Let’s study now the alpha cleavage with carbonyl species and in this case we’ll take
a look at the aldehydes, the ketones, and the carboxylic acids.
So the molecule on the left side could potentially have two different alpha cleavages, one here
on the right side and one on this side.
So we can call this A and B. The molecule in the middle is symmetrical, we’re just going
to pick one of the sides.
Both sides are alpha groups on each side of the carbonyl.
So this will be the alpha carbon because it’s next to the carbonyl and then this is your
alpha bond.
The molecule on the right side also has two different alpha cleavages.
We’re going to represent this side with A and this one with a B. All these ions shown
here have already lost a non-bonding electron, as you can see, therefore there are M+
ions, all three of them.
And so here we have two options, let’s start with this one on the left side, we’ll have
two options for alpha cleavages and we’re going to write both of them.
Option number one, if we cleave the bond shown here as A, we’re going to get the following
fragment and now we’re going to omit the unseen fragments.
We’re just going to draw the one that will produce a cation that can be seen in the
mass spec.
So this will be the product for A and the product for B will be breaking that alpha
bond shown there as B. So I’m going to purposely switch the plus sign so that my radical is
on the left side and I have that long pair there.
So this would be A and B. So remember what I said, whenever you have two radicals next
to each other, they’re always are going to combine and so you will have a triple bonded
specie in both cases.
And so if it’s a triple, notice that you have to keep an inventory of the electrons.
To start you have three electrons and a plus charge and so you end up with a triple bond
so you only have now one lone pair and a plus and since this is a triple bonded ion now,
we have to make it straight.
This is a linear structure at this point and then it’ll be two carbons.
So this will be one and two and so that’s one and two.
So this will be the fragment for A and the fragment for B will be similar but with a
hydrogen attachment.
And so check this out.
If we want to know the molecular mass for this molecule, we’re going to say that we
have C equals three times 12, that’s 36, hydrogens, we have six hydrogens, three, two and one
that’s six, so six times one is equal to six and then we have one oxygen and we multiply
that by 16, one times 16 is 16.
So we do the math here, eight, six, three times is 18, so four plus one is five.
So the molecular weight here is 58.
So for this particular molecule, expect a molecular mass spec that will show like this, you’ll
have a tall fragment of 58, this will be your M+ and you also have a fragment of
57 that has loss a proton.
So the loss of one is a loss of proton.
And so 57 will be the mass to charge ratio for the fragment on the left and this is also
known as the acyl cation.
So this peak will be represented by that and the M+ will be represented by this ion.
So the mass to charge for fragment B will be 16 + 12 + 1.
So it will be 16 plus 12 plus 1, that’s equals to 29.
So B has a mass to charge ratio of 29 because we have one oxygen, one carbon and one hydrogen.
Okay, so the next one only gives us an option, one option, and that is the alpha cleavage
that’s shown there.
Notice that when we did A and B, we’re doing this, so we’re going to split, so this one
goes here, this one goes there, or this one goes there, and this one goes here, all right.
So over here we’ll do the same thing, this electron will go there and this electron will
go to that carbon.
And so we will split that and we’ll have an oxygen with a plus and a radical there.
And this carbon will have a radical here, and then we have an ethyl group.
That plus this ethyl here which has a radical and this will be unseen.
Now we’re going to also have a resonance here because these two electrons on those two radicals
can combine and then it will give us this ion.
So this is your acyl cation, it has a full octet, that’s all for that.
For the next one, we have option A and B, so option A gives us this, if we break on
A this is A and B will be this one.
And so these two will combine, likewise these two will combine to make two new resonance
structures, let’s call this resonance, and so we’ll get a linear structure because it
has a triple bond, one, two carbons, this is your acyl cation, it is seen and you will
have now also this triple bond O with a hydrogen, which this will be your fragment
coming from B. Here the unseen species will be on the left
side A, you will have an OH radical which will be unseen and likewise on the right side
you will have an ethyl group which will be a radical, so this will also be unseen.
