A) 4-methyl octanol
B) 2-methyl decane
C) 4-methyl heptanol
D) 4-mediyL2-octanone
Correct Answer: A
Solution :
Terminal alkenes react rapidly with diborane to form primary trialkyi boranes which on oxidation gives primary alcohols. . \[\underset{4-methyl\,\,octene}{\mathop{C{{H}_{3}}{{(C{{H}_{2}})}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CHC{{H}_{2}}CH}}\,}}\,=C{{H}_{2}}}}\,\xrightarrow{B{{H}_{3}}\,\,or\,\,{{B}_{2}}{{H}_{6}}}\] \[\left( C{{H}_{3}}{{(C{{H}_{2}})}_{3}}-{{\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CHC{{H}_{2}}C{{H}_{2}}.CH}}\,}}\,}_{2}}- \right)\,\,B\] \[\xrightarrow{{{H}_{2}}{{O}_{2}}/NaOH}\underset{4-\,\text{methyl}\,\text{octanol}}{\mathop{C{{H}_{3}}{{(C{{H}_{2}})}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{CHC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH}}\,}}\,}}\,\](In general hydroboration oxidation involve the addition of water according to anti-Markownikoffs rule).
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