How does a hydroboration mechanism actually work?

Let us say that you have 1-propene reacting with $\ce{BH3-THF}$ and then later with hydrogen peroxide and sodium hydroxide. I eventually get 1-propanol. The way it was explained to was that after reacting with $\ce{BH3-THF}$ we have $\ce{CH3CH2CH2BH2}$ and after the second set of reagents we get our product.


But, I never understood the mechanism, the explanation seems forced. $\ce{BH2}$ goes to the less hindered carbon which explains why its at the terminal. And then for some reason, $\ce{BH2}$ gets replaced by $\ce{OH}$. 




I had to write the complete mechanism for you so that you can understand it better. See the image below. I have explained it in the simplest form. Some of the points I would like to discuss before you look at the mechanism are:


1. ***In BH3, Boron is less electronegative than H*** (i.e. hydrogen) , hence in the syn addition I have shown, hydrogen ends up being attached to a more substituted carbon atom.
2. For every bonding orbital we have an anti-bonding orbital which lies exactly opposite to the bonding orbital and ***each bonding-antibonding orbital pair can contain a maximum of 2 electrons in them***.


Ok now lets head towards the mechanism which I wrote exclusively for you:


[![hydroboration-pg 1](https://i.stack.imgur.com/XN2OS.jpg)](https://i.stack.imgur.com/XN2OS.jpg)


[![hydroboration-pg 2](https://i.stack.imgur.com/K4dtV.jpg)](https://i.stack.imgur.com/K4dtV.jpg)


[![hydroboration-pg 3](https://i.stack.imgur.com/Znanf.jpg)](https://i.stack.imgur.com/Znanf.jpg)


If you have any further doubts in the mechanism then you can ask in the comment section below my answer. I hope it helps.


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