Amines Class 12th Chemistry Part Ii CBSE Solution

Class 12th Chemistry Part Ii CBSE Solution
Intext Questions Pg-384
  1. Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii)…
  2. (i) Write structures of different isomeric amines corresponding to the molecular formula,…
Intext Questions Pg-387
  1. Benzene into aniline How will you convert
  2. Benzene into N, N-dimethylaniline How will you convert
  3. Cl-(CH2)4-Cl into hexan-1,6-diamine? How will you convert
Intext Questions Pg-396
  1. C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH Arrange the following in increasing order…
  2. C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 Arrange the following in increasing order of their…
  3. CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. Arrange the following in increasing order…
  4. Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl →…
  5. Write reactions of the final alkylation product of aniline with an excess of methyl iodide…
  6. Write chemical reaction of aniline with benzoyl chloride and write the name of the product…
  7. Write structures of different isomers corresponding to the molecular formula, C3H9N. Write…
Intext Questions Pg-399
  1. 3-Methylaniline into 3-nitrotoluene. Convert
  2. Aniline into 1,3,5 - tribromobenzene Convert
Exercises
  1. (CH3)2CHNH2 Write IUPAC names of the following compounds and classify them into primary,…
  2. CH3(CH2)2NH2 Write IUPAC names of the following compounds and classify them into primary,…
  3. CH3NHCH(CH3)2 Write IUPAC names of the following compounds and classify them into primary,…
  4. (CH3)3CNH2 Write IUPAC names of the following compounds and classify them into primary,…
  5. C6H5NHCH3 Write IUPAC names of the following compounds and classify them into primary,…
  6. (CH3CH2)2NCH3 Write IUPAC names of the following compounds and classify them into primary,…
  7. m-BrC6H4NH2 Write IUPAC names of the following compounds and classify them into primary,…
  8. Methylamine and dimethylamine Give one chemical test to distinguish between the following…
  9. Secondary and tertiary amines Give one chemical test to distinguish between the following…
  10. Ethylamine and aniline Give one chemical test to distinguish between the following pairs…
  11. Aniline and benzylamine Give one chemical test to distinguish between the following pairs…
  12. Aniline and N-methylaniline. Give one chemical test to distinguish between the following…
  13. pKb of aniline is more than that of methylamine. Account for the following:…
  14. Ethylamine is soluble in water whereas aniline is not. Account for the following:…
  15. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.…
  16. Although amino group is o- and p- directing in aromatic electrophilic substitution…
  17. Aniline does not undergo Friedel-Crafts reaction. Account for the following:…
  18. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Account…
  19. Gabriel phthalimide synthesis is preferred for synthesising primary amines. Account for…
  20. In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 Arrange…
  21. In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 Arrange…
  22. In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine (b)…
  23. In decreasing order of basic strength in gas phase: C6H5NH2, (C2H5)2NH, (C2H5)3N and NH3…
  24. In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 Arrange the following:…
  25. In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2 Arrange the…
  26. Ethanoic acid into methanamine How will you convert:
  27. Hexanenitrile into 1-aminopentane How will you convert:
  28. Methanol to ethanoic acid How will you convert:
  29. Ethanamine into methanamine How will you convert:
  30. Ethanoic acid into propanoic acid How will you convert:
  31. Methanamine into ethanamine How will you convert:
  32. Nitromethane into dimethylamine How will you convert:
  33. Propanoic acid into ethanoic acid? How will you convert:
  34. Describe the method for the identification of primary, secondary and tertiary amines. Also…
  35. Carbylamine reaction Write short notes on
  36. Diazotization Write short notes on
  37. Hoffman Bromamide Reaction Write short notes on
  38. Coupling Reaction Write short notes on
  39. Ammonolysis Write short notes on
  40. Acetylation Write short notes on
  41. Gabriel phthalimide reaction Write short notes on
  42. Nitrobenzene to benzoic acid Accomplish the following conversions:…
  43. Benzene to m-bromophenol Accomplish the following conversions:
  44. Benzoic acid to aniline Accomplish the following conversions:
  45. Aniline to 2,4,6-tribromofluorobenzene Accomplish the following conversions:…
  46. Benzyl chloride to 2-phenylethanamine Accomplish the following conversions:…
  47. Chlorobenzene to p-chloroaniline Accomplish the following conversions:…
  48. Aniline to p-bromoaniline Accomplish the following conversions:
  49. Benzamide to toluene Accomplish the following conversions:
  50. Aniline to benzyl alcohol. Accomplish the following conversions:
  51. Give the structures of A, B and C in the following reactions:
  52. Give the structures of A, B and C in the following reactions:
  53. Give the structures of A, B and C in the following reactions:
  54. Give the structures of A, B and C in the following reactions:
  55. Give the structures of A, B and C in the following reactions:
  56. Give the structures of A, B and C in the following reactions:
  57. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’…
  58. C6H5NH2 + CHCl3 + alc.KOH → Complete the following reactions:
  59. C6H5N2Cl + H3PO2 + H2O → Complete the following reactions:
  60. C6H5NH2 + H2SO4(conc.) → Complete the following reactions:
  61. C6H5N2Cl + C2H5OH → Complete the following reactions:
  62. C6H5NH2 + Br2(aq) → Complete the following reactions:
  63. C6H5NH2 + (CH3CO)2O → Complete the following reactions:
  64. C6H5N2Cl Complete the following reactions:
  65. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?…
  66. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.…
  67. 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 2nd 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:

pKb of aniline is more than that of methylamine.


Answer:

Pkb value is the negative logarithm of the basicity constant (Kb) .i.e., pKb = -logKb


Evidently, smaller the value of pKb , 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 pKb 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:

Pkb value is the negative logarithm of the basicity constant (Kb) .i.e., pKb = -logKb


Evidently, smaller the value of pKb , 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 C6H5NHCH3, 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 pkb value, therefore, pKb 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 C2H5NH2, 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: