Q. 255.0( 1 Vote )

An aromatic compo

Answer :


A = C6H5COOH (benzoic acid)


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B = C6H5NH2 (aniline) (formed from Hoffmann bromamide degradation)


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C = C6H5NHCOCH3 (N-phenyl ethanamide also called acetanilide)


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D = C6H5CH2NH2 (benzylamine) (amides on reaction with LiAlH4 yields amines)


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E = 2, 4, 6- tribromoaniline (aniline on reaction with bromine water gives 2, 4, 6- tribromoaniline)


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OR


(a)


(i) Benzene diazonium chloride is reduced to benzene. H3PO2 itself is oxidized to phosphorous acid. Nitrogen formed escapes as gas.


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(ii) This reaction is done to introduce a –CN group on benzene and is known as Sandmeyer’s reaction.


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(iii) This reaction is done to introduce –OH group on benzene thereby forming phenol.


(b) For ethyl substituted amines, order of basicity is 2°>3°>1°.


This order is concluded from considering a subtle interplay of inductive effect, solvation effect and steric hinderance of alkyl group. Therefore, the order of basic strength is C2H5NH2 < (C2H5)3N < (C2H5)2NH


(c) Benzenesulphonyl chloride which is also known as Hinsberg’s Reagent reacts differently with primary and secondary amines.


C6H5SOCl + C6H5NH2 ¾ C6H5SONHC6H5


N- phenylbenzenesulphonamide


The ‘H’ attached to nitrogen in product i.e. sulphonamide is strongly acidic due to presence of strong electron withdrawing sulphonyl group. Hence, the product formed is soluble in alkali.


C6H5SOCl + C6H5NHCH3 ¾ C6H5SON(C6H5)CH3


N,N- Diphenylbenzenesulphonamide


This compound formed does not have any H- atom attached to N- atom. Hence, it is not acidic and product is insoluble n alkali.


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