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Nucleophilic Substitution Reaction

A hydrocarbon...

Question

A hydrocarbon $X$ has molecular formula $C_{5} H_{10}$ . $X$ on treatment with $B_{2} H_{6}$ in $H_{2} O_{2} / N a O H$ gives an optically active $C_{5} H_{12} O$ which on treatment with $C r O_{3} / H C l$ /pyridine gives $C_{5} H_{10} O$ which is still chiral. Which of the following can be a product of reductive ozonolysis of $X$ ?

A

C_{2} H_{5} C H O

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B

C H_{3} C O O C_{2} H_{5}

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C

C H_{3} C O C H_{3}

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D

C H_{3} C O C_{2} H_{5}

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Solution

The correct option is D $C H_{3} C O C_{2} H_{5}$
Degree of unsaturation, DU = $C + 1 - \frac{H}{2} = 5 + 1 - \frac{10}{2} = 1$
DU $= 1$ confirms either one double bond or a cyclic ring. But reaction takes place with $B_{2} H_{6}$ in $H_{2} O_{2} / N a O H$ , it means it is not a cyclic ring but a double bond. Also after oxidation it gives an aldehyde, so the double bond is present at terminal $C$ . Now, as the product after hydroboration and even after formation of aldehyde is optically active, it confirms that the non terminal $C$ with double bond must have only one $H$ after hydroboration step.
Based on these conditions the predicted structure of $X$ and its reactions are as follows:

Ozonolysis of $X$ gives $H C H O$ and $C H_{3} C O C_{2} H_{5}$ .
Hence, (d) is correct.

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