Carbonyl Chemistry:Survey of Reactions and

Mechanisms

Course NotesChemistry 14D

Images and sample reactionstaken from the Chemistry 14DThinkbook for Fall 2004, and

Organic Chemistryby Paula Yurkanis Bruice 4th edition

Carbonyl Chemistry:Survey of Reactions and Mechanisms:

When dealing with Carbonyls, we consider two general mechanism types:

CarbonylsHave X as a Leaving Group Don’t have X as a Leaving GroupCarboxylic Acid (RCO2H) Aldehyde (RCHO)

Acyl Halide (ROX) Ketone (RCOR’)

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

2)… we then get atetrahedralintermediate whosefate is determined bythe presence of aleaving group

1) we start withthe carbonyl...which is attackedby the nucleophile

3b) …If X is NOT a leaving group, the O- accepts a“H+” and the result is an ADDITION REACTION

3a) …If X is a leaving group, then you kick out the leavinggroup and the result is a SUBSTITUTION REACTION.

• Another tetrahedralintermediate is formed.

• The HOCH3+ is

deprotonated by the :Band ultimately, the ketalis formed, as shown in thepink box.

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 areminder that when you produce the hemiacetal or hemiketal, you arehalfway to the acetal/ketal

• A hemiaceta/hemiketal has a carbon attached to an “ether” on one end and an alcohol on theother.

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 theketone, 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 bygiving 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)

NO!! This is not valid because 4 member ringintermediate 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

Now let’s look at an example of an intramolecular formation of hemiacetal as seen in the assembly ofcyclic glucopyranose from acyclic glucose.

• This is an addition reaction where a nucleophilic alcohol attacks the most reactive part of themolecule, 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.

• In step 1) the alcohol attacks the carbonyl carbon, forming the tetrahedral intermediate found instep 2).

• The H2O here serves as a catalyst The OH is underlined in bold yellow; it serves as a protonbus in the reaction by shuttling the proton around. For instance in the second step, itprotonates the negatively charged oxygen in the second step. In the third step, it deprotonatesthe alcohol. Why should we protonate the O- before deprotonating the +OH? In this case, itdoesn’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?

How to speed up the reaction:

We start with thisacyclic glucose

We get these two diastereomers asproducts through identicalmechanisms

1) 2) 3) 4)

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 productside.

• Additionally, acid can protonate the carbonyl oxygen, this increases the electrophilicity of thecarbonyl 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 HO= 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 intermediateo 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 youwant 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 positivecharges on the same molecule in close vicinity to each other which is unfavorable, hence weneed 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 isthen 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 ofwater occurring at the same time another water deprotonates the nitrogen so it can form a newN=C bond, forming the imine.

• The mechanism step shown below CANNOT REALLY HAPPEN in a lab. It can only reallyhappen in an enzyme where the substrate is held tightly. Why can’t it happen? Because it is aTRIMOLECULAR COLLISION, which is unlikely.

How do imines compare with carbonyls?

Carbonyls Imines

Nucleophilicattack at thecarbon

~carbonyl can react because• of δ+ on carbonyl carbon• oxygen can readily accept pair

of electrons on C—O pi bond

O-H2O- ==O +OH2

~ carbon of an imine• has smaller δ+ because nitrogen

is less electronegative• can not accept electrons in the

C—N pi bond as readily asoxygen of a carbonyl

~imine is less readily attacked bynucleophiles than a carbonyl CH3

-NCH3

HO- ==N OH

Electrophilicattack at theOxygen (inthe case ofcarbonyls)orNitrogen (inthe case ofimines)

~carbonyl can react because of twolone pairs on oxygen

+ + HH2O—H ==O ==O + H2O

~imine nitrogen has a lone pair and isless electronegative than oxygen~imines will react more readily withelectrophiles than a carbonyl

CH3

+ HH2O—H ==N ==N+ + CH3

H2O

Enolateformation

~same number of resonance forms asthe imine

~ same number of resonance forms asthe carbonyl

formation ~oxygen heteroatom• more electronegative

:O: :O: O-

-CH2 CH2

HO- H

~nitrogen heteroatom• less electronegative; hence it

can’t stabilize negative chargeas well

~imines won’t form enolates as easilyas 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 PhO=< + 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

PhO=< it’s lone pairs give it resonance therefore it is resistant to nucleophilic O H

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. Inother words, instead of following the black arrow, it would rather follow the pink arrow

Ph PhO=< + CH3O

- O=< NO REACTION OH OCH3

In that case, you might want to change the leaving group from an OH to a halogen such as Cl (i.e. reactwith an acyl halide instead of a carboxylic acid), however since halogens like Chlorine are reactive, werealize that we want to start with a carboxylic acid

Ph PhO=< + CH3O

- O=< Cl OCH3

• The reason that there is some delta+ on the carbonyl carbon is because the more electronegativeoxygen 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 Ph

O== δ+ vs. HO+== δ+ protonating the carbonyl oxygen OH OH makes it more electrophilic

What can we use to protonate the carbonyl oxygen? We can use a strong acid like H2SO4 which doesnothing but protonate!

When do we protonate, and when do we not protonate?~protonation increases electrophilicity of carbonyl carbon, making it more susceptible to nucleophilicattack.

Protonate Doesn’t need to be Protonated• Less reactive carbonyls such as esters

esters and amides• (only the strongest nucleophiles like

LiALH4 and Grignard reagents

• reactive carbonyls like aldehydes andketones

• when the nucleophile is very weak, suchas H2O

So what actually happens

carboxylic acid + alcohol + strong base an ester

Ph CH3OH PhO== δ+ O== OH H2SO4 OCH3

This reaction is called a Fischer esterification where we convert the carboxylic acid into an esterusing 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 atomsare more stable because ofresonance!

B) The proton goes on the ‘alcohol’ of the carboxylic acid

After protonating the carbonyl oxygen, we move on to the step with a yellow star near it. This is theattack 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 theHOCH3 group in the tetrahedral intermediate. Then, in step 4, one of the alcohol groups getsprotonated so that it can leave as water in step 5. The carbocation intermediate in step 6 has aresonance form where oxygen takes the positive charge that was formerly on the carbon, when theoxygen of this resonance structure is deprotonated by the alcohol in step 7, the end product is an esterand a protonated alcohol.

Tetrahedral intermediates that are the same are underlined in light blue to show that the differentmechanisms arrive at the same products. For the E2 mechanism, a bimolecular interaction occurswhere dehydration occurs simultaneously with the oxygen reforming a double bond with the carbonwhile 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 esterTo 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.

Doesn’t increase resonance… butdue to subtle inductive effects,alcohol is more basic

3

4

5

6 7

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, howeverthe alcohol has more difficulty attacking the ester because of the resonance involved OH OR’ R OR R’OHO== R’OH R—C—R in the case of an ester O== R—C—OR” R R ketone OH OH

IV. ESTER HYDROLYSIS WITH A BASEOCH3 NH3 NH3

O== 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 BONDScarbonyl + 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 musthave 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 bondCan we use methyl?H3C—H base H3C

- this doesn’t happen because pka ~ 50 for the hydrogens attachedto the carbon in the methyl, therefore it is hard to take away theproton

But in reality, we don’t need the full negative charge in the Carbon for it to be sufficientlynucleophilic.

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 IHow to use the Grignard reactant? δ- δ+ O- OH 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 strongreagents, 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 donereacting.

Grignard Reactions: Aldehydes/Ketones vs. Esters

Ketone: (or aldehyde, but we’re using a ketone here) δ- δ+ O- OH 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+ O- OCH3 1) >—MgBr * the reaction does not end here! It keeps going!O== δ+ Ph—C— CH2CH3CH3

Ph 2) H3O+

The ester and amide undergo substitution, but when it changes into a ketone, it undergoes an additionprocess and converts into the alcohol

An attack on the ester/amide is SLOWER than the attack on aldehyde/ketoneALKYNE ANION REACTIONS

~ these reactions are a method of adding more than one carbon at a time, wheareas the Grignardreactions 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: CH3 CH3

H3O+ R—C==C- : + O== R—C==C—C—O- R—C==C—C—OH

CH3 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

Nucleophile attackscarbonyl carbon

Eject leavinggroup

There is more Grignard inthe solution, so the newcarbonyl reacts with moreGrignard reactant

No leaving group toeject so addition occurs

~ biggest drawback: if you wanted to onl one add carbon, it is dificult because it’s hard to separate theC==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 makea 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

carbono We need to use a metal that is less electronegative than hydrogeno 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

• we could have used NaBH4 as a proton source but it’s concentration is low compared to thesolvent, 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 isless 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 functionalgroup and leaves the other alone

If we want to react with less reactive carbonyls we have to fine something stronger thanNaBH4.

Net additionoccurs becausenone of thesubstituents are aleaving group

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 furtherdown 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 O H OH 1) LiAlH4

R O- 2) H3O+ R H 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 acido Protonation of a carbonyl requires the strength to be H3O+ or stronger

• Basicity of the nucleophileProtonation may not always be required

• Some nucleophiles like NaBH4, LiAlH4, alkyne anions, organolithiums, and Grignard reactantsare strong bases and will not do well in the presence of a strong acid, therefore, in the presenceof 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 electrophilicityo 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

the reaction is slower without H3O+ because the protonation of the carbonyl oxygen increases theelectrophilicity of the carbonyl carbon, making it easier for the water to attack it

Enzymatic Catalysis catalyst = enzyme• Enzymes catalyze by

o Providing proper orientationo 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 atthe carboxyl end of the phenylalanine (Phe)

H3O+

• No Chymotrypsin = VERY slow reactionWhat does Chymotrypsin do?• Mechanism occurs at the active site• First the imidazole deprotonates the OH

The imidazole deprotonates theOH

Start: the nucleophilic attack onthe carbonyl

Tetrahedral intermediatefragments

Hydrogen bonds stabilizethe transition state

Both acid catalysis andbase catalysis occur inareas near the yellowcircles

Tetrahedral intermediateforms

Hydrogen bonds stillstabilizing the transition state

So Chymotrypsin1) enhances electrophile2) enhances nucleophile3) makes new C—C bonds