EtOH is more acidic than t-BuOH !!
An inductive effect is often cited as the reason why t-BuOH is less acidic than EtOH. In fact in the gas phase, where solvation plays no role, t-BuOH is more acidic than EtOH.
We can apply the concept that, if conjugate base is more stable then the corresponding acid will be more acidic.
In aqueous phase as the conjugate base becomes more hindered, the ability of the solvent to organize itself around the base to partly neutralize the charge by dipole effects or hydrogen bonds becomes increasingly compromised. As a result the conjugate base becomes higher in energy, and the acid becomes weaker.
But what happened in the gas phase where EtOH is less acidic than t-BuOH!!
Have a look at stability of their conjugate base,
It seems that t-BuO— is less stable than EtO— due to +I effect (?) of CH3 groups. What do you think ? CH3 should have +I effect ? Try to recall the order of +I effect of groups. ‘O—’ has greater +I effect than any alkyl group. So CH3 should behave like –I group with respect to O— . In t-BuO— there are three CH3 groups to delocalize the negative charge of O— with respect to only one CH3 group in case of EtO—. So t-BuO— is more stable than EtO— in gas phase. Hence t-BuOH is more acidic than EtOH in gas phase.
These are the cases where we should be careful about the inductive nature of alkyl groups.