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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 conversion from amines

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    $\begingroup$ There's not enough information to answer this unambiguously. A could be benzyl alcohol and B an oxidising agent but there are other possibilities. $\endgroup$
    – Waylander
    Dec 17, 2020 at 9:15
  • $\begingroup$ Or -- as an alternative to the oxidation suggested by @Waylander -- a Grignard reaction of $\ce{PhMgBr}$ and, e.g., ethyl formate. $\endgroup$
    – Buttonwood
    Dec 17, 2020 at 12:56
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    $\begingroup$ @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. $\endgroup$
    – Buttonwood
    Dec 17, 2020 at 13:36
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    $\begingroup$ There still is not enough information to identify A&B if C is acetophenone. The range of possible reactants just got wider. $\endgroup$
    – Waylander
    Dec 17, 2020 at 13:53
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    $\begingroup$ So if the conditions of the first reaction are AlCl3 then we are looking at a Friedel-Crafts reaction with benzene and acetyl hloride $\endgroup$
    – Waylander
    Dec 18, 2020 at 11:29

1 Answer 1

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As @Waylander mentioned in comments, more information is needed to guess A and B.

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.

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