Question

An alcohol (A) on heating with concentrated $H_{2}S{O}_4$ gives alkene (B), which shows geometrical isomerism. The compound (A) is:

• A. $\left(C{H}_3{)}_{2}C\right(OH\left)CH\right(C{H}_3{)}_2$
• B. $\left(C{H}_3{)}_{2}C\right(OH)C{H}_{2}C{H}_3$
• C. $C{H}_{3}C{H}_{2}CH\left(OH\right)C{H}_3$
• D. All of the above

Answer 1

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

Since, alkene shows geometrical isomerism. So, the groups should be different on both sides of double bond.
When 2-butanol is heated with conc. sulfuric acid, it forms 2-butene, which shows geometrical isomerism in the form of cis and trans isomers.
$C{H}_{3}C{H}_{2}CH\left(OH\right)C{H}_3\underset{conc.H_{2}S{O}_4}{\overset{dehydration}{\to }}C{H}_{3}CH=CHC{H}_3$