Question

The treatment of a compound X with $HI{O}_4$ produces $C{H}_{3}CHO$ and $C{H}_{3}C{H}_{2}CHO$. The compound X is :

• A. $C{H}_{3}CH=CHC{H}_{2}C{H}_3$
• B. $C{H}_{3}CH\left(OH\right)CH\left(OH\right)C{H}_{2}C{H}_3$
• C. $C{H}_{3}CH=CHCH=C{H}_2$
• D. $C{H}_{3}CH\left(OH\right)C{H}_{2}CHO$

Answer 1

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

Periodic acid $\left(HI{O}_4\right)$ cleaves 1,2 diols to aldehydes and ketones. So, it will react with 2,3-pentadiol to form ethanal and propanal.
The reaction is:
$C{H}_{3}CH\left(OH\right)CH\left(OH\right)C{H}_{2}C{H}_3\stackrel{HI{O}_4}{\to }C{H}_{3}CHO+C{H}_{3}C{H}_{2}CHO$