A) ethyl alcohol
B) propyl alcohol
C) isobutyl alcohol
D) methyl alcohol
Correct Answer: D
Solution :
Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbons so alcohol of at least two carbon atoms can be formed \[{{H}_{2}}C=C{{H}_{2}}\xrightarrow[({{H}_{2}}S{{O}_{4}})]{{{H}^{+}}}\underset{{{H}_{2}}C}{\overset{{{H}_{3}}C}{\mathop{|}}}\,\xrightarrow{HSO_{4}^{-}}\] \[\underset{C{{H}_{2}}OS{{O}_{3}}H}{\overset{{{H}_{3}}C}{\mathop{|}}}\,\xrightarrow{HOH}H\,\,\,\,\,\,\,\,\,{{\,}_{2}}S{{O}_{4}}+\underset{Ethanol}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,\] \[{{H}_{2}}C=CH-C{{H}_{3}}\] \[\underset{(Markownikoff's\,rule)}{\mathop{\xrightarrow[{{H}_{2}}S{{O}_{4}}]{HOH}}}\,{{H}_{3}}C-\underset{Majorproduct}{\mathop{\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,}}\,H-C{{H}_{3}}\] \[\underset{Minor\,product}{\mathop{+HO-C{{H}_{2}}-C{{H}_{2}}C{{H}_{3}}}}\,\]
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