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1. objectives: Fundamental concepts on basic character of amines. Applications of these concepts in comparing the relative basic strengths of NH 3, various.

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Presentation on theme: "1. objectives: Fundamental concepts on basic character of amines. Applications of these concepts in comparing the relative basic strengths of NH 3, various."— Presentation transcript:

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2 objectives: Fundamental concepts on basic character of amines. Applications of these concepts in comparing the relative basic strengths of NH 3, various alkyl amines, C 6 H 5 NH 2 and other substituted aromatic amines. 2

3  Amines are derivatives of ammonia: NH 3 R – NH 2  Various types of amines are primary (1 0 ), secondary (2 0 ) and tertiary (3 0 ): RNH 2 R 2 NH R 3 N (1 0 ) (2 0 ) (3 0 )  The basic nature of an amine is due to the presence of a lone pair on N, and its ability to donate it; and depends on the stability of the conjugate acid. FUNDAMENTAL CONCEPTS ON BASIC CHARACTER OF AMINES -H +R 3

4  Steric hindrance plays an important role in deciding basic strength.  Electron releasing groups increase the basic strength, whereas electron withdrawing group decrease the basic strength of amines.  A higher value of pK b implies lower basic strength. 4

5 APPLICATION : COMPARISON OF NH 3 AND RNH 2  As the alkyl group has +I – effect, it increases the electron density on N.  Substituted ammonium ion is more stable due to charge dispersal by the alkyl group. So, RNH 2 is more basic than NH 3. 5

6 APPLICATION : COMPARISON OF RNH 2, R 2 NH AND R 3 N IN GASEOUS STATE  As the alkyl groups have +I – effect, it increases the electron density on N.  As these are in the gaseous phase, so only the +I – effect is considered. So, the order of basic nature is: 3 0 > 2 0 > 1 0 > NH 3. 6

7 APPLICATION : COMPARISON OF RNH 2, R 2 NH AND R 3 N IN AQUEOUS STATE  In aqueous solutions, in addition to +I – effect, the solvation effect and steric hindrance are also at work.  As 3 0 amines are least solvated, so their conjugate acids are least stable. So, we expect an order as : 1 0 > 2 0 > 3 0.  But, due to the combination of +I – and solvation effects, secondary amines become most basic. So, the order of basic nature in methyl substituted amines is: 2 0 > 1 0 > 3 0 > NH 3. 7

8 But, this order applies to methyl substituted amines only.  If R– is ethyl (C 2 H 5 –) group, then due to steric hindrance, solvation effect decreases in all the three types of amines. Hence, +I – effect of tertiary amines tends to dominate over the solvation effect of primary amines.  Similarly, in propyl substituted amines, though electron density is increased due to +I – effect, still, due to steric hindrance, ability to donate the electron pair is decreased and hence the order is changed again. Hence, the order of basic nature in ethyl substituted amines is: 2 0 > 3 0 > 1 0 > NH 3. The order of basic nature in propyl substituted amines is: 1 0 > NH 3 > 2 0 > 3 0. Similarly, the order of basic nature in butyl substituted amines is: NH 3 > 1 0 > 2 0 > 3 0. 8

9 APPLICATION : C OMPARISON OF AROMATIC AMINES WITH ALIPHATIC AMINES  It is because the lone pair on nitrogen in aniline is in conjugation with the benzene ring.  Also, anilinium ion (conjugate acid) is less stable than aniline as it has lesser number of resonating structures. pKb of aniline is very high, as compared to that of alkyl amines; i.e., aniline is much less basic than alkyl amines. 9

10  Presence of electron withdrawing group in the benzene ring decreases the basic strength and the presence of electron releasing groups increases the basic strength. NH 3 NO 2 NH 3 CH 3 is less basic than Electron withdrawing group Electron releasing group 10

11 THE ORTHO EFFECT IN AROMATIC AMINES  It has been observed that any group; whether electron withdrawing, or electron releasing, at the ortho position in aniline always decreases its basic strength. This is known as the ORTHO EFFECT.  The reason for the occurence of the ortho effect is steric hindrance. 11

12 QUESTIONS FOR PRACTICE (CLASSROOM EXERCISE)  Predict the increasing order of basic strength in each of the following cases: NH CH 3 – C – NH 2 (I) CH 3 – CH 2 – NH 2 (II) CH 3 O CH 3 – CH – NH 2 (III) CH 3 – C – NH 2 (IV) Answer: IV < II < III < I 12

13 13 NH 2 CH 3 NH 2 CH 3 NH 2 CH 3 (I) (II) (III) (IV) Answer: II < I < III < IV NH 2 NO 2 NH 2 NO 2 NH 2 NO 2 (I) (II) (III) (IV) Answer: II < IV < III < I

14 THANK YOU 14


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