Problem 9:

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.

Solution:

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.

Problem 10:

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.

Solution:

(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

Problem 11:

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.

Solution:

Nature of reagents in the conversion of A to B indicates that the reaction must be Friedal – crafts reaction.