2 - ALDEHYDES AND KETONES
2 - ALDEHYDES, KETONES AND DERIVATIVES Carbonyl chemistry is one of the most important areas of organic chemistry. You have been introduced to the chemistry of carboxylic acids in CHEM*2700, and in this course we will focus on aldehydes and ketones. Hence the compounds that we will study in this chapter will possess a carbonyl group, which has either hydrogens or carbons attached to it.
O
O
O
C
C
R
carbonyl group
R
R
C
H
aldehyde
ketone (R H)
Some significant aldehydes and ketones O H
O
O
H
H
cinnamaldehyde (candy, f ood, drugs)
HO raspberry ketone
O
O helminthosporal (a f ungal toxin)
camphor (liniments, inhalants)
O
O
O
O (-)- enantiomer, spearmint (+)- enantiomer, caraway seed (candy, toothpaste)
biacetyl mar garine) (
muscone (perf umes)
O
OH
O H
H
H
H
O testosterone ma ( le sex hormone)
O
H
progesterone (f emale sex hormone)
CHEM*3750 SCHWAN COURSE NOTES F13
HO O vanillin (f ood flavouring)
Chapter 2 Page 1
The reactivity of aldehydes and ketones is based on two important reactions of them. These species are prone to nucleophilic attack at the carbonyl carbon and one can make a nucleophile on a carbon to the carbonyl group. -
O R
C
R
Nu O C R R
+ Nu -
RCH2
O C
O
base
- C RCH R
R
This chapter will first demonstrate some synthetic routes to aldehydes and ketones, and then will describe some of their vast chemistry based on the two principal modes of reaction shown above. 1. Synthetic Routes to Aldehydes and Ketones (SF10 16.2, 16.4, 16.5) One of the main methods for the preparation of aldehydes and ketones is oxidation of an alcohol as shown in a general sense below.
RCH2OH
O
[O] R
OH R CH R'
H
O
[O] R
R'
The following are some popular routes to aldehydes and ketones and of course include oxidation of an alcohol. Most should be familiar to you. a) Aldehydes Me Me
CHCH2CH2CH2OH
PCC CH2Cl2
Me Me
primary alcohol
O CHCH2CH2 C H aldehyde, 75%
PCC = N+ H ClCrO3 pyridinium chlorochromate
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 2
b) Aldehydes and Ketones O3, CH2Cl2
Zn
-78 °C
HOAc
O H + CH 2O
62%
c) Ketones
Na 2Cr2O7 H2O, H 2SO4
OH menthol
O Ph
menthone, 84%
OH
1. PhMgBr
H Ph
+
H
O
2. H 3O
O
[O]
Ph
Ph
Ph
Grignard chemistry followed by oxidation 1.
OMe
O
OMe
O O
, AlCl 3
93%
2. HCl, H 2O O
Cl O
+ Et 2CuLi
CHEM*3750 SCHWAN COURSE NOTES F13
-78 °C ether
O
Chapter 2 Page 3
A useful method that has selected limitations is the hydration of alkynes. This is achieved with the assistance of mercuric ion Hg+2. When an alkyne is treated with HgSO4/H2SO4/H2O, the elements of water will add across the triple bond. In its simplest form the reaction looks like this:
H C C H
HgSO 4/H2SO4 H2O
H
O C CH3
A limitation arises when one considers the asymmetrically substituted alkyne: one can expect two products!
CH3CH2 C C CH2CH2CH2CH3
HgSO 4/H2SO4
O CH3CH2 CH2 C CH2CH2CH2CH3
H2O +
CH3CH2
O C CH2 CH2CH2CH2CH3
However, if one considers the mechanism of the reaction, which begins with Hg+2 coordinating to the triple bond and creating some cationic character on the triple bond carbons, then water attacks one of the carbons. In the above case, the two alkyl groups can stabilize the cationic charge in a comparable manner, leading to water attack at each carbon and two major products. CH3CH2 C C CH2CH2CH2CH3
Hg
+2
+2
CH3CH2
Hg C+ C+ CH2CH2CH2CH3
OH2
2 products
This mechanistic information does carry within it some synthetic value. What if the two groups on either side of the triple bond differ significantly in their ability to stabilize positive charge? Water will attack at the site bearing more positive charge and that carbon ends up being the site bearing the carbonyl oxygen. Hence terminal alkynes will always be a source of methyl ketones.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 4
+2
H
C C CH2CH2CH2CH3
Hg
+2
H
Hg C+ C+ CH2CH2CH2CH3
which of these carbons bears bear more +ve charge? The more substituted does, according to Markovnikov
O CH3 C CH2CH2CH2CH3 lone product
example:
PhCH2 C C H
HgSO4, H2SO4 H2O
O PhCH2 C CH3
Of course, when the alkyne is symmetrically substituted, then a single product results
CH3CH2 C C CH2CH3
HgSO 4/H2SO4 H2O
O CH3CH2 CH2 C CH2CH3
More examples: O
O +
Cl
AlCl3
CrO3 OH MgBr +
O H
CHEM*3750 SCHWAN COURSE NOTES F13
quench with H2O
Chapter 2 Page 5
O O
HO
O
PCC
O
OH
CrO3
H O
O
+
H
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 6
2. Acidity and Enolization of Aldehydes and Ketones (SF10 3.5-3.7, 18.1, 18.2, 18.3) As stated, one of the origins of the reactivity of aldehydes and ketones is their inherent acidity, which is based on the electronic properties of the carbonyl group. The carbonyl group is an electron-withdrawing group by both induction and resonance. So, hydrogens to the group possess increased acidity simply due to their proximity and the incipient anion can be stabilized through -resonance with the carbonyl group. O
O base
H acidic hydrogen, pKa = ca. 19-20
O-
resonance structures
There is a geometric requirement for facile proton removal in this chemistry. The C-H bond that holds the removable hydrogen must be aligned with C O H the -orbitals of the C=O bond. The required conformation is trivial to achieve in a freely rotating H system, but may not be accessible in cyclic or all aligned, so H is constrained molecules. Without the proper orbital ideal for removal overlap, then the -resonance stabilization of the anion is not possible. The result is that the acidity of the hydrogen is greatly reduced. Bases suitable for deprotonation of an aldehyde or ketone must meet two principal requirements: a) The base must be strong enough to remove the hydrogen. In this regard, any base whose conjugate acid has a pKa greater than 20 will be suitable. Sometimes a weaker base will work, but the reaction must proceed under equilibrium deprotonation conditions. b) The base has to perform selective attack at hydrogen and cannot have properties that will promote nucleophilic attack at the carbonyl carbon. A sterically hindered base usually provides the desired chemoselectivity. LDA, lithium diisopropylamide ([Me2HC]2N-Li+) is often the base of choice. The resonance-stabilized anion that results from these deprotonation reactions is called an enolate, which means that it is the anion of an acid called an enol.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 7
O-
OH
enolate
enol
These species are of course interconvertible through the transfer of one proton. As we will see shortly, the enolate and the enol can be functionalized at the -carbon. Hence one can use an enol or an enolate for derivatizing the compound next to its carbonyl group. Enols can be generated through a reversible acid catalyzed reaction.
O H
+ H O
+
O
H
O
H
+ H
H
H
+
+ H 3O base (= H 2O)
H is much more acidic since the protonated carbonyl is a much better electron sink The methods of forming enols and enolates are fully reversible. So the carbonyl compound can be regenerated from an enol by simply adding acid and isolating the substrate (except in specialized instances). Attempted isolation of the enol from acid solution will provide the aldehyde or ketone. Mechanistically, regeneration of the carbonyl compound can be the reverse of enolate formation, or may involve the enol. -
O
H
+
O
H
O
H Under these conditions, the keto form of the compound is regenerated upon isolation. The equilibrium between enol and keto shown above is called tautomerism, since the compounds are tautomers of one another. The carbonyl compound is usually the thermodynamically favoured form, based on the carbonyl bond strength, but there are exceptions to this rule. A number of equilibrium CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 8
constants have been determined for keto-enol tautomerism. They are calculated as shown, and are of course dependent on the medium in which the measurement is made. If the substrate is neat, the tautomer ratio depends on the origin of the material.
[enol] [keto]
Keq =
The Keq (keto-enol) for acetone in water is 1.5 X 10-7. It should be noted that despite the fact that the ratio can significantly favor the keto form, reactions selective for the enol form can still take place, as we shall see. The relative acidity of the hydrogens next to the carbonyl groups of aldehyde or ketones has both advantages and disadvantages. For one advantage, the acidity allows for deuterium incorporation into the molecule. The protocol involves simple addition of the carbonyl compound to D2O containing a trace amount of acid or base. Any hydrogens in the -position that can adopt the required conformation for enol or enolate formation will become fully deuterated, given sufficient time.
O
O CH3
CH3
-
D2O, OD
D D
H H O (Me)2CH
CH3
CD3
O
+
D2O, D 3O
(Me)2CD
H
H
Another consequence of enol formation, having negative and positive implications, is the ease with which chiral -carbons can invert their stereochemistry. Hence if you have taken the time to make an important, optically pure product possessing stereochemistry at the -carbon, exposure to acid or bases may racemizes the material. For example, optically active 3-phenyl-2-butanone in basic ethanol (r.t.) racemizes within minutes. O Ph
-
CH3
H CH3 (S)-3-Phenyl2-butanone
EtO /EtOH
Ph H3C
-
O
CH3
Achiral, planar
-
EtO /EtOH
O Ph
CH3
H3C H (R)-3-Phenyl2-butanone
isolation racemic ketone
There are instances of advantageous isomerism of -carbon, through their planar form. CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 9
- +
O
O
OK CH2CH=CH2
10% KOH
CH2CH=CH2
CH2CH=CH2
EtOH CH3
CH3
CH3
In the anionic form, the -carbon is planar and in theory can be protonated from either face. However, the configuration of the -carbon, which bears the methyl group, is a constant and that substituent creates a bias in the reprotonation reaction. The allyl and methyl substituents prefer to be trans to one another for steric reasons and trans isomer is produced in >95% yield under these thermodynamic conditions.
The Schwan invaluable mechanism rules. 1. Do not rush mechanistic steps together. 2. Use only reagents and data that have been provided. 3. In aq acid, for every step of the mechanism, the nucleophile is neutral and the electrophile is acidic (e.g., protonated) where possible for enhanced reactivity. It follows that there are no negatively charged entities in acidic solution. 4. In aq base, for every step of the mechanisms, the nucleophile is basic and the electrophile is neutral. It follows that there are no positively charged entities in basic solution. 5. Make sure each mechanistic step is balanced: charge, atoms, electrons 6. Electrons begin every step. 7. Water and alcohols will not do SN2-like substitutions at carbon. 8. In alcohols or water, always use solvent for proton transfers. 9. Know as many pKa’s as possible.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 10
3. Halogenation of Ketones and Aldehydes (SF10 18.3) The following general reaction is the topic of this section.
O +
H
O
acid or
X2
X
base
As noted, halogenation of these carbonyl compounds may proceed under acidic or basic conditions. Under either of these conditions the mechanism for generation of the enol or enolate is the same as previous. The reactive compound then attacks the halogen rather than any other electrophile such as a proton.
O-
via enolate
O
X X
X
-H H
via enol
+
+O H
O
X X
X
Rate studies have shown that initial rates of the halogenation are independent of halogen concentration. This is interpreted to mean that introduction of halogen into the substrate is not rate determining. Furthermore, experiments have shown that the rate is dependent on concentration of carbonyl compound and acid (or base). From this evidence, it is generally believed that enol or enolate formation is rate determining.
O
O Cl2, H
+
70 °C
Cl 85%
Sometimes as shown below, the reaction is autocatalytic. That is, it is very slow until some halogenation occurs which produces acid as a byproduct. The rate of halogenation then accelerates due to the presence of acid. Hence the term autocatalytic, since the reaction provides the means to promote itself.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 11
O
O +
MeOH
Br2
Br
+
HBr
70% Under basic conditions, multiple halogenation is often a problem and therefore basic conditions are not recommended if one is striving for only the monohalogenated product. Note that this reactive character can be used to one’s advantage, as shown below. For the introduction of only one halogen, the reaction is usually performed under acidic conditions since introduction of one halogen into the molecule slows further reaction. For synthetic efficiency, monohalogenation reaction must be performed under acidic conditions.
O
H
+
O
H
+
X
H
In the halogenated form, the electron withdrawing nature of the halogen retards the initial step of enol formation and hence slows the introduction of more halogens.
halogen pulls electron density away from the carbonyl group
Multiple halogenation of methyl ketones provides chemistry that has been used as a functional group identification technique, although spectroscopic methods are now more popular. Multiple halogenation of methyl ketones leads to the haloform reaction and iodine is the more common halogen employed (iodoform reaction). The mechanism involves triple iodination of the methyl group of the ketone under basic conditions.
O R
base I2
O
O R
I
base I2 (2X)
I
R I
I
Then, hydroxide attacks at the carbonyl group to make a tetrahedral intermediate which expels -CI3.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 12
Complete mechanism:
O
-
H
R
-
O OH
-
H
I
-
I
HO
O
OH
R I
I
I
I
H
H
-
OH
I O
I I
I
R
I
I
-
OH
-
R
O
-
O
I
I
I
H
R
O
R
I
I
H
O
R
I
H
R
H
H
O
R
+ OH
HO
-no H's remain; -carbonyl is most electrophilic
-
O R
-
O
I I C I H2O
I I C H I yellow precipitate
Interaction with the solvent provides the products: iodoform and carboxylate. -CI3 is a good leaving group in this case because the three electronegative halogens help to stabilize the negative charge on the carbon atom. The electronegativity also accounts for the multiple halogenations at a single site. Once one halogen is introduced, the remaining hydrogens on that same carbon atom have enhanced acidity. Acidic workup affords the carboxylic acid if desired as opposed to the carboxylate. The iodoform is a bright yellow precipitate that serves as a useful indicator. If a compound is suspected of being a methyl ketone, simply add excess hydroxide and I2 and look for a yellow precipitate for confirmation of the suspected structure. The reaction can also be synthetically useful in that it can be used to achieve the overall conversion of methyl ketone to carboxylic acid.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 13
O
1. Cl2/NaOH
O OH
2. HCl/H 2O 5-methyl-3-hexen-2-one
+ CHCl3
5-methyl-2-pentenoic acid
chloroform
PRACTICE PROBLEMS 3 You can now do Questions 12.4, 16.47, 18.1-18.5 18.20a & 18.33 in SF10 1. Which of the following compounds will give a positive iodoform test. a) 2-pentanone b) 3-pentanone c) hexanal d) cyclohexanone e) 2-methylcyclohexanone f) cyclohexyl methyl ketone 2. For each of the following compounds write reaction equations showing how the compound could be prepared by oxidation of an alcohol, ozonolysis of an alkene and (if possible) hydration of an alkyne. Which of the preparations that you have written would give an unacceptable mixture of isomers? a) b) c) d)
pentanal 3-hexanone 3-heptanone 2-methylbutanal
3. a) Suggest how you could prepare methyl 2-phenylethyl ketone from: i) an alkyne ii) 2-bromoethyl benzene b) Draw the products of the iodoform reaction on methyl 2-phenylethyl ketone. 4. Suggest two sets of reagents that will effect the following reaction.
Cl
O CH3
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 14
SOLUTIONS TO
PRACTICE PROBLEMS 3
Q1. a) 2-pentanone YES b) 3-pentanone NO c) hexanal NO d) cyclohexanone NO e) 2-methylcyclohexanone NO f) cyclohexyl methyl ketone
YES
Q2. a) pentanal PCC
CH2OH H
O C
H
cis or trans isomer
Hg H
O
2
1. O3 2. Zn
C
H
O
+2
H2SO4 H 2O
H
+
O exclusive product
C
H
not likely due to Markovnikov's rule
b) 3-hexanone OH
CrO3 H
O
Hg
+2
H2SO4 H 2O
+
1. O3 2. Zn
cis or trans isomer
c) 3-heptanone
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 15
OH
O
CrO3 H
1. O3
+
2. Zn
Hg
cis or trans isomer
O
+2
+
H2SO4 H 2O
O 50:50 mixture, therefore unacceptable
d) 2-methylbutanal
O
CrO3
HOCH2
H
+
H
1. O3
H
2. Zn
H
cis or trans isomer -use of Hg+2/H2O/H+ on an alkyne is not possible for same reason as in a) above. Q3 a) Hg Ph
Ph
Br
O
+2
Ph
H2SO4 H 2O
Mg ether
CrO3/H
MgBr
Ph
H OH
1. CH3CHO 2. HCl/H2O
+
Ph
b)
O Ph
O
-
1. xs OH I2 + 2. mild H/H2O
Ph
OH
+ CHI3
I I
Q4.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 16
Cl
MgCl
Mg ether
1. CO2 2. HCl/H2O
1. CH3CHO 2. HCl/H2O H
O
CH3
OH CrO3/H
SOCl2 O
OH
+
O Me2CuLi
CH3
Cl
_______________________________________________________ 4. Alkylation Reactions and Enamines (SF10 18.4; 18.9) Generating an enolate from a ketone or aldehyde and quenching it with an alkylating agent affords a method for the synthesis of compounds bearing an alkyl substituent next to the carbonyl group. Simple reactions can occur in some cases. O
O
1. NaH, C 6H6 2. Me 2C=CHCH2Br
CH2CH=CMe2 88%
-
via
O
Br
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 17
-
O
O Li
+
O
LDA
MeI
cold
warm to r.t.
EtO
EtO
Me EtO 93%
+ HN( iPr)2
There can be problems associated with this simple approach. One is that there are two sites for deprotonation when the compound bears -hydrogens on either side of the carbonyl group. For instance, which one of the three (yes three) O different hydrogens on methyl cyclohexanone is most acidic? It can H3 H CH H be challenging to create and maintain a single enolate. Moreover, here can also be problems with multiple H H H H alkylations, depending on the substrate and reaction conditions, because sometimes there are unwanted proton transfers that give a different enolate, even during the course of the reaction. Although some solutions have been established and will be discussed later in the chapter, the problems led to the discovery of alternative protocols to achieve simple efficient alkylation. It is useful to present these as they are still in use today. Uncontrolled alkylations can lead up to five different products, in addition to unreacted starting material!! O
O KOH
O +
Me
MeI O Me
O Me Me
+
O
O + Me e Me M
+
Me
+ Me Me
Me Me
One useful method for overcoming some of the difficulties involves the chemistry of enamines. As their name indicates enamines contain a double bond (ene) and an amine and the name is applied to systems where the two are in conjugation.
N
CHEM*3750 SCHWAN COURSE NOTES F13
-
+
N
Chapter 2 Page 18
Gilbert Stork of Columbia University developed the chemistry shown here and it still bears his name. Enamines are prepared by condensing a secondary amine with a ketone or an aldehyde. If the amine is a primary amine the result is an imine, which is actually the more thermodynamically stable form of an imine/enamine tautomeric equilibrium.
O H
+ R NH R'
R
+
N
R'
(R' = H)
N
R
-H2O enamine
imine
The following mechanism accounts for the enamine formation.
O
+
H /toluene
H
+ H O +
NH
HO
NH
+ N
O
O
O NH O
O H HO+ N
N+
H 2O +
O
O HO
N
+
H
iminium ion
NH
H N + H
O
note with 1 re amines
O
O O
EtNH 2 +
N enamine
N -H2O
via deprotonation of the N at the iminium ion stage
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 19
Water must be driven from the reaction vessel in order to force the equilibrium reaction to completion. This is usually achieved through the use of a Dean Stark apparatus:
evaporated compounds condense, fall into graduated tube and separate. Water stays on lower layer and never return to the flask while the benzene (or toluene) rises to the level where it runs back into the reaction vessel.
reaction mixture of benzene or toluene and maybe water, which also is created by the reaction
As shown above the enamine has nucleophilicity at carbon and it is this atom that is preferentially functionalized in a simple SN2 reaction.
X
X
N
N+ I-
SN2
CH3
CH3 I +
usually X = -CH2CH2- (pyrrolidine) -CH2CH2CH2- (piperidine) -CH2OCH2- (morpholine)
H 3O O
CH3
In this reaction the -carbon of the enamine ( -carbon of the ketone) bears the new substituent. To remove the nitrogen auxiliary and recover the new ketone, the iminium salt is hydrolyzed under aqueous acid conditions. The amine is lost in the acidic medium and can be recycled if desired. This method is preferred over simple deprotonation and alkylation since it minimizes double or multiple alkylation. The
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 20
reaction occurs best with reactive halo compounds such as methyl iodide, benzyl halides, allyl halides, -halocarbonyl compounds and acyl halides. Below is an example of the overall reaction sequence.
O
1. pyrrolidine, H H
O
+
67%
H
2. BrCH2CH3
CH2CH3
+
3. H , H 2O Other examples:
N
1.
O
O Br
O
O
O
+
2. H 3O
N
O
O
1.
O
Cl +
2. H 3O
5. The Aldol Condensation (SF10 19.4-19.5) To this point we have examined the reactions of enols and enolates with a number of reactive electrophiles. One can also react them with the very substrates that provide enols and enolates: CARBONYL COMPOUNDS. Under the proper conditions the carbonyl group is sufficiently reactive to accept electron density from an enol or enolate. The archetypal example in many texts is the base catalyzed reaction of two molecules of acetaldehyde. The name aldol originates from the presence of an aldehyde and an alcohol in the product molecule. This name is used even if ketones are involved in the chemistry.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 21
OH
O 2
10% NaOH, H 2O H
CH3
5 °C
portion that acted as an electrophile
CH3
H
O CH2
H
portion that was enolate
The mechanism for the reaction is straightforward and is the basis of many other reactions yet to be introduced in this chapter. It begins with the reversible generation of an enolate. The enolate then reacts as indicated. Note that all the steps are reversible and sometimes aldol condensations are difficult to complete because of a propensity to revert to starting materials.
CH3
-
-
O
O
O H
H2C
H
CH3
H
O CH2
H
H 2O OH CH3
H
O CH2
H
Sometimes the aldol product will lose water spontaneously to afford a double bond. The conjugation of the double bond with the carbonyl group is the driving force in the dehydration. This can often be achieved simply by heating the reaction mixture during or after the bond forming process. Sometimes making the solution acidic will achieve the same purpose. The term condensation arises from the loss-of-water step. In synthetic chemistry, condensation means: the loss of water or its equivalent. Sometimes when the equilibrium constant for aldol formation is small, the reaction can still be driven by pushing the product through to the dehydrated product, the formation of which proceeds efficiently and is less reversible. Basic conditions:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 22
OH CH3
O
H
O
H
CH
H
H -
CH
CH3
H
+
-
OH
+ H 2O
OH
Often, if the elimination of water does not proceed efficiently under basic condition, one converts the mixture to an acidic one. Under acidic conditions, loss of water is an easier process. Acidic conditions: + H OH2 OH C H3
H
O C H2
+ OH2 CH3
H
H
O
CH H
O
H H OH2
C H3
CH
H
+ H 2O
The whole aldol condensation can also be performed under acidic conditions, where an enol rather than an enolate is the nucleophile. The acid initially induces enol formation by the means shown earlier in this chapter. This enol is a weak nucleophile and will only undergo aldol chemistry when the electron accepting carbonyl group has been protonated. Mechanism for acidic aldol reaction with loss of water:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 23
O CH3
H
+ H O
+
CH3
CH3
OH H CH3
+ H O
OH CH3
CH2
CH3
CH2 CH3
O
+
CH3 enol formation
CH3
OH
H 2O CH3
CH3
O
CH2 CH3
CH3
acidic dehydration as shown earlier CH3 CH3
O
CH
CH3
It is very difficult to isolate the true aldol product under acidic conditions. The reaction usually carries through to the unsaturated material. Both ketones and aldehydes can undergo the aldol condensation. Aldehydes are more reactive since nucleophilic attack at the ketone carbonyl is more sterically hindered. Also, the extra electron donating substituent makes the ketone less electrophilic. Aldol condensations as shown above with two molecules of acetaldehyde are the simplest possible examples. Complications quickly arise when two different aldehydes are used, or when ketones with two sets of -hydrogens are employed. For instance with two different aldehydes, four products are available. The list includes two compounds from self-condensation and two different products from a crossed aldol reaction. OH
O R + O R'
R H
H
R'
H R
+
+
R
O
OH H
R'
CHEM*3750 SCHWAN COURSE NOTES F13
+
O
R
OH
H
OH
O
+
R'
O H
R'
Chapter 2 Page 24
Selectivity can be achieved by choosing one reactant that does not have hydrogens or has -hydrogens but they are of comparatively reduced acidity.
O
O
CHO
OH
NaOH, H 2O
+
25 °C, 4 hr. 100 %
O
CHO
CHO
NaOH, H 2O
+ CH 3CH2CHO
O
CH
C
CH3
72% O CH3
+ (CH3)3C CHO
NaOH, H 2O
O CH
CHC(CH3)3
81% Referring back to page 16, it was mentioned that sometimes one can prepare and maintain a single enolate anion, even when are multiple -hydrogens in the system. Specifically, it may be possible to selectively generate a particular enolate and then quench that enolate with another carbonyl compound (or other electrophile such as an alkylating agent). To begin it is important to know the different between kinetic and thermodynamic enolates. The thermodynamic enolate is the more stable by definition, and represents in this case, the more substituted enolate. This corresponds to the more substituted alkene of two isomers being the more stable and the concept applies in this case, since the enolate is mostly populated by the alkene resonance structure. It follows that the kinetic enolate is the less substituted of two options which can be accessed under mild conditions that make use of the steric differences of the two -hydrogen bearing sites. The best conditions for kinetic deprotonation are: LDA, -78 C in THF. Generally one can expect preferred deprotonation from the 1 over 2 or 3 sites and from the 2 over 3 sites. The reaction is complete and irreversible. See the reaction below. Access to the thermodynamic enolate can be more challenging (e.g., less reliable), but suitable conditions might be Et3N in DMF. DMF = HC(O)NMe2. As indicated, LDA is a reliable method for the complete and irreversible formation of a lithium enolate. With asymmetric ketones, LDA will remove a proton from the least CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 25
sterically hindered position. This is generally believed to be the best method for carrying out crossed aldol condensations. The protocol lends a good amount of certainty to predicting the product in a crossed aldol. O CH3
-
O Li
LDA, THF -78 °C
CH2CH3
CH2
+
CH2CH3 CH3CHO
CH3
-
O
OH CH2
O Li
H 2O CH3
CH2CH3
+
O
CH2
CH2CH3
If a given substrate possesses two carbonyl groups, it is possible for the compound to undergo an intramolecular aldol condensation. This form of the reaction is only suitable for the synthesis of 5, 6, and 7 membered rings. There are some rules that can be applied at this stage. One is a reminder that aldehydes are more electrophilic than ketones. The other is that 5 membered rings will form more readily that 7membered rings while 6-membered rings are the best. Recall that aldol condensations are reversible and there is opportunity to form the most thermodynamically stable product. Intramolecular aldol condensations will virtually always proceed through with loss of water to form the -unsaturated ketone or aldehyde. When deciding the product of this cyclization, an important step in the analysis is determining when water can be readily lost from the aldol. If not, aldol formation often reverses itself.
O
O
O NaOH, H 2O
not
100 °C
O
O O
O NaOH, H 2O 100 °C
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 26
Before we leave the aldol condensation, the role of the carbonyl group in this chemistry should be emphasized. The polarity and resonance ability of the carbonyl group enhances the acidity of the hydrogens to it, allowing for the formation of enolates. That creates the nucleophilic component of the two reactants. For the electrophilic component the carbonyl group is polarized such that the carbon can accept attack by a nucleophile and the oxygen can hold the negative charge. 6. Other Related Condensation Reactions Whereas the Aldol condensation involves the reaction of an enolate of an aldehyde or ketone with another aldehyde or ketone, the Claisen Condensation (SF10 19.2) involves the reaction of an enolate of an ester and a carboxylic acid derivative, usually another ester. One can view the initial steps of the Claisen condensation as analogous to ketone chemistry. Enolates of esters are less acidic than enolates of ketones or aldehydes and can therefore undergo chemistry with less reactive compounds. The archetypal example in this condensation involves the reaction of two molecules of ethyl acetate induced by ethoxide ion. The mechanism of the reaction is shown below. O CH2
-
O
O Na
CH2CH3 CH2
H -
CH3
O
CH2
EtO
-
O
O CH3
-EtO
-
-ketoester
O
CH2CH3
CH3CH2
O Na CH2
O
+
O CH2CH3 CH3
tetrahedral intermediate
-
O CH3CH2
O
CH2CH3
OEt
O CH3CH2
O
+
O
O
O H H
CH3CH2 CH3
O -
O
CH3
H +
H 3O O CH3CH2
O
O H H
CHEM*3750 SCHWAN COURSE NOTES F13
CH3
Chapter 2 Page 27
Although the scheme above shows the -ketoester as the product, it should be realized that under the reaction conditions, the material actually rests as the deprotonated form until acted upon by the addition of acid, which returns the hydrogen. That is, the Claisen reaction mixture must be quenched with acid before isolation of the product. Acetic acid, citric acid or aq. ammonium chloride are often employed. Note that anions derived from the -ketoester have two carbonyl groups available for conjugation. Hence the pKa of the -ketoester is much lower than that of a simple ketone or ester, since anion stabilization is offer by both of these carbonyl groups in a single molecule. Hence, the -ketoester is readily deprotonated by ethoxide. -
O CH3CH2
O CH3CH2
O
O CH2
-
EtO Na CH3
O
O CH
CH3
O
+
CH3CH2
O
O CH
CH3
-
O CH3CH2
O
O CH
CH3
The deprotonation step as shown above is key to the completion of the reaction. The -ketoester anion is essentially inert and furthermore, in anionic form, the species is captured and frozen and cannot succumb to the reversibility of the reaction. The overall reaction only works well when there are 2 or 3 hydrogens on the starting ester. The ketoester drawn above is known by the common name of ethyl acetoacetate and hence the self-condensation of esters is sometimes known as the acetoacetate ester condensation. Example:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 28
-
O
2
1. MeO Na O
Me
2. H
O
+
O Me
O
+
Crossed Claisen condensations are possible when one of the esters does not possess -hydrogens, much like the ideal situation for crossed aldol reactions. O
O O
O
O
+
O
O
O diethyl oxalate
diethyl succinate -
CH3CH3O Na
O
+
O
O
toluene
O
O O
O +
O
90%
-
1. CH3CH3O Na
O O
O
O
+
O
ethanol + 2. H 71%
ethyl benzoate
O
O H
+ O
ethyl formate
CH3
O
-
1. CH3CH3O Na O
ethanol + 2. H
O
+
O
H
O 80%
Note that in each of the three examples above, one of the reacting esters does not contain -hydrogens. Carbonate esters, which also lack -hydrogens, will successfully partake in crossed Claisen reactions. Another reaction that simplifies potentially complicated reactions is the Reformatsky reaction. It begins with an -halo ester * and uses metallic zinc to establish *
-Halo esters can be prepared by the Hell-Volhard-Zellinsky reaction whereby a carboxylic acid is treated with molecular halogen and elemental phosphorus and then water. See SF10 839-841 for details of this preparation. The acid can then simply be esterified.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 29
an ester enolate. In this reaction, the halo ester and zinc are mixed together to create a solution containing a zinc enolate, to which the ketone or aldehyde is added. As with aldol condensations, the product of Reformatsky reactions can be readily converted to the unsaturated material by treatment with acid. This achieves the dehydration process. O Zn
O Br
+
-
O
O (ZnBr)
- + O (ZnBr)
O
CHO
OH
O O 60%
O Br
1. Zn, toluene O
O
2. 3. H
OH O
O
70%
+
If two ester groups are in the same molecule and are separated by 4 or 5 carbon atoms, then one can achieve an intramolecular Claisen condensation. This reaction is called the Dieckmann (SF10 873) condensation. Again there is an archetypal system that exemplifies the reaction.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 30
O
-
H H
O
OMe
O
O
H O
O
-
O
O
O -
O
-
MeO
O
H
O
-
O
MeO O O -
-MeO
O
O
anion, quench w/ acid to obtain product As with the Claisen condensation, the equilibrium is frozen at the desired product by deprotonation of the acidic hydrogen between the carbonyl groups. The Dieckmann condensation is only useful for 5 and 6 membered rings. O CO2Et CO2Et O
EtO
-
O
CHEM*3750 SCHWAN COURSE NOTES F13
H
H
CO2Et
+
O
O
Chapter 2 Page 31
A form of the Claisen Condensation is very prevalent in biological systems: it is vital to the construction of many naturally occurring and biologically important chemicals. In this case the carboxylic acid being attacked is a thiolester (textbooks will tell you thioester), and the sulfur and the remainder of coenzyme A is lost in a potentially reversible step. The nucleophile is not exactly the enolate of an ester, but one that bears an extra -CO2unit. The release of CO2, concurrent with nucleophilic attack, assists the Claisen Condensation over its kinetic barrier. O -
O
O CH2
-
O
O
+ SCoA
O
CH3 CH2 CoAS
H3C
-CO2 SCoA
H3C
SCoA
CH3CH2CH2
SCoA
SCoA steroidal hormones
O more repetitions
CH2
O
reduce, dehydrate, reduce
O
O
-
- SCoA
CH3(CH2CH2)n
SCoA
bile acids terpenes
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 32
PRACTICE PROBLEMS 4 You are capable of doing Questions 18.14, 18.18, 18.19a,b 19.23, 19.1-19.7, 19.1119.12, 19.17-19.18, 19.23-19.30, 19.33-19.34, 19.36, 19.37a-c, 19.38 in SF10.
1. Give the expected products for the following acid catalyzed reactions. a) acetophenone + methylamine b) acetophenone + dimethylamine c) cyclohexanone + aniline (phenylamine) d) cyclohexanone + piperidine 2. Suggest a mechanism, using curly arrows, for the dehydration of the self-aldol condensation product of acetophenone by: a) acid-catalysis
b) base catalysis
3. Draw all the possible aldol condensation combinations if propionaldehyde and methyl ethyl ketone are mixed together with aqueous base.
4. The accompanying experimental observation was made. Draw a mechanism that uses curly arrows and accounts for the transformation.
-
+
O Li O
O + H
MeO -
THF -70
O +
MeO
0°C
+
O Li O
H
5. What reagents would be required to carry out the following series of reactions.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 33
O
O H
H
O H
O H
-
O H
H
OEt H
6. The accompanying compound can be prepared in high yield using a mixed aldol condensation. What starting materials are required?
O Ph
Ph
Ph
Ph
7. a) Draw the key intermediates, or a mechanism showing how the two starting materials can come together using LDA to make the alcohol. Hint: learn the pKa of a typical nitro alkane. b) What reagent would you use to oxidize alcohol to decalin A? c) What would you use isomerize decalin A to decalin B? Why would this reaction occur? O
O H
O
+
using LDA HO
OEt
NO2
O2N H
OH H
H
O
alcohol
OEt
oxidize to a ketone O2N O
H
OH
H H decalin B
O OEt
O2N how? O
OH H
H
H decalin A
O
OEt
FROM PREVIOUS MID-TERM EXAMINATIONS 8. The following two substrates undergo an acid catalyzed aldol condensation to afford a conjugated ketone. Draw the product of the reaction and provide a complete mechanism, including curly arrows, that accounts for its formation.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 34
O
O Ph
CH3
+
+
H
H3O /H2O
9. Considering the equilibrium shown below, address the following questions. a) Drawn a mechanism using curly arrows for the conversion of 1 to 2. Be sure to draw all important intermediates and important resonance structures. b) Which side of the equilibrium is preferred? Provide a very brief answer for your choice. O
OH +
H3O /H2O OH
O
2
1 SOLUTIONS TO
PRACTICE PROBLEMS 4
Q1. O acetophenone =
a)
NCH3
b)
H3C
N
CH3
c)
NPh
d) N
Q2.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 35
a)
O
+ H-OH2
OH Ph
Ph
O H +OH
OH2 Ph
Ph
O
Ph
Ph
H H
b)
O
OH
-
Ph
Ph
O
OH
Ph
Ph
H H
Q3. with propionaldehyde as Nu and ketone as electrophile
O H
O H OH
if elimination occurs
O H
with ketone as Nu and propionaldehyde as electrophil O O
H
if elimination H
OH
occurs
O
H
O O if elimination H HO
H
occurs O
H
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 36
O
if self reactions occurs, aldehyde on itself
H
O H H
H
if elimination occurs
OH
O H H
O
if self reactions occurs, ketone on itself
CH3
O CH3 if elimination CH3
OH
CH3
occurs
O CH3 CH3
O O if elimination CH3 HO
CH3
occurs O
CH3
Q4. Recall that the aldol is a reversible reaction
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 37
-
+
O Li O
+
reverse
MeO
aldol
MeO
Li
O
-
O
+ H
O
forward aldol
H -
+
O Li O
-this reaction goes to the more thermodynamically favoured side of the equilibrium; methoxybenzaldeh yde could be viewed as a more stabilized s.m. compared to benzaldehyde
Q5. O O 1. CH2O
H
O H -
H2SO4
H OH/H2O/
2. PCC O H
O -
O
KOH, Br2 H
EtOH/H
H
+
OEt H
+ CHBr3
Q6. O Ph
Ph + O
Ph
O -
O
OH
Ph
Ph
Ph
Ph
Ph
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 38
Q7. a)
add
O
Li+O-
LDA
O
O
O2N O
OEt
O H
O
O2N
aldol reaction
H
-
+
H O Li
OEt
OEt proton transf er, pKa of nitro alkane is about 10 and alcohol is 16
O2N H
HO
OH
O cyclization
H H
carbanion to carbonyl
OEt
O
O2N Li+ H OH
O OEt
b) PCC or CrO3 would work c) Base should isomerize the ester. The equatorial position is the thermodynamically more stable, so the reaction goes in the direction indicated.
Q8. + H-OH2
O Ph
+O H
CH3
Ph
OH
OH2
CH2
Ph
CH2
H O
+ H-OH2
+O
H
H
H H
O
H
O
OH2 Ph
H
O
H
+O
H Ph
elimination as per earlier in the solutions H
O Ph
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 39
-cyclopropyl H is not as acidic and in any case, it will not eliminate to the conjugated ketone Q9. OH + H -O H2 O
H OH +O
1
H
O H2 +O H
OH + H -O H2 O
OH
H
O H2 O
OH
2 is the m ore therm odynam ically stab becuase of the conjugation of the car with the arom atic. Thus the equilibriu toward2.
_______________________________________________________ 7. Synthetic Applications of Condensation Reactions (SF10 18.5-18.8) a) Acetoacetic Ester Synthesis This synthesis, named after the starting material employed is a valuable means of preparing substituted acetones (methyl ketones). The approach utilizes two key qualities of the starting material, MeC(O)CH2CO2Et. One is that the compound possess a CH2 group with low acidity that can be readily functionalized. The other is that the ester can be converted to an acid, which is then completely removed. The general equation for the overall transformation is as follows, where R = a non-hydrogen group and R' may be hydrogen or another group.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 40
O
O
O O
O
R
R'
O H
O
R
R'
substituted acetone In detail, the first reaction is similar to those you have seen before. One of the acidic hydrogens of the dicarbonyl compound is easily deprotonated by an alkoxide and the anion is captured at carbon by addition of an electrophile. Such chemistry will lead to a mono-substituted acetone. One can repeat the chemistry and introduce a second electrophile and the result would eventually be a di-substituted acetone. O H
O H
-
+
RO M
O
O
O O + H M
pKa = 10.7 -
+
OM O O H -
O
O
O
H
R'
RX
O
+
OM
O H
The next step is alkaline hydrolysis of the ethyl ester, a process called saponification. The solution is quenched with acid to allow for the isolation of the -keto acid. O
O
H
R'
1. NaOH, H2O O
2. HCl, H2O
O
O
H
R'
OH
The last step in the overall transformation has its origin in the structure of the keto acid. -Carbonyl carboxylic acids often adopt a conformation where the acid hydrogen can hydrogen bond in an intramolecular sense with the other carbonyl group. The hydrogen bonding interaction is idealized by the fact that the system forms a six membered ring.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 41
hydrogen bond O
O
H
R'
O
H
O
OH H
R'
O
The hydrogen bonded arrangement allows the -carbonyl carboxylic acids to undergo a thermally induced decarboxylation reaction. The arrows in the diagram below help to assign the re-positioning of the electrons. Only when the system form a 6membered cyclic form is the decarboxylation facilitated, as indicated by the arrows.
O H
H
O R'
O
The decarboxylation is effected by heating the substrate at 100 C. The immediate product of the carboxylation is actually an enol but upon isolation, the material tautomerizes to the ketone.
O H
O R'
OH
heat -CO2
HO
O R'
H
H H
R'
The same chemistry takes place if the target molecule is a di-substituted acetone. The only difference is that after the introduction of the first alkyl group and before the saponification, the substrate is deprotonated again and quenched again with another electrophile. Once the introduction of two groups between the carbonyl is complete then the conversion to acid and subsequent decarboxylation can proceed.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 42
Some examples: O
O
O
O
1. NaOEt, EtOH OEt
O
OEt
2. (Me) 2CHCH2Br
+
2. H 3. heat, -CO 2
O
O
O
1. NaOEt, EtOH O
O
1. NaOH, H 2O
O
O
2. Ph
O
Cl Ph
1. NaOH, H 2O, then H 2. heat, -CO 2
+
O Ph O Sometimes the acid and thermolysis treatments can be addressed simultaneously: O
O
O
1. NaOEt, EtOH O
O O
2. CH3CH2CH2CH2Br
1. NaOH, H 2O 2. H 2O, H 2SO4, heat O
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 43
O
O O
1. NaOEt EtOH
1. NaOEt EtOH
2. MeI
2.
O
O O
Br
1. NaOH, H 2O, then H 2. heat, -CO 2
+
O H
The same chemistry can be performed on a wide range of -keto esters. O
H 1. NaOEt, EtOH CO2Et
O
CH2CH2CH2Br CO2Et
2. excess Br(CH 2)3Br 1. saponification + 2. H 3. heat O
H CH2CH2CH2Br 60%
b) Malonic Ester Synthesis The steps required in the malonic ester synthesis are analogous to those of the acetoacetic ester synthesis. One difference is the starting material, changing the methyl ketone to an ethyl ester creates a difference in the product: a substituted acetic acid derivative is obtained rather than a methyl ketone. Starting material: O
O
O O
vs
CHEM*3750 SCHWAN COURSE NOTES F13
O
O O
Chapter 2 Page 44
Final product:
O
O H
vs R
O R
Overall reaction of malonate synthesis O EtO
O
O EtO
OEt
R
O R'
O OEt
HO
H R
R'
substituted acetic acid Diethyl malonate * is the usual starting material. The anion of it is quenched with an electrophile. The product, also a malonate, is hydrolyzed to a diacid, which is then thermally decarboxylated. The result is the monosubstituted acetic acid. Disubstituted acetic acids are also available and can be obtained by following the same protocol as for disubstituted acetones. Diethyl malonate is slightly less acidic than acetyl acetone, with a pKa of 13.3, but the same chemistry can take place anyway.
CH2(CO2Et)2
1. NaOEt, EtOH 2. allyl bromide
CH2(CO2Et)2
2 eq. NaOEt EtOH
CO2Et 1. NaOH, H 2O C CO Et + 2 2. H 3O H 3. heat
Br(CH2)3Cl
CO2H
C(CO2Et)2 1. KOH, H 2O + 2. H 3O , heat
CO2H cyclobutanecarboxylic acid, 43%
*
1,3-Propanedioic acid is commonly known as malonic acid and so diesters are known as malonates.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 45
CH2(CO2Et)2
CO2Et C CO Et 2 H
1. NaOEt, EtOH 2. Br
1. NaOEt, EtOH 2. EtBr
CO2H C H
CO2Et C CO Et
1. KOH, H 2O
2
+
2. H 3O , heat
Clearly, the prominent structural feature in the acetoacetic esters and the malonates is the methylene group surrounded by two electron-stabilizing functionalities. Although there is a detailed presentation here of the chemistry of the systems with two esters and with a ketone and an ester, other functional groups may also be present. Indeed, a number of groups can serve the role of electron withdrawing groups in the general structure Z-CH2-Z'. Typical ones include of course esters and ketones, but one may also encounter other carbonyl containing compounds as well as nitriles, nitro compounds and sulfur or phosphorus containing groups. The final step of thermal decarboxylation is not always possible. Indeed, sometimes one wishes to maintain both of the groups in the molecule. Furthermore, there may be other methods for removing the particular functional groups once they have served their role.
Me O
O
1 eq. NaOH MeI, H 2O 70%
O
Me Me O
O
O
2 eq. NaOMe, MeI, MeOH 80% c) Robinson Annulation (SF10 19.6) Before introducing the chemistry and overall value associated with the Robinson Annulation, it is important to introduce a very important reaction known as the Michael Addition (SF10 19.7).
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 46
The Michael addition was originally defined as the addition of a malonate to the position of an , -unsaturated ketone, ester or nitrile. This strict definition has subsided and a Michael addition is now defined as the addition of any nucleophile to the position of an alkene bearing one or more electron withdrawing groups. Below are some examples using nucleophiles that have recently been introduced to you. Other carbon nucleophiles can be used in this chemistry as can amines, alcohols and sulfur nucleophiles.Note that the product of these reactions possesses two strong electron withdrawing groups and may well contain one or two carbonyl groups. The Robinson annulation begins with a Michael addition and is completed by an intramolecular aldol condensation. In general, there are other materials that can participate in Michael additions (but not necessarily the full Robinson annulation). Michael donors (nucleophile) O
O
R
-diketone anion
O
R
O
R'
O
Michael acceptors (electrophile)
OR'
-
R2CuLi
R
-keto ester anion dialkyl cuprate
O
conjugated ester OR
O NH2
enamine
N
O R
N C
-
O R
NO2
O
R
-
R'
-keto nitrile anion -nitro ketone anion simple enolate
R2NH or RNH2
amine
ROH
alcohol
RSH
thiol/mercaptan
CHEM*3750 SCHWAN COURSE NOTES F13
conjugated ketone
NO2
conjugated amide conjugated nitro compound
CN
conjugated nitrile
O O S R
conjugated sulfone
O O S OR
conjugated sulfonate
EWG
conjugated, electron deficient alkyne
Chapter 2 Page 47
Mechanism of Michael addition: O
O
O _
OEt O
_
O
H
H
H
O
O
O
O
_
O _
H
O
H
O
_
R O
R
O
O
O H
H O
-OEt + O
_
HOEt
R
O R
O
O
O
H
O
_
acidify O
R
O O
R
+ EtOH
Using the example in the mechanism above, where R = CH3, exposure of that material to base (usually from the original Michael addition reaction mixture), provides a means for the intramolecular aldol reaction. In this particular case, the anion of the methyl group may attack one of two equivalent ketone groups. In many cases there is a lone ketone and the reaction works perfectly well. From this point the regular aldol and its dehydration occur.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 48
O
O H O
O
CH3
O
H
_
base
O
O
CH3
O
O
O
O
CH _ 2
H
protonation & dehydration
O
-
O
O
The thermodynamics of the reaction make the formation of 6 membered ring most favorable. Hence the Robinson annulation is ideal for the formation of cyclohexenones including those fused to other rings. Overall, as stated, the Robinson annulation involves, first a Michael addition and then an aldol condensation.
Michael addition
O
Some examples:
C[H2 O ] CH3 aldol condensation
O
O + CO2Et
methyl vinyl ketone (MVK)
CHEM*3750 SCHWAN COURSE NOTES F13
O cyclohexenone
O
NaOH EtOH CO2Et
Chapter 2 Page 49
O
NaOMe
+
MeO
O
O
O
MeOH
ethyl vinyl ketone
MeO Michael addition product
O
MeO Robinson annulation product
.
. More, general examples of the annulation.
O O
O
O OEt
+
Michael
O OEt O
O Aldol C(O)OEt
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 50
O
O
O
O
Michael
+
O
Aldol
d) Other Conjugate Additions As you know, a number of nucleophiles add to the carbonyl carbon of aldehydes and ketones. The presence of a conjugated double bond creates an additional electrophilic site, and above it was shown that stabilized carbon anions will perform Michael additions on these substrates. Simple carbon nucleophiles such as Grignards will still attack at the C=O group. Organolithium reagents can attack at either electrophilic site and are not synthetically useful for this chemistry. Organocuprate reagents attack in a conjugate fashion and are easily generated using Grignard reagents in the presence of CuI. 1. EtMgBr Et2O + 2. H /H2O
HO Et
1. EtMgBr CuI, Et 2O + 2. H /H2O
O
O
H H Et
A number of other nucleophiles are also useful in conjugate addition reactions. The list includes amines (RNH2, RR’NH), alcohols (ROH), thiols (RSH) and HCN (prepared via KCN/H+). The addition of a nitrile is particularly synthetically useful since it can be converted to a carboxylic acid by hydrolysis.
O
HNMe O MeNH 2
O Ph
KCN/H
O
O
+
NC
CHEM*3750 SCHWAN COURSE NOTES F13
Ph
acid or base hydrolysis
HO
Ph O
Chapter 2 Page 51
A significant example of a Michael addition has recently been found in nature. The Michael reaction is critical to the operation of a rather promising anticancer drug, calicheamicin. In the first step in its operation the trisulfide bond is broken (1). The nucleophilic sulfide then adds in Michael fashion to give enolate 2. This addition changes the shape of the molecule, bringing the ends of the two acetylenes closer together. A cyclization occurs to give a di-radical (3), and this diradical abstracts hydrogen from the cancer cell's DNA, killing it. Calicheamicin depends for its action on the change of shape. Before the Michael reaction, the ends of the two acetylenes are too far apart to cyclize. They are freed to do so when the sulfur adds to the unsaturated carbonyl.
O
NHCOOMe
-
O
SSSMe
S
NHCOOMe
S
-
OR
HO
O
NHCOOMe
OR
HO
OR
HO
2 1
-
O
-
O
NHCOOMe
NHCOOMe
S
S
dead + cancer cells
HO H
OR
DNA
OR
HO
H 3
R = series of sugar moieties
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 52
PRACTICE PROBLEMS 5
You are capable of doing Questions 18.7-18.13, 18.15, 18.23, 18.27b,c,f, 18.30 19.16, 19.35, 19.46, 19.47 and 19.58a in SF10.
1. Why do esters that have only one hydrogen Claisen condensations?
to their carbonyl not work efficiently in
2. Suggest a mechanism, including curly arrows for the following reaction.
O
O OC2H5 + OC2H5
-
CH3CO2C2H5
O
1. C2H5O Na
+
C2H5OH + 2. H3O
CHCO2C2H5 O
3. Draw the product resulting from application of the following sequence of reagents.
O
-
O OC2H5
1. 1 eq.C2H5O Na C2H5OH
+
2. 0.5 eq. Br(CH2)3Br 3. OH/H2O + 4. H3O /H2O,
4. Will the accompanying compound decarboxylate when heated?
HO C
O
O
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 53
FROM PREVIOUS MID-TERM EXAMINATIONS 5. Provide a mechanism for the following transformation: O
-
CH3
O
+
1. EtO Na /EtOH +
CH3
CO2Et
2. mild H3O workup
CO2Et
6. Provide a synthetic route to 4-pentenyl phenyl ketone from diethyl malonate. You may use any reagents and organic compounds to complete your synthesis. Be sure to draw the structure of several of the stable organic intermediates along the way in order to be eligible for part marks.
O EtO
SOLUTIONS TO
O
O OEt
Ph
PRACTICE PROBLEMS 5
Q1. Recall why the Claisen condensation goes to completion. It is because the final step is the deprotonation of the product between their carbonyls. So, if the starting material has only one -H, and that H gets removed to initiate the reaction, then there are not any H’s remaining to be removed in that final step of the reaction. Hence instead of executing that last deprotonation, the alkoxy group will attack the keto part of the product and further reactions will take things back to starting material. Q2.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 54
O CH2 H
-
C2H5O Na
OEt
-
O
+
C2H5OH
O OEt
CH2
OC2H5 OC2H5 O
O
-
O OC2H5
O CH2 OEt OC2H5
H O CH OEt OC2H5
O
O -
+ O
OC2H5 O
-
O
CH OC2H5
OEt
O
CH -
O
O
OEt
OC2H5
O
O CCO 2C 2H5 -
CCO2C2H5
O
O
H +
-
OEt
+
introduce aq. acid at this stage
H OH2
O CHCO2C2H5 O
Q3.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 55
O
O
2
O -
+
2 eq.C2H5O Na
OC2H5
O
O -
O -
OC2H5 +
OC2H5
1 eq. BrCH2CH2CH2Br
O
O
-
O
O
O
-
O O
-
OH/H2O
C2H5O
O
OC2H5
O
O
CH2CH2CH2
CH2CH2CH2
+
H3O /H2O O OH
O
HO
O
O
O
O CH2CH2CH2
CH2CH2CH2
Q4. The reaction will not proceed since that would place a double bond in the bridge head position and the bicyclic nature of the compound does not allow such an arrangement, because of the huge amount of strain associated.
X HO C
O
O
OH
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 56
Q5. This mechanism emphasizes the reversibility of the Claisen/Dieckmann reactions and also the requirement for the existence of some -H’s to prevent the reversibility of the reactions. O
-
CH3
-
EtO O
+
EtO Na /EtOH
O
CH3 OEt
CO2Et
CH3 OEt
EtO -
O
O
H OEt -
O
CH3
EtO EtO
O
OEt H H
O
H
CH3
EtO
OEt
EtO
-
O
CH3
O
O +
O
CH3
EtO H O +
O EtO
-
CH3
O
OEt
CHEM*3750 SCHWAN COURSE NOTES F13
O
+ H OH2
EtO
-
OEt
CH3 O
introduce mild acid at this stage
Chapter 2 Page 57
Q6.
O
O
EtO
OEt +-
1. Na OEt/EtOH 2. Br O
O
EtO
O
O OEt
+-
1. Na OEt/EtOH
OEt Ph
EtO O
O
2. Ph
Cl +-
1. Na OH/H2O +
2. H3O /H2O
Ph
O
O
O
OH Ph
HO -2 CO2 O
note that the introduction of the 3rd carbonyl allows the thermal removal of both carboxylic acid functionalities _______________________________________________________ 8. The Wittig Olefination of Aldehydes and Ketones (SF10 16.10) The most popular method for the conversion of aldehydes and ketone to alkenes is the Wittig Reaction. Generally speaking it involves the reaction of a carbonyl compound with a phosphorus ylide. * The result is an alkene and a phosphine oxide. *
An ylide was originally defined as a compound that possesses a carbon anion directly beside a heteroatomic cation, where the formation of a multiple bond is usually not possible. The definition has eased in recent years and now encompasses any compound bearing adjacent positive and negative charges in situations where a formal multiple bond may or may not be possible. Other Popular methods for olefination chemistry may involve silicon or sulfur atoms rather than a phosphorus atom.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 58
O
+ _ R R"3P C R' ylide
+
+ R"3P O
+
C
R
R' alkene
R"3P O
A specific procedure is usually required to prepare the Wittig reagent. First one treats the halogen in an alkyl halide with a phosphine, usually triphenylphosphine. The result is an alkyltriphenylphosphonium salt. The salt is then treated with a strong base, usually butyllithium in order to generate the ylide.
Ph3P
+
R CH2 X
+ Ph3P CH2R + H
C
Li
-
C6H6
+
H
(n-BuLi)
+ Ph3P CH2R
_ + Ph3P CHR
-
+ X
+ butane + LiX
phosphonium ylide
Deprotonation of the phosphonium salt may also be achieved using alkoxides or NaH. Some ylides are stable. Most often the ylide is prepared in the flask and then is brought together with the carbonyl compound. Under these circumstances, the carbon anion portion of the ylide attacks the carbonyl group. The result is a phosphonium betaine, which readily closes to a cyclic isomer, an oxaphosphetane. To conclude the synthesis, the oxaphosphetane fragments to make the alkene and triphenylphosphine oxide. One of the driving forces of the reaction is the formation of the strong P=O bond.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 59
_ + Ph3P CHR
+
phosphonium ylide
R' R"
R
O
+ Ph3P
R' R" -
O
betaine
fast
R
R'
H
R"
Ph3P=O +
fragmentation
alkene
R
R' R"
Ph3P O oxaphosphetane
The Wittig reaction can be carried out in the presence of ether, ester, halogen and other multiple bond functionalities. It often affords a mixture of geometric isomers about the double bond, although sometimes the reaction is selective. Some examples:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 60
+ PPh3X
NaOMe MeOH
PPh3
O
OC(O)Me H
OC(O)Me
-
HO , H 2O (saponification)
OH
Vitamin A 1
+
H +PPh3
-
+
Bu Li
PPh3
CHO
H
-
O O
+
Ph3P CH O
O CH
O
The Horner-Wadsworth-Emmons (HWE) olefination reaction (SF10 p. 760) follows a path similar to the Wittig olefination. The difference is the nature of the phosphorus containing starting material, in the HWE reaction, one employs a phosphonate ester. The HWE version of the olefination is often preferred since the E- isomer of the alkene is typically formed in high preference or exclusively. The base can be NaH, KOtBu or (Me3Si)2N-Li+.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 61
O EtO P EtO
R
O R EtO P EtO Na+
NaH THF
R is usually a conjugating group e.g, Ar, carbonyl, alkene
H
R' O
R + H
R'
O EtO P - + O Na EtO
The following is a key step in the total synthesis of callipeltoside A, a macrocyclic natural product known to have anti-tumour activity and to protect cell infected with the HIV virus. OR O
OR O
OR MeO
O
+
OR
O H
- + (Me3Si)2N Li . THF -78 °C to rt
O
EtO O P EtO
MeO
O
O H
R = SiMe2tBu Cl Cl
9. Reductive Conversion of C=O to CH2 A final important transformation of aldehydes and ketones is their conversion to methylene groups. The following two reagent systems are observed routinely with ketones and occasionally aldehydes. The Wolff-Kishner approach [There is no SF10 reference] utilizes H2NNH2/high boiling solvent/NaOH/heat while the Clemmensen reduction (SF10 15.9) involves refluxing the ketone in hydrochloric acid with a zinc amalgam [Zn(Hg)]. Clearly the Wolff-Kishner protocol is best for substrates that are not base sensitive while the Clemmensen procedure is preferred when there are no acid sensitive functionalities in the ketone. Each of these procedures are part of the family of reductions known as deoxygenations.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 62
O
H
H2NNH 2, NaOH (HOCH 2CH2)2O heat
H
47%
O Me
O
H2NNH 2, NaOH (HOCH 2CH2)2O H 2O, heat
CHO
Me 69%
CH3 HCl, Zn(Hg) OCH3
heat
OH
OCH3 OH
Both methods are particularly effective in tandem with electrophilic aromatic acylation reactions.
PRACTICE PROBLEMS 6
You are capable of doing Questions 16.7, 16.40, 19.15, 19.19, 19.20 and 19.43, 16.45 in SF10.
1. Name the steps involved in the following sequence and draw the important reaction intermediates. (MVK = methyl vinyl ketone)
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 63
N
1. MVK +
2. H3O
O
2. Draw the product of the sequence below. O Cl
Zn(Hg) HCl
AlCl3
3. Indicate on the arrows directly the reagents that would be required to give the chemistry shown.
O
O O
O
OH
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 64
4. Indicate in each of the boxes accompanying the arrows, provide the full set of conditions, reagents and/or organic molecules that are required to give the chemistry shown. CH2
O
Ph
O
O
O
O
5. Provide a mechanism that accounts for the reaction shown. Be sure to understand why this reaction proceeds.
O
O
-
O
+
1. MeO Na /MeOH +
MeO
2. mild H3O
MeO
O H
6. FROM A PREVIOUS MIDTERM: Draw the product of the treatment shown.
O
1. LDA, -78 °C, THF 2. O O Ph
Ph
3. CH3-Br
7. Write down as many reactions as you can think of for: a) the formation of cyclohexanone and b) the reactions of cyclohexanone CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 65
SOLUTIONS TO
PRACTICE PROBLEMS 6
Q1. N
+
O
Michael addition
-
N+
O
+
H 3O
hydrolysis O
O
OH
O +
intramolecular
N+ H H
aldol
H
O+
+O H
OH
O
HO
Q2.
O Cl AlCl3
Zn(Hg) HCl
Q3.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 66
O 1. O3 2. Zn O O -
O -
OH/H2O
or + H3O /H2O
OH/H2O
or + H3O /H2O
OH
and heat
Q4.
CH2
Ph3P=CH2
Ph
Zn(Hg)/HCl
1. pyrrole/H+/toluene 2. PhC(O)Cl + 3. H3O
O
O
or NH2NH2/KOH HOCH2CH2OH/heat
O -
OH/H2O heat O
O
CHEM*3750 SCHWAN COURSE NOTES F13
-
OH/H2O heat
O
Chapter 2 Page 67
Q5. O
O
O -
MeO
MeO
-
O
-
O
OMe
MeO
O
+
MeO
H
MeO
(or through the mediation of solvent
O MeO
O
-
O H
MeO
O
O OMe H
-
+
H
MeO
MeO
O
-
+ MeO
O H
_
MeO
H H
O
H +
The reaction sequence is fully reversible, but the availability of H's on the product prevent the backwards reaction. The starting material does not have H's between the carbonyls and hence gets attacked at the ketone carbonyl
H2O H
O
O H
MeO
Q6.
O
1. LDA, -78 °C, THF 2. O O Ph
Ph
O Ph
O
Me
O Ph
3. CH3-Br
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 68
Q7. a) Selected Syntheses: O
O
MeO
NaOMe/MeOH
OMe O
O CO2Me 1. OH/H O 2 + 2. H3O 3. , -CO2 -
O 1. O3 2. Zn
b) Some reactions:
O NH2NH2/KOH HOCH2)2, or Clemmenson cond'ns R
O
R'
Ph3P=CRR'
O 2
KOH
O
H 2O
CHEM*3750 SCHWAN COURSE NOTES F13
or other mixed aldol reactions
Chapter 2 Page 69
O
R OH 1. RMgX +
2. mild H3O
O
O
1. Br
O
HO
OEt
OEt
Zn, toluene +
2. mild H3O
O
O R
1. LDA, -78 °C, THF 2. RMgX
O
O Br2, H
O
+
O
Br
O
KOH/H2O
O
NH H+/ toluene/-H2O
N
O R
1. R-X +
2. H3O
_______________________________________________________
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 70
O
O
O
C
C
R
carbonyl group
R
R
C
H
aldehyde
ketone (R H)
Some significant aldehydes and ketones O H
O
O
H
H
cinnamaldehyde (candy, f ood, drugs)
HO raspberry ketone
O
O helminthosporal (a f ungal toxin)
camphor (liniments, inhalants)
O
O
O
O (-)- enantiomer, spearmint (+)- enantiomer, caraway seed (candy, toothpaste)
biacetyl mar garine) (
muscone (perf umes)
O
OH
O H
H
H
H
O testosterone ma ( le sex hormone)
O
H
progesterone (f emale sex hormone)
CHEM*3750 SCHWAN COURSE NOTES F13
HO O vanillin (f ood flavouring)
Chapter 2 Page 1
The reactivity of aldehydes and ketones is based on two important reactions of them. These species are prone to nucleophilic attack at the carbonyl carbon and one can make a nucleophile on a carbon to the carbonyl group. -
O R
C
R
Nu O C R R
+ Nu -
RCH2
O C
O
base
- C RCH R
R
This chapter will first demonstrate some synthetic routes to aldehydes and ketones, and then will describe some of their vast chemistry based on the two principal modes of reaction shown above. 1. Synthetic Routes to Aldehydes and Ketones (SF10 16.2, 16.4, 16.5) One of the main methods for the preparation of aldehydes and ketones is oxidation of an alcohol as shown in a general sense below.
RCH2OH
O
[O] R
OH R CH R'
H
O
[O] R
R'
The following are some popular routes to aldehydes and ketones and of course include oxidation of an alcohol. Most should be familiar to you. a) Aldehydes Me Me
CHCH2CH2CH2OH
PCC CH2Cl2
Me Me
primary alcohol
O CHCH2CH2 C H aldehyde, 75%
PCC = N+ H ClCrO3 pyridinium chlorochromate
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 2
b) Aldehydes and Ketones O3, CH2Cl2
Zn
-78 °C
HOAc
O H + CH 2O
62%
c) Ketones
Na 2Cr2O7 H2O, H 2SO4
OH menthol
O Ph
menthone, 84%
OH
1. PhMgBr
H Ph
+
H
O
2. H 3O
O
[O]
Ph
Ph
Ph
Grignard chemistry followed by oxidation 1.
OMe
O
OMe
O O
, AlCl 3
93%
2. HCl, H 2O O
Cl O
+ Et 2CuLi
CHEM*3750 SCHWAN COURSE NOTES F13
-78 °C ether
O
Chapter 2 Page 3
A useful method that has selected limitations is the hydration of alkynes. This is achieved with the assistance of mercuric ion Hg+2. When an alkyne is treated with HgSO4/H2SO4/H2O, the elements of water will add across the triple bond. In its simplest form the reaction looks like this:
H C C H
HgSO 4/H2SO4 H2O
H
O C CH3
A limitation arises when one considers the asymmetrically substituted alkyne: one can expect two products!
CH3CH2 C C CH2CH2CH2CH3
HgSO 4/H2SO4
O CH3CH2 CH2 C CH2CH2CH2CH3
H2O +
CH3CH2
O C CH2 CH2CH2CH2CH3
However, if one considers the mechanism of the reaction, which begins with Hg+2 coordinating to the triple bond and creating some cationic character on the triple bond carbons, then water attacks one of the carbons. In the above case, the two alkyl groups can stabilize the cationic charge in a comparable manner, leading to water attack at each carbon and two major products. CH3CH2 C C CH2CH2CH2CH3
Hg
+2
+2
CH3CH2
Hg C+ C+ CH2CH2CH2CH3
OH2
2 products
This mechanistic information does carry within it some synthetic value. What if the two groups on either side of the triple bond differ significantly in their ability to stabilize positive charge? Water will attack at the site bearing more positive charge and that carbon ends up being the site bearing the carbonyl oxygen. Hence terminal alkynes will always be a source of methyl ketones.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 4
+2
H
C C CH2CH2CH2CH3
Hg
+2
H
Hg C+ C+ CH2CH2CH2CH3
which of these carbons bears bear more +ve charge? The more substituted does, according to Markovnikov
O CH3 C CH2CH2CH2CH3 lone product
example:
PhCH2 C C H
HgSO4, H2SO4 H2O
O PhCH2 C CH3
Of course, when the alkyne is symmetrically substituted, then a single product results
CH3CH2 C C CH2CH3
HgSO 4/H2SO4 H2O
O CH3CH2 CH2 C CH2CH3
More examples: O
O +
Cl
AlCl3
CrO3 OH MgBr +
O H
CHEM*3750 SCHWAN COURSE NOTES F13
quench with H2O
Chapter 2 Page 5
O O
HO
O
PCC
O
OH
CrO3
H O
O
+
H
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 6
2. Acidity and Enolization of Aldehydes and Ketones (SF10 3.5-3.7, 18.1, 18.2, 18.3) As stated, one of the origins of the reactivity of aldehydes and ketones is their inherent acidity, which is based on the electronic properties of the carbonyl group. The carbonyl group is an electron-withdrawing group by both induction and resonance. So, hydrogens to the group possess increased acidity simply due to their proximity and the incipient anion can be stabilized through -resonance with the carbonyl group. O
O base
H acidic hydrogen, pKa = ca. 19-20
O-
resonance structures
There is a geometric requirement for facile proton removal in this chemistry. The C-H bond that holds the removable hydrogen must be aligned with C O H the -orbitals of the C=O bond. The required conformation is trivial to achieve in a freely rotating H system, but may not be accessible in cyclic or all aligned, so H is constrained molecules. Without the proper orbital ideal for removal overlap, then the -resonance stabilization of the anion is not possible. The result is that the acidity of the hydrogen is greatly reduced. Bases suitable for deprotonation of an aldehyde or ketone must meet two principal requirements: a) The base must be strong enough to remove the hydrogen. In this regard, any base whose conjugate acid has a pKa greater than 20 will be suitable. Sometimes a weaker base will work, but the reaction must proceed under equilibrium deprotonation conditions. b) The base has to perform selective attack at hydrogen and cannot have properties that will promote nucleophilic attack at the carbonyl carbon. A sterically hindered base usually provides the desired chemoselectivity. LDA, lithium diisopropylamide ([Me2HC]2N-Li+) is often the base of choice. The resonance-stabilized anion that results from these deprotonation reactions is called an enolate, which means that it is the anion of an acid called an enol.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 7
O-
OH
enolate
enol
These species are of course interconvertible through the transfer of one proton. As we will see shortly, the enolate and the enol can be functionalized at the -carbon. Hence one can use an enol or an enolate for derivatizing the compound next to its carbonyl group. Enols can be generated through a reversible acid catalyzed reaction.
O H
+ H O
+
O
H
O
H
+ H
H
H
+
+ H 3O base (= H 2O)
H is much more acidic since the protonated carbonyl is a much better electron sink The methods of forming enols and enolates are fully reversible. So the carbonyl compound can be regenerated from an enol by simply adding acid and isolating the substrate (except in specialized instances). Attempted isolation of the enol from acid solution will provide the aldehyde or ketone. Mechanistically, regeneration of the carbonyl compound can be the reverse of enolate formation, or may involve the enol. -
O
H
+
O
H
O
H Under these conditions, the keto form of the compound is regenerated upon isolation. The equilibrium between enol and keto shown above is called tautomerism, since the compounds are tautomers of one another. The carbonyl compound is usually the thermodynamically favoured form, based on the carbonyl bond strength, but there are exceptions to this rule. A number of equilibrium CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 8
constants have been determined for keto-enol tautomerism. They are calculated as shown, and are of course dependent on the medium in which the measurement is made. If the substrate is neat, the tautomer ratio depends on the origin of the material.
[enol] [keto]
Keq =
The Keq (keto-enol) for acetone in water is 1.5 X 10-7. It should be noted that despite the fact that the ratio can significantly favor the keto form, reactions selective for the enol form can still take place, as we shall see. The relative acidity of the hydrogens next to the carbonyl groups of aldehyde or ketones has both advantages and disadvantages. For one advantage, the acidity allows for deuterium incorporation into the molecule. The protocol involves simple addition of the carbonyl compound to D2O containing a trace amount of acid or base. Any hydrogens in the -position that can adopt the required conformation for enol or enolate formation will become fully deuterated, given sufficient time.
O
O CH3
CH3
-
D2O, OD
D D
H H O (Me)2CH
CH3
CD3
O
+
D2O, D 3O
(Me)2CD
H
H
Another consequence of enol formation, having negative and positive implications, is the ease with which chiral -carbons can invert their stereochemistry. Hence if you have taken the time to make an important, optically pure product possessing stereochemistry at the -carbon, exposure to acid or bases may racemizes the material. For example, optically active 3-phenyl-2-butanone in basic ethanol (r.t.) racemizes within minutes. O Ph
-
CH3
H CH3 (S)-3-Phenyl2-butanone
EtO /EtOH
Ph H3C
-
O
CH3
Achiral, planar
-
EtO /EtOH
O Ph
CH3
H3C H (R)-3-Phenyl2-butanone
isolation racemic ketone
There are instances of advantageous isomerism of -carbon, through their planar form. CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 9
- +
O
O
OK CH2CH=CH2
10% KOH
CH2CH=CH2
CH2CH=CH2
EtOH CH3
CH3
CH3
In the anionic form, the -carbon is planar and in theory can be protonated from either face. However, the configuration of the -carbon, which bears the methyl group, is a constant and that substituent creates a bias in the reprotonation reaction. The allyl and methyl substituents prefer to be trans to one another for steric reasons and trans isomer is produced in >95% yield under these thermodynamic conditions.
The Schwan invaluable mechanism rules. 1. Do not rush mechanistic steps together. 2. Use only reagents and data that have been provided. 3. In aq acid, for every step of the mechanism, the nucleophile is neutral and the electrophile is acidic (e.g., protonated) where possible for enhanced reactivity. It follows that there are no negatively charged entities in acidic solution. 4. In aq base, for every step of the mechanisms, the nucleophile is basic and the electrophile is neutral. It follows that there are no positively charged entities in basic solution. 5. Make sure each mechanistic step is balanced: charge, atoms, electrons 6. Electrons begin every step. 7. Water and alcohols will not do SN2-like substitutions at carbon. 8. In alcohols or water, always use solvent for proton transfers. 9. Know as many pKa’s as possible.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 10
3. Halogenation of Ketones and Aldehydes (SF10 18.3) The following general reaction is the topic of this section.
O +
H
O
acid or
X2
X
base
As noted, halogenation of these carbonyl compounds may proceed under acidic or basic conditions. Under either of these conditions the mechanism for generation of the enol or enolate is the same as previous. The reactive compound then attacks the halogen rather than any other electrophile such as a proton.
O-
via enolate
O
X X
X
-H H
via enol
+
+O H
O
X X
X
Rate studies have shown that initial rates of the halogenation are independent of halogen concentration. This is interpreted to mean that introduction of halogen into the substrate is not rate determining. Furthermore, experiments have shown that the rate is dependent on concentration of carbonyl compound and acid (or base). From this evidence, it is generally believed that enol or enolate formation is rate determining.
O
O Cl2, H
+
70 °C
Cl 85%
Sometimes as shown below, the reaction is autocatalytic. That is, it is very slow until some halogenation occurs which produces acid as a byproduct. The rate of halogenation then accelerates due to the presence of acid. Hence the term autocatalytic, since the reaction provides the means to promote itself.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 11
O
O +
MeOH
Br2
Br
+
HBr
70% Under basic conditions, multiple halogenation is often a problem and therefore basic conditions are not recommended if one is striving for only the monohalogenated product. Note that this reactive character can be used to one’s advantage, as shown below. For the introduction of only one halogen, the reaction is usually performed under acidic conditions since introduction of one halogen into the molecule slows further reaction. For synthetic efficiency, monohalogenation reaction must be performed under acidic conditions.
O
H
+
O
H
+
X
H
In the halogenated form, the electron withdrawing nature of the halogen retards the initial step of enol formation and hence slows the introduction of more halogens.
halogen pulls electron density away from the carbonyl group
Multiple halogenation of methyl ketones provides chemistry that has been used as a functional group identification technique, although spectroscopic methods are now more popular. Multiple halogenation of methyl ketones leads to the haloform reaction and iodine is the more common halogen employed (iodoform reaction). The mechanism involves triple iodination of the methyl group of the ketone under basic conditions.
O R
base I2
O
O R
I
base I2 (2X)
I
R I
I
Then, hydroxide attacks at the carbonyl group to make a tetrahedral intermediate which expels -CI3.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 12
Complete mechanism:
O
-
H
R
-
O OH
-
H
I
-
I
HO
O
OH
R I
I
I
I
H
H
-
OH
I O
I I
I
R
I
I
-
OH
-
R
O
-
O
I
I
I
H
R
O
R
I
I
H
O
R
I
H
R
H
H
O
R
+ OH
HO
-no H's remain; -carbonyl is most electrophilic
-
O R
-
O
I I C I H2O
I I C H I yellow precipitate
Interaction with the solvent provides the products: iodoform and carboxylate. -CI3 is a good leaving group in this case because the three electronegative halogens help to stabilize the negative charge on the carbon atom. The electronegativity also accounts for the multiple halogenations at a single site. Once one halogen is introduced, the remaining hydrogens on that same carbon atom have enhanced acidity. Acidic workup affords the carboxylic acid if desired as opposed to the carboxylate. The iodoform is a bright yellow precipitate that serves as a useful indicator. If a compound is suspected of being a methyl ketone, simply add excess hydroxide and I2 and look for a yellow precipitate for confirmation of the suspected structure. The reaction can also be synthetically useful in that it can be used to achieve the overall conversion of methyl ketone to carboxylic acid.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 13
O
1. Cl2/NaOH
O OH
2. HCl/H 2O 5-methyl-3-hexen-2-one
+ CHCl3
5-methyl-2-pentenoic acid
chloroform
PRACTICE PROBLEMS 3 You can now do Questions 12.4, 16.47, 18.1-18.5 18.20a & 18.33 in SF10 1. Which of the following compounds will give a positive iodoform test. a) 2-pentanone b) 3-pentanone c) hexanal d) cyclohexanone e) 2-methylcyclohexanone f) cyclohexyl methyl ketone 2. For each of the following compounds write reaction equations showing how the compound could be prepared by oxidation of an alcohol, ozonolysis of an alkene and (if possible) hydration of an alkyne. Which of the preparations that you have written would give an unacceptable mixture of isomers? a) b) c) d)
pentanal 3-hexanone 3-heptanone 2-methylbutanal
3. a) Suggest how you could prepare methyl 2-phenylethyl ketone from: i) an alkyne ii) 2-bromoethyl benzene b) Draw the products of the iodoform reaction on methyl 2-phenylethyl ketone. 4. Suggest two sets of reagents that will effect the following reaction.
Cl
O CH3
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 14
SOLUTIONS TO
PRACTICE PROBLEMS 3
Q1. a) 2-pentanone YES b) 3-pentanone NO c) hexanal NO d) cyclohexanone NO e) 2-methylcyclohexanone NO f) cyclohexyl methyl ketone
YES
Q2. a) pentanal PCC
CH2OH H
O C
H
cis or trans isomer
Hg H
O
2
1. O3 2. Zn
C
H
O
+2
H2SO4 H 2O
H
+
O exclusive product
C
H
not likely due to Markovnikov's rule
b) 3-hexanone OH
CrO3 H
O
Hg
+2
H2SO4 H 2O
+
1. O3 2. Zn
cis or trans isomer
c) 3-heptanone
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 15
OH
O
CrO3 H
1. O3
+
2. Zn
Hg
cis or trans isomer
O
+2
+
H2SO4 H 2O
O 50:50 mixture, therefore unacceptable
d) 2-methylbutanal
O
CrO3
HOCH2
H
+
H
1. O3
H
2. Zn
H
cis or trans isomer -use of Hg+2/H2O/H+ on an alkyne is not possible for same reason as in a) above. Q3 a) Hg Ph
Ph
Br
O
+2
Ph
H2SO4 H 2O
Mg ether
CrO3/H
MgBr
Ph
H OH
1. CH3CHO 2. HCl/H2O
+
Ph
b)
O Ph
O
-
1. xs OH I2 + 2. mild H/H2O
Ph
OH
+ CHI3
I I
Q4.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 16
Cl
MgCl
Mg ether
1. CO2 2. HCl/H2O
1. CH3CHO 2. HCl/H2O H
O
CH3
OH CrO3/H
SOCl2 O
OH
+
O Me2CuLi
CH3
Cl
_______________________________________________________ 4. Alkylation Reactions and Enamines (SF10 18.4; 18.9) Generating an enolate from a ketone or aldehyde and quenching it with an alkylating agent affords a method for the synthesis of compounds bearing an alkyl substituent next to the carbonyl group. Simple reactions can occur in some cases. O
O
1. NaH, C 6H6 2. Me 2C=CHCH2Br
CH2CH=CMe2 88%
-
via
O
Br
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 17
-
O
O Li
+
O
LDA
MeI
cold
warm to r.t.
EtO
EtO
Me EtO 93%
+ HN( iPr)2
There can be problems associated with this simple approach. One is that there are two sites for deprotonation when the compound bears -hydrogens on either side of the carbonyl group. For instance, which one of the three (yes three) O different hydrogens on methyl cyclohexanone is most acidic? It can H3 H CH H be challenging to create and maintain a single enolate. Moreover, here can also be problems with multiple H H H H alkylations, depending on the substrate and reaction conditions, because sometimes there are unwanted proton transfers that give a different enolate, even during the course of the reaction. Although some solutions have been established and will be discussed later in the chapter, the problems led to the discovery of alternative protocols to achieve simple efficient alkylation. It is useful to present these as they are still in use today. Uncontrolled alkylations can lead up to five different products, in addition to unreacted starting material!! O
O KOH
O +
Me
MeI O Me
O Me Me
+
O
O + Me e Me M
+
Me
+ Me Me
Me Me
One useful method for overcoming some of the difficulties involves the chemistry of enamines. As their name indicates enamines contain a double bond (ene) and an amine and the name is applied to systems where the two are in conjugation.
N
CHEM*3750 SCHWAN COURSE NOTES F13
-
+
N
Chapter 2 Page 18
Gilbert Stork of Columbia University developed the chemistry shown here and it still bears his name. Enamines are prepared by condensing a secondary amine with a ketone or an aldehyde. If the amine is a primary amine the result is an imine, which is actually the more thermodynamically stable form of an imine/enamine tautomeric equilibrium.
O H
+ R NH R'
R
+
N
R'
(R' = H)
N
R
-H2O enamine
imine
The following mechanism accounts for the enamine formation.
O
+
H /toluene
H
+ H O +
NH
HO
NH
+ N
O
O
O NH O
O H HO+ N
N+
H 2O +
O
O HO
N
+
H
iminium ion
NH
H N + H
O
note with 1 re amines
O
O O
EtNH 2 +
N enamine
N -H2O
via deprotonation of the N at the iminium ion stage
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 19
Water must be driven from the reaction vessel in order to force the equilibrium reaction to completion. This is usually achieved through the use of a Dean Stark apparatus:
evaporated compounds condense, fall into graduated tube and separate. Water stays on lower layer and never return to the flask while the benzene (or toluene) rises to the level where it runs back into the reaction vessel.
reaction mixture of benzene or toluene and maybe water, which also is created by the reaction
As shown above the enamine has nucleophilicity at carbon and it is this atom that is preferentially functionalized in a simple SN2 reaction.
X
X
N
N+ I-
SN2
CH3
CH3 I +
usually X = -CH2CH2- (pyrrolidine) -CH2CH2CH2- (piperidine) -CH2OCH2- (morpholine)
H 3O O
CH3
In this reaction the -carbon of the enamine ( -carbon of the ketone) bears the new substituent. To remove the nitrogen auxiliary and recover the new ketone, the iminium salt is hydrolyzed under aqueous acid conditions. The amine is lost in the acidic medium and can be recycled if desired. This method is preferred over simple deprotonation and alkylation since it minimizes double or multiple alkylation. The
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 20
reaction occurs best with reactive halo compounds such as methyl iodide, benzyl halides, allyl halides, -halocarbonyl compounds and acyl halides. Below is an example of the overall reaction sequence.
O
1. pyrrolidine, H H
O
+
67%
H
2. BrCH2CH3
CH2CH3
+
3. H , H 2O Other examples:
N
1.
O
O Br
O
O
O
+
2. H 3O
N
O
O
1.
O
Cl +
2. H 3O
5. The Aldol Condensation (SF10 19.4-19.5) To this point we have examined the reactions of enols and enolates with a number of reactive electrophiles. One can also react them with the very substrates that provide enols and enolates: CARBONYL COMPOUNDS. Under the proper conditions the carbonyl group is sufficiently reactive to accept electron density from an enol or enolate. The archetypal example in many texts is the base catalyzed reaction of two molecules of acetaldehyde. The name aldol originates from the presence of an aldehyde and an alcohol in the product molecule. This name is used even if ketones are involved in the chemistry.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 21
OH
O 2
10% NaOH, H 2O H
CH3
5 °C
portion that acted as an electrophile
CH3
H
O CH2
H
portion that was enolate
The mechanism for the reaction is straightforward and is the basis of many other reactions yet to be introduced in this chapter. It begins with the reversible generation of an enolate. The enolate then reacts as indicated. Note that all the steps are reversible and sometimes aldol condensations are difficult to complete because of a propensity to revert to starting materials.
CH3
-
-
O
O
O H
H2C
H
CH3
H
O CH2
H
H 2O OH CH3
H
O CH2
H
Sometimes the aldol product will lose water spontaneously to afford a double bond. The conjugation of the double bond with the carbonyl group is the driving force in the dehydration. This can often be achieved simply by heating the reaction mixture during or after the bond forming process. Sometimes making the solution acidic will achieve the same purpose. The term condensation arises from the loss-of-water step. In synthetic chemistry, condensation means: the loss of water or its equivalent. Sometimes when the equilibrium constant for aldol formation is small, the reaction can still be driven by pushing the product through to the dehydrated product, the formation of which proceeds efficiently and is less reversible. Basic conditions:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 22
OH CH3
O
H
O
H
CH
H
H -
CH
CH3
H
+
-
OH
+ H 2O
OH
Often, if the elimination of water does not proceed efficiently under basic condition, one converts the mixture to an acidic one. Under acidic conditions, loss of water is an easier process. Acidic conditions: + H OH2 OH C H3
H
O C H2
+ OH2 CH3
H
H
O
CH H
O
H H OH2
C H3
CH
H
+ H 2O
The whole aldol condensation can also be performed under acidic conditions, where an enol rather than an enolate is the nucleophile. The acid initially induces enol formation by the means shown earlier in this chapter. This enol is a weak nucleophile and will only undergo aldol chemistry when the electron accepting carbonyl group has been protonated. Mechanism for acidic aldol reaction with loss of water:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 23
O CH3
H
+ H O
+
CH3
CH3
OH H CH3
+ H O
OH CH3
CH2
CH3
CH2 CH3
O
+
CH3 enol formation
CH3
OH
H 2O CH3
CH3
O
CH2 CH3
CH3
acidic dehydration as shown earlier CH3 CH3
O
CH
CH3
It is very difficult to isolate the true aldol product under acidic conditions. The reaction usually carries through to the unsaturated material. Both ketones and aldehydes can undergo the aldol condensation. Aldehydes are more reactive since nucleophilic attack at the ketone carbonyl is more sterically hindered. Also, the extra electron donating substituent makes the ketone less electrophilic. Aldol condensations as shown above with two molecules of acetaldehyde are the simplest possible examples. Complications quickly arise when two different aldehydes are used, or when ketones with two sets of -hydrogens are employed. For instance with two different aldehydes, four products are available. The list includes two compounds from self-condensation and two different products from a crossed aldol reaction. OH
O R + O R'
R H
H
R'
H R
+
+
R
O
OH H
R'
CHEM*3750 SCHWAN COURSE NOTES F13
+
O
R
OH
H
OH
O
+
R'
O H
R'
Chapter 2 Page 24
Selectivity can be achieved by choosing one reactant that does not have hydrogens or has -hydrogens but they are of comparatively reduced acidity.
O
O
CHO
OH
NaOH, H 2O
+
25 °C, 4 hr. 100 %
O
CHO
CHO
NaOH, H 2O
+ CH 3CH2CHO
O
CH
C
CH3
72% O CH3
+ (CH3)3C CHO
NaOH, H 2O
O CH
CHC(CH3)3
81% Referring back to page 16, it was mentioned that sometimes one can prepare and maintain a single enolate anion, even when are multiple -hydrogens in the system. Specifically, it may be possible to selectively generate a particular enolate and then quench that enolate with another carbonyl compound (or other electrophile such as an alkylating agent). To begin it is important to know the different between kinetic and thermodynamic enolates. The thermodynamic enolate is the more stable by definition, and represents in this case, the more substituted enolate. This corresponds to the more substituted alkene of two isomers being the more stable and the concept applies in this case, since the enolate is mostly populated by the alkene resonance structure. It follows that the kinetic enolate is the less substituted of two options which can be accessed under mild conditions that make use of the steric differences of the two -hydrogen bearing sites. The best conditions for kinetic deprotonation are: LDA, -78 C in THF. Generally one can expect preferred deprotonation from the 1 over 2 or 3 sites and from the 2 over 3 sites. The reaction is complete and irreversible. See the reaction below. Access to the thermodynamic enolate can be more challenging (e.g., less reliable), but suitable conditions might be Et3N in DMF. DMF = HC(O)NMe2. As indicated, LDA is a reliable method for the complete and irreversible formation of a lithium enolate. With asymmetric ketones, LDA will remove a proton from the least CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 25
sterically hindered position. This is generally believed to be the best method for carrying out crossed aldol condensations. The protocol lends a good amount of certainty to predicting the product in a crossed aldol. O CH3
-
O Li
LDA, THF -78 °C
CH2CH3
CH2
+
CH2CH3 CH3CHO
CH3
-
O
OH CH2
O Li
H 2O CH3
CH2CH3
+
O
CH2
CH2CH3
If a given substrate possesses two carbonyl groups, it is possible for the compound to undergo an intramolecular aldol condensation. This form of the reaction is only suitable for the synthesis of 5, 6, and 7 membered rings. There are some rules that can be applied at this stage. One is a reminder that aldehydes are more electrophilic than ketones. The other is that 5 membered rings will form more readily that 7membered rings while 6-membered rings are the best. Recall that aldol condensations are reversible and there is opportunity to form the most thermodynamically stable product. Intramolecular aldol condensations will virtually always proceed through with loss of water to form the -unsaturated ketone or aldehyde. When deciding the product of this cyclization, an important step in the analysis is determining when water can be readily lost from the aldol. If not, aldol formation often reverses itself.
O
O
O NaOH, H 2O
not
100 °C
O
O O
O NaOH, H 2O 100 °C
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 26
Before we leave the aldol condensation, the role of the carbonyl group in this chemistry should be emphasized. The polarity and resonance ability of the carbonyl group enhances the acidity of the hydrogens to it, allowing for the formation of enolates. That creates the nucleophilic component of the two reactants. For the electrophilic component the carbonyl group is polarized such that the carbon can accept attack by a nucleophile and the oxygen can hold the negative charge. 6. Other Related Condensation Reactions Whereas the Aldol condensation involves the reaction of an enolate of an aldehyde or ketone with another aldehyde or ketone, the Claisen Condensation (SF10 19.2) involves the reaction of an enolate of an ester and a carboxylic acid derivative, usually another ester. One can view the initial steps of the Claisen condensation as analogous to ketone chemistry. Enolates of esters are less acidic than enolates of ketones or aldehydes and can therefore undergo chemistry with less reactive compounds. The archetypal example in this condensation involves the reaction of two molecules of ethyl acetate induced by ethoxide ion. The mechanism of the reaction is shown below. O CH2
-
O
O Na
CH2CH3 CH2
H -
CH3
O
CH2
EtO
-
O
O CH3
-EtO
-
-ketoester
O
CH2CH3
CH3CH2
O Na CH2
O
+
O CH2CH3 CH3
tetrahedral intermediate
-
O CH3CH2
O
CH2CH3
OEt
O CH3CH2
O
+
O
O
O H H
CH3CH2 CH3
O -
O
CH3
H +
H 3O O CH3CH2
O
O H H
CHEM*3750 SCHWAN COURSE NOTES F13
CH3
Chapter 2 Page 27
Although the scheme above shows the -ketoester as the product, it should be realized that under the reaction conditions, the material actually rests as the deprotonated form until acted upon by the addition of acid, which returns the hydrogen. That is, the Claisen reaction mixture must be quenched with acid before isolation of the product. Acetic acid, citric acid or aq. ammonium chloride are often employed. Note that anions derived from the -ketoester have two carbonyl groups available for conjugation. Hence the pKa of the -ketoester is much lower than that of a simple ketone or ester, since anion stabilization is offer by both of these carbonyl groups in a single molecule. Hence, the -ketoester is readily deprotonated by ethoxide. -
O CH3CH2
O CH3CH2
O
O CH2
-
EtO Na CH3
O
O CH
CH3
O
+
CH3CH2
O
O CH
CH3
-
O CH3CH2
O
O CH
CH3
The deprotonation step as shown above is key to the completion of the reaction. The -ketoester anion is essentially inert and furthermore, in anionic form, the species is captured and frozen and cannot succumb to the reversibility of the reaction. The overall reaction only works well when there are 2 or 3 hydrogens on the starting ester. The ketoester drawn above is known by the common name of ethyl acetoacetate and hence the self-condensation of esters is sometimes known as the acetoacetate ester condensation. Example:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 28
-
O
2
1. MeO Na O
Me
2. H
O
+
O Me
O
+
Crossed Claisen condensations are possible when one of the esters does not possess -hydrogens, much like the ideal situation for crossed aldol reactions. O
O O
O
O
+
O
O
O diethyl oxalate
diethyl succinate -
CH3CH3O Na
O
+
O
O
toluene
O
O O
O +
O
90%
-
1. CH3CH3O Na
O O
O
O
+
O
ethanol + 2. H 71%
ethyl benzoate
O
O H
+ O
ethyl formate
CH3
O
-
1. CH3CH3O Na O
ethanol + 2. H
O
+
O
H
O 80%
Note that in each of the three examples above, one of the reacting esters does not contain -hydrogens. Carbonate esters, which also lack -hydrogens, will successfully partake in crossed Claisen reactions. Another reaction that simplifies potentially complicated reactions is the Reformatsky reaction. It begins with an -halo ester * and uses metallic zinc to establish *
-Halo esters can be prepared by the Hell-Volhard-Zellinsky reaction whereby a carboxylic acid is treated with molecular halogen and elemental phosphorus and then water. See SF10 839-841 for details of this preparation. The acid can then simply be esterified.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 29
an ester enolate. In this reaction, the halo ester and zinc are mixed together to create a solution containing a zinc enolate, to which the ketone or aldehyde is added. As with aldol condensations, the product of Reformatsky reactions can be readily converted to the unsaturated material by treatment with acid. This achieves the dehydration process. O Zn
O Br
+
-
O
O (ZnBr)
- + O (ZnBr)
O
CHO
OH
O O 60%
O Br
1. Zn, toluene O
O
2. 3. H
OH O
O
70%
+
If two ester groups are in the same molecule and are separated by 4 or 5 carbon atoms, then one can achieve an intramolecular Claisen condensation. This reaction is called the Dieckmann (SF10 873) condensation. Again there is an archetypal system that exemplifies the reaction.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 30
O
-
H H
O
OMe
O
O
H O
O
-
O
O
O -
O
-
MeO
O
H
O
-
O
MeO O O -
-MeO
O
O
anion, quench w/ acid to obtain product As with the Claisen condensation, the equilibrium is frozen at the desired product by deprotonation of the acidic hydrogen between the carbonyl groups. The Dieckmann condensation is only useful for 5 and 6 membered rings. O CO2Et CO2Et O
EtO
-
O
CHEM*3750 SCHWAN COURSE NOTES F13
H
H
CO2Et
+
O
O
Chapter 2 Page 31
A form of the Claisen Condensation is very prevalent in biological systems: it is vital to the construction of many naturally occurring and biologically important chemicals. In this case the carboxylic acid being attacked is a thiolester (textbooks will tell you thioester), and the sulfur and the remainder of coenzyme A is lost in a potentially reversible step. The nucleophile is not exactly the enolate of an ester, but one that bears an extra -CO2unit. The release of CO2, concurrent with nucleophilic attack, assists the Claisen Condensation over its kinetic barrier. O -
O
O CH2
-
O
O
+ SCoA
O
CH3 CH2 CoAS
H3C
-CO2 SCoA
H3C
SCoA
CH3CH2CH2
SCoA
SCoA steroidal hormones
O more repetitions
CH2
O
reduce, dehydrate, reduce
O
O
-
- SCoA
CH3(CH2CH2)n
SCoA
bile acids terpenes
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 32
PRACTICE PROBLEMS 4 You are capable of doing Questions 18.14, 18.18, 18.19a,b 19.23, 19.1-19.7, 19.1119.12, 19.17-19.18, 19.23-19.30, 19.33-19.34, 19.36, 19.37a-c, 19.38 in SF10.
1. Give the expected products for the following acid catalyzed reactions. a) acetophenone + methylamine b) acetophenone + dimethylamine c) cyclohexanone + aniline (phenylamine) d) cyclohexanone + piperidine 2. Suggest a mechanism, using curly arrows, for the dehydration of the self-aldol condensation product of acetophenone by: a) acid-catalysis
b) base catalysis
3. Draw all the possible aldol condensation combinations if propionaldehyde and methyl ethyl ketone are mixed together with aqueous base.
4. The accompanying experimental observation was made. Draw a mechanism that uses curly arrows and accounts for the transformation.
-
+
O Li O
O + H
MeO -
THF -70
O +
MeO
0°C
+
O Li O
H
5. What reagents would be required to carry out the following series of reactions.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 33
O
O H
H
O H
O H
-
O H
H
OEt H
6. The accompanying compound can be prepared in high yield using a mixed aldol condensation. What starting materials are required?
O Ph
Ph
Ph
Ph
7. a) Draw the key intermediates, or a mechanism showing how the two starting materials can come together using LDA to make the alcohol. Hint: learn the pKa of a typical nitro alkane. b) What reagent would you use to oxidize alcohol to decalin A? c) What would you use isomerize decalin A to decalin B? Why would this reaction occur? O
O H
O
+
using LDA HO
OEt
NO2
O2N H
OH H
H
O
alcohol
OEt
oxidize to a ketone O2N O
H
OH
H H decalin B
O OEt
O2N how? O
OH H
H
H decalin A
O
OEt
FROM PREVIOUS MID-TERM EXAMINATIONS 8. The following two substrates undergo an acid catalyzed aldol condensation to afford a conjugated ketone. Draw the product of the reaction and provide a complete mechanism, including curly arrows, that accounts for its formation.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 34
O
O Ph
CH3
+
+
H
H3O /H2O
9. Considering the equilibrium shown below, address the following questions. a) Drawn a mechanism using curly arrows for the conversion of 1 to 2. Be sure to draw all important intermediates and important resonance structures. b) Which side of the equilibrium is preferred? Provide a very brief answer for your choice. O
OH +
H3O /H2O OH
O
2
1 SOLUTIONS TO
PRACTICE PROBLEMS 4
Q1. O acetophenone =
a)
NCH3
b)
H3C
N
CH3
c)
NPh
d) N
Q2.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 35
a)
O
+ H-OH2
OH Ph
Ph
O H +OH
OH2 Ph
Ph
O
Ph
Ph
H H
b)
O
OH
-
Ph
Ph
O
OH
Ph
Ph
H H
Q3. with propionaldehyde as Nu and ketone as electrophile
O H
O H OH
if elimination occurs
O H
with ketone as Nu and propionaldehyde as electrophil O O
H
if elimination H
OH
occurs
O
H
O O if elimination H HO
H
occurs O
H
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 36
O
if self reactions occurs, aldehyde on itself
H
O H H
H
if elimination occurs
OH
O H H
O
if self reactions occurs, ketone on itself
CH3
O CH3 if elimination CH3
OH
CH3
occurs
O CH3 CH3
O O if elimination CH3 HO
CH3
occurs O
CH3
Q4. Recall that the aldol is a reversible reaction
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 37
-
+
O Li O
+
reverse
MeO
aldol
MeO
Li
O
-
O
+ H
O
forward aldol
H -
+
O Li O
-this reaction goes to the more thermodynamically favoured side of the equilibrium; methoxybenzaldeh yde could be viewed as a more stabilized s.m. compared to benzaldehyde
Q5. O O 1. CH2O
H
O H -
H2SO4
H OH/H2O/
2. PCC O H
O -
O
KOH, Br2 H
EtOH/H
H
+
OEt H
+ CHBr3
Q6. O Ph
Ph + O
Ph
O -
O
OH
Ph
Ph
Ph
Ph
Ph
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 38
Q7. a)
add
O
Li+O-
LDA
O
O
O2N O
OEt
O H
O
O2N
aldol reaction
H
-
+
H O Li
OEt
OEt proton transf er, pKa of nitro alkane is about 10 and alcohol is 16
O2N H
HO
OH
O cyclization
H H
carbanion to carbonyl
OEt
O
O2N Li+ H OH
O OEt
b) PCC or CrO3 would work c) Base should isomerize the ester. The equatorial position is the thermodynamically more stable, so the reaction goes in the direction indicated.
Q8. + H-OH2
O Ph
+O H
CH3
Ph
OH
OH2
CH2
Ph
CH2
H O
+ H-OH2
+O
H
H
H H
O
H
O
OH2 Ph
H
O
H
+O
H Ph
elimination as per earlier in the solutions H
O Ph
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 39
-cyclopropyl H is not as acidic and in any case, it will not eliminate to the conjugated ketone Q9. OH + H -O H2 O
H OH +O
1
H
O H2 +O H
OH + H -O H2 O
OH
H
O H2 O
OH
2 is the m ore therm odynam ically stab becuase of the conjugation of the car with the arom atic. Thus the equilibriu toward2.
_______________________________________________________ 7. Synthetic Applications of Condensation Reactions (SF10 18.5-18.8) a) Acetoacetic Ester Synthesis This synthesis, named after the starting material employed is a valuable means of preparing substituted acetones (methyl ketones). The approach utilizes two key qualities of the starting material, MeC(O)CH2CO2Et. One is that the compound possess a CH2 group with low acidity that can be readily functionalized. The other is that the ester can be converted to an acid, which is then completely removed. The general equation for the overall transformation is as follows, where R = a non-hydrogen group and R' may be hydrogen or another group.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 40
O
O
O O
O
R
R'
O H
O
R
R'
substituted acetone In detail, the first reaction is similar to those you have seen before. One of the acidic hydrogens of the dicarbonyl compound is easily deprotonated by an alkoxide and the anion is captured at carbon by addition of an electrophile. Such chemistry will lead to a mono-substituted acetone. One can repeat the chemistry and introduce a second electrophile and the result would eventually be a di-substituted acetone. O H
O H
-
+
RO M
O
O
O O + H M
pKa = 10.7 -
+
OM O O H -
O
O
O
H
R'
RX
O
+
OM
O H
The next step is alkaline hydrolysis of the ethyl ester, a process called saponification. The solution is quenched with acid to allow for the isolation of the -keto acid. O
O
H
R'
1. NaOH, H2O O
2. HCl, H2O
O
O
H
R'
OH
The last step in the overall transformation has its origin in the structure of the keto acid. -Carbonyl carboxylic acids often adopt a conformation where the acid hydrogen can hydrogen bond in an intramolecular sense with the other carbonyl group. The hydrogen bonding interaction is idealized by the fact that the system forms a six membered ring.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 41
hydrogen bond O
O
H
R'
O
H
O
OH H
R'
O
The hydrogen bonded arrangement allows the -carbonyl carboxylic acids to undergo a thermally induced decarboxylation reaction. The arrows in the diagram below help to assign the re-positioning of the electrons. Only when the system form a 6membered cyclic form is the decarboxylation facilitated, as indicated by the arrows.
O H
H
O R'
O
The decarboxylation is effected by heating the substrate at 100 C. The immediate product of the carboxylation is actually an enol but upon isolation, the material tautomerizes to the ketone.
O H
O R'
OH
heat -CO2
HO
O R'
H
H H
R'
The same chemistry takes place if the target molecule is a di-substituted acetone. The only difference is that after the introduction of the first alkyl group and before the saponification, the substrate is deprotonated again and quenched again with another electrophile. Once the introduction of two groups between the carbonyl is complete then the conversion to acid and subsequent decarboxylation can proceed.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 42
Some examples: O
O
O
O
1. NaOEt, EtOH OEt
O
OEt
2. (Me) 2CHCH2Br
+
2. H 3. heat, -CO 2
O
O
O
1. NaOEt, EtOH O
O
1. NaOH, H 2O
O
O
2. Ph
O
Cl Ph
1. NaOH, H 2O, then H 2. heat, -CO 2
+
O Ph O Sometimes the acid and thermolysis treatments can be addressed simultaneously: O
O
O
1. NaOEt, EtOH O
O O
2. CH3CH2CH2CH2Br
1. NaOH, H 2O 2. H 2O, H 2SO4, heat O
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 43
O
O O
1. NaOEt EtOH
1. NaOEt EtOH
2. MeI
2.
O
O O
Br
1. NaOH, H 2O, then H 2. heat, -CO 2
+
O H
The same chemistry can be performed on a wide range of -keto esters. O
H 1. NaOEt, EtOH CO2Et
O
CH2CH2CH2Br CO2Et
2. excess Br(CH 2)3Br 1. saponification + 2. H 3. heat O
H CH2CH2CH2Br 60%
b) Malonic Ester Synthesis The steps required in the malonic ester synthesis are analogous to those of the acetoacetic ester synthesis. One difference is the starting material, changing the methyl ketone to an ethyl ester creates a difference in the product: a substituted acetic acid derivative is obtained rather than a methyl ketone. Starting material: O
O
O O
vs
CHEM*3750 SCHWAN COURSE NOTES F13
O
O O
Chapter 2 Page 44
Final product:
O
O H
vs R
O R
Overall reaction of malonate synthesis O EtO
O
O EtO
OEt
R
O R'
O OEt
HO
H R
R'
substituted acetic acid Diethyl malonate * is the usual starting material. The anion of it is quenched with an electrophile. The product, also a malonate, is hydrolyzed to a diacid, which is then thermally decarboxylated. The result is the monosubstituted acetic acid. Disubstituted acetic acids are also available and can be obtained by following the same protocol as for disubstituted acetones. Diethyl malonate is slightly less acidic than acetyl acetone, with a pKa of 13.3, but the same chemistry can take place anyway.
CH2(CO2Et)2
1. NaOEt, EtOH 2. allyl bromide
CH2(CO2Et)2
2 eq. NaOEt EtOH
CO2Et 1. NaOH, H 2O C CO Et + 2 2. H 3O H 3. heat
Br(CH2)3Cl
CO2H
C(CO2Et)2 1. KOH, H 2O + 2. H 3O , heat
CO2H cyclobutanecarboxylic acid, 43%
*
1,3-Propanedioic acid is commonly known as malonic acid and so diesters are known as malonates.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 45
CH2(CO2Et)2
CO2Et C CO Et 2 H
1. NaOEt, EtOH 2. Br
1. NaOEt, EtOH 2. EtBr
CO2H C H
CO2Et C CO Et
1. KOH, H 2O
2
+
2. H 3O , heat
Clearly, the prominent structural feature in the acetoacetic esters and the malonates is the methylene group surrounded by two electron-stabilizing functionalities. Although there is a detailed presentation here of the chemistry of the systems with two esters and with a ketone and an ester, other functional groups may also be present. Indeed, a number of groups can serve the role of electron withdrawing groups in the general structure Z-CH2-Z'. Typical ones include of course esters and ketones, but one may also encounter other carbonyl containing compounds as well as nitriles, nitro compounds and sulfur or phosphorus containing groups. The final step of thermal decarboxylation is not always possible. Indeed, sometimes one wishes to maintain both of the groups in the molecule. Furthermore, there may be other methods for removing the particular functional groups once they have served their role.
Me O
O
1 eq. NaOH MeI, H 2O 70%
O
Me Me O
O
O
2 eq. NaOMe, MeI, MeOH 80% c) Robinson Annulation (SF10 19.6) Before introducing the chemistry and overall value associated with the Robinson Annulation, it is important to introduce a very important reaction known as the Michael Addition (SF10 19.7).
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 46
The Michael addition was originally defined as the addition of a malonate to the position of an , -unsaturated ketone, ester or nitrile. This strict definition has subsided and a Michael addition is now defined as the addition of any nucleophile to the position of an alkene bearing one or more electron withdrawing groups. Below are some examples using nucleophiles that have recently been introduced to you. Other carbon nucleophiles can be used in this chemistry as can amines, alcohols and sulfur nucleophiles.Note that the product of these reactions possesses two strong electron withdrawing groups and may well contain one or two carbonyl groups. The Robinson annulation begins with a Michael addition and is completed by an intramolecular aldol condensation. In general, there are other materials that can participate in Michael additions (but not necessarily the full Robinson annulation). Michael donors (nucleophile) O
O
R
-diketone anion
O
R
O
R'
O
Michael acceptors (electrophile)
OR'
-
R2CuLi
R
-keto ester anion dialkyl cuprate
O
conjugated ester OR
O NH2
enamine
N
O R
N C
-
O R
NO2
O
R
-
R'
-keto nitrile anion -nitro ketone anion simple enolate
R2NH or RNH2
amine
ROH
alcohol
RSH
thiol/mercaptan
CHEM*3750 SCHWAN COURSE NOTES F13
conjugated ketone
NO2
conjugated amide conjugated nitro compound
CN
conjugated nitrile
O O S R
conjugated sulfone
O O S OR
conjugated sulfonate
EWG
conjugated, electron deficient alkyne
Chapter 2 Page 47
Mechanism of Michael addition: O
O
O _
OEt O
_
O
H
H
H
O
O
O
O
_
O _
H
O
H
O
_
R O
R
O
O
O H
H O
-OEt + O
_
HOEt
R
O R
O
O
O
H
O
_
acidify O
R
O O
R
+ EtOH
Using the example in the mechanism above, where R = CH3, exposure of that material to base (usually from the original Michael addition reaction mixture), provides a means for the intramolecular aldol reaction. In this particular case, the anion of the methyl group may attack one of two equivalent ketone groups. In many cases there is a lone ketone and the reaction works perfectly well. From this point the regular aldol and its dehydration occur.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 48
O
O H O
O
CH3
O
H
_
base
O
O
CH3
O
O
O
O
CH _ 2
H
protonation & dehydration
O
-
O
O
The thermodynamics of the reaction make the formation of 6 membered ring most favorable. Hence the Robinson annulation is ideal for the formation of cyclohexenones including those fused to other rings. Overall, as stated, the Robinson annulation involves, first a Michael addition and then an aldol condensation.
Michael addition
O
Some examples:
C[H2 O ] CH3 aldol condensation
O
O + CO2Et
methyl vinyl ketone (MVK)
CHEM*3750 SCHWAN COURSE NOTES F13
O cyclohexenone
O
NaOH EtOH CO2Et
Chapter 2 Page 49
O
NaOMe
+
MeO
O
O
O
MeOH
ethyl vinyl ketone
MeO Michael addition product
O
MeO Robinson annulation product
.
. More, general examples of the annulation.
O O
O
O OEt
+
Michael
O OEt O
O Aldol C(O)OEt
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 50
O
O
O
O
Michael
+
O
Aldol
d) Other Conjugate Additions As you know, a number of nucleophiles add to the carbonyl carbon of aldehydes and ketones. The presence of a conjugated double bond creates an additional electrophilic site, and above it was shown that stabilized carbon anions will perform Michael additions on these substrates. Simple carbon nucleophiles such as Grignards will still attack at the C=O group. Organolithium reagents can attack at either electrophilic site and are not synthetically useful for this chemistry. Organocuprate reagents attack in a conjugate fashion and are easily generated using Grignard reagents in the presence of CuI. 1. EtMgBr Et2O + 2. H /H2O
HO Et
1. EtMgBr CuI, Et 2O + 2. H /H2O
O
O
H H Et
A number of other nucleophiles are also useful in conjugate addition reactions. The list includes amines (RNH2, RR’NH), alcohols (ROH), thiols (RSH) and HCN (prepared via KCN/H+). The addition of a nitrile is particularly synthetically useful since it can be converted to a carboxylic acid by hydrolysis.
O
HNMe O MeNH 2
O Ph
KCN/H
O
O
+
NC
CHEM*3750 SCHWAN COURSE NOTES F13
Ph
acid or base hydrolysis
HO
Ph O
Chapter 2 Page 51
A significant example of a Michael addition has recently been found in nature. The Michael reaction is critical to the operation of a rather promising anticancer drug, calicheamicin. In the first step in its operation the trisulfide bond is broken (1). The nucleophilic sulfide then adds in Michael fashion to give enolate 2. This addition changes the shape of the molecule, bringing the ends of the two acetylenes closer together. A cyclization occurs to give a di-radical (3), and this diradical abstracts hydrogen from the cancer cell's DNA, killing it. Calicheamicin depends for its action on the change of shape. Before the Michael reaction, the ends of the two acetylenes are too far apart to cyclize. They are freed to do so when the sulfur adds to the unsaturated carbonyl.
O
NHCOOMe
-
O
SSSMe
S
NHCOOMe
S
-
OR
HO
O
NHCOOMe
OR
HO
OR
HO
2 1
-
O
-
O
NHCOOMe
NHCOOMe
S
S
dead + cancer cells
HO H
OR
DNA
OR
HO
H 3
R = series of sugar moieties
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 52
PRACTICE PROBLEMS 5
You are capable of doing Questions 18.7-18.13, 18.15, 18.23, 18.27b,c,f, 18.30 19.16, 19.35, 19.46, 19.47 and 19.58a in SF10.
1. Why do esters that have only one hydrogen Claisen condensations?
to their carbonyl not work efficiently in
2. Suggest a mechanism, including curly arrows for the following reaction.
O
O OC2H5 + OC2H5
-
CH3CO2C2H5
O
1. C2H5O Na
+
C2H5OH + 2. H3O
CHCO2C2H5 O
3. Draw the product resulting from application of the following sequence of reagents.
O
-
O OC2H5
1. 1 eq.C2H5O Na C2H5OH
+
2. 0.5 eq. Br(CH2)3Br 3. OH/H2O + 4. H3O /H2O,
4. Will the accompanying compound decarboxylate when heated?
HO C
O
O
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 53
FROM PREVIOUS MID-TERM EXAMINATIONS 5. Provide a mechanism for the following transformation: O
-
CH3
O
+
1. EtO Na /EtOH +
CH3
CO2Et
2. mild H3O workup
CO2Et
6. Provide a synthetic route to 4-pentenyl phenyl ketone from diethyl malonate. You may use any reagents and organic compounds to complete your synthesis. Be sure to draw the structure of several of the stable organic intermediates along the way in order to be eligible for part marks.
O EtO
SOLUTIONS TO
O
O OEt
Ph
PRACTICE PROBLEMS 5
Q1. Recall why the Claisen condensation goes to completion. It is because the final step is the deprotonation of the product between their carbonyls. So, if the starting material has only one -H, and that H gets removed to initiate the reaction, then there are not any H’s remaining to be removed in that final step of the reaction. Hence instead of executing that last deprotonation, the alkoxy group will attack the keto part of the product and further reactions will take things back to starting material. Q2.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 54
O CH2 H
-
C2H5O Na
OEt
-
O
+
C2H5OH
O OEt
CH2
OC2H5 OC2H5 O
O
-
O OC2H5
O CH2 OEt OC2H5
H O CH OEt OC2H5
O
O -
+ O
OC2H5 O
-
O
CH OC2H5
OEt
O
CH -
O
O
OEt
OC2H5
O
O CCO 2C 2H5 -
CCO2C2H5
O
O
H +
-
OEt
+
introduce aq. acid at this stage
H OH2
O CHCO2C2H5 O
Q3.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 55
O
O
2
O -
+
2 eq.C2H5O Na
OC2H5
O
O -
O -
OC2H5 +
OC2H5
1 eq. BrCH2CH2CH2Br
O
O
-
O
O
O
-
O O
-
OH/H2O
C2H5O
O
OC2H5
O
O
CH2CH2CH2
CH2CH2CH2
+
H3O /H2O O OH
O
HO
O
O
O
O CH2CH2CH2
CH2CH2CH2
Q4. The reaction will not proceed since that would place a double bond in the bridge head position and the bicyclic nature of the compound does not allow such an arrangement, because of the huge amount of strain associated.
X HO C
O
O
OH
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 56
Q5. This mechanism emphasizes the reversibility of the Claisen/Dieckmann reactions and also the requirement for the existence of some -H’s to prevent the reversibility of the reactions. O
-
CH3
-
EtO O
+
EtO Na /EtOH
O
CH3 OEt
CO2Et
CH3 OEt
EtO -
O
O
H OEt -
O
CH3
EtO EtO
O
OEt H H
O
H
CH3
EtO
OEt
EtO
-
O
CH3
O
O +
O
CH3
EtO H O +
O EtO
-
CH3
O
OEt
CHEM*3750 SCHWAN COURSE NOTES F13
O
+ H OH2
EtO
-
OEt
CH3 O
introduce mild acid at this stage
Chapter 2 Page 57
Q6.
O
O
EtO
OEt +-
1. Na OEt/EtOH 2. Br O
O
EtO
O
O OEt
+-
1. Na OEt/EtOH
OEt Ph
EtO O
O
2. Ph
Cl +-
1. Na OH/H2O +
2. H3O /H2O
Ph
O
O
O
OH Ph
HO -2 CO2 O
note that the introduction of the 3rd carbonyl allows the thermal removal of both carboxylic acid functionalities _______________________________________________________ 8. The Wittig Olefination of Aldehydes and Ketones (SF10 16.10) The most popular method for the conversion of aldehydes and ketone to alkenes is the Wittig Reaction. Generally speaking it involves the reaction of a carbonyl compound with a phosphorus ylide. * The result is an alkene and a phosphine oxide. *
An ylide was originally defined as a compound that possesses a carbon anion directly beside a heteroatomic cation, where the formation of a multiple bond is usually not possible. The definition has eased in recent years and now encompasses any compound bearing adjacent positive and negative charges in situations where a formal multiple bond may or may not be possible. Other Popular methods for olefination chemistry may involve silicon or sulfur atoms rather than a phosphorus atom.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 58
O
+ _ R R"3P C R' ylide
+
+ R"3P O
+
C
R
R' alkene
R"3P O
A specific procedure is usually required to prepare the Wittig reagent. First one treats the halogen in an alkyl halide with a phosphine, usually triphenylphosphine. The result is an alkyltriphenylphosphonium salt. The salt is then treated with a strong base, usually butyllithium in order to generate the ylide.
Ph3P
+
R CH2 X
+ Ph3P CH2R + H
C
Li
-
C6H6
+
H
(n-BuLi)
+ Ph3P CH2R
_ + Ph3P CHR
-
+ X
+ butane + LiX
phosphonium ylide
Deprotonation of the phosphonium salt may also be achieved using alkoxides or NaH. Some ylides are stable. Most often the ylide is prepared in the flask and then is brought together with the carbonyl compound. Under these circumstances, the carbon anion portion of the ylide attacks the carbonyl group. The result is a phosphonium betaine, which readily closes to a cyclic isomer, an oxaphosphetane. To conclude the synthesis, the oxaphosphetane fragments to make the alkene and triphenylphosphine oxide. One of the driving forces of the reaction is the formation of the strong P=O bond.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 59
_ + Ph3P CHR
+
phosphonium ylide
R' R"
R
O
+ Ph3P
R' R" -
O
betaine
fast
R
R'
H
R"
Ph3P=O +
fragmentation
alkene
R
R' R"
Ph3P O oxaphosphetane
The Wittig reaction can be carried out in the presence of ether, ester, halogen and other multiple bond functionalities. It often affords a mixture of geometric isomers about the double bond, although sometimes the reaction is selective. Some examples:
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 60
+ PPh3X
NaOMe MeOH
PPh3
O
OC(O)Me H
OC(O)Me
-
HO , H 2O (saponification)
OH
Vitamin A 1
+
H +PPh3
-
+
Bu Li
PPh3
CHO
H
-
O O
+
Ph3P CH O
O CH
O
The Horner-Wadsworth-Emmons (HWE) olefination reaction (SF10 p. 760) follows a path similar to the Wittig olefination. The difference is the nature of the phosphorus containing starting material, in the HWE reaction, one employs a phosphonate ester. The HWE version of the olefination is often preferred since the E- isomer of the alkene is typically formed in high preference or exclusively. The base can be NaH, KOtBu or (Me3Si)2N-Li+.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 61
O EtO P EtO
R
O R EtO P EtO Na+
NaH THF
R is usually a conjugating group e.g, Ar, carbonyl, alkene
H
R' O
R + H
R'
O EtO P - + O Na EtO
The following is a key step in the total synthesis of callipeltoside A, a macrocyclic natural product known to have anti-tumour activity and to protect cell infected with the HIV virus. OR O
OR O
OR MeO
O
+
OR
O H
- + (Me3Si)2N Li . THF -78 °C to rt
O
EtO O P EtO
MeO
O
O H
R = SiMe2tBu Cl Cl
9. Reductive Conversion of C=O to CH2 A final important transformation of aldehydes and ketones is their conversion to methylene groups. The following two reagent systems are observed routinely with ketones and occasionally aldehydes. The Wolff-Kishner approach [There is no SF10 reference] utilizes H2NNH2/high boiling solvent/NaOH/heat while the Clemmensen reduction (SF10 15.9) involves refluxing the ketone in hydrochloric acid with a zinc amalgam [Zn(Hg)]. Clearly the Wolff-Kishner protocol is best for substrates that are not base sensitive while the Clemmensen procedure is preferred when there are no acid sensitive functionalities in the ketone. Each of these procedures are part of the family of reductions known as deoxygenations.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 62
O
H
H2NNH 2, NaOH (HOCH 2CH2)2O heat
H
47%
O Me
O
H2NNH 2, NaOH (HOCH 2CH2)2O H 2O, heat
CHO
Me 69%
CH3 HCl, Zn(Hg) OCH3
heat
OH
OCH3 OH
Both methods are particularly effective in tandem with electrophilic aromatic acylation reactions.
PRACTICE PROBLEMS 6
You are capable of doing Questions 16.7, 16.40, 19.15, 19.19, 19.20 and 19.43, 16.45 in SF10.
1. Name the steps involved in the following sequence and draw the important reaction intermediates. (MVK = methyl vinyl ketone)
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 63
N
1. MVK +
2. H3O
O
2. Draw the product of the sequence below. O Cl
Zn(Hg) HCl
AlCl3
3. Indicate on the arrows directly the reagents that would be required to give the chemistry shown.
O
O O
O
OH
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 64
4. Indicate in each of the boxes accompanying the arrows, provide the full set of conditions, reagents and/or organic molecules that are required to give the chemistry shown. CH2
O
Ph
O
O
O
O
5. Provide a mechanism that accounts for the reaction shown. Be sure to understand why this reaction proceeds.
O
O
-
O
+
1. MeO Na /MeOH +
MeO
2. mild H3O
MeO
O H
6. FROM A PREVIOUS MIDTERM: Draw the product of the treatment shown.
O
1. LDA, -78 °C, THF 2. O O Ph
Ph
3. CH3-Br
7. Write down as many reactions as you can think of for: a) the formation of cyclohexanone and b) the reactions of cyclohexanone CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 65
SOLUTIONS TO
PRACTICE PROBLEMS 6
Q1. N
+
O
Michael addition
-
N+
O
+
H 3O
hydrolysis O
O
OH
O +
intramolecular
N+ H H
aldol
H
O+
+O H
OH
O
HO
Q2.
O Cl AlCl3
Zn(Hg) HCl
Q3.
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 66
O 1. O3 2. Zn O O -
O -
OH/H2O
or + H3O /H2O
OH/H2O
or + H3O /H2O
OH
and heat
Q4.
CH2
Ph3P=CH2
Ph
Zn(Hg)/HCl
1. pyrrole/H+/toluene 2. PhC(O)Cl + 3. H3O
O
O
or NH2NH2/KOH HOCH2CH2OH/heat
O -
OH/H2O heat O
O
CHEM*3750 SCHWAN COURSE NOTES F13
-
OH/H2O heat
O
Chapter 2 Page 67
Q5. O
O
O -
MeO
MeO
-
O
-
O
OMe
MeO
O
+
MeO
H
MeO
(or through the mediation of solvent
O MeO
O
-
O H
MeO
O
O OMe H
-
+
H
MeO
MeO
O
-
+ MeO
O H
_
MeO
H H
O
H +
The reaction sequence is fully reversible, but the availability of H's on the product prevent the backwards reaction. The starting material does not have H's between the carbonyls and hence gets attacked at the ketone carbonyl
H2O H
O
O H
MeO
Q6.
O
1. LDA, -78 °C, THF 2. O O Ph
Ph
O Ph
O
Me
O Ph
3. CH3-Br
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 68
Q7. a) Selected Syntheses: O
O
MeO
NaOMe/MeOH
OMe O
O CO2Me 1. OH/H O 2 + 2. H3O 3. , -CO2 -
O 1. O3 2. Zn
b) Some reactions:
O NH2NH2/KOH HOCH2)2, or Clemmenson cond'ns R
O
R'
Ph3P=CRR'
O 2
KOH
O
H 2O
CHEM*3750 SCHWAN COURSE NOTES F13
or other mixed aldol reactions
Chapter 2 Page 69
O
R OH 1. RMgX +
2. mild H3O
O
O
1. Br
O
HO
OEt
OEt
Zn, toluene +
2. mild H3O
O
O R
1. LDA, -78 °C, THF 2. RMgX
O
O Br2, H
O
+
O
Br
O
KOH/H2O
O
NH H+/ toluene/-H2O
N
O R
1. R-X +
2. H3O
_______________________________________________________
CHEM*3750 SCHWAN COURSE NOTES F13
Chapter 2 Page 70
