Presence of electron withdrawing groups such as nitro group enhances the acidic strength of phenol and the presence of electron releasing groups such as alkyl group decreases the acidic strength of phenol.

\[\underset{\text{2-Ethylpentan-l-ol}}{\mathop{\underset{\begin{smallmatrix} | \\ \,\,\,\,\,\,\,\,\,{{\,}^{1}}C{{H}_{2}}OH \end{smallmatrix}}{\mathop{\overset{5\,\,\,}{\mathop{C{{H}_{3}}}}\,\overset{4\,\,\,}{\mathop{C{{H}_{2}}}}\,\overset{3\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,\overset{2\,\,\,\,}{\mathop{CH}}\,-C{{H}_{2}}C{{H}_{3}}}}\,}}\,\]

When $$conc.$$  $$HI$$  or $$HBr$$  reacts with ether, the corresponding alcohol and alkyl iodide is formed. When there is a case of mixed ethers the halogen atom attaches to the smaller alkyl group, due to steric effect.

Step 2 involves the formation of carbonium ion by the loss of weakly basic \[{{H}_{2}}O\]  molecule. It is slowest step. Step 4 involves the conversion of an unstable ( or intermediate ) into a quite stable product, hence it is fastest step.

The order of reactivity depends upon the stability of the carbocations formed.

Remember that presence of electron-withdrawing group intensifies i.e., destabilises the carbocation thus (i) and (ii) are less stable than (iii). Further (i); is less stable than (ii) because $$–I$$  effect is more pronounced in (i) due to less distance between $$F$$  and positive charge. Thus the stability order of the four carbocations and reactivity of their parent alcohols will be IV > III > II > I

Going backward from $$A$$ and $$B$$

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Due to intramolecular hydrogen bonding this will not be soluble in sodium bicarbonate.