What is the correct relation between acidic strength of primary, secondary and tertiary alcohols?

1°>2°>3°

The correct answer is option 1. 1°>2°>3°.

The correct relation between the acidic strength of primary (1°), secondary (2°), and tertiary (3°) alcohols can be understood by examining the inductive effects and the stability of the resulting alkoxide ion after deprotonation.

When an alcohol loses a proton (H⁺), it forms an alkoxide ion (R-O⁻). The stability of this alkoxide ion influences the acidity of the alcohol. More stable alkoxide ions correspond to stronger acids.

Primary Alcohols (1°):

In primary alcohols (RCH₂OH), the alkyl group (R) is bonded to only one carbon. The inductive effect is minimal, as there is only one alkyl group donating electron density to the oxygen. The alkoxide ion (RCH₂O⁻) is relatively more stable due to less electron-donating inductive effect.

Secondary Alcohols (2°):

In secondary alcohols (R₂CHOH), the carbon attached to the hydroxyl group is connected to two alkyl groups. These two alkyl groups donate more electron density through the inductive effect, making the oxygen more negatively charged. This increased negative charge destabilizes the alkoxide ion (R₂CHO⁻), making the alcohol less acidic compared to primary alcohols.

Tertiary Alcohols (3°):

In tertiary alcohols (R₃COH), the carbon attached to the hydroxyl group is connected to three alkyl groups. The inductive effect is the greatest in this case, with three alkyl groups donating electron density. This leads to a highly destabilized alkoxide ion (R₃CO⁻), making the alcohol the least acidic among the three.

Therefore, the correct order of acidity is: 1° > 2° > 3°

This means primary alcohols are the most acidic, followed by secondary alcohols, and tertiary alcohols are the least acidic.

So, the correct answer is: 1° > 2° > 3°