Question

A substance $C_{4} H_{10} O$ yields on oxidation a compound, $C _{4} H _{8} O$ which gives an oxime and a positive iodoform test. The original substance on treatment with conc. $H _{2} SO _{4}$ gives $C _{4} H _{8}$. The structure of the compound is

• A. $CH _{3} CH _{2} CH _{2} CH _{2} OH$
• B. $CH _{3} CHOHCH _{2} CH _{3}$
• C. $\left( CH _{3}\right)_{3} COH$
• D. $CH _{3} CH _{2}- O - CH _{2} CH _{3}$

Answer 1

Answer: (B)

$C_4H_8\xleftarrow{\text{Conc}H_2SO_2} \,\,\,\xleftarrow {}C_4H_{10}O \xrightarrow {\text{Oxidation}} C_4H_8O(R - COCH_3)$
Thus $C _{4} H _{8} O$ should be $CH _{3} CH _{2} COCH _{3}$,
hence $C _{4} H _{10} O$ should be $CH _{3} CH _{2} CHOH CH _{3}$