The correct order of acidic strength of the following alcohols is

I < I I < I I I < I V

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I V < I I I < I I < I

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I I I < I V < I < I I

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I I I < I I < I V < I

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Solution

The correct option is B $I V < I I I < I I < I$
Acidity of alcohol is due to polar nature of O-H bond.
Acidic strength is inversely proportional to EDGs because after deprotonation the EDGs $(- C H_{3}, - C_{2} H_{5})$ attached to OH in alcohols increases the electron density on oxygen making conjugate base less stable. Thus, the tertiary alcohol will be least acidic among the given compounds because of presence of three EDGs $(- C H_{3})$ .
After tertiary alcohol, secondary alcohol will be second least acidic because of presence of two EDGs.
Now, in between $(- C H_{3} and - C_{2} H_{5})$ , +I effect of $- C_{2} H_{5}$ is more than that of $- C H_{3}$ , hence, $C H_{3} O H$ is the most acidic.
Therefore, the correct order of the acidic strength of the alcohols will be:
$I V < I I I < I I < I$

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