1. IIT-JEE Syllabus
Basicity of aniline and aliphatic amine, preparation from nitro compounds, reaction with nitrous acid, formation and reactions or diazonium salts, and its coupling with phenols, carbylamine reaction.
- Introduction and Nomenclature
2.1 Structure of Amines
Amines are the alkyl (or) aryl derivatives of NH3. The general formula of amine is RNH2,R2NH,R3N, where R is an alkyl (or) aryl group.
In amines one or more Hydrogen atoms of ammonia are replaced by Alkyl or aryl groups. Here nitrogen atom of amines is like that of NH3, it is sp3 hybridised. The three Alkyl groups (or hydrogen atoms) occupy corners of a tetrahedron, one of sp3 orbital occupying the unshared electron pair directed towards the other corner. We say shape of amine as “Trigonal Pyramidal”.
2.2 Classification of Amines
Amines are classified as primary, secondary (or) tertiary according to the number of groups attached to the nitrogen atom. How? If one alkyl group has replaced one hydrogen atom of ammonia, it is primary amine. Similarly, if two hydrogens are replaced, it is secondary amine and if all the three are replaced, it is tertiary amine.
We can also replace all the 4 hydrogens in with alkyl groups to get a quaternary tetra alkyl ammonium ion.
2.3 Nomenclature of Amines
Nomenclature of amines is quite simple. Aliphatic amines are named by naming the alkyl group (or) groups attached to nitrogen , and following there by the word amine.
More complicate amines are often named as prefixing amino – (or-N-methylamino -, N-N, diethyl amino -, etc) to the name of the parent chain.
Aromatic amines – those in which nitrogen is attached to an aromatic ring – are generally named as derivatives of the simplest aromatic amine, aniline.
Salts of amines are generally named by replacing – amine by – ammonium (or – aniline by – anilinium), and adding the name of the anion.
Exercise 1: Give name of the following structures
2.4 Physical Properties
Amines are moderately polar substances; they have boiling points that are higher than those of alkanes but generally lower than alcohols of comparable molecular weight. Molecules of primary and secondary amines can form strong hydrogen bonds to each other and to water. Molecules of tertiary amines can not form hydrogen bonds to each other, but they can form hydrogen bonds to molecules of water or other hydroxylic solvents. As a result, tertiary amines generally boil at lower temperatures than primary and secondary amines of comparable molecular weight.
3.1 Aliphatic Bases
As increasing strength in nitrogenous bases is related to the readiness with which they are prepared to take up protons, and therefore, to the availability of the unshared electron pair on nitrogen, we might expect to see an increase in basic strength on going:
NH3 → RNH2 → R2NH→ R3N, due to the increasing inductive effect of successive alkyl groups making the nitrogen atom more negative. An actual series of amines was found to have related pKa values as follows, however:
It will be seen that the introduction of an alkyl group into ammonia increases the basic strength markedly as expected. The introduction of a second alkyl group further increases the basic strength, but the net effect of introducing the second alkyl group is very much less marked than with the first. The introduction of a third alkyl group to yield a tertiary amine, however, actually decreases the basic strength in both the series quoted. This is due to the fact that the basic strength of an amine in water is determined not only by electron – availability on the nitrogen atom, but also by the extent to which the cation, formed by uptake of a proton, can undergo solvation, and so become stabilized. The more hydrogen atoms attached to nitrogen in the cation, the greater the possibilities of powerful solvation via hydrogen bonding between these and water :
Thus on going along the series, NH3 → RNH2 → R2NH → R3N, the inductive effect will tend to increase the basicity, but progressively less stabilisation of the cation by hydration will occur which will tend to decrease the basicity. The net replacing effect of introducing successive alkyl groups thus becomes progressively smaller, and an actual changeover takes place on going from a secondary to a tertiary amine. If this is the real explanation, no such changeover should be observed if measurements of basicity are made in a solvent in which hydrogen – bonding cannot take place; it has, indeed, been found that in chlorobenzene the order of basicity of the butylamines is
BuNH2 < Bu2NH < Bu3N
Tetralkylammonium salts, e.g. R4N⊕ I–, are known, on treatment with moist silver oxide, AgOH, to yield basic solution comparable in strength with the mineral alkalis. This is readily understandable for the base so obtained, R4N⊕ –OH, is bound to be completely ionised as there is no possibility, as with tertiary amines, etc.,
of reverting to an unionised form. The effect of introducing electron withdrawing groups, e.g. Cl, NO2, close to a basic center is to decrease the basicity, due to their electron withdrawing inductive effect. Thus the amine
is found to be virtually non – basic, due to the three powerfully electron withdrawing CF3 groups. The change is also pronounced with C=O, for not only is the nitrogen atom, with its electron pair, bonded to an electron withdrawing group through an sp2 hybridised carbon atom but an electron withdrawing mesomeric effect can also operate :
Thus amides are found to be only very weakly basic in water [pKa for ethanamide(acetamide) is ≈ 0.5], and if two C=0 groups are present the resultant imides, far from being basic, are often sufficiently acidic to form alkali metal salts, e.g. benzene – 1, 2 – dicarboximide:
Exercise 2: Amines are stronger bases than ammonia. Why?
3.2 Aromatic Bases:
The exact reverse of the above is seen with aniline, which is a very weak base (pKa = 4.62) compared with ammonia (pKa = 9.25) or cyclohexylamine (pKa = 10.68). In aniline the nitrogen atom is again bonded to a sp2 hybridised carbon atom but, more significantly, the unshared electron pair on nitrogen can interact with the delocalised π orbitals of the nucleus:
If aniline is protonated, any such interaction, with resultant stabilisation, in the anilinium cation is prohibited, as the electron pair on N is no longer available :
The aniline molecule is thus stabilised with respect to the anilinium cation, and it is therefore ‘energetically unprofitable’ for aniline to take up a proton ; it thus functions as a base with the utmost reluctance (pKa = 4.62, compared with cyclohexylamine,
pKa = 10.68). The base weakening effect is naturally more pronounced when further phenyl groups are introduced on the nitrogen atom ; thus diphenylamine, Ph2NH, is an extremely weak base (pKa = 0.8), while triphenylamine, Ph3N, is by ordinary standards not basic at all. Introduction of alkyl, e.g. Me, groups, on to the nitrogen atom of aniline results in small increase in pKa :
C6H5NH2 C6H5NHMe C6H5NMe2 MeC6H4NH2
4.62 4.84 5.15 o-4.38
Unlike on such introduction in aliphatic amines this small increase is progressive : suggesting that cation stabilisation through hydrogen – bonded solvation, responsible for the irregular behavior of aliphatic amines, here has less influence on the overall effect. The major determinant of basic strength in alkyl-substituted anilines remains mesomeric stabilisation of the aniline molecule with respect to the cation ; borne out by the irregular effect of introducing Me groups into the o-, m- and p-positions in aniline.
A group with a more powerful (electron – withdrawing) inductive effect, e.g. NO2 is found to have rather more influence. Electron withdrawal is intensified when the nitro group is in the o- or p-position, for the interaction of the unshared pair of the amino nitrogen with the delocalised π orbital sytsem of the benzene nucleus is then enhanced. The neutral molecule is thus stabilised even further with respect to the cation, resulting in further weakening as a base. Thus the nitro – anilines are found to have related pKa values :
The extra base – weakening effect, when the substituent is in the o-position, is due in part to the short distance, over which its inductive effect is operating, and also to direct interaction, both steric and by hydrogen bonding, with the NH2 group. o-Nitroaniline is such a weak base that its salts are largely hydrolysed in aqueous solution, while 2, 4 – dinitroaniline is insoluble in aqueous acids, and 2, 4, 6 – trinitroaniline resembles an amide; it is indeed called picramide and readily undergoes hydrolysis to picric acid (2, 4, 6 – trinitrophenol).
With substituents such as OH and OMe that have unshared electron pairs, an electron – donating, i.e. base strengthening, mesomeric effect can be exerted from the o- and p-, but not from the m-position, with the result that the p-substituted aniline is a stronger base than the corresponding m-compound. The m-compound is a weaker base than aniline itself, due to the electron – withdrawing inductive effect exerted by the oxygen atom in each case. As so often, the effect of the o – substituent remains somewhat anomalous, due to the interaction with the NH2 group by both steric and polar effects. The substituted anilines are found to have related pKa values as follows:
Exercise 3: Aniline is a weak base than ethylamine. Why?
- Preparation of Amines
Now, that we have learnt something about amines, we shall now get into the methods of their preparation.
4.1 From Alkyl halides
Many organic halogen compounds are converted into amines by treatment with aqueous (or) alcoholic solution of ammonia. This reaction is generally carried out either by allowing the reactants to stand together at room temperature (or) by heating them under pressure. Displacement of halogen by NH3 yields the amine salt, from which free amine can be liberated with hydroxide ion.
CH3Cl + NH3 ⎯→ CH3NH2 + Cl– + H2O
The above reaction is a class of substitution reaction, which we know as nucleophilic substitution. Ammonia can act as a nucelophile and it can also act as a base. If ammonia acts a nucleophile substitution takes place,
CH3CH2CH2Br + NH3 ⎯→ CH3CH2CH2NH2 + HBr
And, if ammonia acts as a base, elimination takes place.
It is very evident that primary alkyl halides under go substitution very easily than tertiary alkyl halides, which undergo elimination very easily.
Look at the sequence of reactions below,
The reaction is quite simple and we can convert alkyl halide into all class of amines.
Exercise 4: How can the formation of 2° and 3° amines can be avoided during the preparation of 1° amines by alkylation?
4.2 From nitrogen containing Compounds
Nitro alkanes can be reduced quantitatively to their corresponding amines.
Nitro compound can be reduced in two general ways: (A) by catalytic hydrogenation using molecular hydrogen, or (B) by chemical reduction, usually by a metal and acid. This method cannot be used when the molecule also contains some other easily hydrogenated group, such as a Carbon carbon double bond.
Chemical reduction is most often carried out by adding hydrochloric acid to a mixture of the nitro compound and metal, usually granulated tin or iron.
Alkyl and aryl cyanides can be reduced to their corresponding amines using LiAlH4
Amides can directly be converted into their corresponding amines. This reaction is carried out by treating the amide with a mixture of base and bromine (KOH + Br2). This reaction is called as Hofmann Bromamide reaction. The reaction is as follows,
RCONH2 + Br2 + 4KOH ⎯⎯→ RNH2 + K2CO3 + 2KBr + 2H2O
Here we can see that the amine formed has one carbon less than that of the corresponding amide. Due to the loss of carbon atom, this reaction is also called as Hofmann degradation of amides.
The mechanism of the reaction is as follows:
|Step 1||(one of the hydrogen attached to nitrogen is substituted by a bromine atom)|
|Step 2||(The N-bromamide anion thus formed as a result of proton by base is stabilized by resonance)|
|Step 3||(Bromine leaves and we have an electron deficient nitrogen)|
|Step 4||(There is a shift of alkyl group to the nitrogen)|
|Step 5||(A simple hydrolysis of an imine gives us the amine)|
Apart from this, amides can be dehydrated by P2O5 to their corresponding nitriles and nitriles can then be reduced.
R—C ≡ N R—CH2NH2
By this method you are retaining the number of carbon atoms in both amide and the amine.
4.3 From carbonyl compounds
While studying carbonyl compounds we have seen that carbonyl compounds can be converted into any other functional group. How are we converting carbonyl group into amino group?
See, the following sequence,
CH3CH = O + NH3 ⎯⎯→
The reactions are clear and simple so that, we can get an amine from carbonyl compound just by reductive amination (amination and reduction).
Using this reductive amination we can go from 1° amine to 2° amines. How? Look at the following reaction.
CH3CH2CH = O + ⎯⎯→ CH3CH2CH=NCH2CH3
4.4 Curtius reaction
Amines can be prepared by treating acid chloride with sodium azides the isocyanate thus formed is decomposed with treatment of water and amines are obtained.
4.5 Schmidt Reaction
Hydrozoic acid reacts with carboxylic acid in presence of a mineral acid to give amines.
4.6 By the reduction of an Alkyl Isocyanide
RNC + 4[H] ⎯⎯→ R – NH – CH3
4.7 Preparation of Tertiary Amines
3RX + NH3 R3N + 2HX
Exercise 5: In the presence of base, acyl derivatives of hydroxamic acids undergo the Lossen rearrangement to yield isocyanates or amines.
Write mechanism for the reaction.
- Chemical Reaction
5.1 Basic Nature
Amines turn red litmus blue and also combine with water and mineral acids to form corresponding salts.
R – NH2 + HCl ⎯→ R – –
R–NH2 + H2SO4 ⎯⎯→
When the amine salts are treated with strong bases like NaOH, the parent amines are regenerated.
RN+H3Cl– + OH– ⎯⎯→ RNH2 + H2O + Cl–
Amine salt Amine
(Soluble is water) (insoluble in water)
Further, due to basic character amines react with auric and platinic chlorides in presence of HCl to form double salts.
These double salts decompose on ignition to pure metal, therefore, the formation and decomposition of the double salts is used for determining the molecular weight of amines.
5.2 Acylation (Reaction with Acyl Chlorides or Acid Anhydrides)
Primary and secondary amines can react with acid chlorides or acid anhydrides to form substituted amides.
RNH2 + R′COCl ⎯→ R′CO NHR an N-substituted amide
R2NH + R′COCl ⎯→ R′CO.NR2 an N,N disubstituted amide
5.3 Benzoylation (Schotten Baumann Reaction)
Primary amine reacts with benzoyl chloride to give the acylated product.
(Benzoyl chloride) Benzoyl alkyl amine
5.4 Carbylamine Reaction (Given Only by Primary Amines)
Primary amines when heated with chloroform and alcoholic caustic potash give isocynaides (carbylamines) having very unpleasant smell, which can be easily detected
C2H5NH2 + CHCl3 + 3KOH ⎯→ C2H5NC + 3KCl + 3H2O
Ethylamine Ethyl isocyanide
C6H5 NH2 + CHCl3 + 3KOH ⎯→ C6H5NC + 3KCl + 3H2O
Aniline Phenyl isocyanide
5.5 Action with Aldehyde and Ketone
Both primary aliphatic and aromatic amines react with aldehydes and ketones to form schiff’s bases also called anils.
C2H5NH2 + CH3CHO ⎯→ C2H5N = CHCH3 + H2O
Ethylamine Acetaldehyde Ethylidene ethylamine (Schiff’s base)
5.6 Hofmann Mustard Oil Reaction
Primary amines when warmed with alcoholic carbon disulphide followed by heating with excess of mercuric chloride form isothiocyanates having pungent smell similar to mustard oil.
C6H5NH2 + S = C = S C6H5NCS + 2HCl + HgS
5.7 Reaction with Carbonyl Chloride
This reaction is given only by primary amines.
C2H5 – NH2 + COCl2 ⎯→ C2H5NCO + 2HCl
5.8 Reaction of Quaternary Ammonium Salts:
When a quaternary ammonium hydroxide is heated strongly (125° or higher) it decomposes to yield water, a tertiary amine and an alkene
This reaction is called as the Hofmann elimination. The formation of quaternary ammonium salts followed by an elimination of the kind just described and identification of the alkene and tertiary amine formed was once used in the determination of the structure of complicated amines.
Illustration 1: What is the effect of heat on quaternary bases?
Solution: (C6H5)4+NOH– (C2H5)3N + CH2 = CH2 + H2O
A Hoffman’s elimination occurs if B – H atoms are available and the major product is the least substituted ethylene derivative.
Tetramethyl ammonium hydroxide unable to form an alkene is converted to CH3OH and (CH3)3N. It is particularly useful for cyclic amines.
5.9 Reaction of Amines with Nitrous Acid
The diazonium salts of amines
What are these diazonium salts? Let us look at the name. The name suggests that, the compound has two nitrogen atoms (diazo) and the whole group has a positive charge (ium). There is also an anion to balance it (It is a salt)
So, the possible structure can be
How to prepare them? The preparation is quite simple if we adhere to the experimental conditions.
These diazonium salts are prepared by treating a primary amine with NaNO2 in presence of con. HCl; the temperature being 0°C. (Here the temperature has to be taken care of and if the temperature exceeds 5°C, the reaction will not take place.)
Let us take a case of aliphatic amines,
Mechanism for Diazotization is as follows
The diazonium salts of aliphatic amines are generally unstable and they decompose to give different products.
Thus we can have a wide range of products. Let us now see some thing about aromatic amines
These aryl diazonium salts undergo a variety of displacement reaction. The reactions are simple and are summarised below.
Illustration 2: During diazotisation of aryl amines excess of mineral acid is used. Comment
Solution: It is b’coz high concentration of H+ ions converts aniline to anilinium ion and thus prevents coupling reaction.
Reaction of secondary Amines with Nitrous acid
Secondary amines both aryl and alkyl react with nitrous acid to yield N-nitrosoamines. N-nitrosoamines usually separate from the reaction mixture as oily yellow liquid.
N-nitrosoamines are very powerful carcinogens (cancer causing substances)
Reaction of Tertiary amines with Nitrous acid
When a tertiary aliphatic amine is mixed with nitrous acid, an equilibrium is established among the tertiary amine, its salt, and an N-Nitrosoammonium compound.
Tertiary arylamines react with nitrous acid to form C-nitroso aromatic compound. Nitrosation takes place almost exclusively at the para position if it is open and if not, at the ortho position. The reaction is another example of electrophilic aromatic substitution.
5.10 Coupling Reactions of Arene Diazonium Salts
Arenediazonium ions are weak electrophiles; they react with highly reactive aromatic compounds with phenols and tertiary arylamines to yield azo compound. This electrophilic aromatic substitution is called a diazo coupling reaction occurring mainly at p-position.
Couplings between arenediazonium cations and phenols take place most rapidly in slightly alkaline solution. If the solution is too alkaline (pH > 10), however, the arenediazonium salt itself reacts with hydroxide ion to form a relatively unreactive diazohydroxide or diazotate ion.
Hydrazo compounds are also made as follows:
Diaryl hydrazo compounds undero the benzidine rearrangement
Exercise 6.: A weakly basic solution favours coupling with phenol. Is there any other explanation?
Ring Substitution in Aromatic Amines
The –NH2, – NHR and –NR2 are benzene activating groups through resonance effect of nitrogen where the lone one pair of electron of nitrogen is shifted to the benzene ring making ortho and para, position available for electrophilic attack.
The carbocation formed as intermediate are
The group – NHCOCH3 is less powerful ortho and para director because of the electron-withdrawing character of oxygen makes nitrogen a poor source of electrons. This fact is made use in preparing mono substituted aniline. The –NH2 group is such a powerful activator, that substitution occurs at all available ortho and para positions of aniline. If, however, –NH2 group is converted to –NHCOCH3, the molecule becomes less powerful activator. Hence only mono substitution products are obtained. Finally – NHCOCH3 is converted back to –NH2 by hydrolyzing with acid. This technique is especially used while nitrating aniline as strong oxidizing agent destroys the highly reactive ring.
Exercise 7: Explain the following
- i) It is necessary to acetylate aniline first for preparing bromoaniline.
- ii) While carrying out an electrophilic substitution reaction on aniline. Lewis acid is not used
5.12 Aniline -X rearrangement
Such compounds are not much stable so the group X migrates mainly at p-position.
- Fisher-Hepp rearrangement
- Phenylhydroxylamine – p-aminophenol rearrangement.
Nucleophilic attack by H2O at p – position.
6. Separation of a Mixture of Amines
6.1 Hinsberg’s Method
Treating a mixture of 3 amines with Hinsbergs reagent (benzene sulfonyl chloride) and finally treating the product formed with NaOH can separate the 3 class of amines.
RNH2 + C6H5SO2Cl ⎯→ C6H5– SO2 – NH – R + HCl
(N-alkyl benzene sulfonamides)
(Dissolves in NaOH due to acidic H-attached to Nitrogen)
Tertiary amines do not react with Hinsberg’s reagent.
After reacting with NaOH the aqueous layer and the second layer [Secondary and Tertiary) can be separated by ether. Aqueous layer Hydrolysed with conc. HCl gives primary amine. The ether layer is distilled and tertiary amine is distilled over. Residue hydrolysed with conc. HCl to recover secondary amine.
6.2 Hofmann’s Method
The mixture of amines is treated with diethyloxalate, which forms a solid oxamide with primary amine, a liquid oxime ester with secondary amine. The tertiary amine does not react.
- 7. Test for Amines
Test for Primary Amines (Carbyl Amine Reaction)
When a primary amine is treated with a strong base in presence in chloroform, an isocyanide is formed and this isocyanide thus formed has a very foul smell.
Here attacking electrophile is the dichlorocarbene (:CCl2). The primary amine can be identified with its foul smell.
7.2 Test for Secondary Amines(Libermann Reaction)
The secondary amine is converted into nitrosoamine by treating the amine with nitrous acid. The resultant solutions warmed with phenol and concentrated H2SO4, a brown or red colour is formed at first soon it changes to blue and then to green. The colour changes to red on dilution and further changes to greenish blue on treating with alkali. Tertiary arylamines react with nitrous acid to form o-nitroso aromatic compound.
- Solution to Exercises
Solution 1: i) 2-aminobutane ii) 2-methylpropanamine
Solution 2: Amines have electron-repelling groups, which increase the electron density on nitrogen, while ammonia has no such group.
Solution 3: In aniline the electron pair of nitrogen atom is delocalised due to resonance and hence lesser available for protonation while ethylamine does not undergo resonance.
Solution 4: Use of excess ammonia reduces chances of reaction of 1° amine with alkyl halide to form 2° and 3° amines.
Solution 6: High acidity suppresses the ionisation of C6H5OH to the more reactive C6H5O–. In presence of weak base, C6H5O– is formed and coupling takes place.
Solution 7: i) Amino group, being activating group, activates bromination of aniline and forms tribromoaniline.
- ii) In presence of Lewis acids, aniline undergoes protonation to from anilinium ion, which being m-directing forms unexpected m-substituted derivative.
- Solved Subjective Problems
Problem 1: Arrange the following amines in the increasing order of boiling points giving proper reasoning (A) n-butylamine (B) Ethyldimethylamine (C) diethylamine.
Solution: A > C > B. It is because 1° and 2° amines, unlike 3° amines, form intramolecular H-bonds. A with two H’s available for H – bonding has a higher boiling point than C.
Problem 2: A compound X (C7H7Br) reacts with KCN to give Y (C8H7N). Reduction of Y with LiAlH4 yields Z (C8H11N). Z gives carbylamine reaction, reacts with Hinsberg’s reagent in the presence of aq. KOH to give a clear solution. With NaNO2 and HCl at 0°C (Z) gives a neutral compound which gives red colour with ammonium cerric nitrate. What are X, Y and Z.
Problem 3: Account for the following order of increasing basicity:
RC ≡N: < R′CH = NR < RNH2
nitrile imine amine
Solution: The more s character in the hybrid orbital of the N with the unshared pair of e–s, the less basic is the molecule. The nitrile N using sp HO’s has the most s character, the imine N using sp2 HO’s has the intermediate s character, and the amine N using sp3 HO’s has the least s character.
Problem 4: In terms of s character, explain why NH3 with bond angles of 107° is much more basic than NF3 with bond angles of 103°. The bond angles of are 109°.
Solution: The bond angle of 109° in indicates that N uses pure sp3 HO’s. The bond angle of 107° in NH3 shows that, although N essentially uses sp3 HO’s, it nevertheless has slightly more p character in its bonding HO’s and slightly more s character in its lone pair HO. The bond angle of 103° in NF3 indicates that N has even more p character in its bonding HO’s and even more s character in its lone pair HO. NF3 is less basic than NH3 because its lone pair orbital has more s character.
Problem 5: The C—N—C bond angle in Me3N is 108°.
- a) Describe the hybridization and shape of Me3
- b) Why is an amine of the type R1R2R3N chiral?
- c) Why is it not possible to separate the enantiomers?
Solution: a) N forms three sp3 hybridized σ bonds to the C’s of the Me groups and has a nonbonding electron pair in the fourth sp3 orbital. Me3N has a roughly pyramidal shape.
- b) Because of its pyramidal geometry, such an amine is chiral, the unshared pair being considered a fourth “different” group.
- c) The enantiomers rapidly interconvert; having enough kinetic energy at room temperatures, by a process called nitrogen inversion. For this process H‡ ≈ 6 kcal/ mol (25kJ / mol) and the inversion thus does not involve bond – breaking and subsequent formation. The TS for the inversion is planar, the N being sp2 hybridized, with the unshared pair in the pz
Problem 6: Compare the basicities of
- a) H2C = CHCH2NH2, CH3CH2CH2NH2 and HC ≡ CCH2NH2, and
- b) C6H5CH2NH2, cyclohexyl – CH2NH2 and p-NO2C6H4CH2NH2.
Solution: a) The significant difference among these three bases is the kind of hybrid orbital used by Cβ — the more s character it has, the more electron – withdrawing (by induction) and base weakening it will be. The HO conditions are H2C = CβHCH2NH2(sp2), CH3CβH2CH2NH2(sp3), and HC ≡ CβCH2NH2(sp). The increasing order of electron – attraction is
propargyl > allyl > propyl >, and the decreasing order of basicity is
CH3CH2CH2NH2> H2C = CHCH2NH2 > HC ≡ CCH2NH2
- b) The decreasing order is
The Cβ is cyclohexyl – CH2NH2 uses sp3 HO’s while Cβ in the benzylamines uses sp2 HO’s. The electron withdrawing p-NO2 makes the phenyl ring even more electron–withdrawing and base weakening.
Problem 7: Account for the fact that although N,N—dimethylaniline is only slightly more basic than aniline, 2,6-dimethyl N,N-dimethylaniline is much more basic than
2,6 – dimethylaniline.
Solution: Extended π bonding between the amino N and the ring requires that the σ bonds on N become coplanar with the ring and its ortho bonds. Bulky substituents in the ortho 2,6 – positions sterically hinder the attainment of this geometry and interfere with the base – weakening extended π bonding. This effect is called steric inhibition of resonance.
Problem 8: The reaction of (S) –2– methylbutanamide with Br2 and OH– produces an optically active amine. Give the structure of the product, including its stereochemical designation and the mechanism for its formation.
Solution: The product is (S) sec-butylamine: R migrates with its electron pair to
the electron – deficient , and configuration is retained because C—C
is being broken at the same time that C—N is being formed in
the transition state.
Problem 9: Starting with carbon and hydrogen can we obtain isopropyl amine.
Solution: At first sight this might look impossible. But if you think quietly you can find that this can be done with real ease.
Let us work backwards,
The amine can be obtained from the imine.
The imine, from the ketone,
Now, if you remember, alkynes chapter, you must have converted propyne, to acetone.
CH3 — C ≡ CH
Propyne can be obtained from, acetylene.
CH3 —C ≡ CH CH3—I + HC ≡CH
Acetylene can be obtained from the electrical discharge of graphite rods in presence of hydrogen gas.
C + H2 HC ≡ CH
Now, we have a synthetic route, as follows,
Problem 10: Convert benzoic acid into benzylamine and aniline
Solution: Here we are asked to do two conversions. They are
Here, both the products are amines and one of them is having the same number of carbon atoms as in the acid and another is having one carbon less.
For the other conversion, we can proceed as,
Thus we are effecting the conversions.
Problem 11: Convert Benzene to benzylamine.
Solution: The structure of benzylamine is
This can be obtained by reacting NH3 with benzyl chloride.
Now, let us find a way to prepare benzyl chloride. It can be easily, prepared from toluene by photochemical chlorination.
Now, Friedel – crafts alkylation, can easily obtain Toluene from benzene. The sequence of reaction is thus
Problem 12: Supply the structures A to C
RCH2CH2NH2 A B C
Explain (a) the type of reaction (b) what is leaving group in last step (c) whey does RCH2CH2NH2 not undergo an E2 elimination.
Solution: a) Reaction is E2 (Bimolecular elimination)
- b) NMe3
- c) is a very poor leaving group
Problem 13: An aqueous solution of ethylamine forms a brown ppt. with ferric chloride. Explain
Solution: C2H5NH2 + H2O C2H5N+H3OH– [C2H5NH3]+OH–
FeCl3 + 3OH– ⎯→ + 3Cl–
Problem 14: The amino group in ethyl amine is basic whereas that in acetamide is not basic. Explain.
Solution: In amides the lone pair of electrons on nitrogen atom is delocalized and hence less available for protonation than in amines where no resonance is possible and thus acetamide is a weaker base than ethylamine.
Problem 15: A weakly acidic medium is provided for coupling of benzene diazonium chloride with aniline. Comment.
If the high concentration H+ ions are used during these reactions, then protonation of aniline takes place.
The positive charge on protonated amine exerts –I effect, thus coupling of amine with diazonium salt is not favoured at low pH.
- 10. Assignments (Subjective Problems)
LEVEL – I
- Compound (A) C9H12BrN reacts with benzene sulphonyl chloride in the presence of NaOH to form a white precipitate, insoluble in both acids and bases. When (A) is reacted with KMnO4 under vigorous conditions an acid of neutralization equivalent 83 is produced. The acid forms only one mono-nitro-substitution product. Suggest a structure for A.
- When compound (A) C8H11N is treated with benzene sulphonyl chloride and aq. KOH, no apparent change occurs. Acidification of this mixture gives a clear solution when (A) is treated with NaNO2 and HCl at 0 -5°C and the product is hydrolysed a compound (B) of mole for C2H7N is formed. (B) gives a ppt. with benzene sulphonyl chloride in the presence of aq. KOH. (B) on reaction with NaNO2/HCl at 0°C gives a yellow oily substance (C) which is known to be present in cigarette smoke. (C) is a very powerful carcinogen. What are (A) (B) and (C).
- Compound (A) (C9H11NO) gives a positive Tollen’s Test and is soluble in dil. HCl. It gives no reaction with benzene sulphonyl chloride or with NaNO2 and HCl at 0°C. (A) upon oxidation with KMnO4 gives an acid (B). When (B) is heated with sodalime (C) is formed which reacts with NaNO2 and HCl at 0–5°C. What are (A) (B) & (C)?
- Give structural formulae of compounds A through D
Phthalimide + KOH ⎯→ A
A + CH3 – CH2 – CH2 – Br, heat ⎯→ B
B + H2O, OH– heat ⎯→ C (C3H9N) + D
What are A, B and C
- A BCH2NH⋅CH = CH3
What are A, B, C and the acid D from which A (amide) is obtained?
- Compound (A) C3H9ON when titrated with HCl was found to have a NE of 75 ± 1. A reacted with benzene sulphonyl chloride to give a product that was soluble in dil. NaOH. With excess acetyl chloride (A) was converted to (B) C5H11NO2. When 0.75g of A was treated with excess CH3MgI, 448 ml of CH4 was evolved at STP suggest a structure for (A).
- Compound (A) (C9H14NCl) gives an immediate ppt. with AgNO3. It is very resistant to bromination in either acid or alkaline solution. It is very resistant to heat, nitration and oxidation by KMnO4. Suggest a structure for (A).
- Complete the following series
- RNH2 can be alkylated to form RNHR′ (2° amine) but 3° amine and quarternary ammonium salts are also obtained as side products.
Suggest a way that RNH2 forms only 2° amine?
LEVEL – II
- A pure compound (A) was treated with HBr. The product was a racemate from which optically active compound (B) was obtained. (B) on treatment with NH3 give a further optically active compound (C) which contained 65.7% C, 15.1% H and 19.2% N and had a molecular weight of 73. Treatment of C with HNO2 gave N2 and compound (D) which has oxidized to (E) with acidified K2Cr2O7. (E) reacted with 24-dinitrophenyl hydrazine to give a yellow crystalline substance E gave iodoform test but a negative Tollen’s test. What are (A), (B), (C), (D) and (E).
- A compound (X) has molecular formula of C7H7ON. When reacted with Br2 and KOH. (Y) an aromatic amine is obtained. (Y) give mustard oil reaction and upon diazotisation followed by coupling with phenol (Y) gives an azodye. Write reactions and deduce the structures of (X) and (Y).
- A compound (A) with molecular formula of C8H9NO was boiled with NaOH when an oily liquid (B) was obtained along with a salt (C). Liquid (B) had a fishy smell and gave carbylamine reaction. Salt (C) on heating with soda lime gave methane. Identify (A) and write the reactions.
- An amine contains 31.22% nitrogen. On treatment with nitrous acid, it gives a yellow coloured oily nitroso compound. Give the structure of amine.
- One mole of each of bromo derivative (A) and NH3 react to give one mole of an organic compound (B). (B) reacts with CH3I to give (C), both (B) and (C) react with HNO2 to give compounds (D) and (E), respectively. (D) on oxidation gives 2-methoxy-2-methyl propane. Give structures of (A) to (E) with proper reasoning.
- An organic compound (A) gave the following data in analysis C = 61.1%,
H = 15.25% and N = 23.74%. (A) on heating with HNO2 gives (B). Which contains C = 60%, H = 13.33% and no nitrogen. (B) on careful oxidation gave (C) which shows iodoform test and has molecular weight 58. What are (A), (B) and (C).
- Compound (X) containing chlorine, on treatment with NH3, gives a solid ‘Y’ which is free from chlorine (Y) analysed as C = 49.3% H = 9.59% and N = 19.18% and reacts with Br2 and caustic soda to give a basic compound (Z), (Z) reacts with HNO2 to give ethanol suggest structures for (X), (Y) and (Z).
- An optically active amine (A) is subjected to exhaustive methylation and Hofmann elimination to yield an alkene (B). (B) on ozonolysis gives an equimolar mixture of formaldehyde and butanal. Deduce the structures of (A) & (B). Is there any structural isomer to (A), if yes draw its structure.
- The reaction of n-butylamine with sodium nitrite and hydrochloric acid yields nitrogen and the following mixture; n-butyl alcohol, 25%; sec butyl alcohol, 13%; 1 butene and 2-butene, 37%, n-butyl chloride, 5%; sec butyl chloride, 3% what is the most likely intermediate common to all of these product, and how is it formed.
- A basic volatile nitrogen compound gave a foul smelling gas when treated with CHCl3 and alcoholic KOH. A 0.295 g sample of the substance dissolved in aqueous HCl and treated with NaNO2 solution at 0°C liberated a colourless; odourless gas whose volume correspond to 112 ml at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid, which did not contain nitrogen and which on warming with alkali, and iodine gave yellow precipitate. Identify the original substance. Assume it contains one N atom per molecule.
LEVEL – III
- Compound (A) C10H13ON was insoluble in dil. HCl and NaOH. It was refluxed for 2 hours with 10% NaOH and the alkaline residue was steam distilled. The distillate contained a water insoluble, acid – soluble compound B which readily formed a tribromoderivative. The residue contained an tribromoderivative. The residue was an organic acid which on heating with soda-lime, gave methane. When B was treated with excess C2H5Br, it consumed two moles of C2H5Br. What will be the
structure of (A)?
- When 2.25 gm of an unknown amine was treated with nitrous acid the evolved nitrogen corrected to STP measured 560 ml. The alcohol isolated from reaction mixture gave a positive iodoform reaction. What is the structural formula of the unknown amine?
- A compound (A) which is presented for identification has a molecular weight of 251 and contains 63.7% Br2. Diazotisation of the compound and reduction of the diazonium salt gave a solid substance B containing 67.8% bromine. Only one mono-nitroderivative of (B) can be obtained. What are the formulae of (A) and (B).
- A substance gave the following resultant analysis: 0.25g gave 0.6027g CO2 and 0.339g H2O, 0.2g gave 33.4 cc of N2 at 20°C and 750mm Hg. Its vapour density was 37. Calculate molecular formula of the substance.
- Glycine exists as Zwitter ion but anthranilic acid does not. Comment.
- With the aid of diazonium salts and using PhH, PhMe, and any other needed reagents and solvents, prepare
- a) o-chlorotoluene,
- b) m-chlorotoluene,
- c) 1,3,5 – tribromobenzene,
- d) m-bromochlorobenzene
- e) p-iodotoluene,
- f) p-dinitrobenzene, and
- g) p-cyanobenzoic acid. Do not repeat the synthesis of any intermediate compounds
- A pungent smelling liquid (A) was analysed and found to contain carbon, hydrogen, chlorine and possibly oxygen and upon reaction with aqueous ammonia a neutral compound (B) was formed. When this was treated with bromine and potassium hydroxide a base (C) resulted. In acidic solution substance C reacted with sodium nitrite, giving nitrogen and an alcohol (D). Upon mild oxidation, the alcohol (D) was converted into compound (E) of M.F. C2H4O which gave a positive test with ammoniacal silver nitrate solution. Describe the chemistry involved in these reactions and identify the compounds (A) (B) (C) (D) and (E).
- One mole of each of bromo derivative (A) and NH3 react to give one mole of an organic compound (B). (B) reacts with CH3I to give (C). Both (B) and (C) react with HNO2 to give compounds (D) and (E), respectively. (D) on oxidation and subsequent decarboxylation gives 2-methoxy -2-methyl propane. Give structures of (A) to (E) with proper reasoning.
- An organic compound (A) (C4H11N) on reacting with HNO2 gave a compound (B) (C4H10O). Compound (B) on mild oxidation gave (C) (C4H8O). Compound (C) on drastic oxidation gave (D) (C2H4O2). Compound (D) on treatment with PCl5 followed by NH3 gave (C2H5NO) which on treatment with Br2/NaOH gave (F) (CH5N). Deduce the structures of (A) (B) (C) (D) (E) & (F) with reasoning. Write down the reactions involved.
- Reaction of a compound (A) (C11H14) with trioxygen and then water gave (B) (C8H8O) and (C) (C3H6O) in equimolar amounts. (C) was unaffected by ammoniacal silver nitrate, but (B) reacted with this reagent to give (D) (C8H8O2). When (D) was treated successively with sulphur dichloride oxide and ammonia, (E) (C8H8NO) was formed. E reacted with bromine and sodium hydroxide to give (F) (C7H9N), a much weaker base than methylamine. Identify the compounds (A) to (F), as far as is possible and comment on the reactions. Suggest further experiments (i) to confirm the identification of (C) and (ii) to be carried out on compound (D) to complete the identification of compounds (D) to (F).
- Assignments (Objective Problems)
LEVEL – I
- RN ≡ C + HgO ⎯→ A + Hg2O. A is
(A) RNH2 (B) RCONH2
(C) R – NCO (D) RCOOH
- . Identify A and B
(A) RCHO, RCOOH (B) RNH2, RCOOH
(C) RN2+, Cl–, R – H (D) RNCO, RNH2
- . Identify A
(A) C6H5 – CHO (B) C6H5 – COOH
(C) C6H5 – CH2OH (D) None
- H2O + A
(A) RC ≡ N (B) RN ≡ C
(C) RC = N = O (D) R – N = C = O
- . A, B, C are
(A) CH4 + N2 + CO2 (B) CO2 + H2 + N2
(C) C2H6 + N2 + H2 (D) CO2 + H2 + N2
- Aniline on treatment with sodium hypochlorite gives
(A) p-aminophenol (B) Phenol
(C) Sodium salt of aniline (D) Anilinium chloride
- On heating aniline with fuming sulphuric acid at 180°C, the compound formed will be
(A) Aniline disulphate (B) Aniline 2, 4, 6-trisulphonic acid
(C) Sulphanilic acid (D) None
- . A is
(A) CH3CONH2 (B) CH3COCONH2
(C) CH3CH2CONH2 (D) CH3CH(NH2)COOH
(A) Cyclohexanoic acid (B) Adipic acid
(C) Cyclohexcanal (D) Cyclohexane
- The structural form of a compound C6H11N (A) that is optically active, dissolves in dil. HCl and releases N2 with nitrous acid is
|(C)||C6H5 – CH(NH2)CH3||(D)||C6H5 – NHC2H5|
- Acetamide is treated with the following reagents. Which one of these would give methylamine?
(A) PCl5 (B) NaOH + Br2
(C) Soda lime (D) Hot conc. HSO4
- Correct order of increasing basicity
(A) NH3 < C6H5NH2 < (C2H5)2NH < C2H5NH2 < (C2H5)3N
(B) C6H5NH2 < NH3 < (C2H5)3N < (C2H5)2NH < NH3
(C) C6H5NH2 < NH3 < (C2H5)3N < (C2H5)2NH < C2H5NH2
(D) C6H5NH2 < (C2H5)3N < NH3 < C2H5NH2 < (C2H5)2NH
- Which of the following does not reduce ArNO2 to ArNH2
(A) Fe/HCl (B) Zn/HCl
(C) LiAlH4 (D) SnCl2 in HCl
- In the reaction PhSO2Cl + EtNH2 ⎯⎯→ A B C
Then C will be
(A) PhSO2NHEt (B) PhSO2NEt2
(C) PhSO2OH (D) None
- For the reaction : C6H5OH + H2C=O + R2NH ⎯⎯→ P
Here P will be
LEVEL – II
- The end product of the following sequence will be
(A) Methylamine (B) Ethyl amine
(C) Methyl cyanide (D) Ammonium acetate
- How many primary amines are possible with formula C4H11N
(A) 1 (B) 2
(C) 3 (D) 4
- An organic compound (A) C6H4N2O4 is insoluble both in dil. HCl and base. Its dipole moment is zero. Deduce structure of A.
- Dinitrobenzene when oxidized with an alkaline solution of K3[Fe(CN)6] yields.
(A) 2,4-dinitrophenol (B) m-nitroaniline
(C) Both (A) and (B) (D) None
- When aniline is heated with glacial acetic acid in the presence of anhydrous ZnCl2 the product formed is
(A) Acetamide (B) Acetanilide
(C) Phenyl acetamide (D) Chlorobenzene
- Which of the following on heating with propionamide will form ethylamine
(A) LiAlH4 (B) KOH + Br2
(C) Na + Alcohol (D) Sn/HCl
- When phenol and aniline are heated at 260°C in presence of ZnCl2, it yields
(A) Diphenyl amine (B) o-amino phenol
(C) p-aminophenol (D) o-diaminobenzene
- Aniline is treated with NaNO2/HCl at 0°C followed by alkaline β-napthol solution
(A) Brilliant red dye (B) Blue coloured solution
(C) Purple precipitate (D) Yellow coloured complex
- Ethylamine with nitrosyl chloride gives
(A) Ethyl chloride (B) Ethyl alcohol
(C) Ethyl nitride (D) Nitro ethane
- Iso-electric point for glycine corresponds to pH, corresponding to which
(A) Glycine molecules move towards cathode
(B) Glycine molecules move towards anode
(C) No net movement takes place
(D) None of the above
- Identify Z in the sequence
C6H5NH2 X Y Z
(A) C6H5NH2COOH (B) C6H5COOH
(C) C6H5NHCH3 (D) C6H5CH2NH2
- The major product (70-80%) of the reaction between m – dinitrobenzene with
- Acetoaldoxime reacts with P2O5 to give
(A) Methyl cyanide (B) Methyl cyanate
(C) Ethyl cyanide (D) None of these
- What is the end product in the following sequence of reactions?
C2H5NH2 A B C
(A) Ethyl cyanide (B) Ethyl amine
(C) Methyl amine (D) Acetamide
(A) p-Bromoaniline (B) 2,4,6-Tribromoaniline
(C) Nitrobenzene (D) m-Bromoaniline
- Answers (Objective Assignment)
LEVEL – I
- C 2. D
- D 4. B
- B 6. A
- C 8. D
- B 10. C
- B 12. C
- C 14. B
LEVEL – II
- B 2. D
- C 4. A
- B 6. B
- A 8. A
- A 10. C
- B 12. B
- A 14. B