Question

Decreasing order of reactivity of HX in the reaction $ROH+HX\to RX+H_{2}O$

• A. $HI>HBr>HCl>HF$
• B. $HBr>HCl>Hl>HF$
• C. $HCl>HBr>Hl>HF$
• D. $HF>HBr>HCl>HI$

Answer 1

The correct option is A $HI>HBr>HCl>HF$

When alcohols react with halogen acids, alkyl halides are formed. Generally, reaction takes place via $S{N}_1$ mechanism.
In the first step protonation of alcohol takes place. In the second step carbocation is formed. In the third step attack of nucleophile occurs.

As we know, fluorine is the most electronegative element, so HF bond will be stronger and formation of $F^{-}$ from HF will be difficult. The electronegativity order is as follows:
$F>Cl>Br>I$. Hence $I^{-}$ will release easily from HI. Hence, the correct order of reactivity will be $HI>HBr>HCl>HF$ which is option (a).