//M0//QN1//SUB//DL0//EQ

What are amine compounds? Explain with an example.

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Amines can be considered as derivatives of ammonia.

The compounds obtained when one, two or three hydrogen atoms in ammonia are displaced by an alkyl or aryl group are called amine compounds.

e.g. CH3 – NH2, C6H5 – NH2, CH3 – NH – CH3, CH3 – N etc...

//M0//QN2//SUB//DL0//EQ

Explain the structure of amine compounds by drawing a diagram.

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Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons.

Nitrogen orbitals in amines are therefore, sp3 hybridized and the geometry of amines is pyramidal.

Each of the three sp3 hybridized orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines.

The fourth orbital of nitrogen in all amines contains an unshared pair of electrons.

The angle C–N–E (where E is C or H) is less than 109.5°.

For instance, it is 108° in case of trimethylamine as in the figure below:

//M0//QN3//SUB//DL0//EQ

Explain the classification of amine compounds.

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Amine compounds are classified as primary (1°), secondary (2°) and tertiary (3°) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule.

If one hydrogen atom of ammonia is replaced by R or Ar, we get RNH2 or ArNH2, a primary amine (1°).

If two hydrogen atoms of ammonia or one hydrogen atom of R–NH2 are replaced by another alkyl/aryl (R') group we get secondary amine (2°) R–NH–R'.

The second alkyl/aryl group may be same or different.

Replacement of another hydrogen atom by alkyl/aryl group leads to the formation of tertiary amine (3°).

NH2 → RNH2 → →

Primary (1°) secondary (2°) tertiary (3°)

Amines are said to be 'simple' when all the alkyl or aryl groups are the same, and 'mixed' when they are different.

//M0//QN4//SUB//DL0//EQ

Give IUPAC name of the following compounds.

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(i) (ii) CH3 – CH2 – NHCOCH3

Ans. (i)

N, N – Dimethyl propan-2-amine.

(ii) CH3 – CH2 – NH – COCH3

N – Ethylethanamide.

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Write the IUPAC names of the compounds as shown below. If possible, write their common names and types of amines.

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//M0//QN6//SUB//DL0//EQ

Write a note on formation of amine compounds by reduction of nitro compounds.

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Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in acidic medium.

Nitroalkanes can also be similarly reduced to the corresponding alkanamines.

(i)

(ii)

Reduction with iron scrap and hydrochloric acid is preferred because FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction.

Fe + 2HCl → H2 + FeCl2

FeCl2 + H2O → Fe(OH)2 + 2HCl

Thus, only a small amount of hydrochloric acid is required to initiate the reaction.

//M3//QN7//SN//DL0//EQ

Describe the formation of amine compounds by ammonolysis of alkyl halide compounds.OR Write short notes on: Ammonolysis

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The carbon–halogen bond in alkyl or benzyl halides can be easily cleaved by nucleophile.

Hence, an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which
the halogen atom is replaced by an amino
(–NH2) group.

This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis.

The reaction is carried out in a sealed tube at 373 K.

The primary amine obtained which behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt.

The free amine can be obtained from the ammonium salt by treatment with a strong base:

Ammonolysis has the disadvantage of yielding a mixture of primary, secondary and tertiary amines and also a quaternary ammonium salt.

However, primary amine is obtained as a major product by taking large excess of ammonia.

The order of reactivity of halides with amines is RI > RBr > RCl.

//M0//QN8//SUB//DL0//EQ

Discuss the formation of amine compounds by reduction of nitrile and amide compounds.

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Nitriles on reduction with lithium aluminium hydride (LiAlH4) or catalytic hydrogenation produce primary amines. This reaction is used for ascent of amine series, i.e., for preparation of amines containing one carbon atom more than the starting amine.

R – C ≡ N R – CH2 – NH2

e.g.

The amides on reduction with lithium aluminium hydride yield amines.

R – CH2 – NH2

e.g.

//M3//QN9//SUB//DL0//EQ

Give following conversion in three steps: Ethanoic acid into Methanol.

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//M3//QN10//SN//DL0//EQ

Write short note on: Gabriel phthalimide synthesis.

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Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines can not be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

//M2//QN11//SN//DL0//EQ

Explain the formation of amines by Hofmann bromamide degradation reaction and state the features of this process. OR Write short notes on: Hofmann's bromamide  reaction.

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Hofmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom.

The amine so formed contains one carbon less than that present in the amide.

Example:(1)

CH3 – NH2 + Na2CO3 + 2NaBr + 2H2O

Example:(2)

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State the general physical properties of amine compounds.

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//M0//QN13//SUB//DL0

Write note on solubility of amine compounds. Explain which of the amine and alcohol compounds has higher solubility? Why?

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//M0//QN14//SUB//DL0//EQ

Discuss the effect of hydrogen bonding on the boiling point of isomeric amine compounds.

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Primary and secondary amines are engaged in inter molecular association due to hydrogen bonding between nitrogen on one and hydrogen of another molecule.

This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it.

Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation.

Therefore, the order of boiling points of isomeric amines is as follows:

Primary > Secondary > Tertiary

Intermolecular hydrogen bonding in primary amines is shown in figure.

//M0//QN15//SUB//DL0//EQ

Explain the basic characterization of amine compounds with the process equation and state the importance of the process.

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Amines, being basic nature react with acids to form salts.

Amine salts on treatment with a base like NaOH, regenerate the parent amine.

Amine salts are soluble in water but insoluble in organic solvents like ether. This reaction is the basis for the separation of amines from the non basic organic compounds insoluble in water.

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Arrange the following in increasing order of their basic strength.

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//M0//QN17//SUB//DL0//EQ

Explain the basicity of alkanamine compounds relative to ammonia in gaseous state based on inductive effect of the alkyl group.

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Let us consider the reaction of an alkanamine and ammonia with a proton to compare their basicity.

Due to the electron releasing nature of alkyl group a pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid.

Moreover, the substituted ammonium ion formed from the amine gets stabilized due to dispersal of the positive charge by the +I effect of the alkyl group.

Hence, alkylamines are stronger bases than ammonia.

Thus, the basic nature of aliphatic amines should increase with increase in the number of alkyl groups.

The order of basicity of amines in gaseous phase follows the expected order:

tertiary amine > secondary amine > primary amine > ammonia

//M2//QN18//SUB//DL0//EQ

State the factors that determine the basicity of alkyl amine compounds in aqueous medium and explain the order of basicity of alkyl amine compounds based on these factors.

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There is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in aqueous state.

In the aqueous phase the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group (+I) but also by solvation with water molecules.

The greater the size of the ion, lesser will be the solvation and the less stabilized is the ion.

The order of stability of ions are as follows:

Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base.

Thus, the order of basicity of aliphatic amines should be primary > secondary > tertiary which is opposite to the inductive effect based order.

Secondly, when the alkyl group is small like –CH3 group there is no steric hinderance to H-bonding.

In case the alkyl group is bigger than CH3 group there is steric hinderance to H-bonding.

Therefore, the change of nature of the alkyl group e.g. from –CH3 to –C2H5 results in change of the order of basic strength.

The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows:

(C2H5)2 NH > (C2H5)3 N > C2H5NH2 > NH3

(CH3)2 NH > CH3NH2 > (CH3)3 N > NH3

//M0//QN19//SUB//DL0//EQ

Describe the basic strength of aryl amine compounds relative to ammonia.

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In aniline or other arylamines, the –NH2 group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation.

Five resonance Hybrid structures of aniline are as following:

On the other hand, anilinium cation obtained by accepting a proton can have only two resonating structures.

We know that greater the number of resonating structures greater is the stability, thus aniline is more stable than anilinium ion.

Hence, the proton acceptability or the basic nature of aniline or other aromatic amines would be less than that ammonia.

In case of substituted aniline, it is observed that electron releasing groups like –OCH3, –CH3 increase basic strength where as electron withdrawing groups like –NO2, –SO3H, –COOH, –X decrease it.

//M0//QN20//SUB//DL0//EQ

Explain the alkylation process of amine compounds with example.

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The primary amine can react with alkyl halide, to form secondary and tertiary amines, and finally quaternary ammonium salt.

The order of reactivity of halides with amines is RI > RBr > RCI

//M3//QN21//SN//DL0//EQ//PYQ

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“Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. This reaction is known as acylation.”

This reaction is the replacement of hydrogen atom of –NH2 and of group by the acyl group.

The reaction is carried out in the presence of a base stronger than the amine like pyridine, which removes the formed HCl and shifts the equilibrium to the right hand side.

Amines also react with benzoyl chloride (C6H5COCl) this reaction is known as benzoylation.

//M3//QN22//SUB//DL0//EQ

Give conversion in three steps: Nitrobenzene into chlorobenzene.

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Nitrobenzene into chlorobenzene

//M2//QN23//SN//DL0//EQ

Explain the carbylamine reaction of amine compounds. OR Write short notes on: carbylamine reaction.

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Aliphatic and aromatic primary amines on heating with chloroform and Ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances.

Secondary and tertiary amines do not show this reaction.

This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

R – NH2 + CHCl3 + 3KOH R – NC + 3KCl + 3H2O

e.g. (1)

(2)

//M3//QN24//SUB//DL0//EQ

Explain the reaction of primary aliphatic amine and aromatic amine compounds with nitrous acid with necessary equations.Or Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

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(i) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts, which are unstable and quantitatively liberate nitrogen gas along with alcohols. Quantitative evolution of nitrogen is used in estimation of amino acids and proteins.

R – NH2 + HNO2

[R – N2+CI –] ROH + N2 + HCI

(ii) Aromatic amines react with nitrous acid at low temperatures to form diazonium salts.

A very important class of compounds used for synthesis of a variety of aromatic compounds.

//M3//QN25//SUB//DL0//EQ

Write a note on the reaction of amine compounds with arylsulphonyl chloride and state the utility of this reaction.Or Describe the method for the determination of primary, secondary and tertiary amines. Write the chemical equations involved.

Or Explain with equations how primary, secondary and tertiary amine compounds can be separated using the Hinsberg reaction.

//X

The reaction of amine compounds with arylsulphonyl chloride (benzenesulphonyl chloride, C6H5SO2Cl) is known as the Hinsberg test.

This test is used to distinguish and separate primary, secondary, and tertiary amines.

(a) Reaction with Primary Amine

A primary amine reacts with benzenesulphonyl chloride to form N-substituted benzenesulphonamide, which is soluble in alkali.

The sulphonamide formed contains a hydrogen atom attached to nitrogen, which is acidic due to the electron-withdrawing sulphonyl group.

Hence, it dissolves in aqueous alkali.

(b) Reaction with Secondary Amine

A secondary amine reacts with benzenesulphonyl chloride to form N, N-disubstituted benzene-sulphonamide, which is insoluble in alkali.

The product does not contain any hydrogen atom attached to nitrogen, so it is not acidic and hence insoluble in alkali.

(c) Reaction with Tertiary Amine

A tertiary amine does not react with benzenesulphonyl chloride.

However, it dissolves in dilute acids due to the formation of an ammonium salt.

The different behaviour of primary, secondary, and tertiary amines towards benzenesulphonyl chloride is useful in:

• Identifying the type of amine

• Separating mixtures of amines

At present, p-toluenesulphonyl chloride is commonly used instead of benzenesulphonyl chloride.

//M3//QN26//SUB//DL0//EQ

Explain the bromination of Aniline. State what is done if mono substituted derivative of aniline is to be formed by this process.

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Aniline reacts with bromine water at room temperature to give a white precipitate of
2, 4, 6-tribromoaniline.

Aromatic amine compounds are very reactive towards electrophilic substitution reactions.

If we have to prepare monosubstituted aniline derivative than we have to control the activating effect of –NH2 group.

This can be done by protecting the –NH2 group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.

The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below:

Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore activating effect of –NHCOCH3 group is less than that of amino group.

//M3//QN27//SUB//DL0//EQ

Explain nitration of aniline. Explain what is done by this process to obtain the para nitro derivative as the major product.

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Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.

However by protecting the –NH2 group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the P-nitro derivative can be obtained as the major product.

//M0//QN28//SN//DL0//EQ

Short note: Sulphonation of Aniline.

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Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid of 453–473K produces P-amino benzene sulphonic acid commonly known as sulphanilic acid as the major product.

Diazonium Salts

//M2//QN29//SUB//DL0//EQ

Give the general formula of diazonium salt. State the possible negative ions present in this salt and explain with an example how diazonium salts are named.

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The diazonium salts have the general formula RN2+X–, where R stands for an aryl group and X– ion may be Cl–, Br–, HSO4–, BF4– etc.

They are named by suffixing diazonium to the name of the parent hydrocarbon from which they are formed, followed by the name of anion such as chloride, hydrogensulphate, etc.

The group is called diazonium group.

For example, is named as benzene diazonium chloride and is known as benzenediazonium hydrogen sulphate.

//M2//QN30//SUB//DL0//EQ

Give the resonance structure of arenediazonium ion.

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Resonance structure of arenediazonium ion is as follows.

//M2//QN31//SN//DL0//EQ

Q.31 Write a note on the formation of benzene diazonium chloride. OR Write short notes on: Diazotisation

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Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278 K.

Nitrous acid is produced in the reaction
mixture by the reaction of sodium nitrite with hydrochloric acid.

The conversion of primary aromatic amines into diazonium salts is known as diazotization.

Due to its instability, the diazonium salt is not generally stored and is used immediately after its preparation.

C6H5NH2 + NaNO2 + 2HCl

C6H5N+2Cl– + NaCl + 2H2O

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Describe the physical properties of benzenediazonium salts.

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//M0//QN33//SUB//DL0

State the types of chemical reaction of diazonium salts.

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//M0//QN34//SUB//DL0//EQ

Explain the displacement equation of diazonium group of benzene diazonium salt by iodide ion and chloride ion.

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Replacement by iodide ion: Iodine is not easily introduced into the benzene ring directly, but when the diazonium salt solution is treated with potassium iodide, iodo benzene is formed.

Ar2C + KI → ArI + KCl + N2

Replacement by fluoride ion: When arenediazonium chloride is treated with fluoroboric acid, arene diazonium fluoroborate is precipitated which on heating decomposes to yield aryl fluroide.

Ar2C + HBF4 → Ar-2B4 Ar-F + BF3 + N2

//M2//QN35//SUB//DL0//EQ

Explain the displacement of diazonium group of benzene diazonium salt by hydrogen with equation. OR Explain the reduction process of benzene diazonium salt by giving an equation.

//X

Certain mild reducing agents like hypophosphorous acid (phosphinic acid) or ethanol reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanol, respectively.

Ar2C + H3PO2 + H2O → ArH + N2 + H3PO3 + HCl

Ar2C + CH3CH2OH → ArH + N2 + CH3CHO + HCl

//M0//QN36//SN//DL0//EQ

Explain the displacement of diazonium group of benzene diazonium salt by halide or cyanide ion with equation. OR Short note on: Sandmeyer reaction and Gattermann reaction.

//X

Replacement by halide or cyanide ion:

The Cl–, Br– and CN– nucleophiles can easily be introduced in benzene ring in the presence of Cu(I) ion.

This reaction is called Sandmeyer reaction.

Alternatively, chlorine or bromine can also be introduced in the benzene ring by treating the diazonium salt solution with corresponding halogen acid in the presence of copper powder. This is referred as Gattermann reaction.

The yield in Sandmeyer reaction is found to be better than Gattermann reaction.

//M3//QN37//SUB//DL0//EQ

Explain the displacement equation of the diazonium group of benzene diazonium salt by the hydroxyl group as well as by the nitro group.

//X

Replacement by hydroxyl group:

If the temperature of the diazonium salt solution is allowed to rise upto 283 K, the salt gets hydrolyzed to phenol.

Ar2C + H2O → ArOH + N2 + HCl

Replacement by nitrogroup: When diazonium fluroborate is heated with aqueous sodium nitrite solution in the presence of copper, the diazonium group is replaced by –NO2 group.

//M3//QN38//SN//DL0//EQ

Explain the coupling processes of benzene diazonium chloride with equation. OR Write short notes on: Coupling reactions.

//X

The azo products obtained have an extended conjugate system having both the aromatic rings joined through the –N = N– bond.

These compounds are often coloured and are used as dyes.

Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-Hydroxyazo benzene.

This type of reaction is known as coupling reaction. Similarly, the reaction of diazonium salt with aniline yields p-Aminoazobenzene. This is an example of electrophilic substitution reaction.

//M3//QN39//SUB//DL0

Identify X, Y, Z from the following reactions (Write formula)

//X

//M2//QN40//SUB//DL0

State the importance of diazonium salts in the synthesis of aromatic compounds.

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//M4//QN41//SUB//DL0//EQ

Give the structures of A and B in the following reactions.

//X

(i) CH3CH2Br A B (ii) C6H5NO2 A B

Ans. (i) CH3CH2Br CH3CH2CN CH3CH2CHO

Bromo ethane Propane nitrile Propanal

(ii) C6H5NO2 C6H5NH2

Nitrobenzene Aniline N - Phenylethanamide

Class 12 Chemistry (Part 2) 016