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Standard XII

Chemistry

Physical Properties of Alkenes

An organic co...

Question

An organic compound has molecular formula A $(C_{7} H_{10})$ and known to decolourized $B r_{2}$ -water solution.A on reduction with $H_{2} / P t$ produces B $(C_{7} H_{14})$ . B on monochlorination with $C l_{2} / h v$ produces four isomeric monochloro derivatives $C_{7} H_{13} C l$ . A on ozonolysis followed by work-up with Zn-dimethyl sulphide affords two and only two products, one being methanal and other C $(C_{6} H_{8} O_{3})$ . C is oxidised with acidic solution of dochromate to yields D $(C_{6} H_{8} O_{5})$ which is soluble in $N a H C O_{3}$ . D is reduced with $N a B H_{4}$ to yield E $(C_{6} H_{10} O_{3})$ 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 $B r_{2}$ water
Step 2- Ozonolysis of $A$ in the presence of $Z n - M e_{2} S$ (Stops the oxidation upto aldehyde) will result in $C$
Step 3- Oxidation with strong oxidising agent $K_{2} C r_{2} O_{7}$ will results in compound $D$ , that will oxidise the aldehyde part to $C O O H$
Step 4- $N a B H_{4}$ can't reduce the $C O O H$ so it will reduce the ketone part and convert it to alcohol.

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