The correct sequence of decreasing acidity is:

(C H_{3})_{3} C O H > (C H_{3})_{2} C H O H > C_{2} H_{5} O H > C H_{3} O H

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C H_{3} O H > C_{2} H_{5} O H > (C H_{3})_{2} C H O H > (C H_{3})_{3} C O H

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C_{2} H_{5} O H > C H_{3} O H > (C H_{3})_{3} C O H > (C H_{3})_{2} C H O H

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(C H_{3})_{2} C H O H > (C H_{3})_{3} C O H > C_{2} H_{5} O H > C H_{3} O H

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Solution

The correct option is D $C H_{3} O H > C_{2} H_{5} O H > (C H_{3})_{2} C H O H > (C H_{3})_{3} C O H$
The acidity of alcohols depends on the stability of the corresponding conjugate base i.e. the alkoxide ion $R O^{-}$ . The stability of the base is determined by the stabilization of the negative charge on the conjugate base. Thereby, higher is the stabilization of $R O^{-}$ more is the acidity of $R O H$ .
Alkyl group has an electron donating effect (+ Inductive effect), thus will increase the electron density on the conjugate base (alkoxide) and will destabilise it. Thus primary alcohols having least +I effect will be more acidic than secondary alcohols. Similarly, secondary alcohols are more acidic than tertiary alcohols. Therefore the order of decreasing acidity will be:
$C H_{3} O H > C_{2} H_{5} O H > (C H_{3})_{2} C H O H > (C H_{3})_{3} C O H$

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C H_{3} C H_{2} O H + H C l a n h . Z n C l_{2} - ----- \to

C H_{3} C H_{2} O H + H C l \to

(C H_{3})_{3} C O H + H C l \to

(C H_{3})_{2} C H O H + H C l a n h . Z n C l_{2} - ----- \to

B (m i x) c o n c . H I \leftarrow ---- - (C H_{3})_{3} C - O - C H_{3} a n h y d H I - ----- \to A (m i x)

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