Learning Objectives

After studying this unit, the student would be able to

• classify the amines according to their degree of substitution

• tell about the shape and hybridisation of amines

• explain the methods of preparation of amines

• describe the physical properties of amines

• tell the effect of various substituents on basicity of amines

• describe the various chemical reactions of amines

• distinguish between the three classes of amines, primary, secondary and tertiary

• discuss the preparation of diazonium salts

• understand the importance of diazonium salts in the preparation of other organic compounds

• describe the utility of diazonium salts in preparing azo dyes and acid-base indicators

• explain the uses of amines

Amines

Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by alkyl or aryl groups.

Nitrogen atom of amines is trivalent and contains an unshared pair of electrons in its fourth orbital. Hybridisation is $s{p}^3$ and shape is pyramidal.

Classification : Amines are classified according to their degree of substitution at nitrogen atom. Thus amines are designated as primary $(1^{\circ }$ ), secondary $(2^{\circ }$ ), or tertiary $(3^{\circ }$ ) depending on whether one, two or all three hydrogen atoms in ammonia are substituted by alkyl or aryl groups. For example

Primary $\left(1^{\circ }\right)-{\mathrm{RNH}}_2$

Secondary $\left(2^0\right)-{\mathrm{R}}_2\mathrm{NH}$

Tertiary $\left(3^{\circ }\right)-{\mathrm{R}}_3\mathrm{N}$

The R group may be aryl or alkyl group and they may be identical or different.

Secondary and tertiary amines having different kind of alkyl groups are known as N-substituted primary amines where the longest alkyl chain attached to nitrogen is taken as the basic amine, for example,

Preparation of amines

1. Reduction of nitro compounds

$\underset{\text{ Nitroethane }}{{\mathrm{CH}}_3{\mathrm{CH}}_2-{\mathrm{NO}}_2+3{\mathrm{H}}_2}\stackrel{\text{ Raney Ni or Pt. }}{\to }\underset{\text{ Ethylamine }}{{\mathrm{CH}}_3{\mathrm{CH}}_2-{\mathrm{NH}}_2}+2{\mathrm{H}}_2\mathrm{O}$

$\underset{\text{Nitroethane}}{{\mathrm{CH}}_3{\mathrm{CH}}_2-}{\mathrm{NO}}_2\stackrel{\text{Sn/HCl}}{\to }\underset{\text{Ethylamine }}{{\mathrm{CH}}_3{\mathrm{CH}}_2-{\mathrm{NH}}_2}+2{\mathrm{H}}_{2}0$

2. Hofmann’s Ammonolysis of alkyl halides

${\mathrm{RNH}}_2+\mathrm{RX}⟶{\mathrm{R}}_2\mathrm{NH}\stackrel{\mathrm{RX}}{\to }{\mathrm{R}}_3\mathrm{N}\stackrel{\mathrm{RX}}{\to }{\mathrm{R}}_4{\mathrm{N}}^{+}{\mathrm{X}}^{-}$

Hofmann’s ammonolysis of alkyl halides usually gives a mixture of primary, secondary and tertiary amines along with some quaternary ammonium halides, because the replacement of hydrogens on $\mathrm{N}$ atom in ammonia does not stop at the first stage. The process of alkylation on $\mathrm{N}$ atom continues to form a secondary and a tertiary amine and finally a quaternary salt. A good yield of primary amine may still be obtained by taking ammonia in large excess over alkyl halide. Reactivity of halides with amines is $\mathrm{RI}>\mathrm{RBr}>\mathrm{RCl}$

Limitation

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

3. From nitriles and Isonitriles

Aromatic nitriles are also reduced similarly

b) Reduction of isocyanides or carbylamines

$\begin{array}{rl}& \underset{\text{ Ethyl isocyanide }}{{\mathrm{CH}}_3{\mathrm{CH}}_2}-\mathrm{N}\equiv \mathrm{C}\underset{\text{ or }{\mathrm{LiAlH}}_4\text{/ether }}{\overset{{\mathrm{H}}_2/\text{ Raney }\mathrm{Ni}\text{ or }\mathrm{Pd}}{\to }}\underset{\text{ Ethylmethylamine (secondary amine) }}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NHCH}}_3}\\ & \end{array}$

4. Reduction of amides

Substituted amides give secondary amines

5. Gabriel phthalimide reaction

Limitations

a) Only primary amines can be synthesized by this method.

b) Aromatic primary amines such as aniline, toluidines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution reaction with potassium phthalimide under mild conditions.

6. Hofmann degradation of primary amides (Hofmann bromamide reaction)

$\underset{\text{Propionamide}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CONH}}_2}+{\mathrm{Br}}_2+4\mathrm{KOH}⟶\underset{\text{Ethylamine}}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2}+{\mathrm{K}}_2{\mathrm{CO}}_3+2\mathrm{KBr}+2{\mathrm{H}}_2\mathrm{O}$

This reaction provides a method for stepping down the series, since the amine produced contains one carbon atom less than the starting amide.

7. Reduction of Oximes

$\underset{\text{ Acetaldoxime }}{{\mathrm{CH}}_3\mathrm{CH}=\mathrm{NOH}}\underset{\text{ or Na/ }{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}{\overset{{\mathrm{LiAlH}}_4\text{/ether }}{\to }}\underset{\text{ Ethylamine }}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2}$

8. By reductive amination of aldehydes and ketones

$\underset{\text{Acetaldehyde}}{{\mathrm{CH}}_3-\mathrm{CH}=\mathrm{O}}+\underset{\text{Ammonia}}{{\mathrm{NH}}_3}\underset{-{\mathrm{H}}_2\mathrm{O}}{\overset{\mathrm{\Delta }}{\to }}\underset{\text{Acetaldimine}}{[{\mathrm{CH}}_3-\mathrm{CH}=\mathrm{NH}]}\underset{\text{ or }\mathrm{NaBH},\mathrm{CN}}{\overset{\mathrm{H}/\mathrm{Ni}}{\to }}\underset{\text{Ethylamine}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{NH}}_2}$

Physical Properties

Physical state : Amines containing one to three carbon atoms are colourless gases at room temperature and have fishy odour. Amines with four or more carbon atoms are volatile liquids and the higher amines are solid. Amines are usually colourless but due to oxidation on keeping, they become coloured.

Boiling Points : The boiling points of primary and secondary amines are lower than those of alcohols of comparable molecular mass because of lesser degree of intermolecular association. This is because hydrogen bonds in amines are less strong than in alcohols because of lower electronegativity of nitrogen than oxygen. The boiling points of tertiary amines are near to those of alkanes of comparable molecular mass because they do not form hydrogen bonds since they do not have a hydrogen bonded to nitrogen. Thus the boiling points of isomeric amines are in the order primary $>$ secondary $>$ tertiary.

Solubility : All the lower amines (containing less than six carbon atoms) are soluble in water because all the three classes of amines are capable of forming hydrogen bonds with water molecules. The solubility in water decreases with increasing size of the alkyl group. Amines are soluble in non-polar solvents like benzene, ether etc. Aromatic amines are insoluble in water

Chemical Reactions of Amines

1. Basic character of amines :

All the three classes of amines are basic in nature because of their unshared electron pair. The three amines react reversibly with water to form alkyl substituted ammonium ions and hydroxide ions.

They also react with acids to form salts.

${\mathrm{RNH}}_2+\mathrm{HX}⟶\underset{\text{Alkylammonium halide }}{{\mathrm{RN}}^{+}{\mathrm{H}}_3{\mathrm{X}}^{-}}$

The measure of the basic strength can be understood in terms of ${\mathrm{K}}_b$ and ${\mathrm{pK}}_b.\mathrm{K}$ (equilibrium constant) can be expressed as $\mathrm{K}={\displaystyle \frac{\stackrel{+}{{\mathrm{RNH}}_3][\mathrm{OH}]}}{\left[{\mathrm{RNH}}_2\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}}$ or $\mathrm{K}\left[{\mathrm{H}}_2\mathrm{O}\right]={\displaystyle \frac{\stackrel{+}{\left[{\mathrm{RNH}}_3\right]\left[{\mathrm{OH}}^{-}\right]}}{\left[{\mathrm{RNH}}_2\right]}}$

$\text{ or }{\mathrm{K}}_{\mathrm{b}}={\displaystyle \frac{\stackrel{+}{\left[{\mathrm{RNH}}_3\right][\mathrm{OH}]}}{\left[{\mathrm{RNH}}_2\right]}}$

or ${\mathrm{pK}}_{\mathrm{b}}=-\log {\mathrm{K}}_{\mathrm{b}}$

Larger the value of ${\mathrm{K}}_b$, stronger would be the base and smaller the value of ${\mathrm{pK}}_b$, stronger the base.

Effect of Structure on Basicity - All aliphatic amines are stronger bases than ammonia. This is because of the availability of the unshared electron pair on the nitrogen. Alkyl groups are electron releasing and thus electrons will shift towards the nitrogen making the unshared electron pair on nitrogen more readily available. However the basicity order for three classes of amines is, $\mathrm{R}-{\mathrm{NH}}_2<{\mathrm{R}}_2-\mathrm{NH}>{\mathrm{R}}_3\mathrm{N}$.

Dialkyl amine having two electron releasing alkyl groups will be a strong base than alkyl amine (primary amine) because of the enhanced electron availability due to inductive effect of alkyl groups. Accordingly trialkyl amine having three alkyl groups should be more basic but it is less basic in aqueous solution. This is because in the aqueous phase, the substituted ammonium cations get stabilized by solvation with water molecules apart from stabilization by electron releasing alkyl groups. The order of stability of the ions are dependent on the extent of $\mathrm{H}$-bonding and is as follows

According to this, the order of basicity is $1^{\circ }>2^{\circ }>3^{\circ }$, which is opposite to the basicity due to inductive effect. Also in trialkylamine, the overcrowding of the three alkyl groups causes a steric effect which retards the protonation of nitrogen resulting in lower base strength of trialkylamine in water. All effects i.e. inductive effect, steric effect and solvation effect put together give the above order in aqueous phase. Whereas in gas phase, the basicity of amines follows the expected order i.e. $3^{\circ }>2^{\circ }>1^{\circ }>{\mathrm{NH}}_3$ because only inductive effect plays a role in gas phase.

Arylamines are weaker base than ammonia since the aromatic ring is electron withdrawing and thus reduces the electron density on nitrogen as shown,

In substituted anilines, the electron releasing groups, $-\mathrm{R}$ (alkyl), $-\mathrm{OR},-{\mathrm{NH}}_2$ increase the basic character since they increase the electron availability on nitrogen. Whereas electron withdrawing groups such as $-{\mathrm{NO}}_2,-\mathrm{COOH},-\mathrm{X}$ (halogen), $-{\mathrm{C}}_{-}^{-}$, etc decrease the basicity since they withdraw electrons and make the electron pair less available for donation.

2. Alkylation

3. Acylation -

Reaction with acid chlorides and acid anhydrides. The amine may be aliphatic or aromatic, primary or secondary amines. A tertiary amine lacking any hydrogen on nitrogen, does not react at all.

Benzoylation of compounds containing an active hydrogen atom such as alcohols, phenols and amines with benzoyl chloride in presence of dilute aqueous $\mathrm{NaOH}$ solution is called Schotten Baumann reaction.

4. Reaction with aldehydes and ketones

5. Reaction with grignard reagent

6. Reaction with carbon disulphide

This reaction is called Hofmann mustard oil reaction and is used as a test for primary amines.

7. Reaction with Nitrous Acid-Primary, secondary and tertiary amines react differently with nitrous acid.

Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which are unstable and liberate nitrogen gas thus forming the corresponding alcohol. Nitrous acid which is produced in situ is unstable and thus reaction is carried at low temperature.

$${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2+{\displaystyle \frac{{\mathrm{HNO}}_2}{{({\mathrm{NaNO}}_2+\mathrm{HCl})}^2}}{\mathrm{CH}}_3{\mathrm{CH}}_2-\stackrel{+}{{\mathrm{N}}_2}{\mathrm{Cl}}^{-}⟶{\mathrm{CH}}_3{\mathrm{CH}}_2-\mathrm{OH}+{\mathrm{N}}_2$$

Aromatic primary amines react with nitrous acid at low temperature to form diazonium salts, which is an important class of compounds used for various synthetic purposes. The reaction also forms the basis for azo dye test.

This reaction of converting aromatic primary amines into diazonium salts by treatment with a cold [273-278 K] solution of nitrous acid is called diazotisation.

Reactions given by aromatic amines only.

8. Reaction with phosgene or carbonyl chloride

$\underset{\text{Aniline}}{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2}+{\mathrm{COCl}}_2\underset{-\mathrm{HCl}}{\overset{}{\to }}{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{NH}-\mathrm{COCl}\underset{-\mathrm{HCl}}{\overset{\mathrm{\Delta }}{\to }}\underset{\text{Phenyl isocyanate}}{{\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{N}=\mathrm{C}=0}$

9. Electrophilic substitution reactions

$-{\mathrm{NH}}_2$ group strongly activates the aromatic ring through delocalization of the lone pair of electrons of the $\mathrm{N}$-atom over the aromatic ring. As a result, electron density increases more at 0 - and p-positions as compared to m-positions. Therefore, the $-{\mathrm{NH}}_2$ group directs the incoming group to 0 - and p-positions, i.e ${\mathrm{NH}}_2$ is an $0-$ - p-directing group. Due to the strong activating effect of the ${\mathrm{NH}}_2$ group, aromatic amines undergo electrophilic substitution reactions readily and it is difficult to stop the reaction at the monosubstitution stage.

i) Halogenation

If a monohalogenated derivative is required, the amino group is first acetylated and then halogenation of the ring is carried out. After halogenation, the acetyl group is removed by hydrolysis and the monohalogenated amine is obtained.

The lone pair on nitrogen enters into resonance with oxygen of carbonyl group thus making it less available for donation to benzene ring.

ii) Nitration

The reason for formation of large amount of unexpected m-nitroaniline is that under strongly acidic conditions of nitration, most of the aniline is converted into anilinium ion and since $-{\mathrm{NH}}_3{}^{+}$is a m-directing group, so an unexpected large amount of $\mathrm{m}$-nitroaniline is obtained. To produce only 0 - and p-product during nitration of aniline, first protect the amino group by acetylation followed by nitration. The acetyl group is finally removed by hydrolysis to give a mixture of 0 - and $p$-nitroanilines.

iii) Sulphonation

Sulphanilic acid contains both an acidic $\left({\mathrm{SO}}_3\mathrm{H}\right)$ as well as a basic $\left({\mathrm{NH}}_2\right)$ group. It exists as an internal salt or zwitterion. Due to zwitterion character, sulphanilic acid has high melting point and is insoluble in water and organic solvents.

iv) Friedel-Crafts reactions

Aniline is a Lewis base while ${\mathrm{AlCl}}_3$ is a Lewis acid. They combine with each other to form a salt.

Due to the presence of a positive charge on $\mathrm{N}$-atom in the salt, it acts as a strong electronwithdrawing group. As a result, it reduces the electron density in the benzene ring and hence aniline does not undergo Friedel Crafts (alkylation or acylation) reaction.

10. Oxidation Reactions : Aromatic amines get easily oxidized on exposure to air forming complex coloured products. Aniline on oxidation with sodium dichromate and sulphuric acid gives a black dye known as aniline black. Though on controlled oxidation of aniline, p-benzoquinone is formed

Tests to distinguish between $1^{\circ },2^{\circ }$ and $3^{\circ }$ Amines

1. Carbylamine test or Isocyanide test: Both aliphatic and aromatic primary amines on heating with chloroform in presence of alcoholic $\mathrm{KOH}$ form carbylamines or isocyanides having extremely unpleasent smell.

2. Hinsberg Test

The reaction of primary and secondary amines with benzenesulphonyl chloride forms the basis of this test. In this test, the given amine is treated with benzene sulphonyl chloride (Hinsberg reagent) in the presence of cold aqueous $\mathrm{NaOH}$ and then the resulting mixture is acidified.

a) A primary amine gives a clear solution which on acidification gives an insoluble $\mathrm{N}$-alkylbenzene sulphonamide

b) A secondary amine gives an insoluble N, N-dialkylbenzenesulphonamide which remains unaffected on addition of acid.

c) A teriary amine does not react at all. Therefore, it remains insoluble in the alkaline solution but dissolves on acidification to give a clear solution.

3. Azo dye test (only for $1^{\circ }$ aromatic amines): Aromatic primary amines can be distinguished from aliphatic $1^{\circ }$ amines by azo dye test. Dissolve the $1^{\circ }$ amine in dil. $\mathrm{HCl}$ and cool it to $273-278\mathrm{K}$ and then treat it with ice-cold solution of ${\mathrm{HNO}}_2({\mathrm{NaNO}}_2+$ dil. $\mathrm{HCl})$ at $273-278\mathrm{K}$. The resulting solution is then added to a cold alkaline solution of 2-naphthol ( $\beta$-naphthol). Appearance of an orange or yellow dye confirms the presence of an aromatic $1^{\circ }$ amine.

4. Libermann nitrosoamine test for secondary amines

a) Secondary amines (aliphatic and aromatic) react with nitrous acid to give yellow oily compounds called nitrosoamines and the reaction is called nitrosation. (Replacement of the available hydrogen on nitrogen by nitroso group)

b) When nitrosoamine is warmed with a little phenol and small amount of sulphuric acid, it produces a green solution which turns deep blue or violet on treatment with $\mathrm{NaOH}$.

5. Aliphatic tertiary amines $\left({\mathrm{R}}_3\mathrm{N}\right)$ have no hydrogen on nitrogen and thus do not react with nitrous acid. They simply dissolve in nitrous acid by forming the trialkyl ammonium salt of nitrous acid.

${\mathrm{R}}_3\mathrm{N}+\mathrm{HO}-\mathrm{N}=\mathrm{O}⟶{[{\mathrm{R}}_3\mathrm{N}-\mathrm{H}]}^{+}\bar{\mathrm{O}}-\mathrm{N}=0$

Diazonium Salts

Aromatic diazonium salts are more stable than aliphatic diazonium salts due to resonance.

Because of their instability, diazonium salts cannot be stored and are used immediately after preparation.

Preparation

This process of conversion of a primary aromatic amine into its diazonium salt is called diazotisation.

General formula is ${\mathrm{RN}}_2^{+}{X}^{-}$where $\mathrm{R}$ is alkyl or aryl group and ${\mathrm{X}}^{-}={\mathrm{Cl}}^{-},{\mathrm{Br}}^{-},{\mathrm{NO}}_3^{-}$or any other anion.

Name: They are named by adding the word diazonium to the parent aromatic compound followed by the name of anion. For example, ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2{}^{+}{\mathrm{Cl}}^{-}$is benzene diazonium chloride, $\mathrm{p}-{\mathrm{ClC}}_6{\mathrm{H}}_4{\mathrm{N}}_2{}^{+}{\mathrm{HSO}}_4^{-}$is $\mathrm{p}$-chlorobenzene diazonium hydrogen sulphate.

Physical Properties: Dry diazonium salts are crystalline solids. They are readily soluble in water. They are unstable and may explode in dry state. Thus they are not isolated and are used in the solutions in which they are prepared.

Reactions of diazonium salt

1. Sandmeyer reaction

2. Gattermann reaction

3. Reaction with $\mathrm{KI}$

4. Balz Schiemann reaction

5. Reaction with ${\mathrm{H}}_2\mathrm{O}$

6. Formation of benzene

$\underset{\text{Arenediazonium salt}}{{\mathrm{ArN}}_2{}^{+}}{\mathrm{X}}^{-}+{\mathrm{H}}_3{\mathrm{PO}}_2+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{Cu}}^{+}}{\to }\underset{\text{Arene}}{\mathrm{Ar}-\mathrm{H}}+\underset{\text{Phosphorus acid}}{{\mathrm{H}}_3{\mathrm{PO}}_3}+\mathrm{HX}+{\mathrm{N}}_2\uparrow$

$\underset{\text{Arenediazonium salt}}{\mathrm{Ar}-\stackrel{+}{\mathrm{N}}\equiv {\mathrm{NX}}^{-}}+\underset{\text{Ethanol}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}}\stackrel{\mathrm{\Delta }}{\to }\underset{\text{Arene}}{\mathrm{Ar}-\mathrm{H}}+{\mathrm{N}}_2+\mathrm{HX}+{\mathrm{CH}}_3\mathrm{CHO}$

7. Formation of nitrobenzene

8. Coupling reactions

• In coupling reactions, nitrogen is retained in the product.

• Diazonium salts only react with aromatic compounds having strong electron releasing groups.

• Substitution usually occurs at para position to the activating group.

• Phenols couple in mildly alkaline medium and amines in mildly acidic medium.

• These products are called azo compounds.

• Azo compounds are brightly coloured, yellow, orange, red, blue etc and are of importance as dyes.

• Azo compounds are also used as acid - base indicators.

Uses of Amines

1. Aliphatic amines are used as solvents, intermediates in drug manufacture and also as reagents in organic synthesis.

2. Quaternary ammonium salts, also derived from long chain aliphatic tertiary amines are widely used as detergents.

3. Aromatic amines are also used as intermediates for the synthesis of dyes, drugs, textiles and photographic developers.

4. Aromatic amines serve as the starting material for diazotization reaction to give diazonium salts. These diazonium salts are used for the synthesis of many compounds.

Solved Examples

Question 1- p-Chloroaniline and anilinium hydrochloride can be distinguished by

(a) Sandmeyer reaction

(b) ${\mathrm{NaHCO}}_3$

(c) ${\mathrm{AgNO}}_3$

(d) Carbylamine test

Question 2- A positive carbylamine test is given by

(a) $\mathrm{N},\mathrm{N}$-dimethylaniline

(b) 2,4-dimethylaniline

(c) $\mathrm{N}-$ methylaniline

(d) p-methylbenzylamine

Question 3- An organic compound $\mathrm{A}$ on treatment with ${\mathrm{NH}}_3$ gives ’ $\mathrm{B}$ ’ which on heating gives ’ $\mathrm{C}$ ‘. ’ $\mathrm{C}$ ’ when treated with ${\mathrm{Br}}_2$ in the presence of $\mathrm{KOH}$ produces ethylamine. Compound ’ $A$ ’ is

(a) ${\mathrm{CH}}_3\mathrm{COOH}$

(b) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{COOH}$

(c) ${\mathrm{CH}}_3-\mathrm{CHCOOH}$
$\text{ | }$
${\mathrm{CH}}_3$

(d) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{COOH}$

Show Answer

Answer- is (d) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{COOH}$

Question 4- In the chemical reaction

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\underset{\mathrm{HCl},278\mathrm{K}}{\overset{{\mathrm{NaNO}}_2}{\to }}\mathrm{A}\underset{\mathrm{\Delta }}{\overset{\mathrm{CuCN}}{\to }}\mathrm{B}$

The compounds A and B are :

(a) Nitrobenzene and chlorobenzene

(b) Phenol and bromobenzene

(c) Flourobenzene and phenol

(d) Benzene diazonium chloride & benzonitrile

Question 5- An organic compound $\left[{\mathrm{C}}_3{\mathrm{H}}_9\mathrm{N}\right][\mathrm{A}]$ when treated with nitrous acid gave an alcohol and ${\mathrm{N}}_2$ gas was evolved. $[A]$ on warming with ${\mathrm{CHCl}}_3$ and caustic potash gave $[B]$ which on reduction gave isopropylmethylamine. Predict the structure of $[\mathrm{A}]$ :

Show Answer

Answer- is (a)

Question 6- Predict the product

Show Answer

Answer- is (c)

Practice Questions

Question 1- Carbylamine test is performed in alc. $\mathrm{KOH}$ by heating a mixture of

(a) Chloroform and silver powder

(b) Trihalogenated methane and a primary amine

(c) An alkyl halide and a primary amine

(d) An alkyl cyanide and a primary amine

Question 2- Which of the following will not undergo diazotization?

(a) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2$

(b) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{NH}}_2$

(c) ${\mathrm{CH}}_3{\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{NH}}_2$

(d) ${\mathrm{NO}}_2{\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{NH}}_2$

Question 3- Benzamide can be converted to benzyl amine using

(a) ${\mathrm{Br}}_2/\mathrm{KOH}$

(b) ${\mathrm{PCl}}_5$

(c) ${\mathrm{LiAlH}}_4$

(d) ${\mathrm{HN}}_3$

Question 4- The compound which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosamine is

(a) methylamine

(b) ethylamine

(c) diethylamine

(d) triethylamine

Question 5- ${\mathrm{CH}}_3{\mathrm{NH}}_2+{\mathrm{CHCl}}_3+\mathrm{KOH}\to$ Nitrogen containing compound $+\mathrm{KCl}+{\mathrm{H}}_2\mathrm{O}$. Nitrogen containing compound is

(a) ${\mathrm{CH}}_2\mathrm{CN}$

(b) ${\mathrm{CH}}_3{\mathrm{NHCH}}_3$

(c) ${\mathrm{CH}}_3-\bar{\mathrm{N}}\equiv \stackrel{+}{\mathrm{C}}$

(d) ${\mathrm{CH}}_3-\stackrel{+}{\mathrm{N}}\equiv \bar{\mathrm{C}}$

Question 6- Acetamide is treated separately with the following reagents. Which of these would give methylamine?

(a) ${\mathrm{PCl}}_5$

(b) Sodalime

(c) $\mathrm{NaOH}+{\mathrm{Br}}_2$

(d) Hotconc. ${\mathrm{H}}_2{\mathrm{SO}}_4$

Question 7- In the chemical reactions,

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\underset{\mathrm{HCl},278\mathrm{K}}{\overset{{\mathrm{NaNO}}_2}{\to }}A\stackrel{{\mathrm{HBF}}_4}{\to }B$

The compounds A and B respectively are

(a) Nitrobenzene and chlorobenzene

(b) Nitrobenzene and flourobenzene

(c) Phenol and benzene

(d) Benzene diazonium chloride and flouro benzene

Question 8- Which of the following amines cannot be prepared by Gabriel’s synthesis

(a) Butylamine

(b) Isopropylamine

(c) $\mathrm{N}-$ phenylethylamine

(d) Benzylamine

Question 9- The strongest base among the following is

Question 10- In the chemical reaction

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2+{\mathrm{CHCl}}_3+3\mathrm{KOH}\to [\mathrm{A}]+[\mathrm{B}]+3{\mathrm{H}}_2\mathrm{O}$

the compound $[A]$ and $[B]$ are respectively

(a) ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{NC}$ and $3\mathrm{KCl}$

(b) ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{CN}$ and $3\mathrm{KCl}$

(c) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CONH}}_2$ and $3\mathrm{KCl}$

(d) ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{NC}$ and ${\mathrm{K}}_2{\mathrm{CO}}_3$

Question 11- Predict the product

Question 12- The product of the following reaction is:-

Question 13- In the following reaction

The structure of the major product $\mathrm{X}$ is

Question 14- Amongst the compounds given, the one that would form a brilliant coloured dye on treatment with ${\mathrm{NaNO}}_2$ in dil. $\mathrm{HCl}$ followed by addition to an alkaline solution of $\beta$-naphthol is

Question 15-

’ $X$ ’ in the above reaction is

(a) ${\mathrm{H}}_3{\mathrm{PO}}_2$

(b) ${\mathrm{H}}_3{\mathrm{PO}}_3$

(c) ${\mathrm{H}}_3{\mathrm{PO}}_4$

(d) ${\left({\mathrm{HPO}}_3\right)}_2$

Question 16- Benzene diazonium chloride on reaction with phenol in weakly basic medium gives

(a) diphenylether

(b) p-hydroxyazobenzene

(c) chlorobenzene

(d) benzene

Question 17- Chlorobenzene can be prepared by reacting aniline with

(a) hydrochloric acid

(b) cuprous chloride

(c) chlorine in presence of anhyd ${\mathrm{AICl}}_3$

(d) nitrous acid followed by heating with cuprous chloride

Question 18- The major product of the following reaction is

Question 19- The final product in the following reaction sequence is

p-chloroaniline $\underset{0-5^{\circ }\mathrm{Cl}}{\overset{{\mathrm{NaNO}}_2/\mathrm{HCl}}{\to }}$ ? $\stackrel{\mathrm{KCN}}{\to }$ ? $\stackrel{{\mathrm{LiAlH}}_4}{\to }$ ?

(a) p-chlorobenzamide

(b) p-chlorophenol

(c) p-chlorobenzylamine

(d) p-chlorobenzylalcohol