NCERT Solutions for Class 12 Chemistry Chapter 13 - Organic Compounds containing Nitrogen (Amines and Diazonium Salts)

Question 1:

Write chemical equations for the following reactions:
(a) Reaction of ethanolic NH3 with C2H5Cl.
(b) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of CH3Cl.

Answer:

Question 2:

Write chemical equations for the following conversions:

Answer:

Question 3:

Write structures and IUPAC names of

  1. the amide which gives propanamine by Hoffmann bromamide reaction
  2. the amine produced by the Hoffmann degradation of benzamide.
Answer:

(i) Since propanamine contains 3 carbon atoms, the amide molecule must contain four carbon atoms.

Question 4:

Arrange the following in the decreasing order of their basic strength:
C6H5NH2, C2H5NH2, (C2H5)2NH, NH3

Answer:

The decreasing order of basic strength is:
(C2H5)2 NH > C2H5NH2 > NH3> C6H5NH2

Question 5:

Classify the following amines as primary, secondary and tertiary:

Answer:
  1. Primary (1°)
  2. Tertiary(3°)
  3. Primary (1°)
  4. Secondary (2°)
Question 6:
  1. Write structures of different isomeric amines corresponding to the molecular formula, C4H11N.
  2. Write IUPAC names of all the isomers.
  3. What type of isomerism is exhibited by different pairs of amines?
Answer:

(i), (ii)
Primary amines

Question 7:

How will you convert

  1. Benzene into aniline ?
  2. Benzene into N, N-dimethylaniline
  3. Cl-(CH2)6-Cl into hexane-1,6-diamine?
Answer:

Question 8:

Arrange the following in increasing order of their basic strength:

Answer:

Question 9:

Complete the following acid-base reactions and name the products:

Answer:

Question 10:

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

Answer:

Question 11:

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

Answer:

Question 12:

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:

Four structural isomers are possible

Question 13:

Convert
(i) 3-Methylaniline into 3-nitrotoluene (ii) Aniline into 1,3,5-tribromobenzene.

Answer:

Question 14:

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

Answer:
  1. Propan-2-amine (1°)
  2. Propan-1-amine (1°)
  3. N-methylpropan-2-amine (2°)
  4. 2-Methylpropan-2-amine (1°)
  5. N-Methylbenzenamine(2°)
  6. N-Ethyl-N-methylethanamine (3°)
  7. 3-Bromobenzenamine or 3-Bromoaniline (1°)
Question 15:

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

  1. Methylamine and dimethylamine
  2. Secondary and tertiary amines
  3. Ethylamine and aniline
  4. Aniline and benzylamine
  5. Aniline and N-methylaniline
Answer:

(i) These can be distinguished by carbylamine test. When heated with alcoholic solution of KOH and chloroform, methylamine gives foul smell of methyl isocyanide.

Question 16:

Account for the following:

  1. pKb of aniline is more than that of methylamine.
  2. Ethylamine is soluble in water, whereas aniline is not.
  3. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
  4. Although amino group is o-and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
  5. Aniline does not undergo Friedel Crafts reaction.
  6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
  7. Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer:

(i) In aniline, the lone pair of electrons on N atom is delocalized over the benzene ring. As a result, electron density on the nitrogen decreases. On the other hand, in CH3NH2, +I effect of CH3 group increases the electron density on N atom. Therefore, aniline is less basic than methylamine and hence pKb of aniline is higher than that of methylamine.

(ii) Ethylamine dissolves in water due to intermolecular hydrogen bonding as shown below:

However, because of large hydrophobic part (i.e., hydrocarbon part) of aniline, the extent of hydrogen bonding is less and therefore, aniline is insoluble in water.

(iii) Methylamine is more basic than water and therefore, accepts a proton from water forming OH– ions

(iv) Under strongly acidic conditions of nitration (in the presence of a mixture of conc. HNO3 + H2SO4), aniline gets protonated and is converted into anilinium ion having –NH3+ group. This group is deactivating group and is m-directing. So, the nitration of aniline gives o, p-nitroaniline (mainly p-product) while the nitration of anilinium ion gives m-nitroaniline.

Question 17:

Arrange the following :

Answer:

Question 18:

How will you convert

  1. Ethanoic acid into methanamine
  2. Hexanenitrile into 1-aminopentane,
  3. Methanol to ethanoic acid,
  4. Ethanamine into methanamine,
  5. Ethanoic acid into propanoic acid, (vi) Methanamine into ethanamine,
  6. Nitromethane into dimethylamine,
  7. Propanoic acid into ethanoic acid ?
Answer:

Question 19:

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

Answer:

The primary, secondary and tertiary amines can be distinguished by the following tests :

Question 20:

Write short notes on the following :

  1. Carbylamine reaction
  2. Diazotisation
  3. Hoffmann’s bromamide reaction
  4. Coupling reaction
  5. Ammonolysis
  6. Acetylation
  7. Gabriel phthalimide synthesis
Answer:

(i) Carbylamine reaction. When a primary amine is warmed with chloroform and alcoholic potash, an alkyl isocyanide (carbylamine) is formed which gives very foul smell.

This reaction is not given by secondary and tertiary amines. Therefore, this reaction provides an excellent test for the identification of a primary amine.

(ii) Diazotisation. The reaction of aniline or other aromatic amines with nitrous acid at 0–5°C to form diazonium salts is called diazotisation. Nitrous acid needed for this reaction is prepared in situ by the action of dil. HCl on NaNO2.

(iii) From amides by Hoffmann’s degradation method. Primary amines can be prepared from amides by treatment with Br2 and KOH solution. The amine formed contains one carbon atom less than the parent amide.

(iv) Coupling reaction. Aromatic amines react with diazonium salts to form azo compounds in acidic medium called dyes. The reaction is known as coupling or diazo reaction. For example, aniline couples with benzene diazonium chloride to form diazo amino benzene which ultimately changes to p-amino azo benzene on warming with a small quantity of hydrochloric acid.

(v) Ammonolysis of alkyl halides. The hydrogen atoms in ammonia can be substituted by alkyl groups by heating alkyl halides with ammonia.

The primary amine so formed reacts further with alkyl halides to give secondary and tertiary amines and quaternary salts.

(vi) Acylation. Primary and secondary amines (which contain replaceable hydrogen atoms) react with acid chlorides to form substituted amides. For example,

Since tertiary amines do not contain replaceable hydrogen atom, they donot react with acetyl chloride.

(vii) Gabriel’s phthalimide synthesis. This method is used for preparing only primary amines. In this method, phthalimide is treated with alcoholic KOH to give potassium phthalimide, which is treated with alkyl halide or benzyl halide to form N-alkyl or aryl phthalimide. The hydrolysis of N-alkyl phthalimide with 20% HCl under pressure or refluxing with NaOH gives primary amine.

Phthalic acid can again be converted into phthalimide and is used again and again. This method is very useful because it gives pure amines. Aryl halides cannot be converted to aryl amines by Gabriel synthesis because they do not undergo nucleophilic substitution with potassium phthalimide.

Question 21:

Accomplish the following conversions.

  1. Nitrobenzene to benzoic acid
  2. Benzene to m-bromophenol
  3. Benzoic acid to aniline
  4. Aniline to 2,4,6-tribromofluorobenzene
  5. Benzyl chloride to 2-phenylethanamine
  6. Chlorobenzene to p-chloroaniline
  7. Aniline to p-bromoaniline
  8. Benzamide to toluene
  9. Aniline to benzyl alcohol
Answer:

Question 22:

Give the structures of A, B and C in the following reactions

Answer:

Question 23:

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:
  1. Since the compound C of molecular formula C6H7N is formed from B on treatment with Br2 and KOH (Hoffmann bromamide reaction), therefore, the compound ‘B’ must be an amide and ‘C’ must be an amine. The only aromatic amine having molecular formula C6H7N is C6H5NH2 (aniline).
  2. Since ‘C’ is aniline, the amide from which is formed by must be benzamide (C6H5CONH2).

Question 24:

Complete the following reactions :

Answer:

Question 25:

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

Answer:

Aromatic amines such as aniline cannot be prepared by Gabriel phthalimide reaction because it requires the treatment of potassium phthalimide with C6H5Cl or C6H5Br. Since aryl halides donot undergo nucleophilic substitution reaction under ordinary conditions, therefore, the reaction does not occur. Hence, aniline cannot be prepared by this method.

Question 26:

How do aromatic and aliphatic primary amines react with nitrous acid ?

Answer:

Aromatic primary amines react with HNO2 at 273–278 K to form aromatic diazonium salts.

Aliphatic primary amines also react with HNO2 at 273–278 K to form aliphatic diazonium salts. However, these are unstable even at low temperature and therefore, decompose readily to form alcohols (generally predominate) and N2 is evolved.

Question 27:

Give plausible explanation for each of the following :

  1. why are amines less acidic than alcohols of comparable molecular masses?
  2. why are primary amines higher boiling than tertiary amines ?
  3. why are aliphatic amines stronger bases than aromatic amines ?
Answer:
  1. Loss of proton from amines give amide ion whereas loss of a proton from alcohol gives an alkoxide ion.

    Since O is more electronegative than N, therefore, RO can accommodate the –ve charge more easily than RNH. Consequently, RO is more stable than RNH. Thus, alcohols are more acidic than amines.
  2. Primary amines (RNH2) have two hydrogen atoms on the N atom and therefore, form intermolecular hydrogen bonding.

    Tertiary amines (R3N) donot have hydrogen atoms on the N atom and therefore, these donot form hydrogen bonds. As a result of hydrogen bonding in primary amines, they have higher boiling points than tertiary amines of comparable molecular mass. For example, b.p. of n-butylamine is 351 K while that of tert-butylamine is 319 K.
  3. Both arylamines and alkylamines are basic in nature due to the presence of lone pair on N-atom. But arylamines are less basic than alkylamines. For example, aniline is less basic than ethylamine as shown by Kb values :

Ethylamine : Kb = 4.7 × 10–4 Aniline : Kb = 4.2 × 10–10
The less basic character of aniline can be explained on the basis of aromatic ring present in aniline. Aniline can have the following resonating structures :

It is clear from the above resonating structures that three of these (III, IV and V) acquire some positive charge on N atom. As a result, the pair of electrons become less available for protonation. Hence, aniline is less basic than ethyl amine in which there is no such resonance.

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