Mechanism in formation of acetals

and my questions are
1. which one is right?
2. does the protonation by H2SO4 and free proton both mean the protonation by H3O+?
3. I know sp2 hybridized C is more electronegative, but at the same time it's losing e- to oxygen through pi bond. is C=O still less polarized than C-O even considering pi bond?
4. in answer #2, it's saying like free proton(H3O+?) is less acidic than hydrogen attached to positively charged oxygen(ROH2+?). how is this right?

Hope somebody can help me understand it !!

The first answer is wrong, and the second answer is somewhat true (alcohols are more polar than carbonyls, but not for this reason), but unhelpful.

You use protonated ethanol because it is the most likely acid present.

Let me provide an analogy.

Sulfuric acid is a strong acid. When sulfuric acid is dissolved in water, it ionizes (reacts with water)to an extent that only the tiniest fraction of $\ce{H2SO4}$ is left:

$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$

In an aqueous solution, its fair to say that the concentration of hydronium cation outweighs the concentration of free sulfuric acid by quite a lot. If you were drawing an acid-catalyzed mechanism in water, would you use $\ce{H2SO4}$ or would you use $\ce{H3O+}$?

This reaction is done in ethanol as the solvent. The basicity of ethanol is pretty similar to water, and we have a lot of ethanol, so we can expect the sulfuric acid to ionize to a great extent. These acid-base equilibrium shenanigans reaction would occur regardless of whatever else we added to the mix.

$$\ce{CH3CH2OH + H2SO4 -> CH3CH2OH2+ + HSO4-}$$

Now, if we are doing an acid-catalyzed reaction in ethanol using a strong acid, which species should we use as the proton source when drawing a mechanism: $\ce{H2SO4}$ or $\ce{CH3CH2OH2+}$?

Note that this is an example of the "leveling effect". The strongest acid that can exist in your solution is the conjugate acid of your solvent, and the strongest base that can exist in your solvent is the conjugate base of your solvent. Anything stronger reacts with your solvent.

• Thank you! Now that I saw your answer I realize I was asking so simple thing T_T. – NK Yu Apr 4 '16 at 4:07
• @NKYu: You may consider it simple, but go easy on yourself. The fine details of this issue are overlooked by many organic chemistry students (and instructors). I find myself often taking the lazy way out and just using $\ce{H+}$. By the way, if you are new to the Stack Exchange experience, you can upvote good answers and accept answers that were most useful to you for your question. – Ben Norris Apr 5 '16 at 16:54