Question
An alcohol $$X$$ on heating with concentrated $${H_2}S{O_4}$$ gives an alkene $$Y$$ which can show geometrical isomerism. The alcohol $$X$$ is
A.
$$C{H_3}C{H_2}CH\left( {OH} \right)C{H_3}$$
B.
$${\left( {C{H_3}} \right)_2}C\left( {OH} \right)CH{\left( {C{H_3}} \right)_2}$$
C.
$${\left( {C{H_3}} \right)_3}C\left( {OH} \right)$$
D.
$${\left( {C{H_3}} \right)_2}C\left( {OH} \right)C{H_2}C{H_3}$$
Answer :
$$C{H_3}C{H_2}CH\left( {OH} \right)C{H_3}$$
Solution :
\[\underset{\left( X \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}CH\left( OH \right)C{{H}_{3}}}}\,\] \[\xrightarrow[{{H}_{2}}S{{O}_{4}}]{\text{Conc}\text{.}}\underset{\left( Y \right)}{\mathop{C{{H}_{3}}CH=CHC{{H}_{3}}}}\,\]
\[\underset{\left( X \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}CH\left( OH \right)C{{H}_{3}}}}\,\] \[\xrightarrow[{{H}_{2}}S{{O}_{4}}]{\text{Conc}\text{.}}\underset{\left( Y \right)}{\mathop{C{{H}_{3}}CH=CHC{{H}_{3}}}}\,\]