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Question 1: Compare the basicities of
a) H2C = CHCH2NH2, CH3CH2CH2NH2 and HCº CCH2NH2, and
b) C6H5CH2NH2, cyclohexyl – CH2NH2 and p-NO2C6H4CH2NH2.
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.
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
(b) CH3 – (CH2)5 – CONH2
(c) CH3 –C(CH3)2- CH2 – CH2 – CONH2
(d) O – CH3 – C6H4NH2
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 Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A is
(a) o-toluic acid
(c) o-bromo toluene
(d) m-toluic acid
Solution: The compound is o-toluic acid, and the reactions involved are
Hence, the correct option is A.