Question

The correct gas phase acidity order for alcohols \(FCH_2OH\), \(CH_3OH\), \(ClCH_2OH\) and \(BrCH_2OH\) is

• A. \(ClCH_2OH > BrCH_2OH > CH_3OH > FCH_2OH\)
• B. \(FCH_2OH > BrCH_2OH > ClCH_2OH > CH_3OH\)
• C. \(CH_3OH > FCH_2OH > BrCH_2OH > ClCH_2OH\)
• D. \(ClCH_2OH > BrCH_2OH > FCH_2OH > CH_3OH\)

Hint:

In gas phase acidity, conjugate base stabilization depends not only on inductive effect but also on polarizability of the substituent.

Answer 1

The Correct Option is B

Step 1: Understand gas phase acidity.
Gas phase acidity depends on how easily an alcohol loses \(H^+\) to form its conjugate base.
\[ ROH \rightarrow RO^- + H^+ \]

Step 2: Relate acidity with conjugate base stability.
Greater stability of the conjugate base means greater acidity of the parent alcohol.
\[ \text{More stable } RO^- \Rightarrow \text{more acidic } ROH \]

Step 3: Compare methanol.
For \(CH_3OH\), the \(CH_3\) group has \(+I\) effect, which destabilizes the conjugate base \(CH_3O^-\).
Therefore, \(CH_3OH\) is the least acidic among the given alcohols.

Step 4: Effect of halogen substitution.
In \(XCH_2OH\), the halogen atom withdraws electron density through the \(-I\) effect and stabilizes the conjugate base.
Thus, haloalcohols are more acidic than methanol.

Step 5: Compare halogen effects in gas phase.
In the gas phase, both electronegativity and polarizability affect conjugate base stabilization.
Fluorine gives the strongest electron-withdrawing effect, so \(FCH_2OH\) is the most acidic.

Step 6: Compare \(BrCH_2OH\) and \(ClCH_2OH\).
In the gas phase, bromine stabilizes the negative charge better than chlorine due to greater polarizability.
Hence,
\[ BrCH_2OH > ClCH_2OH \]

Step 7: Final order.
Therefore, the correct gas phase acidity order is
\[ FCH_2OH > BrCH_2OH > ClCH_2OH > CH_3OH \]
\[ \boxed{FCH_2OH > BrCH_2OH > ClCH_2OH > CH_3OH} \]
Hence, the correct answer is option (B).