00:01
Hello, today we are doing problem 9 .46, and this problem asks us to draw the organic products formed when propanol is treated with each of the following reagent.
00:12
So we'll start from a and work our way to i.
00:15
So a, we have propanol reacting with a strong acid.
00:18
The strong acid in this case is sulfuric acid.
00:21
So as you guys know, i like just to rewrite my strong acid is ha.
00:25
And you know a strong acid fully dissociates in solution to h plus and a minus.
00:31
So when you see an alcohol with a strong acid, we know that a dehydration reaction will occur.
00:39
So as before, as every single problem, you need to identify what is your nucleophile, what is your electrofile? nuclophile being the source of electrons, something that's very electron -rich.
00:48
The only thing that's electron -rich here is your oxygen alcohol.
00:52
So you know that's going to nucleophilically attack your hydrogen to pick up a proton and get protonated into oh2 plus.
01:04
Now we know that oh2 plus is not favorable.
01:08
Oxion does not like positive charge.
01:09
It likes to be neutral or negative.
01:11
So this will want to fall off.
01:13
But in order to fall off, we will use our a minus, which is free in solution, to allow our dehydration reaction to a good.
01:24
So remember dehydration, you go from sp3 carbons to sp2, so you lose a degree of saturation.
01:31
So essentially from an alkan, you go to an alken.
01:35
So the way you do that is by picking up one of these beta protons.
01:39
So beta proton, meaning something that is actually adjacent to your leaving group.
01:45
And actually, sorry, it's not these ones that are the beta proton.
01:49
These are the beta protons here.
01:51
So if you see here, this alcohol group was attached to this, which is called your alpha proton, and right adjacent to that is this carbon here, which is your beta proton.
02:05
So your conjugate acid will pick up the protons from that beta carbon.
02:15
So if i were to just erase this, i can show that our conjugate base, our conjugate acid, sorry.
02:26
We'll pick up a proton making a bond.
02:30
When you make a bond, you need to break a bond.
02:32
So sigma bond between the hydrogen and carbon will break, forming our al -keen, making a new bond.
02:38
When you make a bond, you need to break a bond.
02:40
So that's sigma bond attached to our water will break to give our dehydration product and free water.
02:52
Plus h -2 -0.
02:55
Plus we reform our strong acid.
02:59
And because we reform our acid, we know that this is an acid -catalyzed reaction because we reform the catalyst.
03:07
So next for letter b, we see that we have the propanol reacted with sodium hydride.
03:13
Sodium -hydrate dissociates to sodium -plus and h -minus, sodium -plus being our spectator ion, h -minus being the hydride.
03:21
You can imagine that h -minus is very reactive and it would want to pick up another hydrogen to become h -2 gas.
03:28
So the only hydrogen that is labile, labol meaning that it's susceptible to nucleophilic attack or to be ripped off, is this hydrogen attached to our alcohol.
03:40
So we make a bond between the two hydrants forming our h2 gas and therefore the bond between o and h will break, giving those two electrons of that sigma bond onto oxygen's nucleus to make a formal negatively charged o minus plus h2 gas.
03:59
And now if we were to continue this reaction, this o minus can act as a strong nucleophile to attack any electrophile that we add into the mixture.
04:10
So below here, we see that we have a primary alcohol with a strong acid that will associate into h plus and cl minus.
04:21
So as similar to the first one, we know that we are going to protonate our alcohol because of its nucleophilic properties.
04:33
And now because we know we have a primary alcohol originally, we know that this is going to occur through an s &2 mechanism in which our nucleophile will backside attack our electrophilic carbon, making a new bond.
04:55
And when you make a bond, you need to break a bond.
04:59
So now that we broke that bond, we can draw our substitution product plus our h2o.
05:16
Next, we have another strong acid, so this is exactly going to be the same thing as what we did before.
05:24
Hbr dissociates in the solution to h plus and br minus.
05:28
Oxygen having nucleophilic properties will attack this proton to protonate itself...
