Question

An alcohol gave Lucas test in about 5 minutes. When the alcohol was treated with hot concentrated $H_{2}S{O}_4$, it gave an alkene of molecular formula $C_4{H}_8$ which on ozonolysis gives single product with molecular formula $C_2{H}_{4}O$. The structure of alcohol is:

• A. $C{H}_{3}CH\left(OH\right)C{H}_{2}C{H}_{2}C{H}_3$
• B. $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH$
• C. $C{H}_{3}CH\left(OH\right)C{H}_{2}C{H}_3$
• D. $(C{H}_3{)}_{3}C-OH$

Answer 1

The correct option is C $C{H}_{3}CH\left(OH\right)C{H}_{2}C{H}_3$

$C{H}_{3}C{H}_2-CH\left(OH\right)-C{H}_3\stackrel{H^{+}}{\to }C{H}_3-CH=CH-C{H}_3\stackrel{O_3,Zn/H_{2}O}{\to }2C{H}_3-CHO$

As Lucas test is given by compound in about 5 minutes, it is a secondary alcohol.

After ozonolysis only one product is obtained confirms that the alkene form was symmetric and that is possible if and only if alcohol was secondary.

Hence, the alcohol is $C{H}_{3}CH\left(OH\right)C{H}_{2}C{H}_3$.