“The chief commercial use of amines is as intermediates in the synthesis of medicines and fibres” .
Amines constitute an important class of organic compounds derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl/aryl group(s). In nature, they occur among proteins, vitamins, alkaloids and hormones. Synthetic examples include polymers, dyestuffs and drugs. Two biologically active compounds, namely adrenaline and ephedrine, both containing secondary amino group, are used to increase blood pressure. Novocain, a synthetic amino compound, is used as an anaesthetic in dentistry. Benadryl, a well known antihistaminic drug also contains tertiary amino group. Quaternary ammonium salts are used as surfactants. Diazonium salts are intermediates in the preparation of a variety of aromatic compounds including dyes. In this Unit, you will learn about amines and diazonium salts.
13.1 Structure of Amines
Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp3 hybridised and the geometry of amines is pyramidal. Each of the three sp3 hybridised 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. Due to the presence of 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 shown in Fig. 13.1.
Amines 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, what would you get? You get R-NHR’, secondary amine. 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. Amines are said to be ‘simple’ when all the alkyl or aryl groups are the same, and ‘mixed’ when they are different.
In common system, an aliphatic amine is named by prefixing alkyl group to amine, i.e., alkylamine as one word (e.g., methylamine). In secondary and tertiary amines, when two or more groups are the same, the prefix di or tri is appended before the name of alkyl group. In IUPAC system, amines are named as alkanamines, derived by replacement of ‘e’ of alkane by the word amine. For example, CH3NH2 is named as methanamine. In case, more than one amino group is present at different positions in the parent chain, their positions are specified by giving numbers to the carbon atoms bearing –NH2 groups and suitable prefix such as di, tri, etc. is attached to the amine. The letter ‘e’ of the suffix of the hydrocarbon part is retained. For example, H2N–CH2–CH2–NH2 is named as ethane-1, 2-diamine.
In arylamines, –NH2 group is directly attached to the benzene ring. C6H5NH2 is the simplest example of arylamine. In common system, it is known as aniline. It is also an accepted IUPAC name. While naming arylamines according to IUPAC system, suffix ‘e’ of arene is replaced by ‘amine’. Thus in IUPAC system, C6H5–NH2 is named as benzenamine. Common and IUPAC names of some alkylamines and arylamines are given in Table 13.1.
13.2 (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?
13.4 Preparation of Amines
Amines are prepared by the following methods:
1. Reduction of nitro compounds
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.
Reduction with iron scrap and hydrochloric acid is preferred because FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction. Thus, only a small amount of hydrochloric acid is required to initiate the reaction.
2. Ammonolysis of alkyl halides
You have read (Unit 10, Class XII) that the carbon – halogen bond in alkyl or benzyl halides can be easily cleaved by a 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 thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt.
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 &amp;amp;amp;amp;amp;amp;amp;gt; RBr &amp;amp;amp;amp;amp;amp;amp;gt;RCl.
Write chemical equations for the following reactions:
(i) Reaction of ethanolic NH3 with C2H5Cl.
(ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of CH3Cl.
3. Reduction of nitriles
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.
5. Gabriel phthalimide synthesis
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 cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
6. Hoffmann bromamide degradation reaction
Hoffmann 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.
Write structures and IUPAC names of
(i) the amide which gives propanamine by Hoffmann bromamide reaction.
(ii) the amine produced by the Hoffmann degradation of benzamide.
(i) Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. Structure and IUPAC name of the starting amide with four carbon atoms are given below:
13.3 How will you convert
(i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) Cl–(CH2)4–Cl into hexan-1,6-diamine?
13.5 Physical Properties
The lower aliphatic amines are gases with fishy odour. Primary amines with three or more carbon atoms are liquid and still higher ones are solid. Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation.
Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part. Higher amines are essentially insoluble in water. Considering the electronegativity of nitrogen of amine and oxygen of alcohol as 3.0 and 3.5 respectively, you can predict the pattern of solubility of amines and alcohols in water. Out of butan-1-ol and butan-1-amine, which will be more soluble in water and why? Amines are soluble in organic solvents like alcohol, ether and benzene. You may remember that alcohols are more polar than amines and form stronger intermolecular hydrogen bonds than amines.
Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of 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 &amp;amp;amp;amp;amp;amp;amp;gt; Secondary &amp;amp;amp;amp;amp;amp;amp;gt; Tertiary
Intermolecular hydrogen bonding in primary amines is shown in Fig. 13.2.
Boiling points of amines, alcohols and alkanes of almost the same molar mass are shown in Table 13.2.
13.6 Chemical Reaction
Difference in electronegativity between nitrogen and hydrogen atoms and the presence of unshared pair of electrons over the nitrogen atom makes amines reactive. The number of hydrogen atoms attached to nitrogen atom also decides the course of reaction of amines; that is why primary (–NH2), and tertiary amines differ in many reactions. Moreover, amines behave as nucleophiles due to the presence of unshared electron pair. Some of the reactions of amines are described below:
1. Basic character of amines
Amines, being basic in nature, react with acids to form salts.
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.
The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base. Basic character of amines can be better understood in terms of their Kb and pKb values as explained below:
K = [R - N+H3][OH−] / [R - NH2[H2O]
or K[H2O]= [R - N+H3][OH−] / [R - NH2]
pKb = –log Kb
Larger the value of Kb or smaller the value of pKb, stronger is the base. The pKb values of few amines are given in Table 13.3.
|Name of amine||pKb|
pKb value of ammonia is 4.75. Aliphatic amines are stronger bases than ammonia due to +I effect of alkyl groups leading to high electron density on the nitrogen atom. Their pKb values lie in the range of 3 to 4.22. On the other hand, aromatic amines are weaker bases than ammonia due to the electron withdrawing nature of the aryl group.
You may find some discrepancies while trying to interpret the Kb values of amines on the basis of +I or –I effect of the substituents present in amines. Besides inductive effect, there are other effects like solvation effect, steric hinderance, etc., which affect the basic strength of amines. Just ponder over. You may get the answer in the following paragraphs.
Structure-basicity relationship of amines
Basicity of amines is related to their structure. Basic character of an amine depends upon the ease of formation of the cation by accepting a proton from the acid. The more stable the cation is relative to the amine, more basic is the amine.
Due to the electron releasing nature of alkyl group, it (R) 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 stabilised 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. This trend is followed in the gaseous phase. The order of basicity of amines in the gaseous phase follows the expected order: tertiary amine &amp;amp;amp;amp;amp;amp;amp;gt; secondary amine &amp;amp;amp;amp;amp;amp;amp;gt; primary amine &amp;amp;amp;amp;amp;amp;amp;gt; NH3. The trend is not regular in the aqueous state as evident by their pKb values given in Table 13.3. 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 stabilised is the ion. The order of stability of ions are as follows:
Decreasing order of extent of H-bonding in water and order of stability of ions by solvation.
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 &amp;amp;amp;amp;amp;amp;amp;gt; secondary &amp;amp;amp;amp;amp;amp;amp;gt; tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like –CH3 group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than CH3 group, there will be 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. Thus, 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 the aqueous state. The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows:
(C2H5)2NH &amp;amp;amp;amp;amp;amp;amp;gt; (C2H5)3N &amp;amp;amp;amp;amp;amp;amp;gt; C2H5NH2 &amp;amp;amp;amp;amp;amp;amp;gt; NH3
(CH3)2NH &amp;amp;amp;amp;amp;amp;amp;gt; CH3NH2 &amp;amp;amp;amp;amp;amp;amp;gt; (CH3)3N &amp;amp;amp;amp;amp;amp;amp;gt; NH3
(b) Arylamines versus ammonia
pKb value of aniline is quite high. Why is it so? It is because 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. If you write different resonating structures of aniline, you will find that aniline is a resonance hybrid of the following five structures.
We know that greater the number of resonating structures, greater is the stability. Thus you can infer that aniline (five resonating structures) 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 of ammonia. In case of substituted aniline, it is observed that electron releasing groups like –OCH3, –CH3 increase basic strength whereas electron withdrawing groups like –NO2, –SO3H, –COOH, –X decrease it.
Example 13.4 Arrange the following in decreasing order of their basic strength:
C6H5NH2, C2H5NH2, (C2H5)2NH, NH3
The decreasing order of basic strength of the above amines and ammonia follows the following order:
(C2H5)2NH &amp;amp;amp;amp;amp;amp;amp;gt; C2H5NH2 &amp;amp;amp;amp;amp;amp;amp;gt; NH3 &amp;amp;amp;amp;amp;amp;amp;gt; C6H5NH2
Amines undergo alkylation on reaction with alkyl halides (refer Unit 10, Class XII).
Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. This reaction is known as acylation. You can consider this reaction as the replacement of hydrogen atom of –NH2 or &amp;amp;amp;amp;amp;amp;amp;gt;N–H group by the acyl group. The products obtained by acylation reaction are known as amides. The reaction is carried out in the presence of a base stronger than the amine, like pyridine, which removes HCl so formed and shifts the equilibrium to the right hand side.
Amines also react with benzoyl chloride (C6H5COCl). This reaction is known as benzoylation.
CH3NH2 +C6H5COCl → CH3NHCOC6H5 + HCl
Methanamine Benzoyl chloride N − Methylbenzamide
What do you think is the product of the reaction of amines with carboxylic acids ? They form salts with amines at room temperature.
4. Carbylamine reaction
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.
5. Reaction with nitrous acid
Three classes of amines react differently with nitrous acid which is prepared in situ from a mineral acid and sodium nitrite.
(a) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas quantitatively and alcohols. Quantitative evolution of nitrogen is used in estimation of amino acids and proteins.
(b) Aromatic amines react with nitrous acid at low temperatures (273-278 K) to form diazonium salts, a very important class of compounds used for synthesis of a variety of aromatic compounds discussed in Section 13.7.
Secondary and tertiary amines react with nitrous acid in a different manner.
6. Reaction with arylsulphonyl chloride
Benzenesulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides.
The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.
Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.
(c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines. However, these days benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride.
7. Electrophilic substitution
You have read earlier that aniline is a resonance hybrid of five structures. Where do you find the maximum electron density in these structures? Ortho- and para-positions to the –NH2 group become centres of high electron density. Thus –NH2 group is ortho and para directing and a powerful activating group.
The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their very high reactivity. Substitution tends to occur at ortho- and para-positions. If we have to prepare monosubstituted aniline derivative, how can the activating effect of –NH2 group be controlled ? 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.
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.
(b) Nitration: 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.
(c) Sulphonation: Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid, as the major product.
Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction.
3.4 Arrange the following in increasing order of their basic strength:
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2.
13.5 Complete the following acid-base reactions and name the products:
(i) CH3CH2CH2NH2 + HCl →
(ii) (C2H5)3N + HCl →
13.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
13.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
13.8 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.
II. DIAZONIUM SALTS
The diazonium salts have the general formula RN+2X where R stands for an aryl group and X– ion may be Cl– Br–, HSO−4 , BF–4 , 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 N+2 group is called diazonium group. For example, C6H5N2+Cl– is named as benzenediazonium chloride and C6H5N2HSO4 is known as benzenediazonium hydrogensulphate.
Primary aliphatic amines form highly unstable alkyldiazonium salts (refer to Section 13.6). Primary aromatic amines form arenediazonium salts which are stable for a short time in solution at low temperatures (273-278 K). The stability of arenediazonium ion is explained on the basis of resonance.
13.7 Method of preparation of Diazonium Salts
Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278K. 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 diazotisation. Due to its instability, the diazonium salt is not generally stored and is used immediately after its preparation.
13.8 Physical Properties
Benzenediazonium chloride is a colourless crystalline solid. It is readily soluble in water and is stable in cold but reacts with water when warmed. It decomposes easily in the dry state. Benzenediazonium fluoroborate is water insoluble and stable at room temperature.
13.9 Chemical Reaction
The reactions of diazonium salts can be broadly divided into two categories, namely (A) reactions involving displacement of nitrogen and (B) reactions involving retention of diazo group.
A. Reactions involving displacement of nitrogen Diazonium group being a very good leaving group, is substituted by other groups such as Cl– , Br– , I– , CN– and OH– which displace nitrogen from the aromatic ring. The nitrogen formed escapes from the reaction mixture as a gas.
1. Replacement by halide or cyanide ion: The Cl– , Br– and CN– nucleophiles can easily be introduced in the 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 Gatterman reaction.
The yield in Sandmeyer reaction is found to be better than Gattermann reaction.
4. Replacement by H: Certain mild reducing agents like hypophosphorous acid (phosphinic acid) or ethanol reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanal, respectively.
B. Reactions involving retention of diazo group coupling reactions
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-hydroxyazobenzene. 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.
13.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds
From the above reactions, it is clear that the diazonium salts are very good intermediates for the introduction of –F, –Cl, –Br, –I, –CN, –OH, –NO2 groups into the aromatic ring.
Aryl fluorides and iodides cannot be prepared by direct halogenation. The cyano group cannot be introduced by nucleophilic substitution of chlorine in chlorobenzene but cyanobenzene can be easily obtained from diazonium salt.
Thus, the replacement of diazo group by other groups is helpful in preparing those substituted aromatic compounds which cannot be prepared by direct substitution in benzene or substituted benzene.
(i) 3-Methylaniline into 3-nitrotoluene.
(ii) Aniline into 1,3,5 – tribromobenzene.
Amines can be considered as derivatives of ammonia obtained by replacement of hydrogen atoms with alkyl or aryl groups. Replacement of one hydrogen atom of ammonia gives rise to structure of the type R-NH2, known as primary amine. Secondary amines are characterised by the structure R2NH or R-NHR’ and tertiary amines by R3N, RNR′R′′ or R2NR′. Secondary and tertiary amines are known as simple amines if the alkyl or aryl groups are the same and mixed amines if the groups are different. Like ammonia, all the three types of amines have one unshared electron pair on nitrogen atom due to which they behave as Lewis bases.
Amines are usually formed from nitro compounds, halides, amides, imides, etc. They exhibit hydrogen bonding which influence their physical properties. In alkylamines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. In aromatic amines, electron releasing and withdrawing groups, respectively increase and decrease their basic character. Aniline is a weaker base than ammonia. Reactions of amines are governed by availability of the unshared pair of electrons on nitrogen. Influence of the number of hydrogen atoms at nitrogen atom on the type of reactions and nature of products is responsible for identification and distinction between primary, secondary and tertiary amines. p-Toluenesulphonyl chloride is used for the identification of primary, secondary and tertiary amines. Presence of amino group in aromatic ring enhances reactivity of the aromatic amines. Reactivity of aromatic amines can be controlled by acylation process, i.e., by treating with acetyl chloride or acetic anhydride. Tertiary amines like trimethylamine are used as insect attractants.
Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes by reductive removal of the diazo group. Coupling reaction of ary
13.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
13.2 Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline.
13.3 Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
13.4 Arrange the following:
(i) In decreasing order of the pKb values:
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2
(ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2
(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2.
(iv) In decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
(v) In increasing order of boiling point:
C2H5OH, (CH3)2NH, C2H5NH2
(vi) In increasing order of solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2.
13.5 How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid?
13.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
13.7 Write short notes on the following:
(i) Carbylamine reaction
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(vii) Gabriel phthalimide synthesis.
13.8 Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol.
13.9 Give the structures of A, B and C in the following reactions:
13.10 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.
13.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
13.14 Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answers to Some Intext Questions
13.4 (i) C6H5NH2 &amp;amp;amp;amp;amp;amp;amp;lt; NH3 &amp;amp;amp;amp;amp;amp;amp;lt; C6H5CH2NH2 &amp;amp;amp;amp;amp;amp;amp;lt; C2H5NH2 &amp;amp;amp;amp;amp;amp;amp;lt; (C2H5)2NH
(ii) C6H5NH2 &amp;amp;amp;amp;amp;amp;amp;lt; C2H5NH2 &amp;amp;amp;amp;amp;amp;amp;lt; (C2H5)3N &amp;amp;amp;amp;amp;amp;amp;lt; (C2H5)2NH
(iii) C6H5NH2 &amp;amp;amp;amp;amp;amp;amp;lt; C6H5CH2NH2 &amp;amp;amp;amp;amp;amp;amp;lt; (CH3)3N &amp;amp;amp;amp;amp;amp;amp;lt; CH3NH2 &amp;amp;amp;amp;amp;amp;amp;lt; (CH3)2NH
Some Useful Links
I. Multiple Choice Questions (Type-I)
1. Which of the following is a 3° amine?
2. The correct IUPAC name for CH2==CHCH2 NHCH3 is
3. Amongst the following, the strongest base in aqueous medium is ____________.
(iii) (CH3)2 NH
4. Which of the following is the weakest Brönsted base?
5. Benzylamine may be alkylated as shown in the following equation :
Which of the following alkylhalides is best suited for this reaction through SN1 mechanism?
6. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?
(i) H2 (excess)/Pt
(ii) LiAlH4 in ether
(iii) Fe and HCl
(iv) Sn and HCl
7. In order to prepare a 1° amine from an alkyl halide with simultaneous addition of one CH2 group in the carbon chain, the reagent used as source of nitrogen is ___________.
(i) Sodium amide, NaNH2
(ii) Sodium azide, NaN3
(iii) Potassium cyanide, KCN
(iv) Potassium phthalimide, C6H4(CO)2N–K+
8. The source of nitrogen in Gabriel synthesis of amines is _____________.
(i) Sodium azide, NaN3
(ii) Sodium nitrite, NaNO2
(iii) Potassium cyanide, KCN
(iv) Potassium phthalimide, C6H4(CO)2N–K+
9. Amongst the given set of reactants, the most appropriate for preparing 2° amine is _____.
(i) 2° R—Br + NH3
(ii) 2° R—Br + NaCN followed by H2/Pt
(iii) 1° R—NH2 + RCHO followed by H2/Pt
(iv) 1° R—Br (2 mol) + potassium phthalimide followed by H3O+/heat
10. The best reagent for converting 2–phenylpropanamide into 2-phenylpropanamine is _____.
(i) excess H2
(ii) Br2 in aqueous NaOH
(iii) iodine in the presence of red phosphorus
(iv) LiAlH4 in ether
11. The best reagent for converting, 2-phenylpropanamide into 1- phenylethanamine is ____.
(i) excess H2/Pt
12. Hoffmann Bromamide Degradation reaction is shown by __________.
13. The correct increasing order of basic strength for the following compounds is _________.
(i) II &amp;amp;amp;amp;amp;amp;amp;lt; III &amp;amp;amp;amp;amp;amp;amp;lt; I
(ii) III &amp;amp;amp;amp;amp;amp;amp;lt; I &amp;amp;amp;amp;amp;amp;amp;lt; II
(iii) III &amp;amp;amp;amp;amp;amp;amp;lt; II &amp;amp;amp;amp;amp;amp;amp;lt; I
(iv) II &amp;amp;amp;amp;amp;amp;amp;lt; I &amp;amp;amp;amp;amp;amp;amp;lt; III
14. Methylamine reacts with HNO2 to form _________.
15. The gas evolved when methylamine reacts with nitrous acid is __________.
16. In the nitration of benzene using a mixture of conc. H2SO4 and conc. HNO3, the species which initiates the reaction is __________.
17. Reduction of aromatic nitro compounds using Fe and HCl gives __________.
(i) aromatic oxime
(ii) aromatic hydrocarbon
(iii) aromatic primary amine
(iv) aromatic amide
18. The most reactive amine towards dilute hydrochloric acid is ___________.
19. Acid anhydrides on reaction with primary amines give ____________.
(iii) secondary amine
(i) Sandmeyer reaction
(ii) Gatterman reaction
(iii) Claisen reaction
(iv) Carbylamine reaction
21. Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is
(i) Hoffmann Bromamide reaction
(ii) Gabriel phthalimide synthesis
(iii) Sandmeyer reaction
(iv) Reaction with NH3
22. Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride.
23. Which of the following compounds is the weakest Brönsted base?
24. Among the following amines, the strongest Brönsted base is __________.
25. The correct decreasing order of basic strength of the following species is _______.
H2O, NH3, OH–, NH2–
(i) NH2– &amp;amp;amp;amp;amp;amp;amp;gt; OH– &amp;amp;amp;amp;amp;amp;amp;gt; NH3 &amp;amp;amp;amp;amp;amp;amp;gt; H2O
(ii) OH– &amp;amp;amp;amp;amp;amp;amp;gt; NH2– &amp;amp;amp;amp;amp;amp;amp;gt; H2O &amp;amp;amp;amp;amp;amp;amp;gt; NH3
(iii) NH3 &amp;amp;amp;amp;amp;amp;amp;gt; H2O &amp;amp;amp;amp;amp;amp;amp;gt; NH2– &amp;amp;amp;amp;amp;amp;amp;gt; OH–
(iv) H2O &amp;amp;amp;amp;amp;amp;amp;gt; NH3 &amp;amp;amp;amp;amp;amp;amp;gt; OH– &amp;amp;amp;amp;amp;amp;amp;gt; NH2–
26. Which of the following should be most volatile?
(I) CH3CH2CH2NH2 (II) (CH3)3N (III) (IV) CH3CH2CH3
27. Which of the following methods of preparation of amines will give same number of carbon atoms in the chain of amines as in the reactant?
(i) Reaction of nitrite with LiAlH4.
(ii) Reaction of amide with LiAlH4 followed by treatment with water.
(iii) Heating alkylhalide with potassium salt of phthalimide followed by hydrolysis.
(iv) Treatment of amide with bromine in aqueous solution of sodium hydroxide.
II. Multiple Choice Questions (Type-II)
Note : In the following questions two or more options may be correct.
28. Which of the following cannot be prepared by Sandmeyer’s reaction?
29. Reduction of nitrobenzene by which of the following reagent gives aniline?
30. Which of the following species are involved in the carbylamine test?
(iv) NaNO2 + HCl
31. The reagents that can be used to convert benzenediazonium chloride to benzene are __________.
32. The product of the following reaction is __________.
33. Arenium ion involved in the bromination of aniline is __________.
34. Which of the following amines can be prepared by Gabriel synthesis.
(i) Isobutyl amine
35. Which of the following reactions are correct?
36. Under which of the following reaction conditions, aniline gives p-nitro derivative as the major product?
(i) Acetyl chloride/pyridine followed by reaction with conc. H2SO4 + conc. HNO3.
(ii) Acetic anyhdride/pyridine followed by conc. H2SO4 + conc. HNO3.
(iii) Dil. HCl followed by reaction with conc. H2SO4 + conc. HNO3.
(iv) Reaction with conc. HNO3 + conc.H2SO4.
37. Which of the following reactions belong to electrophilic aromatic substitution?
(i) Bromination of acetanilide
(ii) Coupling reaction of aryldiazonium salts
(iii) Diazotisation of aniline
(iv) Acylation of aniline
III. Short Answer Type
38. What is the role of HNO3 in the nitrating mixture used for nitration of benzene?
39. Why is NH2 group of aniline acetylated before carrying out nitration?
40. What is the product when C6H5CH2NH2 reacts with HNO2?
41. What is the best reagent to convert nitrile to primary amine?
42. Give the structure of ‘A’ in the following reaction.
43. What is Hinsberg reagent?
44. Why is benzene diazonium chloride not stored and is used immediately after its preparation?
45. Why does acetylation of —NH2 group of aniline reduce its activating effect?
46. Explain why MeNH2 is stronger base than MeOH?
47. What is the role of pyridine in the acylation reaction of amines?
48. Under what reaction conditions (acidic/basic), the coupling reaction of aryldiazonium chloride with aniline is carried out?
49. Predict the product of reaction of aniline with bromine in non-polar solvent such as CS2.
50. Arrange the following compounds in increasing order of dipole moment.
CH3CH2CH3, CH3CH2NH2, CH3CH2OH
51. What is the structure and IUPAC name of the compound, allyl amine?
54. A primary amine, RNH2 can be reacted with CH3—X to get secondary amine, R—NHCH3 but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH2 forms only 2° amine?
55. Complete the following reaction.
56. Why is aniline soluble in aqueous HCl?
57. Suggest a route by which the following conversion can be accomplished.
58. Identify A and B in the following reaction.
59. How will you carry out the following conversions?
(i) toluene → p-toluidine
(ii) p-toluidine diazonium chloride → p-toluic acid
60. Write following conversions:
(i) nitrobenzene → acetanilide (ii) acetanilide → p-nitroaniline
61. A solution contains 1 g mol. each of p-toluene diazonium chloride and pnitrophenyl diazonium chloride. To this 1 g mol. of alkaline solution of phenol is added. Predict the major product. Explain your answer.
62. How will you bring out the following conversion?
63. How will you carry out the following conversion?
64. How will you carry out the following conversion?
65. How will you carry out the following conversions?
IV. Matching Type
Note : Match the items of Column I and Column II in the following questions.
66. Match the reactions given in Column I with the statements given in Column II.
|Column I||Column II|
|(i)||Ammonolysis||(a)||Amine with lesser number of carbon atoms|
|(ii)||Gabriel phthalimide synthesis||(b)||Detection test for primary amines.|
|(iii)||Hoffmann Bromamide reaction||(c)||Reaction of phthalimide with KOH and R—X|
|(iv)||Carbylamine reaction||(d)||Reaction of alkylhalides with NH3|
67. Match the compounds given in Column I with the items given in Column II.
|Column I||Column II|
|(i)||Benzene sulphonyl chloride||(a)||Zwitter ion|
|(ii)||Sulphanilic acid||(b)||Hinsberg reagent|
|(iii)||Alkyl diazonium salts||(c)||Dyes|
|(iv)||Aryl diazonium salts||(d)||Conversion to alcohols|
V. Assertion and Reason Type
Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
68. Assertion : Acylation of amines gives a monosubstituted product whereas alkylation of amines gives polysubstituted product.
Reason : Acyl group sterically hinders the approach of further acyl groups.
69. Assertion : Hoffmann’s bromamide reaction is given by primary amines.
Reason : Primary amines are more basic than secondary amines.
70. Assertion : N-Ethylbenzene sulphonamide is soluble in alkali.
Reason : Hydrogen attached to nitrogen in sulphonamide is strongly acidic.
71. Assertion : N, N-Diethylbenzene sulphonamide is insoluble in alkali.
Reason : Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.
72. Assertion : Only a small amount of HCl is required in the reduction of nitro compounds with iron scrap and HCl in the presence of steam.
Reason : FeCl2 formed gets hydrolysed to release HCl during the reaction.
73. Assertion : Aromatic 1° amines can be prepared by Gabriel Phthalimide Synthesis.
Reason : Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.
74. Assertion : Acetanilide is less basic than aniline.
Reason : Acetylation of aniline results in decrease of electron density on nitrogen.
VI. Long Answer Type
75. A hydrocarbon ‘A’, (C4H8) on reaction with HCl gives a compound ‘B’, (C4H9Cl), which on reaction with 1 mol of NH3 gives compound ‘C’, (C4H11N). On reacting with NaNO2 and HCl followed by treatment with water, compound ‘C’ yields
an optically active alcohol, ‘D’. Ozonolysis of ‘A’ gives 2 mols of acetaldehyde. Identify compounds ‘A’ to ‘D’. Explain the reactions involved.
76. A colourless substance ‘A’ (C6H7N) is sparingly soluble in water and gives a water soluble compound ‘B’ on treating with mineral acid. On reacting with CHCl3 and alcoholic potash ‘A’ produces an obnoxious smell due to the formation of compound ‘C’. Reaction of ‘A’ with benzenesulphonyl chloride gives compound ‘D’ which is soluble in alkali. With NaNO2 and HCl, ‘A’ forms compound ‘E’ which reacts with phenol in alkaline medium to give an orange dye ‘F’. Identify compounds ‘A’ to ‘F’.
77. Predict the reagent or the product in the following reaction sequence.
I. Multiple Choice Questions (Type-I)
1. (ii) 2. (iv) 3. (iii) 4. (i) 5. (iii) 6. (ii)
7. (iii) 8. (iv) 9. (iii) 10. (iv) 11. (ii) 12. (ii)
13. (iv) 14. (iii) 15. (ii) 16. (iii) 17. (iii) 18. (ii)
19. (i) 20. (ii) 21. (ii) 22. (iv) 23. (iii) 24. (iv)
25. (i) 26. (ii) 27. (iii)
II. Multiple Choice Questions (Type-II)
28. (iii), (iv) 29. (i), (ii), (iii) 30. (i), (ii) 31. (ii), (iii)
32. (i), (ii) 33. (i), (ii), (iii) 34. (i), (ii) 35. (i), (iii)
36. (i), (ii) 37. (i), (ii)
III. Short Answer Type
38. HNO3 acts as a base in the nitrating mixture and provides the electrophile, NO2+ .
39. See NCERT textbook for Class XII.
41. Reduction of nitriles with sodium/alcohol or LiAlH4 gives primary amine.
43. Benzene sulphonylchloride.
44. Benzene diazonium chloride is very unstable.
45. See NCERT textbook for Class XII.
46. Nitrogen is less electronegative than oxygen therefore lone pair of electrons on nitrogen is readily available for donation. Hence, MeNH2 is more basic than MeOH.
47. Pyridine and other bases are used to remove the side product i.e. HCl from the reaction mixture.
48. Reaction is done in mild basic conditions.
49. A mixture of 2-bromoaniline and 4-bromoaniline is formed.
50. CH3CH2CH3 &amp;amp;amp;amp;amp;amp;amp;lt; CH3CH2NH2 &amp;amp;amp;amp;amp;amp;amp;lt; CH3CH2OH
51. CH2== CH—CH2—NH2, prop-2-en-1-amine
52. N, N-Dimethylbenzenamine
53. Z is an aliphatic amine which gives a solid insoluble in base. This implies that reaction with C6H5SO2Cl must give a product without any replaceable hydrogen attached to nitrogen. In other words, the amine must be a secondary amine. i.e. Z is ethylmethylamine.
Carbylamine reaction is shown by 1° amine only which results in the replacement of two hydrogen atoms attached to nitrogen atom of NH2 group by one carbon atom. On catalytic reduction the isocyanide will give a secondary amine with one methyl group.
55. The reaction exhibits azo-coupling of phenols. In mild alkaline conditions phenol moiety participates in the azo-coupling and para position of phenol is occupied.
61. This reaction is an example of electrophilic aromatic substitution. In alkaline medium, phenol generates phenoxide ion which is more electronrich than phenol and hence more reactive for electrophilic attack. The electrophile in this reaction is aryldiazonium cation. Stronger the electrophile faster is the reaction. p-Nitrophenyldiazonium cation is a
stronger electrophile than p-toluene diazonium cation. Therefore, it couples preferentially with phenol.
(ii) Conversion (A) given below is same as in part (i) given above after that reaction (B) can be carried out.
IV. Matching Type
66. (i) → (d) (ii) → (c) (iii) → (a) (iv) → (b)
67. (i) → (b) (ii) → (a) (iii) → (d) (iv) → (c)
V. Assertion and Reason Type
68. (iii) 69. (iii) 70. (iv) 71. (ii) 72. (iv) 73. (i) 74. (iv)
VI. Long Answer Type
Since products of ozonolysis of compound ‘A’ are CH3 — CH == O and O == CH—CH3. The compound ‘A’ is CH3—CH== CH—CH3
On the basis of structure of ‘A’ reactions can be explained as follows :