Give the decreasing order of reactivity of the following compounds with $H B r$ .

III > IV > II > I

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III > II > IV > I

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III> II > I > IV

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II > III > IV > I

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Solution

The correct option is B III> II > I > IV
The decreasing order of reactivity of the following compounds with HBr is $I I I > I I > I > I V$ .
The $- O H$ group of alcohol is protonated with $H B r$ followed by loss of water molecule to form carbocation.
$A r - C H_{2} - O H + H^{+} \to A r - C H_{2} - O H_{2}^{+}$
$A r - C H_{2} - O H_{2}^{+} \to A r - C H_{2}^{+} + H_{2} O$
Electron donating substituents (such as methyl and methoxy group in para position) on benzene ring stabilizes the carbocation and decreases the reactivity.
Electron withdrawing substituent (such as methoxy group in meta position) on benzene ring destabilizes the carbocation and decreases the reactivity.

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(c) IV > V > III > II > I
(d) V > IV > III > II > I

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Give the decreasing order of reactivity of the following compounds with HBr.

Give the decreasing order of reactivity of the following compounds with $H B r$ .