A) ethyl alcohol
B) iso-propyl alcohol
C) propyl alcohol
D) tert-pentyl alcohol
Correct Answer: B
Solution :
Secondary alcohols upon oxidation give ketones with the same number of carbon atoms. Thus, A is iso-propyl alcohol and the reactions are as \[\underset{{}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}}}\,\xrightarrow[{}]{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}}\] \[\underset{propanone}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}}}\,\xrightarrow[\begin{smallmatrix} Vigorous \\ oxidation \end{smallmatrix}]{[O]}C{{H}_{3}}COOH\]You need to login to perform this action.
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