Solved Examples on Amines

Question 1 . The sequence not correct for basic strength of compounds are  

(A) Dimethylamine > methylamine > trimethylamine in aqueous solution  

(B) diethylamine > triethylamine > ethylamine in aqueous solution  

(C) methyl amine > pyridine > aniline  

(D) aniline > pyrole > pyridine  

Solution. Pyridine is more basic than aniline.  

Hence (D) is not correct.  

Question 2: An aqueous solution of ethylamine gives a red precipitate with ferric chloride. Explain.  

Solution. Ethyl amine acts as a base in water to produce appreciable OH^- ions which then reacts with Fe⁺³ ions to give a red precipitate of ferric hydroxide  

C₂H₅NH₂ + + H₂O = C₂H₅NH₃⁺ + OH^-  

 Fe⁺³ + 3OH^– ————> Fe(OH)₃  

Question 3: p-Methoxyaniline is stronger base than aniline but p – nitroaniline is a weaker base than aniline. Explain.  

Solution:Methoxy group is electron repelling and thus intensifies the negative charge on N atom of NH₂ group to increase the electron pair donating tendency whereas nitro group is electron attracting and thus intensifies positive charge on N atom of – NH₂ group to decrease the electron donating nature.  

Question 4: Which of the following is a better nucleophilie, aniline or anilinium ion and why?  

Solution: Aniline. The presence of postive charge on anilinium ion C₆H₅NH₃⁺ reduce the tendency to donate lone pair of electrons.    

Question 5:   Give name of the following structures

Solution:       

i)    2-aminobutane

ii)   2-methylpropanamine

iii)   N-methyl-2-aminopropane                

Question 6: 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.

Solution 7: Convert Benzene to benzylamine.

Solution:  The structure of benzylamine is 

\

Question 8: 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.

 So,

For the other conversion, we can proceed as,

Question  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.

Now, if you remember, alkynes chapter, you must have converted propyne, to acetone.

Acetylene can be obtained from the electrical discharge of graphite rods in presence of hydrogen gas.

Question  10:     The C—N—C bond angle in Me₃N is 108°.

a)   Describe the hybridization and shape of Me₃N.

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 s 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 p_z orbital.

Question 11:      Account for the following order of increasing basicity:

RCN: < 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.

Question 12:   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  NH₄⁺ are 109°.

Solution:  The bond angle of 109° in NH₄⁺  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.

Question 13:   Place the isomeric compounds ethylenediammine(A), n-buytlamine (B) and diethyl amine(C) in order of decreasing boiling points and give an expalantion.

Solution:        B > C > A

1° and 2° amines, unlike 3° amines, form intramolecular H-bonds, B with two H’s available for HG-bonding has a higher boiling than C.

Question 14:   a)   Why is an amine of the type R¹R²R³N chiral?

b)   Why is it not possible to separate the enantiomers?

Solution:  a)   Because of the pyramidal geometry, the amine is chiral, the unshared pair being considered as fourth different group.

 b)   The enantiomers rapidly interconvert, having enough kinetic energy at room temperature by a process called nitrogen inversion. For this process 4 kcal/mole (25 kJ/mole) 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 2p_z orbital.

Question 15:  a)   In general the order of base strength as measured by K_b values for aqueous solution of aliphatic amines is R₂NH > RNH₂ < R₃N > NH₃. Explain the order

 b)   What order would you expect to observe in the gas phase.

Solution: a)   Two effects, induction and rotation, determine the K_b of an alkyl amine. Inducting, alkyl groups, being electron releasing, increase the electron density on nitrogen, making the amine more basic (larger K_b). Consequently, they stabilize the positive charge on the conjugate acid, making the ammonium ion less acidic. [It is better to use bases for comparisons rather than the conjugate acids, as was done for carboxylic acids. In terms of induction alone, increasing the number of R’s should increase the basicity. Solvation through H-bonding with H₂O is more important in the conjugate acid than in the free amine because the ammonium ion has a +ve charge and a greater number of H’s. In terms of solvation alone, the more H’s in the ammonium ion the more it is stabilized through H-bonding and the equilibrium shifts more to the rightward. As R’s replace H’s H-bonding deadlines and basicity of amines decrease. Hence the non-uniformity of K values.

b)   Free from solvation effects only induction prevails and the order is R₃N > R₂NH > RNH_­2 > NH₃.

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