Leah here from leah4sci.com and In this video, I will take a break from specific mechanisms
and focus on an important phenomenon that arises in many reactions.
And this is the concept of the hydride or the alkyl shift.
A hydride is a hydrogen atom along with two electrons and a negative charge and an alkyl
shift is the shift of an entire R group.
Having done many reactions, you will recognize that the rate or speed of a reaction depends
on the stability of the intermediates.
If you have a reaction that undergoes a stable intermediate, the reaction will happen quickly
but if the reaction undergoes an unstable or less stable intermediate, it will happen
more slowly.
A common example of this is the idea of a carbocation formation.
If you have a leaving group on a tertiary carbon, this will give you a tertiary carbocation
which is very stable.
When you have a leaving group on a secondary carbon, you will form a secondary carbocation
which is stable but not as stable.
And finally, when you have a leaving group on a primary carbon, the carbocation that
would form is so unstable that this actually will not be seen in reactions.
For more in depth explanation of carbocation stability, go ahead and watch my carbocation
video.
Assuming you understand the concept, let's move on.
At times, you will come across a reaction where the carbocation that forms is the more
stable of possibilities.
For example, in this hydrohalogenation of an alkene, the reaction will start out as
a standard alkene reaction with the Markovnikov or secondary carbocation forming as a result
of the pi bond attacking the hydrogen.
I get the hydrogen on the terminal carbon and a positive charge on the secondary carbon.
However, this carbocation is located next to a tertiary carbon which would ultimately
form a more stable carbocation.
So how can I get the carbocation in the tertiary position if the double bond is only on a primary
and on a secondary carbon?
And this is where the concept of the hydride shift comes in.
The hydrogen is bound to the carbon with two electrons.
In a hydride shift, I get the entire hydrogen plus electrons jumping over to the less stable
position thus creating a deficiency in the tertiary carbon.
A quick formal charge on the tertiary carbon shows that what started out as a carbon with
four bonds now has only three and therefore a positive charge forms on the most stable
carbon of this molecule.
The movement of this hydrogen with its electrons is called a hydride shift.
And since the carbocation that forms is a lot more stable, this will happen rather quickly.
And when the bromide ion in solution comes to attack, it winds up attacking at the tertiary
rather than in the secondary position.
The final product for this molecule is a bromine added in a tertiary position despite the fact
that the double bond only started out between a primary and a secondary carbon.
Let's look at another example.
For this reaction, let's look at the acid catalyzed hydration.
As with any alkene addition, the pi bond will reach out with its electrons to grab a hydrogen
and collapse these electrons on to the water molecule.
In addition to the H2O forming, I also have two potential carbocation intermediates where
the double bond broke.
I have the option to place the hydrogen in two positions to give me two different carbocation
intermediates.
The tricky aspect is that both intermediates have secondary carbocations.
And so in order to predict the product, you have to think one step further.
On the left, I have a secondary carbocation directly near a tertiary position with a free
hydrogen.
On the right, the carbocation only has secondary carbon neighbors.
And so while it will form, it would be a minor product while the structure on the left will
be the major product.
And the reason for that is the hydrogen with its lone pair of electrons can perform a hydride
shift and take the place of the secondary carbocation providing a deficiency in the
tertiary position and therefore a stable tertiary carbocation intermediate.
Jumping ahead a few steps, I wind up with an alcohol in the tertiary position directly
neighboring the secondary carbon which originally held my double bond.
What if instead you're faced with a double bond next to a quaternary carbon or carbon
having four groups attached to it and no hydrogen to perform a hydride shift?
The reaction starts off the same way where the electrons grab the hydrogen and collapse
the bonding electrons on to the chlorine.
In addition to a Cl minus in solution, the hydrogen adds to the Markovnikov position
giving me a secondary carbocation.
In this situation, because the secondary carbocation is directly neighboring a quaternary carbon
and the fact that there are no hydrogens on those quaternary carbon, we will shift the
next best group.
These lines here represent methyl groups attached to the quaternary carbon by a pair of electrons.
In an alkyl shift specifically here, a methyl shift, the entire CH3 group with its lone
pair of electrons will jump over towards the secondary carbocation to free up a space at
the quaternary now tertiary position.
This gives me a methyl group in the secondary now tertiary position and a carbocation in
the newly formed tertiary position.
This happens rather quickly.
And so, when the chloride ion comes to attack, it winds up attacking the tertiary carbocation
to give me a final product where the chlorine is added in the tertiary position.
When you see these reactions, it helps to understand the mechanism but you don't always
have to go through every step to give the reaction.
If you recognize your double bond is located next to a carbon that is more substituted,
try to envision what can possibly shift to give you a more stable intermediate and place
your product there directly.
For more information about organic chemistry reactions, join my mailing list at Leah4Sci.com/OrganicChemistry