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Solved Examples on Organic Compounds Containing Nitrogen

Question: 1, Compare the basicities of

a)  H2C = CHCH2NH2, CH3CH2CH2NH2 and HC º 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 Cb — the more s character it has, the more electron -  withdrawing (by induction) and base weakening it will be. The HO conditions are H2C = CbHCH2NH2(sp2), CH3CbH2CH2NH2(sp3), and HC º CbCH2NH2(sp). The increasing order of electron – attraction is  propargyl > allyl > propyl >, and the decreasing order of basicity is CH3CH2CH2NH2> H2C = CHCH2NH2 > HC º CCH2NH2

b) The decreasing order is

The Cb is cyclohexyl – CH2NH2 uses sp3 HO’s while Cb in the benzylamines uses sp2 HO’s. The electron withdrawing p-NO2 makes the phenyl ring even more electron–withdrawing and base weakening.

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Question: 2, A compound X with seven carbon atoms on treatment with Br2 and KOH gives Y. Y gives carbylamine test and upon diazotisation and coupling with phenol gives azodye. X is

(A)     C6H5CONH2

(B)     CH­3 – (CH2)5 – CONH2

(C)     CH3 –C(CH3)2- CH­2 – CH2 – CONH2

(D)     O – CH3 – C6H4NH2

Answer: A

Solution: Since Y gives coupling reaction after diazotisation it suggest that Y can be aniline or benzener ring substituted aniline. Since Y has been obtained from Hoffmann bromamide it means has – CO NH2 group with benzene ring. Hence it is C6H5CONH2.Hence, the correct option is  (A).

Question: 3, A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with...

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