Question

Why does H bind to the first Carbon, i know that H binds to the C that already has more H's attached to it. But i still dont get it because the first Carbon has only two hydrogens?

Answer 1

Well, to see why `"HCl"` adds Markovnikov (hydrogen on the less-substituted carbon, or the carbon with more hydrogens in other words), let's look at the transition state of this reaction.

As the alkene donates its `pi` electrons to grab a proton from `"HCl"`, which is a strong acid, we have to consider which carbocation is more stable in the transition state: the one with carbon-1 (terminal) or with carbon-2 (non-terminal).

The more stable carbon can more easily hold a positive charge, and the OTHER carbon thus gets protonated.

Both carbons are `sp^2` hybridized, having only three electron groups. However, carbon-1 has two hydrogens, while carbon-2 has no hydrogens. Hm... hinthint.

CARBOCATIONS PRESENT

A primary (`1^@`) carbocation is written as `"R"-"CH"_2^((+))`.

By definition, carbon-1 would have become a primary carbocation intermediate---just stare at it for a while and you should see that the `"R"` group is everything to the left of carbon-1.

A tertiary (`3^@`) carbocation is written as `"R"_1"R"_2"R"_3"C"^((+))`.

In this case, looking at carbon-2, your `"R"` groups are `"H"_3"C"`, `"H"_3"C"-"H"_2"C"`, and `"CH"_2`. Therefore, carbon-2 is by definition going to become a tertiary carbocation intermediate.

CARBOCATION STABILITY

Okay, so which is more stable? If you recall, your professor should have talked about hyperconjugation, where the surrounding `"C"-"H"` bonding electron pairs interact with the empty `p` orbital of the electropositive carbon.

This interaction distributes electron density from the `"C"-"H"` bonding pair into the empty `p` orbital, stabilizing the molecule.

(It's like distributing an even amount of water in multiple glasses instead of pouring it all into one glass.)

Since there are three `"R"` groups surrounding the tertiary electropositive carbon (carbon-2), instead of just one (like in the primary carbocation intermediate), the tertiary carbocation intermediate is more stable.

In fact, we can make this statement about the carbocation stabilities:

`3^@ > 2^@ > 1^@ > "methyl"`

CONCLUSIONS

So, carbon-2, being tertiary, is more favorably positively charged, and thus, the proton must go on carbon-1 to prevent it from forming a less-stable intermediate. The carbon that the proton does not go onto becomes positively charged!

This correlates with your idea that the proton normally goes onto the carbon with more hydrogens---carbon-2 has no hydrogens directly attached to it at all. Carbon-1 has two.

So even by that logic (which, though simpler, is also correct), carbon-1 should indeed be protonated via Markovnikov addition.