Answer to Question #16774 in Organic Chemistry for kenneth mcdonald
An optically active compound A, C9H10O, forms a condensation product when treated with hydroxylamine. TREATMENT OF COMPOUND A, with warm ammoniacal silver nitrate results in the production of metallic silver.Reduction of compound A.with lithium aluminium hydride in ether followed by acidification gives an optically active compound B, C9H12O , which on treatment with concentrated sulphuric acid , is converted to a product C,C9H10 , ozonolysis of C, followed by decomposition of the ozonides with water gives formaldehyde and a compound D, C8H8O , which reacts with iodine in aqueous sodium hydroxide on warming to give idoform. Subsequent acidification of the reaction mixture gives a precipitate of benzioc acid.deduce the structures of the compounds A, B,C, and D explain why compound A& B are optically active why does A lose its optical activity on treatment with cold dilute alcoholic potassium hydroxide.
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