The C-O bonding electrons could then attack the positively charged oxygen
Not exactly sure what you're saying here, but I think the language could be tightened up a bit.
If you're talking about the mechanism by which the alcohol becomes protonated, then a lone pair on the oxygen belonging to the alcohol moves to attack a proton on the acid. The lone pair on the oxygen forms a sigma bond to the proton, and the electrons in the proton-acid bond move on to the acid to form the conjugate base. The oxygen of the alcohol is now sigma bonded to two protons and a carbon, and thus has a positive charge.
I think you're talking about the subsequent substitution reaction, though. Here, the electrons in the carbon-oxygen bond of the C-O
+H
2 group can heterolytically cleave and both electrons can move on to the oxygen atom. This leaves a carbocation and neutral water (I think this is what you were saying). This is the rate determining step of the S
N1 reaction. It's not an 'attack,' just a heterolytic cleavage.
But bottom line, both of you are right. To protonate an alcohol, you need an acid with a lower pk
a than the protonated alcohol. A strong base will
deprotonate an alcohol, making it an even worse leaving group (O
2-) than it already is (O
-)