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>> Solved Problems on Chemical Bonding Part III
An aromatic compound A on treatment with CHCl3 and KOH gives B & C, both of which, in turn give the same compound D when distilled with Zn dust. Oxidation of D yields E of formula C7H6O2. The sodium salt of E on heating with soda lime gives F which can also be obtained by distilling A with Zn dust. Identify A, B, C, D, E and F.
Molecular formula of (E) is C7H6O2 and reaction of its sodium salt with soda lime (decarboxylation) to form (F) indicates that (E) and (F) should be C6H5COOH and C6H6 respectively. Since (F) is also obtained from (A) by reaction with Zn dust, it indicates that (A) should be phenol. Nature of (A) as phenol is confirmed by the fact that it explains all the given reactions.
Five isomeric para – disubstituted aromatic compounds A to E with molecular formula C8H8O2 were given for identification. Based on the following observations, give structures of the compounds?
(i) Both A and B form a silver mirror with Tollen’s reagents; B gives a positive test with FeCl3 solution also.
(ii) C gives positive iodoform test.
(iii) D is readily extracted in NaHCO3 solution.
(iv) E on acid hydrolysis gives 1, 4 dihydroxy benzene.
(A) and (B) gives tollen’s test. Hence must contain aldehyde group. But (B) shows phenolic test also. Hence (A) & (B) are
(C) shows Iodoform test show it must contain group. Therefore (C) is
(D) shows presence of carboxylic group. Hence (D) and (E) are
An organic compound A, C8H4O3 in dry benzene in the presence of anhydrous AlCl3gives compound B. The compound B on treatment with PCl5, followed by reaction with H2/Pd (BaSO4) gives compound C, which on reaction with hydrazine gives cyclised compound D (C14H10N2). Identify A, B, C and D.
Nature of reagents in the conversion of A to B indicates that the reaction must be Friedal – crafts reaction.