Okay today we've got another mechanism challenge we have our starting material and we have this product
We have the reaction conditions so we know everything that's involved, but we want to know how it's going to happen
So we got to write every single electron pushing arrow exactly. How is this going to proceed?
So take a moment and give it a shot
[Think Fast and correct]
[C' mon you can do it]
[Time is going to be up in nearly 8 secs]
[6.......5........4.........3........2.........1....................]
Okay, taking a look here we see some very interesting structural features
We've got these two five-membered rings
But then on the product we've got this five-membered ring which appears to have [uh] remained intact
But then this has become a six-membered ring. So this five-membered ring is now a six-membered ring and there's an additional six-membered ring
[Uh] and nitrogen is part of both of those two six-membered rings. So definitely we're gonna look over here to this azide
And that's definitely gonna be doing uh some some chemistry
So the first thing we have to do is understand exactly what n3 is
so what exactly is n3 we've got n
n
N like that. Uh, and so this one has a minus charge
And this one has a plus charge
So that's what this functional group actually is when we see n3 and so what that means is that this nitrogen
Is actually right. It's only got two bonds. It's got two bonds and two lone pairs
So if it makes an additional if it acts as a nucleophile and makes an additional covalent bond
It's going to go to three bonds in one lone pair, which is ideal. So this is ready to attack something
And it just so happens that we've got a carbonyl nearby
So let us attack this carbonyl
And we're going to kick that up there
and
in doing so we are making uh, we are making a a a seven membered ring, right we're making
this is
one two three, four five six
Seven so let's go ahead and draw that seven membered ring
Whoops, we're going to want the pen tool
so
That is going to be n right there
And the rest of the azide is still there
This one's still got a positive charge we've got a methyl up here
And then we've got boom boom boom boom
We've got o minus we've got a methyl here
And we've got another five-membered ring there
And that is a methyl
And that's
ch2
OH
so
Not too not too tricky here. Right just nitrogen attacked the carbonyl popped open the carbonyl formed that new seven membered ring now
Here's where the real interesting chemistry happens because we know that we can we can close up that carbonyl
But we're going to have to pop something open, right we're going to if this carbon accepts this pi bond again
It's got three sigma bonds here
One of them is going to have to break we could break the one that just formed and go back to where he went
That's not very interesting. So what if we could open up something else now?
Where is another uh portion of the molecule that could be a driving force here?
We've got this nitrogen with a with a positive charge
So watch this we can satisfy all of the formal charges here in one fell swoop if we close that up there
And then let's take this sigma bond and kick that over here
So this is going to detach from we're going to this carbon is going to detach from this carbon
And is instead going to coordinate to this nitrogen atom?
And the reason that this is thermodynamically favorable is that this can then uh kick that off
And we are going to go straight to the product losing nitrogen gas
So nitrogen gas is lost and look at what we've done. We have we have made this ring one
uh one
One carbon smaller and we have made one unit smaller. I should say and then this uh
And then this ring has become one atom larger
so the seven membered ring has condensed and the five-membered ring has ex has expanded and that's great because
We're going to uh, two
Six-membered rings which are more stable than seven and five-membered rings remember six-membered rings?
Uh, all the carbons are perfectly, uh sp3 tetrahedral
109.5 degree bond angles
So that is the optimal ring size
So the two driving factors here are number one going from seven and five to six and six
And two we get to neutralize this nitrogen atom, uh and kick off n2
Which by the way also is entropically favorable
So those are essentially the three factors that are driving this
And we're just going to go straight to the product and that's all it is. So
A pretty short one, but just two interesting things or really one interesting step
But just number one and minus from the azide attacks the carbonyl kicks that open and then obviously the interesting step
is that the carbonyl reforms and then this uh
This carbon will shift and coordinate to this nitrogen instead so that we go from seven and five to six and six
so that is today's uh
mechanism challenge
Thanks for watching guys subscribe to my channel for more tutorials support me on patreon so I can keep making content
And as always, feel free to email me ProfessorDaveExplains@gmail.com
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