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 $$conc.\,{H_2}S{O_4}$$ gives $${C_4}{H_8}.$$ The structure of the compound is
A.
$$C{H_3}C{H_2}C{H_2}C{H_2}OH$$
B.
$$C{H_3}CHOHC{H_2}C{H_3}$$
C.
$${\left( {C{H_3}} \right)_3}COH$$
D.
$$C{H_3}C{H_2} - O - C{H_2}C{H_3}$$
Answer :
$$C{H_3}CHOHC{H_2}C{H_3}$$
Solution :
\[{{C}_{4}}{{H}_{8}}\xleftarrow[\left( -{{H}_{2}}O \right)]{Conc.\,{{H}_{2}}S{{O}_{4}}}{{C}_{4}}{{H}_{10}}O\xrightarrow{\text{Oxidation}}{{C}_{4}}{{H}_{8}}O\left( R-COC{{H}_{3}} \right)\]
Thus \[{{C}_{4}}{{H}_{8}}O\] should be \[C{{H}_{3}}C{{H}_{2}}COC{{H}_{3}},\] hence \[{{C}_{4}}{{H}_{10}}O\] should be \[C{{H}_{3}}C{{H}_{2}}CHOHC{{H}_{3}}\]