Arrange the following alcohols in order of reactivity towards gaseous HBr:
A
II>III>IV>I
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B
I>IV>II>III
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C
I>II>III>IV
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D
III>II>IV>I
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Solution
The correct option is BI>IV>II>III The reaction with gaseous HBr and alcohols involve the formation of carbocations. Thus, the stable the intermediate carbocation, the faster the reaction will proceed. We can see the corresponding carbocations and observe that they all are 2° carbocations. The stability factor will be determined by the -I group (−F). -I groups destabilise carbocations and thus III will be least stable due to the presence of two −F groups. IV will be comparatively stabler than II as the −F group is located farther away from the positively charged carbon. Hence, the correct order of reactivity will be: I>IV>II>III Thus, Option B is correct.