3B Chemistry Lecture 11 Thursday February 26th, 2009
UV spectroscopy with regards to aldehydes and ketones:
2 peaks observed in a UV spectrum for that particular
UV spectrum. If you were to plot this out, you would get
your wavelength and absorbance; the intensity of the
peak means more is happening or more is being
absorbed. For this molecule, what we are going to see is
the following graph. If you were able to look to the left,
you would actually see another peak to the left. But the
cutoff is at 200; that is where the UV lamps start
working. The intensity of the two peaks are very
different. If you were to just look at the numbers, and
not look at the intensities, then you should remember :
you would only see anything on a TLC plate is when the
molecule is conjugated otherwise you wouldnt see it.
But there is an absorbance at 280 nm, which is in the
realm of the UV light. The intensity of that peak is very
small so it wouldnt show up on the TLC plate for our
peak. Pi to pi* is the transition and it is 190 nm. It means
an electron goes up to the next level. The nb to pi*
transition is a lone pair (non bonding) electrons going to
the LUMO. It is the lowest energy transition in that
system so it has the longest wavelength of light and it
happens to be in the UV spectrum even though it has a
low intensity. There is a name for that transition, and it
is called symmetry forbidden. The pi to pi* is called
symmetry allowed. Symmetry forbidden simply refers to
the fact that the lone pairs are perpendicular to the pi
system. Electrons cannot do a perpendicular hop. P
orbitals are always supposed to be parallel; non bonding
orbitals to a pi system are symmetry forbidden and does
not occur very easily. But pi to pi* is a normal transition
going from the HOMO to the LUMO. Even though there
is a transition, it is so weak we usually dont see it.
Thats why everyday ordinary ketones dont show up
under a UV lamp in the lab.
Once we remove 2 hydrogens from the molecule and
make it a conjugated system, then we know that the
levels are going to get closer to each other. The gaps
have shrunk and the numbers include: 219 nm going
from pi to pi*, and 324nm going from nb to pi*. If you
look at the UV spectrum, you will see a peak at 219 and
the point is that the symmetry allowed transition is now
in the UV spectrum. You can now see it and it will show
up as a dark spot on your TLC plate.
Ketones/Aldehydes/ Carbonyl Chemistry: Our most
common method for synthesizing aldehydes and ketones
are oxidizing alcohols to the corresponding carbonyls.
For a secondary alcohol, take chromium trioxide and
sulfuric acid, or you can use PCC. PCC with a
secondary alcohol ( no special conditions needed) will
give us a ketone. Primary alcohols: to go to the aldehyde
we need to use PCC and no water. The reason is if you
try to do this oxidation with PCC and you havent gone
through extra precautions to eliminate water from the
system, then you will end up with the carboxylic acid.
Primary alcohols dont like to stop at the aldehyde; they
like to go all the way to the carboxylic acid. The
chromium trixode and sulfuric acid does the same thing.
MnO2 : will oxidize to a ketone. If you use PCC with 2
hydroxy groups, you will get 2 ketones. If you react the
same molecule with MnO2, then you will only oxidize
one of the alcohols: the difference is that MnO2 only
oxidizes allylic alcohols. It wont react with everyday
ordinary secondary or primary alcohols that are not
allylic. It has to do with the unique CH bond strength of
allylic hydrogens: they are very weak. We have to break
that bond and react. That is another example of a
chemoselective reagent. It also oxidizes benzylic
alcohols (which are attached to benzyllic carbons).
MnO2 with primary allylic or benzyllics will oxidize it
to the aldehyde with water or not.
Ozonolysis: cleaves C-C double bonds to give 2
carbonyls. In terms of the mechanism, you only need to
know the first step. Ozonolysis is good for cleaving
cyclic alkenes, because both carbonyls we get are still
part of the same molecule. Ozonolysis will not work on
aromatic rings. The benzene ring stays untouched.
Ozonolysis is another method for making aldehydes and
ketones.
Friedel Crafts Accylation reaction makes aromatic
ketones. You cant make aldehydes using this method
though. Any acid chloride you make can which is stableyou can do a friedel crafts acyllation with. Oxidation of
alcohols, ozonolysis, and friedel crafts acylation: these
are the three methods for making aldehydes and ketones.
Reactivity of the carbonyl functional group: For
complete reductions all the way to the CH3 group: we
know 3 ways of doing this already including: hydrozene,
base, and heat. Another one is Zn(Hg), and HCl. Then
we have Paladium on carbon, and ammonium formate
which only works of R or R =Aryl (aromatic ring). If
you use palladium over carbon and ammonium formate,
then you will only reduce the ketone next to the aromatic
ring. However, if you use Zn(Hg) and HCl, then you will
reduce all of them. If you use hydrozene, base, and heat,
then you will reduce the ketone but if you use Zn(Hg),
and HCl, then you will get the following. The reason you
get it is because the 2 reducing conditions include one is
basic and one is acidic. A functional group may be
sensitive to acid but not sensitive to base so if we try to
do something somewhere else in the molecule under
acidic conditions, we might do things at other sites in the
molecule. But if we do it under basic conditions, then we
can leave it alone. That molecule plus HCl and water
will give the tertbutoxide plus methanol. We would
protonate the oxygen, and then the proton can even fall
back off so it is in equilibrium; one thing that can
certainly happen to relieve that positive charge is to lose
the tert butyl carbocation which is particularly stable.
Water will react with that. This is considered an acid
sensitive functional group. We can have functional
groups that are not tolerant to certain conditions. Those
are methods to reducing a carbonyl completely to the
CH3 group. All we are doing is replacing a C-O double
bond with 2 hydrogens.
Reduction with H- where we stop at the alcohol: sodium
borohydride and methanol will give the alcohol. If you
dont have any protons present in the course of the
reaction you will get an O- up there and youwill have to
purposely add a source of protons later on. That is 2
steps instead of just one if you use methanol or ethanol
or something as one of your reagents in the first palce.
We take the same molecule and react it with the other
reducing agent we know: Lithium aluminum hyride: it is
a 2 step process. We usually use H30+ as our second
step. One step is better than two. The consequences of
the fact that the carbonyl group has 2 faces: behind the
plane of the paper and the one in the front: in some cases
you will get the same product.
Reduction using a carbon bearing nucleophile, namely, a
Grignard or alkyl lithium. If you see R-Br/Cl/I, R-Alkyl,
R-Aryl, etc and magnesium, then we know that that
makes our Grignard reagent. We know that Grignard and
alkyl lithiums are strong bases so we cant have acids.
Our solvent needs to be free of water or any other source
of protons. We can also use these to add aldehydes and
ketones. R- will be our Grignard/Alkyl lithium and this
is an addition reaction. We need to work it up with
aqueous acid afterwards. It is a 2 step synthesis of
alcohols and we have made a C-C bond. It allows us to
extend the chain of carbons in a molecule by having a
reaction of an aldehyde or ketone with a carbon bearing
nucleophile. If we start with an aldehyde which is not
formaldehyde, then we will get a secondary alchol.
(Formaldehyde will give primary alcohols). If we start
with a cycloheptanone, well get a tertiary alcohol. We
can create primary, secondary, or tertiary alcohols.
Some predict the product examples: We will use
magnesium bromide and H3O+ . We are not limited to
alkyls: we have phenyl lithium. Acid base reactions are
always faster than addition reactions, so we are going to
get some benzene as well as our other product. The
phenyl lithium is the carbanion, pulled off the proton,
made the benzene and then we get an O-. When reacted
with H3O+, it was protonated. We can protect an OH
group, or we if we have an inexpensive reagent, then we
can use 2 equivalents of it. One of them would make O-,
and the other would react with the carbonyl. For the next
example, you would get the expected molecule plus the
benzene which protected the OH group. It made a
tertiary alcohol. Next, we get magnesium reacting first
with the iodine. Just do the mechanism stepwise in 3
steps.