A) using excess of\[PB{{r}_{3}}\]
B) choosing low temperature for reaction
C) converting alcohol into sulphuric or sulphonic acid
D) converting alcohol into carboxylate esters
Correct Answer: C
Solution :
\[\underset{3-pentanol}{\mathop{C{{H}_{3}}-C{{H}_{2}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}C{{H}_{3}}}}\,\xrightarrow{{}}\] \[\underset{\downarrow rearrangement}{\mathop{C{{H}_{3}}C{{H}_{2}}-\overset{+}{\mathop{C}}\,H-C{{H}_{2}}C{{H}_{3}}}}\,\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}\overset{\oplus }{\mathop{C}}\,HC{{H}_{3}}\] Hence, it give 2- and 3-bromopentane, but when alcohol is converted into sulphonic acid and then bromination takes place a single compound is formed. \[C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}C{{H}_{3}}+HOS{{O}_{3}}H\] \[\xrightarrow{{}}{{(C{{H}_{3}}C{{H}_{2}})}_{2}}\underset{B{{r}^{-}}\downarrow PB{{r}_{3}}}{\mathop{CH.OS{{O}_{3}}H}}\,\] \[\underset{3-bromopentane}{\mathop{C{{H}_{3}}C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{3}}}}\,+HSO_{4}^{-}\] In this reaction during the formation of sulphonate\[CO\]bond does not break. Hence, sulphonate group is converted into bromide group without rearrangement.
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