Question

Identify the correct trend of acidic strength for the given alcohols

• A. \((i)\gt (iii)\gt (ii)\)
• B. \((i)\gt (ii)\gt (iii)\)
• C. \((iii)\gt (i)\gt (ii)\)
• D. \((iii)\gt (ii)\gt (i)\)

Hint:

The acidity of alcohols decreases with increasing alkyl substitution because alkyl groups show a \(+I\) effect and destabilize the alkoxide ion. \[ 1^\circ \text{ alcohol} \gt 2^\circ \text{ alcohol} \gt 3^\circ \text{ alcohol} \] in acidic strength.

Answer 1

The Correct Option is B

Step 1: Recall the acidic nature of alcohols.
Alcohols behave as weak acids and can lose a proton to form alkoxide ions. \[ ROH \rightleftharpoons RO^- + H^+ \] The acidity of an alcohol depends on the stability of the corresponding alkoxide ion.

Step 2: Effect of alkyl groups on acidity.
Alkyl groups exhibit a \(+I\) (electron-releasing) effect.
This electron-donating effect increases the electron density on the oxygen atom of the alkoxide ion and destabilizes it.
As the number of alkyl groups attached to the carbon bearing the \(-OH\) group increases, the stability of the alkoxide ion decreases.
Therefore, acidity decreases.

Step 3: Compare the three alcohols.
For the primary alcohol \((i)\), \[ RCH_2OH \] there is only one alkyl group exerting the \(+I\) effect. Hence, its alkoxide ion is relatively more stable.
For the secondary alcohol \((ii)\), \[ R_2CHOH \] two alkyl groups exert the \(+I\) effect, reducing acidity.
For the tertiary alcohol \((iii)\), \[ R_3COH \] three alkyl groups strongly donate electrons, making the alkoxide ion least stable and therefore least acidic.
Thus, \[ \text{Primary alcohol} \gt \text{Secondary alcohol} \gt \text{Tertiary alcohol} \]

Step 4: Final conclusion.
Hence, the correct order of acidic strength is \[ \boxed{(i)\gt (ii)\gt (iii)} \] Therefore, the correct option is \[ \boxed{(2)} \]