# Formation of Acetal from Hemiacetal?

In the reaction mechanism of formation of Acetal from Hemiacetal, why doesn't H+ attack the Oxygen of OR Group instead of the OH?

Since O in OR group has +I effect from two alkyl groups, shouldn't it donate its lp more easily?

• Of course. The point is, you can protonate on OR. But after you protonate on OR, what is the next step? Either ROH is kicked out (which brings you back to a carbonyl), or the proton falls back off and you got back to square one. That's what we sometimes call an unproductive step: it most certainly can happen, but it doesn't get you to the product which you want to get. Only by protonating on OH can you eventually get to an acetal. If the equilibrium is indeed biased towards acetal, then the system will get there eventually, and it has to get there by protonating OH, not OR. @user93057 – orthocresol May 2 at 8:16
• The difference in protonatability between the two Oxygens due to +I effect is not enough that the OH is never protonated. Some is, and as that is the only pathway by which the reaction can proceed, that is what appears in the reaction mechanism diagram. – Waylander May 2 at 8:56

The acetal reaction is performed in $$\ce{H+/ROH}$$ condition (e.g., $$\ce{H+/CH3OH}$$). The general mechanism can be found in any Organic Chemistry textbook. The reaction is reversible. The first step is protonation of aldehyde (or ketone) carbonyl oxygen by an acid ($$\ce{H+}$$):
The hemiketal has two oxygen molecules, $$\ce{OH}$$ and $$\ce{OR}$$, both of which are susceptible to protonate. However, if $$\ce{OR}$$ is protonate, reaction would go backward. Of cause, it protonates because the reaction is reversible (see the scheme). Yet, more favored is $$\ce{OH}$$, which would subsequently eliminate as a $$\ce{H2O}$$ molecule. The resultant positively charged carbonyl compound undergoes second nucleophilic attack by another $$\ce{CH3OH}$$ molecule to give protonated acetal, which proceeds to give expected acetal by deprotonation. Since we are using the alcohol as the reaction solvent, forward reaction is favored by multiple times.
OP's suggestion of protonation of $$\ce{OH}$$ group is legitimate, but reaction goes backward for a moment, but forward reaction favorably proceed to the end (Recall Fischer esterification). If you want back your aldehyde (or ketone), you can make backward reaction favored by using water as your solvent ($$\ce{H+/H2O}$$).