Leah here from Leah4Sci.com. In this video, we'll discuss the what and how behind the
alkene reaction mechanisms.
In this video, I will take you through the mechanism of the acid catalyzed hydration
of alkenes.
There are a number of different reactions that will put an alcohol group on to an alkene
and the most common one is the acid catalyzed hydration. This is typically accomplished
by using a strong acid such as sulfuric acid or H2SO4.
Now, just because we're saying that sulfuric acid is the reagent, this doesn't mean this
is the only thing in solution. Recall from general chemistry that acid strength is measured
on a pH scale where pH is negative log of the H plus concentration.
And while you don't have to perform this map in organic chemistry, it's important to remember
the concept. If I tell you that I have a really string acid where the pH is equal to one this
simply means that I have one dissociated acid or one H plus for every ten molecules of water.
And if I have an acid with a pH of 2, still a very strong acid, that means I have one
H plus for every 102 or 100 water molecules. This really means that you have your H2SO4
in a very large quantity of water and that means water will always be available for the
reaction.
Let's look at the specifics of how the H2SO4 reacts with the water. If I do a standard
dissociation reaction, H2SO4 will break down to give me and H plus and then HSO4 minus.
Now, the HSO4 minus is a spectator ion so we won't really look at it but what we do
care about is the H plus.
When an H plus ion is placed in a solution of water, the electronegative oxygen will
act as a necleophile and reach out to grab the H plus. This will give me a molecule of
H3O plus or the hydronium acid catalyst. Hydronium is simply a water molecule with an H plus
attached to it since H plus is not going to be free floating in your solution.
Now, anytime you see acid catalyzed reaction regardless of what the acid is, if this reaction
is aqueous then you always be able to use an H3O plus in your reaction rather than showing
the acid.
Taking that into account, we can now start the mechanism of acid catalyzed hydration
by simply placing H3O plus over the reaction arrow and assuming that something like H2SO4
is in the pod [0:03:04] [Phonetic]. I like to write H3O plus as a hydrogen bound to an
OH2 because this shows me the hydrogen that is about to come off and makes it easier to
draw out the reaction.
Starting with the symmetrical 2-butene, the pi electrons will act as the nucleophile and
the positive hydronium will act as the electrophile. As in the previous alkene reactions, the pi
electrons reach out and grab the hydrogen. This causes the electrons between hydrogen
and oxygen to collapse on to the oxygen giving it back its lone pair.
The intermediate for this reaction will have my hydrogen on either one of the carbon atoms
given that I started with a symmetrical for molecule. And a carbocation forming on the
second carbon that used to hold the double bond. Also formed in this reaction, I have
my water molecule now with its two lone pairs of electrons and no charge.
For this intermediate stuff, the positive charge makes the carbocation my electrophile
and the lone pairs of electrons on the electronegative oxygen makes water the nucleophile. This is
a distinction to recognize in any acid solution. If the charges of this solution are between
neutral and positive then anything even partially negative will be a nucleophile in the solution.
And because this is an acidic solution, make sure never ever to create negative charges
in the solution.
Going back to the reaction, the nucleophilic electrons on the water will reach out and
attack the positive carbon giving me another intermediate for this reaction where I have
my oxygen bound to the carbon and bound to two hydrogens. This oxygen now has just one
lone pair and a formal charge of plus one.
To get rid of this positive charge, I will bring in another water molecule which will
reach out with its lone pair of electrons to grab one of the partially positive hydrogens
and return the lone pair of electrons to the oxygen. This gives me my final product which
is a two butanol having an OH where the carbocation was and then H3O plus which is my acid catalyst
regenerated.
Recall that the definition of a catalyst is something that will speed up the reaction
and will not get used up in the reaction itself. So if my hydronium takes part of the mechanism,
then the hydronium has to be regenerated in the last set of the mechanism as I have here.
Now, let's look at an example starting with an asymmetrical alkene for my reactants. In
this reaction of methylcyclohexane with an acid catalyst, I will start the reaction in
the same way. The nucleophilic pi electrons of the alkene will reach out and grab the
hydrogen causing the electrons between hydrogen and oxygen to collapse on to the oxygen.
If I look at the structure formed by breaking the double bonds, let's name the starting
materials so we can keep track of the steps. We'll start at the tertiary carbon 1, 2, 3,
4, 5, 6. After breaking the double bond, I have the option to place the hydrogen in the
tertiary position giving me a carbocation in the secondary position or the hydrogen
secondary in the carbocation tertiary. And this is because when you break the carbon
to carbon double bond, one of the carbons will get the hydrogen, the other one will
be the [0:07:04] [indiscernible] bonds and therefore have a formal charge of plus one.
Comparing the stability of the carbocation, I have a secondary carbocation versus a tertiary
carbocation. Since the secondary carbocation has fewer carbon neighbors therefore is less
stabilized and the tertiary carbon having more carbon neighbors is more stabilized.
The secondary will not form and my intermediate will have a tertiary carbocation with a hydrogen
placed in the secondary position.
Another water molecule in solution will then reach out with its lone pair of electrons
and attack the carbocation giving me the oxonium intermediate or positive oxygen intermediate.
And finally, another water molecule in solution will reach out and grab that extra hydrogen
giving oxygen back its lone pair of electrons. This gives me a final product where the alcohol
is placed on the most substituted carbon and the hydronium ion regenerated as my catalyst
in solution. This placement of the alcohol in the tertiary position is in accordance
with Markovnikov's Rule which tells you to place the carbocation intermediate in the
most stable position.
In future videos, I will address hydration mechanisms that don't necessarily follow Markovnikov's
Rule due to the catalyst or reagents being used in the mechanism.
I will be happy to answer any additional mechanism questions in my Weekly Review Sessions live
online and from the comfort of your home. I also offer one-on-one private tutoring.
Visit my website for more information, http://Leah4Sci.com/OrganicChemistry.