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Question

An organic compound has molecular formula A (C7H10) and known to decolourized Br2-water solution.A on reduction with H2/Pt produces B (C7H14). B on monochlorination with Cl2/hv produces four isomeric monochloro derivatives C7H13Cl. A on ozonolysis followed by work-up with Zn-dimethyl sulphide affords two and only two products, one being methanal and other C (C6H8O3). C is oxidised with acidic solution of dochromate to yields D (C6H8O5) which is soluble in NaHCO3. D is reduced with NaBH4 to yield E (C6H10O3) which may be resolved into enantiomers. Identify A to E.

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Solution

Step 1- A has multiple bond as it gives the unsaturation test with Br2 water
Step 2- Ozonolysis of A in the presence of ZnMe2S (Stops the oxidation upto aldehyde) will result in C
Step 3- Oxidation with strong oxidising agent K2Cr2O7 will results in compound D, that will oxidise the aldehyde part to COOH
Step 4- NaBH4 can't reduce the COOH so it will reduce the ketone part and convert it to alcohol.

1011964_239330_ans_f77317b0ea264c7b8e20d4abe11497da.PNG

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