In the given diol,which of the following stat | vedclass

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(A) The acidity of an alcohol depends on the stability of the resulting alkoxide ion. $C_2$ is a tertiary $(3^{\circ})$ carbon,while $C_5$ is a second

Vedclass · Accepted Answer

$OH$ at $C_2$ is more basic than that at $C_5$

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(A) The acidity of an alcohol depends on the stability of the resulting alkoxide ion. $C_2$ is a tertiary $(3^{\circ})$ carbon,while $C_5$ is a secondary $(2^{\circ})$ carbon. In the $OH$ group at $C_2$,the electron-donating inductive effect ($+I$ effect) of the three alkyl groups attached to $C_2$ destabilizes the negative charge on the oxygen atom of the alkoxide ion. In contrast,the $OH$ group at $C_5$ is attached to a secondary carbon,which has a weaker $+I$ effect compared to the tertiary carbon. Therefore,the alkoxide ion formed from the $OH$ at $C_5$ is more stable than the one formed from the $OH$ at $C_2$. Since greater stability of the conjugate base implies higher acidity,the $OH$ group at $C_5$ is more acidic than the $OH$ group at $C_2$. Conversely,the $OH$ group at $C_2$ is more basic than the $OH$ group at $C_5$ because the conjugate base of the $C_2$ alcohol is less stable and thus more reactive (more basic).

In the given diol,which of the following statements is correct regarding the acidity of the hydroxyl groups at $C_2$ and $C_5$?

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