# Conversions in amines

We have to find A, B, C, D, E, F, G, H. I got C is acetophenone and D is benzoic acid and further I got what are E,F,G,H I am just a bit confused in A and B please any hints..

Edits : C is acetophenone and the reagent in first reaction where A and B are reactants is AlCl3 not H2O

• There's not enough information to answer this unambiguously. A could be benzyl alcohol and B an oxidising agent but there are other possibilities. Dec 17 '20 at 9:15
• Or -- as an alternative to the oxidation suggested by @Waylander -- a Grignard reaction of $\ce{PhMgBr}$ and, e.g., ethyl formate. Dec 17 '20 at 12:56
• @Rover If the entry about «2,4-DNP» is about 2,4-dinitrophenylhydrazine, then this may react on either aldehyde, or cetone (reference). This then requires a cross check in a database (e.g., Reaxys) or/and table (e.g. CRC Handbook Of Tables For Organic Compound Identification) if your diagram is supported by experimental evidence. We got colored hands preparing derivative hydrazones, picrates, etc. Dec 17 '20 at 13:36
• There still is not enough information to identify A&B if C is acetophenone. The range of possible reactants just got wider. Dec 17 '20 at 13:53
• So if the conditions of the first reaction are AlCl3 then we are looking at a Friedel-Crafts reaction with benzene and acetyl hloride Dec 18 '20 at 11:29

Since, in the question other products have not been mentioned, I am writing the answers for them with reasoning.

1) Since on reaction of D with soda lime gives benzene, which is a decarboxylating reaction, we guess that D should be benzoic acid or sodium benzoate as you have correctly guessed.

2) Now we see that C on reaction with $$\ce{NaOH}+\ce{Br_2}$$ gives benzoic acid, and as the mentioned reagents are reagents for iodoform reaction, we conclude that C must have been a methyl ketone and hence acetophenone which also gives 2,4-DNP test.

3) Now D i.e, benzoic acid on heating with $$\ce{NH_3}$$ gives benzamide which is a condensation reaction in which water condenses out. So, E is benzamide.

4) Benzamide on reaction with $$\ce{NaOH}+\ce{Br_2}$$ will give aniline, which is the Hoffmann Bromamide degradation reaction. So,F is aniline.

5) Aniline on reaction with $$\ce{CHCl_3}+\ce{KOH}$$ will give phenyl isocynaide which is the carbylamine reaction.

6) Ozonolysis of phenyl isocyanide gives phenyl isocyanate. So, H is phenyl isocyanate.

Were there any options in the question from where this has been asked? If there were, it would make it easy to guess A and B.