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Aktuel tid:0:00Samlet varighed:10:05

Voiceover: Before we
get into the reactivity of aldehydes and ketones, lets first review the
bonding in a carbonyl. A carbonyl is the carbon
double bonded to the oxygen, so lets focus then on
this carbon right here on the formaldehyde molecule. Lets find the hybridization
stage of this carbon. So I'm going to draw an arrow to this. And to find the hybridization state, one way to do it is to think
about the steric number. Where the steric number is the number of sigma bonds plus the number
of lone pairs of electrons. So to that carbon, let's count
up some sigma bonds here. So we have a sigma bond to this hydrogen, a sigma bond to this hydrogen, and in our double bond here, one of those is a sigma bond and one of those is a pi bond. So we have a total of three sigma bonds. So three sigma bonds and zero lone pairs of electrons gives us a
steric number of three, which we know means it must
have three hybrid orbitals. And so this carbon is sp-two hybridized. So if I'm going to go ahead
and draw that carbon over here. So that carbon is sp-two hybridized, which means it has three
sp-two hybrid orbitals. And we go ahead and put those
three sp-two hybrid orbitals in here like that, alright. And we know that carbon has
un-hybridized p orbitals. And we go ahead and draw in that un-hybridized p orbital right here. Next lets think about those hydrogens. So these hydrogens ... Let me go ahead and put them in red here. So these hydrogens right here are bonded to that carbonyl carbon. Those have an electron in s orbital, which we know is spherically shaped, so I can put a s orbital in here. And the overlap of course
would be sigma bond. And so I have those
sigma bonds right there. Next lets look at the hybridization
of the carbonyl oxygen. So same idea. Number of sigma bonds plus number of lone pairs of electrons. So there is one sigma bond between the oxygen and the carbon, and then we have two
lone pairs of electrons. So one sigma bond and two
lone pairs of electrons gives us a steric number of three, which means that oxygen must
be sp-two hybridized as well. So the oxygen has three
sp-two hybrid orbitals. So let me go ahead and draw those. So put in the oxygen right here. Oxygen has three sp-two hybrid orbitals. Let me go ahead and draw in those sp-two hybrid orbitals. So there is one, and I
have these two over here. And so the lone pairs of
electrons on the oxygen ... One lone pair is going into
this sp-two hybrid orbital. The other is going to go into
this sp-two hybrid orbital. And then we have an overlap
right here for this carbon, so that is of course the sigma bond between the carbon and the oxygen. If the oxygen is sp-two hybridized, it must also have an
un-hybridized p orbital. So I'm going to draw in
the un-hybridized p orbital on the oxygen here like that. And then we can see that the pi bond comes from the side by side
overlap of our p orbitals. And so let me go ahead and
highlight the pi bond over here. So and that double bond again on the dot structure for formaldehyde. One of those is a sigma bond
and one of those is a pi bond. And over here we can
see that on the right. So this represents the
bonding of the carbonyl. And that is going to be
important when we think about things like molecular geometry. So if the carbon is sp-two hybridized, then we know that these atoms
lie on the same plane here, and we have bond angles
close to 120 degrees. We will talk more about
that in a few minutes. Next lets think about the
polarization of that carbonyl. So once again we look at the ... Lets go down to this
generic aldehyde here, and then we have this carbonyl carbon attached to this oxygen. Oxygen is more
electronegative than carbon, so it's going to withdraw
some electron density. We show the polarization
with this arrow here. The arrow points in the
direction of the electrons, so the electrons are going to be pulled closer to the oxygen, and so the oxygen is going to get a little
bit partially negative. So we draw a partial negative sign here. The oxygen is withdrawing
some electron density from my carbonyl carbon right here, so my carbonyl carbon is going to be partially positive like that. And for an aldehyde ... We know that alkyl groups
are electron donating. So our group right here on the left, you can think about that as being a little bit electron donating. Which means our group is going to donate some electron density. So I will draw an arrow showing
that some electron density is being donated by the R group. So remember when we did [carbo-cat-ines], and we put alkyl groups
on our [carbo-cat-ines]. The more alkyl groups we had, the more the [carbo-cat-ine]
was stabilized. So here we have one R
group donating a little bit of electron density
attempting to stabilize that partially positive
charge on the carbonyl carbon. Lets go over here to the ketone, and we have a similar situation. We have once again the oxygen withdrawing some electron density
from our carbonyl carbon. So we have a partial
negative on our oxygen, and our carbonyl carbon
gets a partial positive. But this time we have two R groups. So this R group on the left can donate some electron density. This R group on the right can donate some electron density. And once again, think
about the [carbo-cat-ines]. The more alkyl groups we had the more our full positive charge on our [carbo-cat-ine] was stabilized. Similar idea here. The more R groups you have, the more you stabilize the partial positive charge on your carbonyl. And so because of that, ketones are a little bit
more stable than aldehydes just thinking about the polarization. So there is more to polarization in an aldehyde carbonyl than in a ketone. Alright, so now lets put
these ideas together. Let's think about a
nucleophilic addition reaction to a carbonyl, and so
I'm going to go ahead and draw a ketone down here. And I know that the geometry around the carbon ketone is trigonal planar. Because we talked about
the bonding already. So let me go ahead and draw this in. So we are going to have ... Lets make a R prime group
coming out add some space and make a R group going
away from this space. And we know that these all are
on the same plane like that. So I'm drawing in my plane here. We also know that there is a
polarization of the carbonyl. So the oxygen is more negative and the carbon is a little
bit positive like that. So the carbon is partially positive. That means that ones electrons, it's electrophilic, and
that is extremely important when you are talking about reactions. Nucleophilic additions to carbonyls. So a nucleophile is going to come along. So lets go ahead a draw
in a nucleophile here. So lets go ahead and make it a negatively charged
nucleophile like that. So the nucleophile is
going to be attracted to positive things, right? Opposite charges attract. And so the nucleophile is going to attack the partially positive
carbonyl carbon like that. And when it does so, it
is going to perform a bond and therefore kick off these pi electrons here off onto this oxygen. So lets go ahead and draw the results of our nucleophilic attack here. So we are going to show a bond form between the carbon an the nucleophile. So let me go ahead and
highlight those electrons here. So these electrons right
here on the nucleophile have now formed a bond between the nucleophile and the carbon. And, lets see. We still have our oxygen. Our oxygen used to have two
lone pairs of electrons, but it picked up another
lone pair of electrons, so negative one a formal charge now, and then we have our R groups. So lets go ahead and draw
our R groups over here. So we have R prime and then
we have R over here like that. and so this will be an intermediate. And if we think about the
geometry of this carbon. Let me go ahead a label it here. So the geometry of this carbon now, if I calculate the steric
number I have four sigma bonds. And so four sigma bonds
means a steric number for four hybrid orbitals, so this carbon must be
sp-3 hybridized now. So a sp-3 hybridized carbon. And so therefore your bond angle is closer to 109 degrees. So in particular we are going to think about these R groups here. and so this angle right in here is somewhere around 109 degrees. So I'm going to go ahead and write that approximately 109 degrees. And so the bond angle has changed. So go back over here to
this situation on the left. This carbon right here
was sp-two hybridized, and therefore everything was planar, and so these bond angles
were approximately equal at approximately 120 degrees. And so we've gone from
approximately 120 degrees to 109 degrees for our intermediate here. So over here on the right
our sp-3 hybridized carbon has tetrahedral geometry so we call this our tetrahedral intermediate. So let me go ahead and write that. This is our tetrahedral intermediate, and it is an alk oxide and ion. And so now we can compare aldehydes and ketones
in terms of reactivity. And so the first factor is the
polarization of the carbonyl. So we've already seen that aldehydes are more polarized than ketones, and so therefore the carbonyl carbon is a little more positive. And that means the nucleophile can attack that positive charge a little bit more, and so that's one reason why aldehydes are more reactive than ketones. The polarization of the carbonyl. Another reason has to do
with steric hindrance. So when you think about a ketone, and lets say that we have some big bulky R groups here on this ketone, so these bulky R groups might prevent the nucleophile from attacking. So it turns out there is an optimum angle for the nucleophile to
attack the carbonyl carbon, and if you have bulky R groups
they might prevent that. They also would interfere
with the formation of the tetrahedral
intermediate because if I had big bulky R groups and I'm
changing the bond angle from 120 degrees,
approximately 120 degrees over here on the left to 109 degrees, those bulky R groups have to
be closer together in space and they would of course repel. And there is some steric
hindrance on formation of your tetrahedral intermediate as well. So for those two reasons we observe aldehydes to be more
reactive than ketones.