question_answer

1-propanol can be easily obtained from propene \[C{{H}_{3}}-CH=C{{H}_{2}}\], by oxidation in the presence of

A)  \[{{H}_{2}}S{{O}_{4}}\]                

B)  \[{{H}_{2}}O\]

C)  \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}}\]             

D)  None of these

Correct Answer: C

Solution :

                 1-propanol can be obtained only by hydroboration-oxidation. \[\underset{propene}{\mathop{6C{{H}_{3}}CH=C{{H}_{2}}}}\,+{{B}_{2}}{{H}_{6}}\xrightarrow{{}}\underset{tri-n-propyi\text{ }borane}{\mathop{2{{(C{{H}_{3}}C{{H}_{2}}C{{H}_{2}})}_{3}}B}}\,\]                                                 \[\underset{1-propanol}{\mathop{3C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH}}\,+\underset{boric\text{ }acid}{\mathop{2B{{(OH)}_{3}}}}\,\] This reaction takes place by anti- Markownikoff s rule. By the addition of water 2-propanol is obtained. (According to Markownikoffs rule) \[C{{H}_{3}}CH=C{{H}_{2}}+H.\,OH\xrightarrow{{{H}^{+}}}\] \[C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  OH  \\  2-propanol \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}\] When propene reacts with \[{{H}_{2}}S{{O}_{4}}\], 2-propanol is obtained. (According to Markownikoffs rule) \[C{{H}_{3}}CH=C{{H}_{2}}+{{H}^{+}}-HSO_{4}^{-}\xrightarrow{{}}\]                                 \[C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  HS{{O}_{4}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}\xrightarrow{{{H}_{2}}O}\]                                 \[C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  OH \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}+{{H}_{2}}S{{O}_{4}}\]                                 2-propanol Note - According to Markownikoffs rule the addition of a reagent to an unsymmetrical alkene takes place in such a way that the negative part of the reagent will be attached to the carbon atom which contains lesser number of hydrogen atom.