low temperature(<272-278K) forming carbocation and nitrogen gas.

Hence, diazonium salts of aromatic amine are much more stable than aliphatic diazonium salts.

Note: Carbocation is an ion in which carbon atom consists of positive charge

Question 19.
Account for the following:

Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Answer:
Gabriel phthalimide synthesis is a very convenient method for the preparation of pure aliphatic amines ( especially primary amines) Phthalimide on treatment with ethanolic KOH gives potassium phthalimide which on heating with a suitable alkyl hallide gives N-substituted phthalimides. These upon susbsequent hydrolysis with dil.HCl under pressure or with alkali gives primary amines.

Step 1: Phthalimide is treated with KOH to form potassium phthalimide

Step 2: Potassium phthalimde is treated with suitable alkyl hallide to form N-substituted phthalimides.

Step 3: N-substituted phthalimides undergoes hydrolysis in the prsence of dil HCl or with alkali(NaOH) to give primary amines.

Overall reaction:

∴ Gabriel phthalimide synthesis results in the formation of primary(1° amine) only. Secondary or tertiary amines are not formed through this synthesis. Hence, Gabriel phthalimide synthesis preferred for the formation of primary amines only.

Question 20.
Arrange the following:

In decreasing order of the pKb values:

C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

Answer:
Pk_b value is the negative logarithm of the basicity constant (K_b) .i.e., pK_b = -logK_b

Evidently, smaller the value of pK_b , stronger is the base (strong tendency to donate electrons)

Aliphatic amines(R-NH₂) are more basic(tendency to donate electrons) than aromatic amines(C₆H₅NH₂) because of the following reasons:

⇒ In aliphatic amines, alkyl groups are present. Alkyl groups are electron releasing groups, hence they increase the elecron density of N-atom and thus is easily available to donate electrons. This poperty makes aliphatic amines more basic.

⇒ In aromatic amines, aryl group is present. Aromatic amine shows resonance:

As a result of resonance, the lone pair of electrons on the nitrogen atom gets delocalized over benzene ring. As a result, electron density on the nitrogen decreases and thus is less easily available to donate electrons making it less basic.

Hence aliphatic amines(R-NH₂) are more basic than aromatic amines(C₆H₅NH₂)

Now, in C₂H₅NH₂, one ethyl group(alkyl) is present and in (C₂H₅)₂NH₂ two ethyl groups are present. As we know that more alkyl groups are

present, more basic will be the amine.

Hence, (C₂H₅)₂NH₂ is more basic than the C₂H₅NH₂

Now in C₆H₅NH₂, due to delocalisation of lone pair of electrons of the N-atom over the benzene ring, makes it less basic than (C₂H₅)₂NH₂ and C₂H₅NH₂

Now, in C₆H₅NHCH_3, due to presence of CH₃ group, makes it more basic than C₆H₅NH₂ but less basic than (C₂H₅)₂NH₂ and C₂H₅NH₂ due to the presence of aromatic ring which is responsible for the delocalisation of lone pair of electrons of N-atom over the benzene ring.

Combining all these facts, the relative basic strength of these four amines decrease in the order:

(C₂H₅)₂NH₂ > C₂H₅NH₂ > C₆H₅NHCH₃ > C₆H₅NH₂

Since, a stronger base has a lower pk_b value, therefore, pK_b values decrease in the reverse order:

C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH₂

Question 21.
Arrange the following:

In increasing order of basic strength:

C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂

Answer:
In (C₂H₅)₂NH₂, two ethyl groups are present and in CH₃NH₂ one methyl group is present. As we know that more alkyl groups are

present, more basic will be the amine. Hence, (C₂H₅)₂NH₂ is more basic than the CH₃NH₂

Now in C₆H₅NH₂, due to delocalisation of lone pair of electrons of the N-atom over the benzene ring (decreases the electron density of N-atom) makes it less basic than (C₂H₅)₂NH₂ and CH₃NH₂

Now, in C₆H₅N(CH₃)₂ due to presence of two CH₃ groups (increases the electron density of N-atom) makes it more basic than C₆H₅NH₂ but less basic than (C₂H₅)₂NH₂ and CH₃NH₂ due to the presence of aromatic ring which is responsible for the delocalisation of lone pair of electrons of N-atom over the benzene ring (decreases the electron density of N-atom)

Combining all these facts, the relative basic strength of these four amines increases in the order:

C₆H₅NH₂ < C₆H₅NHCH₃ < CH₃NH₂ < (C₂H₅)₂NH

Question 22.
Arrange the following:

In increasing order of basic strength:

(a) Aniline, p-nitroaniline and p-toluidine

(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂

Answer:
(a) Electron-donating groups such as –CH₃, -OCH₃, -NH₂, etc. increase the basicity and electron-withdrawing groups such as –NO₂, -CN, -SO₃H, -COOH, -X(halogen), etc. decrease the basicity of amines.

Explanation: Electron-donating groups releases electrons, stabilizes the conjugate acid (cation) and thus increases the basic strength.

Electron-withdrawing groups withdraws electrons, destabilizes the conjugate acid (cation) and thus decreases the basic strength.

In p-nitroaniline, NO₂ group is present. As we know that NO₂ group is an electron withdrawing group which decreases the basic strength of amine. In p-toluidine, CH₃ group is present and as we know that CH₃ group is an electron donating(releasing) group which increases the basic strength of amine.

Hence, p-toluidine is more basic than p-nitroaniline

Now in C₆H₅NH₂ (aniline), due to delocalisation of lone pair of electrons of the N-atom over the benzene ring (decreases the electron density of N-atom) makes it less basic than p-toluidine but more basic than p-nitroaniline (NO₂ group is present which decreases the density of aniline)

Combining all these facts, the relative basic strength of these three amines increases in the order:

p-nitroaniline< aniline < p-toluidine

(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂

In C₆H₅NH₂ and C₆H₅NHCH₃, the N-atom is directly attached to the aromatic ring. Hence, they both show resonance in which delocalisation of lone pair of electrons N-atom takes palce. As a result the electron density of N-atom decreases. Hence both are weaker bases. However in C₆H₅NHCH₃, CH₃ group(electron withdrawing) is present which increases the overall density of electrons.

Hence, C₆H₅NHCH₃ is more basic than C₆H₅NH₂

In C₆H₅CH₂NH₂, the N-atom is not directly attached to the aromatic ring. As a result, it does not show resonance. There is no effect on the electron density of lone pair of electrons of N-atom. Hence, it is more basic than C₆H₅NH₂ and C₆H₅NHCH₃

Combining all these facts, the relative basic strength of these three amines increases in the order:

C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂

Question 23.
Arrange the following:

In decreasing order of basic strength in gas phase:

C₆H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃

Answer:
Amines in gas phase or in non-aqueous solvents , there is no solvation effect(getting influenced by the solvent).It means that the stabilization of conjugate acid formed due to the formation of hydrogen bonding are absent.

Hence, the basic strength of amines depends only on the +I effect of the alkyl groups. Alkyl groups are electron releasing groups, they release electron to the nitrogen in amine and increase the overall electron density of electrons and thus is easily available to donate electron. This property makes it more basic. As s result, more alkyl groups are attached, the higher the +I effect. Hence, the higher the +I effect, stronger is the base(high tendency to accept electrons)

In (C₂H₅)₃N, 3 alkyl groups are present, hence it is more basic. In (C₂H₅)₂NH, 2 alkyl groups are present, hence it is less basic than (C₂H₅)₃N. In C₂H₅NH_2, only one alkyl group is present, hence it is less basic than (C₂H₅)₃N and (C₂H₅)₂NH. In NH₃, no alkyl group is present, so there is no +I effect. Hence it is less basic among all the amines.

Combining all these facts, the relative basic strength of these four amines decreases in the order:

(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃

Question 24.
Arrange the following:

In increasing order of boiling point:

C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂

Answer:
As we know that boiling point of compounds depend upon the formation of H-bonding. Amines have higher boiling points than hydrocarbons of simple molecular masses. This is due to the reason that amines being polar, form intermolecular H-bonding(except tertiary amine which do not have hydrogen atom linked to N-atom, i.e., R₃N)

Further, since the electronegativity (tendency to attract a shared pair of electrons) of nitrogen in amine is lower (3.0) than that of oxygen (3.5) in alcohol, therefore, amines form weaker H-bonds than electronegative oxygen atom.

Hence, C₂H₅OH has higher boiling point than (CH₃)₂NH and C₂H₅NH₂. Because in C₂H₅OH, the electronegativity of O-atom is higher than H-atom which makes strong intermolecular H-bonding whereas in ( CH₃)₂NH and C₂H₅NH₂, the electronegativity of N-atom is higher than H-atom but lower than O-atom in C₂H₅OH which makes weak intermolecular H-bonding.

Further, since, the extent of H-bonding depends upon the number of H-atoms on the N-atom. More the no. of H-atoms linked to nitrogen, the higher the boiling point. Since in(CH₃)₂NH have one hydrogen atom and in C₂H₅NH₂ have two H-atoms linked to nitrogen, therefore,

C₂H₅NH₂ has higher boiling point than (CH₃)₂NH.

Combining all these facts, the boiling point of the given three compounds increases in the order:

(CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

Question 25.
Arrange the following:

In increasing order of solubility in water:

C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂

Answer:
All the three classes of aliphatic amines form H-bonds with water. As the size of alkyl group increases, the solubility decreases due to a corresponding increase in the hydrophobic part (hydrocarbon part) of the molecule. On the other hand, aromatic amines are insoluble in water due to presence of larger hydrocarbon part (C₆H₅-group) which tends to retard the formation of H-bonds.

Now, among the given compounds, C₆H₅NH₂ is insoluble in water due to presence of C₆H₅-group (hydrocarbon part). In (C₂H₅)₂NH, two alkyl groups are present and in C₂H₅NH₂ only one alkyl group is present, hence C₂H₅NH₂ is more soluble in water than (C₂H₅)₂NH.

Combining all these facts, solubility of the given three compounds increases in the order:

C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

Question 26.
How will you convert:

Ethanoic acid into methanamine

Answer:

Explanation: To convert ethanoic acid into methanamine (CH₃NH₂), we need ethanamide (amide group-RCONH₂) and we can get ethanamide from ethanoyl chloride (acid chloride group- RCOCl).

Step-1: Convert ethanoic acid into ethanoyl chloride

Ethanoic acid (CH₃COOH) reacts with thionyl chloride (SOCl₂) or PCl₅ or PCl₃ to form ethanoyl chloride (CH₃COCl) by the replacement of OH group by Cl atom.

Step-2 Convert ethanoyl chloride into ethanamide

Ethanoyl chloride (CH₃COCl) reacts with ammonia (in excess) to form ethanamide (CH₃CONH₂) by removal of NH₄Cl

Step-3: Convert ethanamide into methanamine (Hoffman Bromamide reaction)

Ethanamide (CH₃CONH₂) is treated with an aqueous or ethanolic solution of potassium hydroxide (KOH) and bromine (Br₂), it gives ethanamine (CH₃NH₂-final product)

Note: Hoffman Bromamide Reaction is a reaction in which a primary amide is treated with an aqueous KOH or NaOH and bromine, it gives a primary amine which has one carbon atom less than the original amide.

Question 27.
How will you convert:

Hexanenitrile into 1-aminopentane

Answer:

Explanation: The structure of Hexanenitrile is CH₃CH₂CH₂CH₂CH₂CN or C₅H₁₁CN. To convert Hexanenitrile into 1-aminopentane, first we need hexanamide (amide group-RCONH₂) and we can get hexanamide from hexanoyl chloride (acid chloride group – RCOCl) and we can get hexanoyl chloride from hexanoic acid.

Step 1: Convert Hexanenitrile into Hexanoic acid

Hexanenitrile (C₅H₁₁CN) undergoes hydrolysis to form Hexanoic acid (C₅H₁₁COOH)

Step 2: Convert Hexanoic acid into hexanoyl chloride

Hexanoic acid (C₅H₁₁COOH) reacts with thionyl chloride (SOCl₂) or PCl₅ or PCl₃ to form hexanoyl chloride (C₅H₁₁COCl) by the replacement of OH group by Cl atom.

Step 3: Convert Hexanoyl chloride into hexanamide

Hexanoyl chloride (C₅H₁₁COCl) reacts with excess ammonia to form hexanamide (C₅H₁₁CONH₂) by the removal of NH₄Cl.

Step 4: Convert hexanamide into 1- amino pentane by Hoffman Bromamide reaction)

Hexanamide (C₅H₁₁CONH₂) is treated with an aqueous or ethanolic solution of potassium hydroxide (KOH) and bromine (Br₂), it gives 1-aminopentane (final product) which has one carbon atom less than the hexanamide.

Question 28.
How will you convert:

Methanol to ethanoic acid

Answer:

Explanation: The structure of methanol is CH₃OH. To convert methanol to ethanoic acid (the number of carbon atoms is increasing from one carbon atom to two carbon atoms) we need ethanenitrile and we can get ethanenitrile from methyl chloride.

Note: If the number of carbon atoms is increasing, we need a nitrile group and if the number of atoms is decreasing, we need an amide group (Hoffman Bromamide reaction)

Step 1: Convert methanol into methyl chloride

Methanol (CH₃OH) reacts with thionyl chloride (SOCl₂) or PCl₅ or PCl₃ to form methyl chloride (CH₃Cl) by the replacement of OH group by Cl atom.

Step 2: Convert methyl chloride into ethanenitrile

Methyl chloride (CH₃Cl) reacts with ethanolic NaCN/ KCN to form ethanenitrile (CH₃CN) by the removal of NaCl / KCl

Step 3: Convert ethanenitrile into ethanoic acid

Ethanenitrile (CH₃CN) undergoes hydrolysis to form ethanoic acid (final product) which has one carbon atom more than the ethanenitrile.

Question 29.
How will you convert:

Ethanamine into methanamine

Answer:
Ethanamine into methanamine (ethylamine to methylamine)

Explanation: The structure of ethanamine is CH₃CH₂NH₂. To convert ethanamine into methanamine (the number of carbon atoms is decreasing from two carbon atoms to one carbon atom), first we need ethanamide ( amide group- RCONH2) and we can get acetamide from acetic acid. By oxidation of ethanol, we can get ethanoic acid.

Step 1: Convert Ethanamine into ethanol with the help of diazonium salt

Ethanamine (CH₃CH₂NH₂)reacts with NaNO₂ and HCl to give Diazonium salt (R--N₂⁺Cl^-) which undergoes hydrolysis to form ethanol.

Note: The diazonium salts or diazonium compounds are the class of organic compounds with general formula R−N₂⁺X⁻ where X is an organic or inorganic anion (for example, Cl^–, Br^–, BF₄^–, etc.) and R is an alkyl or aryl group. Hence, they have two nitrogen atoms with one being charged. Example - Benzenediazonium chloride (C₆H₅N₂⁺Cl^–)

Step 2: Convert ethanol to ethanoic acid by oxidation

Ethanol (CH₃CH₂OH) undergoes oxidation in the presence of strong oxidizing agent KMNO₄ to form ethanoic acid (CH₃COOH)

Step 3: Convert ethanoic acid into ethanamide (amide group)

Ethanoic acid (CH₃COOH) is treated with ammonia (in excess) to form ethanamide (CH₃CONH₂)

Step 4: Convert ethanamide into methanamine by Hoffman Bromamide Reaction

Ethanamide (CH₃CONH₂) is treated with alc.NaOH or KOH in the presence of bromine, it gives methanamine (CH₃NH₂ - final product) which has one carbon atom less than the ethanamide.

Question 30.
How will you convert:

Ethanoic acid into propanoic acid

Answer:

Explanation: To convert ethanoic acid into propanoic acid (the number of carbon atoms are increasing from two carbon atoms to three carbon atoms), we need propionitrile (CH₃CH₂CN) and we can get propionitrile from ethyl chloride. To get ethyl chloride, we need ethanol which can be formed by the reduction of ethanoic acid.

Step 1: Convert ethanoic acid into ethanol by reduction

Ethanoic acid (CH₃COOH ) undergoes reduction in the presence of lithium aluminium hydride (LiAlH₄) to form ethanol (CH₃CH₂OH)

Step 2: Convert ethanol into ethyl chloride

Ethanol (CH₃CH₂OH ) reacts with PCl₅ to form ethyl chloride (CH₃CH₂Cl) by the replacement of OH group by Cl atom.

Step 3: Convert ethyl chloride into ethyl cyanide/ propionitrile

Ethyl chloride (CH₃CH₂Cl) reacts with ethanolic NaCN / KCN to give ethyl cyanide/ propionitrile (CH₃CH₂CN)

Step 4: Convert ethyl cyanide/ propionitrile into propanoic acid

Ethyl cyanide (CH₃CH₂CN) undergoes hydrolysis to form a propanoic acid (CH₃CH₂COOH-final product) which has one carbon atom more than the propionitrile.

Question 31.
How will you convert:

Methanamine into ethanamine

Answer:

Explanation: The structure of methanamine is CH₃NH₂ and the structure of ethanamine is CH₃CH₂NH₂. To convert methanamine into ethanamine (the number of carbon atoms are increasing from one carbon atom two carbon atoms) so we need ethanenitrile which we can get from ethyl chloride. We can obtain ethyl chloride from alcohol which can be obtained from diazonium salt (R--N₂⁺Cl^-)

Step 1: Convert methyl amine to methanol

Methyl amine (CH₃NH₂) is first treated with HNO₂ and HCl, which gives a fresh diazonium salt (R--N₂⁺Cl^-) and then diazonium salt undergoes hydrolysis to form methanol.

Step 2: Convert methanol to methyl chloride

Methanol is treated with PCl₅ or thionyl chloride, gives ethanenitrile (CH₃Cl) by the replacement of OH atom by Cl atom.

Step 3: Convert methyl chloride to ethanenitrile

Methyl chloride (CH₃Cl) is treated with ethanolic NaCN or KCN, gives ethanenitrile (CH₃CN)

Step 4: Convert ethanenitrile to ethanamine

Ethanenitrile (CH₃CN) undergoes a reduction in the presence of sodium and ethyl alcohol to form ethanamine (CH₃CH₂NH₂ – final product) which has one carbon atom more than the ethanenitrile.

Question 32.
How will you convert:

Nitromethane into dimethylamine

Answer:

Explanation: The structure of Nitromethane is CH₃NO₂ and structure of dimethylamine (CH₃NHCH₃), as dimethylamine is a secondary amine so to obtain this, we need methyl isocyanide (CH₃NC) which we can get from methanamine through carbylamine reaction.

Step 1: Convert nitromethane to methanamine

Nitromethane (CH₃NO₂) undergoes reduction in the presence of Sn and HCl to form methanamine (CH₃NH₂)

Step 2: Convert methanamine to methyl isocyanide by Carbylamine reaction

Methanamine (CH₃NH₂) is heated with alcoholic potassium hydroxide and chloroform, the methyl isocyanide (CH₃NC) is formed.

Note: Carbylamine reaction is given only by primary amines

Primary amines when heated with chloroform and alcoholic potassium hydroxide give isocynaides (carbylamines) having very unpleasant smell, which can be easily detected.

Step 3: Convert methyl isocyanide to diethylamine

Methyl isocyanide (CH₃NC) undergoes reduction in the presence of sodium and ethyl alcohol to form diethylamine (CH₃NHCH₃-final product)

Question 33.
How will you convert:

Propanoic acid into ethanoic acid?

Answer:
Propanoic acid to Ethanoic acid/ acetic acid

Explanation: The structure of propanoic acid is CH₃CH₂COOH and the structure of ethanoic acid is CH₃COOH. To convert propanoic acid to ethanoic acid (the number of carbon atoms are decreasing), we need propionamide (amide group-RCONH₂). Then propionamide undergoes Hoffman Bromamide reaction to form methylamine. Then methylamine formed can be converted to ethanol to form ethanoic acid by oxidation.

Step 1: Convert propanoic acid to propionamide

Propanoic acid (CH₃CH₂COOH) reacts with ammonia (in excess) to form propionamide (CH₃CH₂CONH₂)

Step 2: Convert propionamide to ethyl amine by Hoffman Bromamide reaction

Propionamide (CH₃CH₂CONH₂) reacts with potassium hydroxide and bromine to form ethylamine (CH₃CH₂NH₂) which has one carbon atom less than the propionamide (Hoffman Bromamide reaction)

Step 3: Convert ethyl amine to methanol by forming diazonium salt

Ethyl amine (CH₃CH₂NH₂) reacts with NaNO₂ and HCl to form diazonium salt (R--N₂⁺Cl^-) then diazonium salt undergoes hydrolysis to from ethanol (CH₃CH₂OH)

Step 4: Convert ethanol to ethanoic acid by oxidation

Ethanol (CH₃CH₂OH) undergoes oxidation in the presence of strong oxidizing agent KMNO₄ to form ethanoic acid (CH₃COOH-final product)

Question 34.
Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reaction involved.

Answer:
Primary amine: A primary (1°) amine is an amine that has the following general structural formula.

R= alkyl or aryl group

Secondary amine: A secondary (2°) amine is an amine that has the following general structural formula.

R¹ and R²= alkyl or aryl group

Tertiary amine: A tertiary amine is an amine that has the following structure

R¹, R² and R³ are alkyl or aryl groups

Identification of Primary, Secondary and Tertiary amines

Primary, secondary and tertiary amines can be identified by the following test:

Hinsberg’s test: This is an excellent test for the identification of primary, secondary and tertiary amines. In this test, the amine is shaken with benzenesulphonyl chloride ( Hinsberg’s reagent) in the presence of an excess of aqueous KOH solution when

(i) A primary amine gives a clear solution which on acidification gives an N-alkylbenzene sulphonamide which is soluble in alkali.

Due to the presence of strong electron withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as a proton. So it is acidic and dissolves in alkali.

(ii) A secondary amine reacts with Hinsberg's reagent to give a sulphonamide which is soluble in water.

There is no H-atom attached to the N-atom in the sulphonamide Therefore it is not acidic and soluble in alkali.

(iii) A Tertiary amine does not react with Hinsberg’s reagent at all

Question 35.
Write short notes on

Carbylamine reaction

Answer:
Carbylamine reaction is given only by primary amines.Primary amines when heated with chloroform and alcoholic potassium hydroxide give isocyanides (carbylamines) having a very unpleasant smell, which can be easily detected.

Question 36.
Write short notes on

Diazotization

Answer:

Aromatic primary amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) at low temperatures 273-278K to form Diazonium salts (Ar-N₂⁺Cl^-) This process of conversion of a primary aromatic amine into its diazonium salt is called Diazotization.

Question 37.
Write short notes on

Hoffman Bromamide Reaction

Answer:

Hoffman Bromamide Reaction is a reaction in which a primary amide is treated with an aqueous KOH or NaOH and bromine, it gives a primary amine which has one carbon atom less than the original amide. This reaction involves the migration of alkyl or aryl group from the carbonyl carbon atom amide to the nitrogen atom.

For example:

Question 38.
Write short notes on

Coupling Reaction

Answer:
The reaction of joining two aromatic rings through the --N=N—bond is known as coupling reaction.

Arenediazonium salts (Ar-N₂⁺Cl^-) such as benzene diazonium salts (C₆H₅-N₂⁺Cl^-) react with phenol or aromatic amine (C₆H₅-NH₂) to form coloured Azo compounds.

The coupling reaction is an example of electrophilic substitution reaction in which the diazonium cation with the positive charge on the terminal nitrogen acts as the electrophile while the electron rich compounds such as phenols and amines act as the nucleophiles.

Note: Electrophilic substitution reaction is a chemical reaction in which an electrophile (electron loving) displaces a group in a compound. For example:

Question 39.
Write short notes on

Ammonolysis

Answer:

When an alkyl or benzyl hallide allowed to react with an ethanolic solution of ammonia, it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by amino (-NH₂ group) The process of cleavage of the C-X bond by ammonia is called Ammonolysis.

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is formed.

Though primary amine is produced as the major product, this process produces a mixture of primary, secondary and tertiary amines and also quaternary ammonium salts as shown below:

This method cannot be used for the preparation of arylamines (aniline) since aryl halides are much less reactive than alkyl halides towards nucleophilic substitution reaction.

Note: Nucleophilic substitution reaction is the reaction of an electron pair donor (Nu, the nucleophile) with the electron pair acceptor (the electrophile) For example:

Question 40.
Write short notes on

Acetylation

Answer:
Acetylation is the process of introducing an acetyl group into a molecule.

With aliphatic amines: Primary and secondary amines( but not tertiary amines because they do not contain an H-atom on the N-atom) undergo nucleophilic substitution reaction when treated with acid chlorides to form N- substituted amides.

This reaction involves the replacement of Hydrogen atom of NH₂ or NH group of acetyl group which in turn leads to the production of amides.

With aromatic amine: Acylation of aromatic amines is usually carried out in the presence of a catalyst. For example, acetylation with acetyl chloride is carried out in the presence of a base stronger than the amide, like pyridine, which removes HCl formed during the reaction and shifts the equilibrium in the forward direction.

Note: Pyridine is a basic heterocyclic organic compound with the chemical formula C₅H₅N.

Question 41.
Write short notes on

Gabriel phthalimide reaction

Answer:
This is a very convenient method for the preparation of pure aliphatic primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phthalimide which on heating with a suitable alkyl halides gives an N- substituted phthalimides followed by the subsequent hydrolysis to give corresponding primary amines.

Note: Aromatic primary amines such as aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution reaction with potassium phthalimide under mild conditions

Question 42.
Accomplish the following conversions:

Nitrobenzene to benzoic acid

Answer:
First nitrobenzene is reduced to aniline by reduction with hydrogen gas with palladium catalyst followed by nitration and substitution and final hydrolysis by acid by a proton yields benzoic acid.

Question 43.
Accomplish the following conversions:

Benzene to m-bromophenol

Answer:
Benzene by nitration with conc. hydrochloric acid and nitrous acid gives nitrobenzene which on reaction with bromine liquid gives m-bromobenzene.
Upon reduction with tin and acid followed by heating with diluted hydrochloric acid gives m-brmophenol.

Question 44.
Accomplish the following conversions:

Benzoic acid to aniline

Answer:
Benzoic acid, reacted with sulphonyl chloride undergoes reaction, followed by ammine and bromine liquid in presence of sodium hydroxide gives aniline. This reaction is called as Hoffman bromamide degradation.

Question 45.
Accomplish the following conversions:

Aniline to 2,4,6-tribromofluorobenzene

Answer:
Aniline reaction with bromine water, followed by sodium nitrite gives 2,4,6- tribromodiazoniumchloride, which is very reactive gives 2,4,6-tribrmofluorobenzene upon treatment with hydrofluoroburic acid.

Question 46.
Accomplish the following conversions:

Benzyl chloride to 2-phenylethanamine

Answer:
Benzyl chloride reacted with alcoholic NaCN and reduction gives 3-phenylethananmine.

Question 47.
Accomplish the following conversions:

Chlorobenzene to p-chloroaniline

Answer:
Chlorobenzene upon nitration with nitronium ion and para product obtained undergoes reduction to give p-chloroaniline.

Question 48.
Accomplish the following conversions:

Aniline to p-bromoaniline

Answer:
As aniline is a very activating group, it is first reacted with anhydride to make it less activating , which on reaction with bromine in acetic acid , followed by acid hydrolysis gives p-bromoaniline.

Question 49.
Accomplish the following conversions:

Benzamide to toluene

Answer:
Benzamide undergoes Hoffman bromamide degradation to give aniline upon treatment with sodium nitrite gives benzenediazonium chloride, reacts with phosphoric acid , followed by friedal crafts reaction with methyl chloride in solvent gives toulene.

Question 50.
Accomplish the following conversions:

Aniline to benzyl alcohol.

Answer:
Aniline reacts with sodium nitrite following by substitution reaction with KCN, by acid hydrolysis gives benzoic acid which upon treatment with reducing agent gives benzyl alcohol.

Question 51.
Give the structures of A, B and C in the following reactions:

Answer:
Ethyl iodide reacts with NaCN gives a substitution reactions to give propanitrile upon partial hydrolysis gives B , upon reaction with sodium hydroxide gives C.

Question 52.
Give the structures of A, B and C in the following reactions:

Answer:
Benzenediazoniumchloride gives nucleophilic substitution reactions gives A, upon hydrolysis the CN ion is replaced by OH ion which is less better leaving group gives B upon heating with ammmonia gives C.

Question 53.
Give the structures of A, B and C in the following reactions:

Answer:
Ethylbromide gives nucleophilic substitution reactions gives B ,upon reduction gives B followed by reacting with nitrous acid ,i.e oxidation gives propanol.

Question 54.
Give the structures of A, B and C in the following reactions:

Answer:
Nitrobenzene upon reduction with iron/acid gives A , reacting with sodium nitrite gives benzenediazonium chloride , followed by complete Hydrolysis gives phenol.

Question 55.
Give the structures of A, B and C in the following reactions:

Answer:
Acetic acid upon heating with ammmonia gives A , which is an amide. This amide reacts with NaOBr which extracts the carbonyl carbon giving B , followed by reacting with sodium nitrite gives methanol.

Question 56.
Give the structures of A, B and C in the following reactions:

Answer:
Nitrobenzene upon reduction with iron/acid mixture gives A, followed by oxidation with nitrous acid gives B and reacting with phenol undergoes addition reaction to give C.

Question 57.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.

Answer:
It is given that compound 'C' having the molecular formula,C₆H₇N is formed by heating compound 'B' with Br₂and KOH. This is a Hoffmann bromamide degradation reaction(in which isicyanide compound is formed). Therefore, compound 'B' is an amide and compound 'C' is an amine. The only amine having the molecular formula,C₆H₇N is aniline, .
Therefore, compound 'B' (from which 'C' is formed) must be benzamide, .

Further, benzamide is formed by heating compound 'A' with aqueous ammonia. Therefore, compound 'A' must be benzoic acid.The given reactions can be explained with the help of the following equations:

Benzoic acid

Question 58.
Complete the following reactions:

C₆H₅NH₂ + CHCl₃ + alc.KOH →

Answer:
It is a carbylamine reaction in which a isocyanide compound is formed along with side products of potassium chloride.Basically the name of reaction is given is due to formation of a foul smelling compound called as isocyanide.

Question 59.
Complete the following reactions:

C₆H₅N₂Cl + H₃PO₂ + H₂O →

Answer:
Benzenediazonium chloride is a very reactive compound which oxidises hypophosphorous acid to hypophosphoric acid and the reactant is reduced to benzene.

Question 60.
Complete the following reactions:

C₆H₅NH₂ + H₂SO₄(conc.) →

Answer:
aniline undergoes sulphonation to anillium hydrogensulphate.

Question 61.
Complete the following reactions:

C₆H₅N₂Cl + C₂H₅OH →

Answer:
aniline is very activating group which undergoes reaction to give ortho and para product. But in acidic medium aniline accquires positive charge on N atom which is metadirecting group and hence the product.

Question 62.
Complete the following reactions:

C₆H₅NH₂ + Br₂(aq) →

Answer:
It is clubbing reaction of a very reactive compound undergoing addition to give benzene.

Question 63.
Complete the following reactions:

C₆H₅NH₂ + (CH₃CO)₂O →

Answer:
As mentioned aniline is avery activating group, so to reduce its activity it is reacted with anhydride to give n-phenylethanamine.

Question 64.
Complete the following reactions:

C₆H₅N₂Cl

Answer:
Hydrofluroburic acid reacts qith benzenediazonium chloride along with sodium nitrite to give a reduced product as Nitrobenzene.

Question 65.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer:
Gabrielphthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (S_N2) of alkyl halides by the anion formed by the phthalimide.But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Therefore aromatic primary amines can not be formed by gabriel phthalimide process.

Question 66.
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Answer:
(i) Aromatic amines react with nitrous acid (prepared in situ fromNaNO₂ and a mineral acid such as HCl) at 273 - 278 K to form stable aromatic diazonium salts i.e., NaCl and water.This reaction is widely used for preparation of variety of compounds.

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO₂ and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which is very reactive, which further produce alcohol and HCl with the evolution of nitrogen gas.

Question 67.
Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

(ii) Why do primary amines have higher boiling point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Answer:
i) Amines undergo protonation to give amide ion.
R-NH₂ --> R-NH + H⁺
Similarly, alcohol loses a proton to give alkoxide ion.
R-OH --> R-O + H⁺
In an amide ion, the negative charge is on the N-atom whereas, an alkoxide ion, the negative charge is on the O-atom. Since O is more electronegative than N, O can accommodate the negative charge more easily than N. As a result, the amide ion is less stable than the alkoxide ion. Hence, amines are less acidic than alcohols of comparable molecular masses.

ii) In a molecule of a tertiary amine, there are no H−atoms whereas, in primary amines, two hydrogen atoms are present. Due to the presence of H−atoms, primary amines undergo extensive intermolecular H−bonding.

As a result, extra energy is required to separate the molecules of primary amines. Hence, primary amines have higher boiling points than tertiary amines.
iii)Due to the −R effect of the benzene ring, the electrons on the N- atom are less available in case of aromatic amines. Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily. This explains why aliphatic amines are stronger bases than aromatic amines.