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Carbonyl Chemistry:
Survey of Reactions and
Mechanisms
Course Notes
Chemistry 14D
Images and sample reactions
taken from the Chemistry 14D
Thinkbook for Fall 2004, and
Organic Chemistry
by Paula Yurkanis Bruice 4th edition
Carbonyl Chemistry:
Survey of Reactions and Mechanisms:
When dealing with Carbonyls, we consider two general mechanism types:
3a) …If X is a leaving group, then you kick out the leaving
group and the result is a SUBSTITUTION REACTION.
1) we start with
the carbonyl...
which is attacked
by the nucleophile
2)… we then get a
tetrahedral
intermediate whose
fate is determined by
the presence of a
leaving group
3b) …If X is NOT a leaving group, the O- accepts a
“H+” and the result is an ADDITION REACTION
Carbonyls
Have X as a Leaving Group
Carboxylic Acid (RCO2H)
Don’t have X as a Leaving Group
Aldehyde (RCHO)
Acyl Halide (ROX)
Acid Anhydride (RCO2COR’)
Ester (RCO2R’)
Amide (RCONR’2)
Nitrile (RCN) (doesn’t look like a Carbonyl because it doesn’t
have a C=O but it reacts very similarly)
H3C—C=N
Ketone (RCOR’)
C= O Survey of Reactions
I. HEMIACETAL/ACETAL and
HEMIKETAL/KETAL FORMATION
Aldehyde  Hemiacetal  Acetal
Ketone  Hemiketal  Ketal
• Ketone/aldehyde + alcohol = hemiacetal/hemiketal after one equivalent of alcohol
= acetal/ketal after two equivalents of alcohol
*hemiacetal, hemiketal, and half all start with the letter h, this is a
reminder that when you produce the hemiacetal or hemiketal, you are
halfway to the acetal/ketal
• A hemiaceta/hemiketal has a carbon attached to an “ether” on one end and an alcohol on the
other.
o i.e. RO—C—OH
• An acetal/ketal has a carbon that is attached to two “ethers”. If you take a hemiacetal and
replace the OH with an OR’ group then you get an acetal.
o i.e. RO—C—OR’
First, let us look at a generic mechanism for the formation of an acetal or ketal. Here, we start with the
ketone, so we will be forming a hemiketal, and then a ketal.
• In step 1) the oxygen is protonated. This makes the carbonyl carbon more electrophilic by
giving it a greater partial positive charge
• Then the first equivalent of alcohol attacks the electrophile in step 2)
• Then, the tetrahedral intermediate is formed, when this is deprotonated by the :B, the hemiketal
is formed as seen inside the orange box
• The mechanism doesn’t stop here… the remaining alcohol group on the molecule is protonated
by H—B+, to form H2O as a leaving group, with this, a double bond is formed.
• Then, the second equivalent of alcohol attacks the electrophilic carbonyl carbon
•
•
1)
2)
Another tetrahedral
intermediate is formed.
The HOCH3+ is
deprotonated by the :B
and ultimately, the ketal
is formed, as shown in the
pink box.
Now let’s look at an example of an intramolecular formation of hemiacetal as seen in the assembly of
cyclic glucopyranose from acyclic glucose.
We start with this
acyclic glucose
We get these two diastereomers as
products through identical
mechanisms
•
This is an addition reaction where a nucleophilic alcohol attacks the most reactive part of the
molecule, which is the aldehyde.
Since we are focusing on the aldehyde, let the rest of the glucose “loop” structure be denoted
•
by
Mechanism: * remember that this whole process is reversible.
•
1)
•
•
•
2)
3)
4)
In step 1) the alcohol attacks the carbonyl carbon, forming the tetrahedral intermediate found in
step 2).
The H2O here serves as a catalyst The OH is underlined in bold yellow; it serves as a proton
bus in the reaction by shuttling the proton around. For instance in the second step, it
protonates the negatively charged oxygen in the second step. In the third step, it deprotonates
the alcohol. Why should we protonate the O- before deprotonating the +OH? In this case, it
doesn’t really matter what the sequence of the proton transfer is.
The final product is formed and water is regenerated in step 4).
Is the following a valid mechanism step? NO!! This is not valid because 4 member ring
intermediate has ring strain!
• There is a 1,3 H- shift
• This does not happen because of the nature of
the orbitals
• This process if forbidden
How to speed up the reaction:
Why are hemiacetal/acetal and hemiketal/ketal formation faster in the presence of an acid?
• Because alcohol is a poor nucleophile , it usually helps to have an acid catalyst. Protonating
the alkoxide (RO-) of the tetrahedral intermediate shifts the equilibrium towards the product
side.
• Additionally, acid can protonate the carbonyl oxygen, this increases the electrophilicity of the
carbonyl group.
Why are hemiacetal/acetal and hemiketal/ketal formation faster in the presence of a base?
• Deprotonating the tetrahedral intermediate’s oxonium ion (R2OH+) makes the deprotonated
atom a poorer leaving group.
• The deprotonation also converts the weakly nucleophilic alcohol into a stronger alkoxide
nucleophile.
How would this reaction occur in a living cell?
• Inside a living cell, the proton shuttle is more likely to be an enzyme containing a sufficiently
acidic proton and the lone pair would function as the base.
II. IMINE FORMATION
Aldehyde/Ketone + Primary Amine = Imine
•
•
•
Imine: is a compound with a C=N bond
Imine formation requires small amount of catalytic acid
An example of this is the reaction of retinal & opsin to form rhodopsin
In the mechanism, we will use: RNH2 as the opsin and
H
O=
as the retinal
ret
mechanism for imine formation:
• First, the amine attacks the carbonyl carbon as seen in step 1
• Then, the alkoxide (O-) gains a proton (2a) and the Nitrogen loses a proton (2b) to form a
neutral tetrahedral intermediate
o The equilibrium favors the tetrahedral intermediate with nitrogen protonated because
nitrogen is more basic than oxygen, it can be forced towards the imine by
 removing water
 Precipitation of imine produced
• At this point in the reaction, you might be wondering, why not protonate the Oxygen since you
want water to leave? As you can see in 3 this structure has a positive charge on the nitrogen,
protonating the oxygen would put a positive charge on the oxygen. This would lead to positive
charges on the same molecule in close vicinity to each other which is unfavorable, hence we
need to deprotonate the positively charged nitrogen first in 3.
• Now, in 4 we protonate the alcohol
• Starting with step 5 we see that the water leaves and a carbocation intermediate is formed that
has resonance with the nitrogen, once this is protonated, it forms the protonated imine which is
then deprotonated to yield the imine. This proceeds in an E1 like fashion
•
The reaction can also proceed in an E2 like fashion that was see in 5b, with the elimination of
water occurring at the same time another water deprotonates the nitrogen so it can form a new
N=C bond, forming the imine.
•
The mechanism step shown below CANNOT REALLY HAPPEN in a lab. It can only really
happen in an enzyme where the substrate is held tightly. Why can’t it happen? Because it is a
TRIMOLECULAR COLLISION, which is unlikely.
How do imines compare with carbonyls?
Carbonyls
~carbonyl can react because
Nucleophilic
• of δ+ on carbonyl carbon
attack at the
• oxygen can readily accept pair
carbon
of electrons on C—O pi bond
OH2O-
==O

+OH2
Electrophilic
attack at the ~carbonyl can react because of two
Oxygen (in
lone pairs on oxygen
the case of
+
carbonyls)
+ H
or
H2O—H
==O  ==O
Nitrogen (in
+
the case of
H2O
imines)
Enolate
formation
~same number of resonance forms as
the imine
Imines
~ carbon of an imine
• has smaller δ+ because nitrogen
is less electronegative
• can not accept electrons in the
C—N pi bond as readily as
oxygen of a carbonyl
~imine is less readily attacked by
nucleophiles than a carbonyl
CH3
NCH3
HO==N 
OH
~imine nitrogen has a lone pair and is
less electronegative than oxygen
~imines will react more readily with
electrophiles than a carbonyl
+
H2O—H
CH3
==N 
H
==N+
+
CH3
H2O
~ same number of resonance forms as
the carbonyl
formation
~oxygen heteroatom
• more electronegative
:O:
:O:
O-CH2
CH2
HO- H
~nitrogen heteroatom
• less electronegative; hence it
can’t stabilize negative charge
as well
~imines won’t form enolates as easily
as carbonyls
~need a strong base like LDA
III. FISCHER ESTERIFICIATION
Carboxylic Acid + Alcohol + Strong Acid  Ester
Suppose you want to make an Ester… what would you use?
If we react
Ph
Ph
O=<
+ CH3OH  O=<
OH
OCH3
a carboxylic acid with an alcohol…. there is NO REACTION.
You might be thinking that perhaps carboxylic acid is not a good enough electrophile
Ph
O=<
it’s lone pairs give it resonance therefore it is resistant to nucleophilic
OH
You might want to deprotonate CH3OH to make it CH3O- but this doesn’t work because it would
rather grab the acidic proton from the carboxylic acid instead of attacking the carbonyl carbon. In
other words, instead of following the black arrow, it would rather follow the pink arrow
Ph
O=<
OH
+ CH3O-  O=<
Ph
NO REACTION
OCH3
In that case, you might want to change the leaving group from an OH to a halogen such as Cl (i.e. react
with an acyl halide instead of a carboxylic acid), however since halogens like Chlorine are reactive, we
realize that we want to start with a carboxylic acid
Ph
O=<
Cl
•
-
+ CH3O
Ph
 O=<
OCH3
The reason that there is some delta+ on the carbonyl carbon is because the more electronegative
oxygen is taking some electron density away from the carbon, in a carboxylic acid.
Ph
O== δ+  this δ+ is not very big because of electron donation from resonance
OH
So how do we make the carbonyl carbon more electrophilic? We can protonate the carbonyl oxygen,
this would make the oxygen more electron deficient and make the δ+ bigger
Ph
O== δ+
Ph
+
vs.
HO ==
δ+
OH
protonating the carbonyl oxygen
OH makes it more electrophilic
What can we use to protonate the carbonyl oxygen? We can use a strong acid like H2SO4 which does
nothing but protonate!
When do we protonate, and when do we not protonate?
~protonation increases electrophilicity of carbonyl carbon, making it more susceptible to nucleophilic
attack.
Protonate
Doesn’t need to be Protonated
• Less reactive carbonyls such as esters
• reactive carbonyls like aldehydes and
esters and amides
ketones
• (only the strongest nucleophiles like
• when the nucleophile is very weak, such
LiALH4 and Grignard reagents
as H2O
So what actually happens
carboxylic acid + alcohol + strong base  an ester
Ph
O== δ+
CH3OH
Ph
O==
OH
H2SO4
OCH3
This reaction is called a Fischer esterification where we convert the carboxylic acid into an ester
using a small alcohol and a strong acid.
Fischer Esterification Mechanism:
CH3OH
H—OSO3H

CH3O+H2
~ first we protonate the alcohol using strong acid
Now one, of two possibilities can happen, either
A) the proton goes on the carbonyl oxygen
This is better! The oxygen atoms
are more stable because of
resonance!
B) The proton goes on the ‘alcohol’ of the carboxylic acid
Doesn’t increase resonance… but
due to subtle inductive effects,
alcohol is more basic
After protonating the carbonyl oxygen, we move on to the step with a yellow star near it. This is the
attack of the alcohol on the carbonyl carbon.
Then, we see in step 3 that the alcohol grabs the hydrogen from the positively charge oxygen in the
HOCH3 group in the tetrahedral intermediate. Then, in step 4, one of the alcohol groups gets
protonated so that it can leave as water in step 5. The carbocation intermediate in step 6 has a
resonance form where oxygen takes the positive charge that was formerly on the carbon, when the
oxygen of this resonance structure is deprotonated by the alcohol in step 7, the end product is an ester
and a protonated alcohol.
3
4
5
6
7
Tetrahedral intermediates that are the same are underlined in light blue to show that the different
mechanisms arrive at the same products. For the E2 mechanism, a bimolecular interaction occurs
where dehydration occurs simultaneously with the oxygen reforming a double bond with the carbon
while an alcohol deprotonates it.
In the above reaction, the carbonyl is protonated after methanol because methanol is a stronger base.
In real life:
There isn’t much stability difference between a carboxylic Acid and an ester
To drive the reaction towards the ester, we can use Le Chatelier’s Principle
• To go towards the ester, use a large amount of alcohol or remove the water as it is formed
• To go towards the alcohol, add a lot of water
• Note that adding a lot of acid does not influence the position of the equilibrium, you just have
to make sure that the acid is strong (sufficiently acidic) to protonate the carbonyl or alcohol.
In the reaction where:
“H+”
PhCO2H + CH3OH
PhCO2CH3 + H2O
In the case of an aldehyde or a ketone, it’s easier for the alcohol to attack the carbonyl carbon, however
the alcohol has more difficulty attacking the ester because of the resonance involved
OH
OR’
R
OR R’OH
O==
R’OH R—C—R
in the case of an ester  O==
R—C—OR”
R
R
ketone
OH
OH
IV.
O==
ESTER HYDROLYSIS WITH A BASE
OCH3
NH3
NH3
O==
In this reaction, a new C—N bond is formed.
First, the nitrogen of the NH3 attacks the carbonyl carbon, forming a tetrahedral intermediate.
Afterwards, the OCH3 leaves and the new C—N bond is formed
V.
FORMING NEW C—C BONDS
carbonyl + Grignard reagent/alkyne  a new C—C bond
GRIGNARD REACTIONS
carbonyl + Grignard reagent  alcohol and a new C—C bond
In order for carbon (C) to be nucleophilic enough to attack the electrophilic carbonyl carbon, it must
have a sufficiently negative charge.
i.e R3C-
:
functions as the nucleophile
H3C-
:
this is the carbon anion or “carbanion”
In these reactions we want to form a new C—C bond
Can we use methyl?
H3C—H
base
H3Cthis doesn’t happen because pka ~ 50 for the hydrogens attached
to the carbon in the methyl, therefore it is hard to take away the
proton
But in reality, we don’t need the full negative charge in the Carbon for it to be sufficiently
nucleophilic.
To give Carbon a δ- then it has to be attached to something that is less electronegative than it
• Carbon is more electronegative than any metal
• Metals generally have low electronegativity
δ- δ+
H3C—M
electronegativity: less than 2.5 make a C—Metal bond
Electronegativity: 2.5
Usually M = a metal. This metal is usually Magnesium (Mg)
Hence, Grignard reactants are organometallic, this means that they have a Carbon—Metal bond
Why Magnesium?
Mg prefers to be in a 2+ oxidation state base
How do you prepare the Grignard reactant?
H3C— Br Mg
H3C—MgBr
the solvent has to be an ether because it doesn’t have any
Cl ether
lone pairs to donate
I
How to use the Grignard reactant?
δ- δ+
OOH
Ph H3C—MgBr
H3O+
O== δ+
Ph—C—H
Ph—C—H
H
ether
CH3
CH3
Overall Grignard Reaction
1. Mg, ether
OH Ph
2. carbonyl
CH3Br
3. H3O+
Ph
CH
~this makes a new C—C bond
~ very versatile reaction
~ can use a wide variety of reactants, the only real limitataion is that because Grignards are such strong
reagents, you can’t make them in a reasonable acidic solution
~ can make a lot of alcohols, can use alcohols in the reaction
~ definitely cannot stick this in water, or any acid for that matter, until the Grignard is completely done
reacting.
Grignard Reactions: Aldehydes/Ketones vs. Esters
Ketone: (or aldehyde, but we’re using a ketone here)
δ- δ+
OOH
1) H3C—MgBr
H3O+
O== δ+
Ph—C—H
Ph—C—H
2) H3O+
CH3
CH3
We have seen that aldehydes and ketones undergo addition…
Do esters and ketones undergo substitution ? Yes, but the net reaction is an addition reaction
δ- δ+
OOCH3 1) >—MgBr
* the reaction does not end here! It keeps going!
O== δ+
Ph—C— CH2CH3CH3
Ph
2) H3O+
There is more Grignard in
No leaving group to
the solution, so the new
eject so addition occurs
Eject leaving
carbonyl reacts with more
Nucleophile attacks group
Grignard reactant
carbonyl carbon
The ester and amide undergo substitution, but when it changes into a ketone, it undergoes an addition
process and converts into the alcohol
An attack on the ester/amide is SLOWER than the attack on aldehyde/ketone
ALKYNE ANION REACTIONS
~ these reactions are a method of adding more than one carbon at a time, wheareas the Grignard
reactions were for one or more carbons.
R—C==C—H
Na+ NH2 -
R—C==C-:
This H has pKa of 25
This proton is much more acidic than the one in methane, this has to do with the
Hybridization of carbon
Mechanism:
R—C==C- :
CH3
+ O==
R—C==C—C—OCH3
H3O+
CH3
R—C==C—C—OH
CH3
~this reaction is rather different in that it can make all kinds of different alkyne anions
~the alcohol still has the C==C bond
~ biggest drawback: if you wanted to onl one add carbon, it is dificult because it’s hard to separate the
C==C bond so that only one carbon attaches.
V.
METAL HYDRIDE REACTIONS
Metal + Carbonyl = alcohol
With the Grignard reactants we have created new C—C bonds, what happens when you want to make
a new C—H bond?
• We use metals with a hydrogen attached to it
•
•
Why?
We want a hydride ion -H: that has a negative charge large enough to attack the carbonyl
carbon
o We need to use a metal that is less electronegative than hydrogen
o 2 are really useful for this process
 NaBH4 a borohydride
H
it isn’t very polar but it has a negative charge that it
H—B —H would like to get rid of
*the blue bond or any bond for that matter, can be seen essentially
H
as “H-“
 LiAlH4
Mechanism for reactions with NaBH4 or LiAlH4
Net addition
occurs because
none of the
substituents are a
leaving group
•
we could have used NaBH4 as a proton source but it’s concentration is low compared to the
solvent, so we use CH3OH instead
Although NaBH4 is a good source of creating a new C—H bond, it does not react with anything that is
less reactive than an aldehyde or a ketone
O
O
OH
O
NaBH4
alcohol
OCH3
OCH3
•
NaBH4 is chemoselective in the above reaction, this means that it operates on one functional
group and leaves the other alone
If we want to react with less reactive carbonyls we have to fine something stronger thanNaBH4.
H—BH3
BH3
in order for it to be more reactive, you have to make B less willing to have a
a negative charge
~looking down at the same row as Boron, we find that we can use Aluminum (Al) which is further
down than Boron
• This is less electronegative so it is more inclined to dump the hydrogen
• More reactive source of hydride
• Good enough to attack esters, anhydrides, amides, etc.
• HOWEVER YOU CANNOT USE A PROTIC SOLVENT due to it’s high reactivity
So our new reactant is:
Li+ H — AlH3
LiAlH4 reacts similarly to NaBH4 except that it is stronger!
• For aldehydes and ketones: reaction is simply addition because there’s no leaving group
• For other carbonyls with a leaving group, you get two new C—H bonds when reacting with
LiAlH4
• Look at the following mechanism
•
•
•
It reacts just like NaBH4 except that since there’s a leaving group, substitution occurs before
net addition is achieved.
LiAlH4 can also react with amides to form terminal amines by getting rid of the carbonyl
O
1) LiAlH4
R—CH2NR2
R
NR2
2) H3O+
LiAlH4 is so strong that it can even react with a carboxylate ion
O
R
O
O-
1) LiAlH4
2) H3O+
R
H
H
OH
Ph
H
Overall things to keep in mind when doing Carbonyl Chemistry
In terms of protonating the Carbonyl with an acid:
Must take into account
• Strength of the acid
o Protonation of a carbonyl requires the strength to be H3O+ or stronger
• Basicity of the nucleophile
Protonation may not always be required
• Some nucleophiles like NaBH4, LiAlH4, alkyne anions, organolithiums, and Grignard reactants
are strong bases and will not do well in the presence of a strong acid, therefore, in the presence
of these reactants, do not protonate the carbonyl with a strong acid.
What happens if a Catalyst is Present?
(See Thinkbook Lecture Supplement: Carbonyl Reaction Catalysis)
General Base Catalysis catalyst = base
~ use a base to remove a hydrogen from the nucleophile
~ Nuc: is a stronger nucleophile than Nuc—H
•
Example: Peptide Hydrolysis
the pink line indicates where the bond is broken when the base attacks the carbonyl carbon
• Hydrolysis is faster with OH- than without it
o Without OH- the nucleophile would be H2O
General Acid Catalysis catalyst= acid
• Acid speeds up the reaction
o By enhancing carbonyl electrophilicity
o It shifts the equilibrium towards the tetrahedral intermediate, a neutral tetrahedral
intermediate is less likely to kick out the leaving group
• Strong acids provide “H+”, some examples are :
o H2SO4
o H3O+
o ROH2+
• Example: Peptide hydrolysis
H3O+
the reaction is slower without H3O+ because the protonation of the carbonyl oxygen increases the
electrophilicity of the carbonyl carbon, making it easier for the water to attack it
Enzymatic Catalysis catalyst = enzyme
• Enzymes catalyze by
o Providing proper orientation
o Stabilizing the transition state (with factors such as hydrogen bonding, and other
stabilizing features)
• Example: Peptide Hydrolysis with Chymotrypsin
o Chymotrypsin is a serine protease, it effects the selective hydrolysis of the peptide at
the carboxyl end of the phenylalanine (Phe)
•
No Chymotrypsin = VERY slow reaction
What does Chymotrypsin do?
• Mechanism occurs at the active site
• First the imidazole deprotonates the OH
The imidazole deprotonates the
OH
Start: the nucleophilic attack on
the carbonyl
Tetrahedral intermediate
fragments
Hydrogen bonds stabilize
the transition state
Both acid catalysis and
base catalysis occur in
areas near the yellow
circles
Tetrahedral intermediate
forms
Hydrogen bonds still
stabilizing the transition state
So Chymotrypsin
1) enhances electrophile
2) enhances nucleophile
3) makes new C—C bonds