Say you're working on this mechanism and find yourself with a carbocation intermediate. Are we
good, or can I do a carbocation rearrangement to make it even more stable. Methyl shift,
or Hydride shift? That's exactly what we're going to cover in this leah4sci video. Reactions
are all about stability. If a reaction has an unstable intermediate with very high energy,
it's not going to happen as quickly delaying your product. But if I find a way to make
that intermediate more stable. it's more likely to form, therefore faster to form giving me a quicker
product. Carbocation rearrangement is about taking a high energy unstable intermediate and making
it more stable to get a faster reaction. The 3 most common carbocation rearrangements include
the HYDRIDE SHIFT, the ALKYL or METHYL SHIFT, and the RING EXPANSION, which can then undergo
another hydride shift as I explain in the video linked below. Let's quickly review carbocation
stability using the pencil trick taught in this video. We identify this as a tertiary carbocation
because the carbocation has three lines coming out of it to other carbons. The secondary carbocation
only has two and the primary carbocation has just one. Having substituents on a carbocation is like
providing moral support that helps stabilize that positive charge. And so the tertiary carbocation
is more stable than the secondary, which is more stable than the primary and that would
be more stable than a methyl carbocation. In fact, primary and methyl carbocations are so unstable,
you'll rarely see them in reactions. Hydride shift is the movement of a hydride, a hydrogen atom and
two electrons. The electrons coming from the bond holding it to carbon. In our initial example,
we saw a secondary carbocation. But if I can somehow turn that into a tertiary carbocation,
it would be so much more stable. This is easily accomplished by grabbing that now visible hydrogen
and its electrons. Remember we want to move the entire hydride group towards that carbocation.
None of the other hydrogens are involved in this reaction and so we'll ignore them,
but we will show the two electrons from that initial bond that moved over, now attached to
the secondary position. The tertiary carbon looks exactly the same as before the hydride shift,
only now we know it's missing its invisible hydrogen leaving it with a formal charge of
plus one. The carbocation itself didn't move. This is where a lot of students get confused. Instead,
the secondary carbon has a positive charge because it has a lack, it's missing an extra bond,
carbon has an incomplete octet with a formal charge of plus one. When I fill in the second
invisible hydrogen, you can see that carbon now has four bonds and a complete octet. So it no
longer has a carbocation because it is no longer lacking. The tertiary carbon on the other hand,
had a complete octet. It was happy, but when we take away that hydrogen and look
at what's left on carbon, we have only three bonds and an incomplete octet. So this carbon
now has a carbocation simply because it lost a hydrogen. The second thing to keep in mind is you
can only shift to the next position. Since this carbon is directly attached to the carbocation,
I'm allowed to move the hydrogen one space over. But if I had a carbocation here, I'm not allowed
to move the hydrogen because that's more than one position. Now that you get the basics here's the
trick to recognize. A secondary carbocation near a tertiary carbon will give you a hydride shift.
Are you with me so far? If yes, make sure to give this video a thumbs up, subscribe to my channel
and hit that bell notification icon so you don't miss out on any new videos and let's continue.
An alkyl shift as the name implies is the movement of an alkyl or an R-group. A carbon chain,
for example, a methyl shift. Take this secondary carbocation. Secondary is kinda sort of stable,
but tertiary would be even more stable. So I look over to the nearby carbon, primary,
no good, quaternary, definitely more substituted, but I don't have a hydrogen so I grab the next
best thing. I’ll grab the smallest R-group that I can and move that over instead. In this case,
we'll mark it as the green methyl group. The entire CH3 group moves over including the
electrons that bind it to carbon. This gives me a nearly unrecognizable structure. Because I started
with a quaternary and secondary carbon, and now I have a tertiary and another tertiary carbon, and a
carbocation on the position that lost the methyl group. Let's not forget the invisible hydrogen
which helps us understand why this carbon is no longer positive. It has four bonds and a complete
octet. This carbon had no invisible hydrogen because it had 4 bonds to carbon, now has 3
bonds to carbon with an incomplete octet, a formal charge of plus one and a more stable tertiary
carbocation. What if you're looking to do an alkyl shift? and you find yourself with different
options. In this case, the quaternary carbon has a methyl group, an ethyl group, and an isopropyl
group. The smaller the group, the easier it is to move. In this case, the methyl group moves over,
once again giving me a more substituted tertiary carbocation. Now that we know the basics,
here's the trick to look out for. A secondary carbocation near a quaternary carbon will give
you an alkyl shift, ideally a methyl shift. Let's see how these look in actual reactions starting
with the acid-catalyzed hydration of this alkene which I teach in more detail in the alkene series
linked below. H2SO4 gives me H+ in solution which is grabbed by the pi bond. Following Markovnikov's
rule, hydrogen adds at the end and a carbocation forms on the secondary more substituted carbon. I
immediately recognize a secondary carbocation near a tertiary carbon and so I add in my invisible
hydrogen allowing me to show the mechanism for the hydride shift. I'll only show the hydrogen here
because I did a hydride shift and in the future let it be invisible like the rest. And I have
my tertiary carbocation at the more substituted position. H2SO4 implies aqueous, meaning water
is the solvent, and one of these molecules will reach out and attack the carbocation, giving me
an oxonium intermediate. This gets deprotonated by another solvent molecule for a final neutral
alcohol. To get this without the mechanism, recognize that acid-catalyzed hydration of an
alkene requires a carbocation intermediate at the more stable position, and a secondary carbocation
near a tertiary carbon will automatically result in a hydride shift which is where the water will
add giving me the final alcohol. Next we'll look at an SN1 reaction where this molecule
reacts with water in a mild acidic solution. I cover substitution and elimination in detail in
the series linked below. The mechanism begins when bromine, the leaving group breaks away giving me a
secondary carbocation sitting next to a quaternary carbon. This would give me a more stable
carbocation but without a hydrogen to move, I’ll grab an entire methyl group and shift it over.
This results in a tertiary carbocation at the former quaternary carbon. Once again, water comes
in to attack the carbocation forming an oxonium which we deprotonate with another water molecule
for a final tertiary alcohol. Look at that, an alcohol on the carbon near where the leaving group
departed. And the shortcut, recognizing that an SN1 reaction with a carbocation intermediate will
form on the secondary carbon near a quaternary carbon automatically resulting in a methyl shift,
a carbocation at the more stable position, and an OH group adding at the more stable carbocation
position. Now take a look at this molecule where my alkene has a primary and secondary carbon,
one of which would form a carbocation, near a tertiary carbon that is also on
a constrained ring. Yes we'll have a carbocation rearrangement but we're also going to have a ring
expansion which is what I cover in the video linked below at leah4sci.com/carbocation.