Lecture 4

Question 1

Chapter 16 recap

Answer 1

Go to the introduction slides and follow the instructions there.

Okay, everything on that paper you WILL be tested on trust, matter fact everything in this lecture you also will be tested on so pay close attention, and don’t get lazy.

I also recommend to fully go through lecture 5 and 6 of organic 2 but mainly focus on the highlighted part.

Question 2

What is the Baeyer–Villiger oxidation?

Answer 2

It is a reaction where aldehydes and ketones react with the conjugate base of a peroxyacid to form carboxylic acids and esters. It is an oxidation reaction because the number of C¬O bonds increases. A particularly good reagent for a Baeyer-Villiger oxidation is peroxytrifluoroacetate ion (Shown in written notes).

Question 3

If the two alkyl substituents attached to the carbonyl group of the ketone are different, then on what side of the carbonyl carbon is the oxygen inserted?

Answer 3

To answer this question, we need to look at the mechanism of the reaction.

Look at the wall

Question 4

What is a ylide?

Answer 4

a ylide is a compound with opposite charges on adjacent covalently bonded atoms that have complete octets.

The phosphonium ylide can be written in the doubly bonded form because phosphorus can have more than eight valence electrons. This is shown in the digital notes

Question 5

How is a phosphonium ylide made?

Answer 5

Look at the wall.

Question 6

What is The Wittig reaction?

Answer 6

It is a reaction where an aldehyde or a ketone reacts with a phosphonium ylide (“ILL-id”) to form an alkene. This reaction interchanges the doubly bonded oxygen of the carbonyl compound with the doubly bonded carbon group of the phosphonium ylide as shown in digital notes.

The Wittig reaction is a concerted [2 + 2] cycloaddition reaction. It is called a [2 + 2] cycloaddition reaction because, of the four
electrons involved in the cyclic transition state, two come from the carbonyl group and two come from the ylide.

Question 7

What is the mechanism for The Wittig reaction?

Answer 7

Look at the wall

Question 8

What is the importance of Wittig’s reaction?

Answer 8

1. The Wittig reaction is a powerful way to make an alkene because the reaction is completely regioselective—only one alkene is formed.
2. The Wittig reaction is the best way to make a terminal alkene because other methods form a terminal alkene only as a minor product. (shown in digital notes).

Question 9

What is a limitation of Wittig’s reaction?

Answer 9

A limitation of the Wittig reaction is that when it is used for the synthesis of an internal alkene, a mixture of E and Z stereoisomers is generally formed.

Question 10

What should be considered when synthesizing an alkene using a Wittig reaction? (Retrosynthetic Analysis)

Answer 10

The first thing you must do is decide which
part of the alkene should come from the carbonyl compound and which part should come from the ylide. If both sets of carbonyl compound and ylide are available, the better choice is the set that requires the less sterically hindered alkyl halide for the synthesis of the ylide via an SN2 reaction.

For the synthesis of 3-ethyl-3-hexene, for example, it is better to use a three-carbon alkyl halide for the ylide and a five-carbon carbonyl compound than a five-carbon alkyl halide for the ylide and a three-carbon carbonyl compound, because it is easier to form a ylide from a primary alkyl halide (1-bromopropane) than from a secondary alkyl halide (3-bromopentane).

Question 11

Random note i remebered (srry)

Answer 11

The NaBH4 is less reactive than the LiAlH4 since:

1. Al is larger (leading to a smaller overlap with the hydrogen)
2. Al and H have a larger electronegative difference (Al is less electronegative)

Both these make the H more electrophilic and more likely to react.

Question 12

What are a,b-unsaturated carbonyl compounds?

Answer 12

They are carbonyl compounds that have a double bond between the alpha and beta positions.

Question 13

What is interesting about the a,b-unsaturated carbonyl compounds?

Answer 13

The resonance contributors for an a,b-unsaturated carbonyl compound show that the molecule has two electrophilic sites: the carbonyl carbon and the b-carbon (This is shown in written notes).

This means that a nucleophile can add either to the carbonyl carbon or to the b-carbon.

Question 14

What is direct addition and conjugate addition?

Answer 14

direct addition or 1,2-addition is the Nucleophilic addition to the carbonyl carbon.

Conjugate addition or 1,4-addition is the nucleophilic addition to the b-carbon. It is called 1,4 because it occurs at the 1- and 4-positions (1 is oxygen 4 is the b carbon). The initial product of 1,4-addition is an enol, which tautomerizes to a ketone or to an aldehyde (Shown in written notes).

Whether the product obtained from nucleophilic addition to an a,b-unsaturated aldehyde or ketone is the direct addition product or the conjugate addition product depends on the nature of the nucleophile and the structure of the carbonyl compound

Question 15

What is the nature of conjugate and direct addition?

Answer 15

• Addition to the b-carbon (conjugate addition) is generally irreversible.

-Addition to the carbonyl carbon (direct addition) can be reversible or irreversible

Question 16

When is a reaction under kinetic and or thermodynamic control?

Answer 16

When two competing reactions are both irreversible, the reaction is under kinetic control, and when one or both of the reactions is reversible, the reaction is under thermodynamic control

Question 17

What are the conditions that allow for conjugate or direct addition?

Answer 17

• Conjugate addition:

1. Nucleophiles that are weak bases
   form conjugate addition products.
2. Nucleophiles that are strong based form conjugate addition products with
   less reactive carbonyl groups.

• Direct addition:

1. Nucleophiles that are strong bases
   form direct addition products with
   reactive carbonyl groups.

Question 18

Why do nucleophiles that are weak bases form conjugate addition products?

Answer 18

When the nucleophile is a weak base, such as a halide ion, a cyanide ion, a thiol, an alcohol, or an amine (IMP GROUPS), then direct addition is reversible, because a weak base is a good leaving group. Therefore, the reaction is under thermodynamic control.

The reaction that prevails when the reaction is under thermodynamic control is the one that forms the more stable product. The conjugate addition product is always the more stable product because it retains the very stable carbonyl group. Therefore, weak bases form conjugate
addition products.

This is depicted in the digital notes.

Question 19

Why do nucleophiles that are strong bases
form direct addition products with
reactive carbonyl groups but form
conjugate addition products with
less reactive carbonyl groups?

Answer 19

When the nucleophile is a strong base, such as a Grignard reagent or a hydride ion, then direct addition is irreversible. Now, because the two competing reactions are both irreversible, the reaction is under kinetic control.

The reaction that prevails when the reaction is under kinetic control is the one that is faster.

Therefore, the product depends on the reactivity of the carbonyl group. Compounds with reactive carbonyl groups form primarily direct addition products because for those compounds, direct addition is faster. Compounds with less reactive carbonyl groups (High steric hindrance) form conjugate addition products because for those compounds, conjugate addition is faster.

Question 20

What addition reaction does a,b-unsaturated aldehydes and ketones undergo with strong bases?

Answer 20

Aldehydes have more reactive carbonyl groups than ketones do, so aldehydes form primarily direct addition products with hydride ions and Grignard reagents. Compared with aldehydes, ketones form less of the direct addition product and more of the conjugate addition product, because ketones are more sterically hindered and, therefore, less reactive than
aldehydes.

Mental note: in the example they give for the ketone 51% was the direct addition product and 49% was the conjugate addition product. For aldehyde, it was 97% direct addition product

Question 21

What are Hard and Soft Electrophiles and Nucleophiles?

Answer 21

Electrophiles and nucleophiles can be classified as either hard or soft. Hard electrophiles and nucleophiles are more polarized than soft ones.

Hard nucleophiles (strong bases) are like chihuahuas, they are charged, small and aggressive and they typically undergo direct addition, while soft nucleophiles (weak bases) are less aggressive, and are uncharged or have delocalized charge.

Hard nucleophiles prefer to react with hard electrophiles, and soft nucleophiles prefer to react with soft electrophiles.

Question 22

What addition reaction does Grignard Reagents and Organocuprates bring about?

Answer 22

Grignard reagents are strong bases and, therefore, add irreversibly to carbonyl
groups. Thus, the reaction is under kinetic control. If the carbonyl compound is reactive, the reaction with the Grignard reagent will form the direct addition product. If, however, the rate of direct addition is slowed down by steric hindrance, reaction with the Grignard reagent will form the conjugate addition product because conjugate addition then becomes the faster reaction.

Only conjugate addition occurs when organocuprates react with a,b-unsaturated aldehydes and ketones, due to high steric hindrance of the base and it being generally weak. An example of an organocuprate is shown in the notes.

Question 23

What is an important note and ig a summary of this

Answer 23

Okay, first the summary because I feel the important info might have slipped:

So if the base and the carbonyl compound have steric hindrance and the base is a soft nucleophile then it will undergo conjugate addition.

If the base and the carbonyl compound have little steric hindrance and the base is strong then it will undergo direct addition.

Anything in between can occur each way. Be smart with it (y3ani if you see an organocuprate know it is conjugate addition).

A good slide for the lecture is shown in the notes.

The important note is that a base is less nucleophilic the bulkier it is.

Question 24

What is the reaction of nucleophiles with a,b-unsaturated carboxylic acid derivatives?

Answer 24

Nucleophiles react with a,b-unsaturated carboxylic acid derivatives with reactive carbonyl groups, such as acyl chlorides, at the carbonyl group, forming nucleophilic acyl substitution products. Conjugate addition products are formed from the reaction of nucleophiles with less reactive
carbonyl groups, such as esters and amides.

Notice that a,b-unsaturated carboxylic acid derivatives undergo nucleophilic acyl
substitution rather than direct nucleophilic addition because they have a group that can be replaced by a nucleophile. (SO direct addition reaction never happens)

Question 25

What is the general chemistry of a C-H bond?

Answer 25

Hydrogen and carbon have similar electronegativities, which means that the two atoms share the electrons that bond them together almost equally. Consequently, a hydrogen bonded to a carbon is usually not acidic. This is particularly true for hydrogens bonded to sp3 carbon because these carbons are the most similar to hydrogen in electronegativity. For example, the pKa of ethane is greater than 60.

Question 26

What is the case when the hydrogen bonded to an sp3 carbon is acidic?

Answer 26

When the sp3 carbon is adjacent to a carbonyl carbon, it is considered to be acidic.

Question 27

What is a carbon acid?

Answer 27

A compound that contains relatively acidic hydrogen bonded to an sp3 carbon is called a carbon acid.

The pKa values of some carbon acids are shown in Table 17.1

Question 28

Why is the hydrogen bonded to an a-carbon more acidic than hydrogens bonded to other sp3 carbons?

Answer 28

A hydrogen bonded to an a-carbon is more acidic than hydrogens bonded to other sp3 carbons because the base formed when a proton is removed from an a-carbon is relatively stable and the more stable the base, the stronger its conjugate acid

Question 29

Why is the base formed by removing a proton from an a-carbon more stable than bases formed by removing a proton from other sp3 carbons?

Answer 29

To understand this we make a comparison between the base formed by ethane and the base formed from a a-carbon:

Why is the base formed by removing a proton from an a-carbon more stable than bases formed by removing a proton from other sp3 carbons?

In contrast, when a proton is removed from an a-carbon, two factors combine to increase the stability of the base that is formed. First, the electrons left behind when the proton is removed are delocalized, and electron delocalization increases stability. More importantly, the
electrons are delocalized onto an oxygen, an atom that is better able to accommodate them because it is more electronegative than carbon. This is shown in digital notes.

note that Nitroalkanes, nitriles, and N,N-disubstituted amides also have a relatively acidic a-hydrogen because, in each case, the electrons left behind when the proton is removed can be delocalized onto an atom that is more electronegative than carbon.

Question 30

Following the reason why a base formed from an a-carbon is more stable. Would ketones/aldehydes have a more acidic hydrogen than carbon?

Answer 30

Yes they would. This is because the electrons left behind when a proton is removed from the a-carbon of an ester are not as readily delocalized onto the carbonyl oxygen as they would be in an aldehyde or a ketone. This is because the oxygen of the OR group of the ester also has a lone pair that can be delocalized onto the carbonyl oxygen. Thus, the lone pair on carbon and the lone pair on oxygen compete for delocalization onto the same oxygen.

This is depicted in digital notes.

Question 31

Why are a-hydrogens bonded to carbons flanked by two carbonyl groups generally more acidic?

Answer 31

They are more acidic because the electrons left behind when the proton is removed can be delocalized onto two oxygens.

Question 32

What are b-diketones and b-keto esters?

Answer 32

a b-diketone is a ketone that has a second keto group at the b-position

a b-keto ester is a ketone that has a ester group at the b-position.

b-Diketones have lower pKa values than b-keto esters because, as we saw, electrons
are more readily delocalized onto the carbonyl oxygen of a ketone than they are onto the carbonyl oxygen of an ester.

This is also depicted in the notes.

Question 33

What is the fraction of tautomers for a b-diketone?

Answer 33

15% enol tautomer and 85% keto tautomer

This is much much greater than usual. This is because the enol tautomer is stabilized both by intramolecular hydrogen bonding and by conjugation of the carbon-carbon double bond with the second carbonyl group. This is shown in digital notes.

Question 34

Which tautomer is more stable in phenol?

Answer 34

In phenol, its enol tautomer is more stable than its keto tautomer because the enol
tautomer is aromatic, but the keto tautomer is not. This is shown in the notes

Usually, the keto tautomer is more stable so this is very unusual.

Question 35

What is the mechanism for the acid and based catalyzed mechanism of Keto–Enol Interconversion?

Answer 35

Some repetition occurs here specifically with the acid Acid-Catalyzed Keto–Enol Interconversion. Look at the wall ;)

Notice 2 things:

• The way the equilibria arrows are shown indicates that the intermediate is always less stable than the enol and keto tautomers
• The steps are reversed in the base- and acid-catalyzed interconversions. In the base-catalyzed reaction, the base removes a proton from an a-carbon in the first step and the oxygen is protonated in the second step. In the acid-catalyzed reaction, the oxygen is protonated in the first step and the proton is removed from the a-carbon in the second step.

Question 36

What happens when Br2, Cl2, or I2 is added to a solution of an a-carbon aldehyde or ketone?

Answer 36

What happens depends on the conditions. This reaction can be catalyzed by either an acid or a base. In the acid-catalyzed reaction, the halogen replaces one of the hydrogens. In the base-catalyzed reaction, the halogen replaces all of the a-hydrogens.

These reactions are known as a-substitution reactions because one electrophile (Br+) is substituted for
another (H+) on the a-carbon.

Question 37

Why is only one a-hydrogen substituted in acid-catalyzed conditions but two a-hydrogens are substituted in base-catalyzed conditions?

Answer 37

In basic conditions, each successive halogenation is more rapid than the previous one because the electron-withdrawing halogen atom increases the acidity of the remaining a-hydrogen (allowing for the base catalyst to remove it more easily). This is why all of the a-hydrogens are replaced by halogens.

Under acidic conditions, on the other hand, each successive halogenation is slower than the previous one because the electron-withdrawing halogen atom decreases the basicity of the carbonyl
oxygen, thereby making protonation of the carbonyl oxygen (the first step in the acid-catalyzed reaction) less favourable.

Question 38

How is keto-enol interconversion related to a-substitution?

Answer 38

Actually, keto–enol interconversion is an a-substitution reaction in which hydrogen serves as both the electrophile that
is removed from the a-carbon and the electrophile that is added to the a-carbon when the enol or enolate ion reverts back to the keto tautomer.

In my opinion, this statement is very far-fetched

Question 39

How can we halogenate the a-carbon of carboxylic acids?

Answer 39

When we try to halogenate the a-carbon of carboxylic acids we are into that problem where the carboxylic acid cannot undergo substitution reactions at the a-carbon because a base removes a proton from the OH group instead of from the a-carbon, since the OH group is more acidic.

If, however, a carboxylic acid is treated with PBr3 and Br2, the a-carbon will be brominated. This halogenation reaction is called the Hell–Volhard–Zelinski reaction or, more simply, the HVZ reaction which is covered in lecture 3. when you examine the reaction, that a-substitution occurs because an acyl bromide, rather than a carboxylic acid, is the compound that undergoes a-substitution. Do the mechanism (ACTUALLY DO IT BECAUSE IF YOU ARE NOT ABLE TO WE MIGHT ASWELL FAIL)

Question 40

How can we replace the a-halogen of Carbonyl Compounds?

Answer 40

A halogen attached to the a-carbon of a carbonyl compound can be replaced only by a poor nucleophile (a weak base). A strong base will form a,b-unsaturated carbonyl compound by removing a proton from the b-carbon and eliminating the bromide ion. Recall that a weak base favours substitution and a strong base favours elimination.

Question 41

What does the amount of carbonyl compound converted to an enolate ion depend on?

Answer 41

The amount of carbonyl compound converted to an enolate ion depends on the pKa of the carbonyl compound and the particular base used to remove the a-hydrogen.

Question 42

What is the base generally used to remove an a-hydrogen and why?

Answer 42

LDA (lithium diisopropylamide) is generally used to remove an a-hydrogen.

This is because the product acid from the reaction (diisopropylamine, or DIA) is a much weaker acid than the reactant acid (the ketone). Whereas if we use other bases like hydroxide the product acid is water which is a stronger acid than the starting ketone (Recall that the equilibrium of an acid-base reaction favours dissociation of the strong acid and formation of the weak acid).

This is why LDA is the base of choice for those reactions that require the carbonyl compound to be completely converted to an enolate ion before it reacts with an electrophile.

This is all depicted in the digital notes

Question 43

How is LDA prepared?

Answer 43

LDA is easily prepared by adding butyllithium to diisopropylamine in THF at -78 °C (that is, at the temperature of a dry ice/acetone bath.) Reaction shown in DIGITAL notes.

Question 44

Why might it be problematic to use LDA?

Answer 44

sing a nitrogen base to form an enolate ion can be a problem because a nitrogen base can also react as a nucleophile and add to the carbonyl carbon (Section 16.8). However, the two bulky isopropyl substituents attached to the nitrogen of LDA make it difficult for the nitrogen to get close enough to the carbonyl carbon to react with it. Consequently, LDA is a strong base but a poor nucleophile—that is, it removes an a-hydrogen much faster than it adds to a carbonyl carbon.

Question 45

Why is it important to alkylate the a-carbon?

Answer 45

Putting an alkyl group on the a-carbon of a carbonyl compound is an important reaction because it gives us another way to form a carbon-carbon bond.

Question 46

What is the mechanism for alkylating the a-carbon?

Answer 46

Look at the wall. Note: Although an enolate ion has two resonance contributors, for the sake of simplicity, only the resonance contributor with the negative charge on carbon is shown in many of the reactions, still in the exam draw both. So this way you can get intuition. If there are tautomers, recognise them, it will probably be the key to the question.

Question 47

What happens if we use a weaker base such as hydroxide when alkylating a carbon?

Answer 47

As mentioned before, the weaker base of OH will form only a small amount of the enolate ion at equilibrium (Section 17.6). Therefore, most of the hydroxide will be present when the alkyl halide is added to the reaction mixture. As a result, the major product of the reaction will be an alcohol
formed from an SN2 reaction of hydroxide ion with the alkyl halide, and the alkylated ketone will be a minor product.

Question 48

What happens when you alkylate unsymmetrical ketones?

Answer 48

If the ketone is unsymmetrical and has hydrogens on both a-carbons, two monoalkylated products can be obtained because either a-carbon can be alkylated.

One product will be formed from a kinetic enolate ion and one product will be formed from a thermodynamic enolate ion. The relative amounts of the two products depend on the reaction conditions

Question 49

What is the kinetic enolate ion and when will its product be the major one?

Answer 49

Recall back to the beginning of the lecture and redefine what kinetic control is.

The kinetic enolate ion is the one that is generally formed faster. The a-hydrogen that is removed to make this enolate ion is more accessible to the base and it is slightly more acidic. Because an enolate ion is formed faster, the product formed from it will be the major product if the reaction is carried out under conditions (LDA at -78 °C) that cause the reaction to be irreversible.

Question 50

What is the thermodynamic enolate ion and when will its product be the major one?

Answer 50

The thermodynamic enolate ion is the one that is more stable since it has a more stable double bond (most likely due to more alkyl substitution). Therefore the product it forms is the major one if the reaction is carried out under conditions that cause enolate ion formation to be reversible (LDA at 0 °C).

Question 51

How can we alkylate the a-carbon using an enamine intermediate?

Answer 51

It is very straightforward since enamines react with electrophiles in the same way that enolate ions do. aldehydes and ketones can be alkylated at the a-carbon by first forming an enamine of the carbonyl compound. This is shown on the wall.

Note that In addition to using an enamine to alkylate the a-carbon of an aldehyde or ketone, it can also be used to acylate the a-carbon.

Question 52

What is the main advantage of using enamine intermediate?

Answer 52

The main advantage to using an enamine intermediate to alkylate an aldehyde or a ketone is that it forms a monoalkylated product without having to use a strong
base (LDA).

Question 53

What are the two possible ways of alkylating the B-carbon?

Answer 53

1. Via enamine
2. Michael reaction

Question 54

How can you alkylate the B-carbon using an enamine?

Answer 54

Enamines, which are relatively weak bases, attach the a-carbon of an aldehyde or ketone (that they were derived from) to the
b-carbon of an a,b-unsaturated carbonyl compound via conjugate addition. The product is a 1,5-dicarbonyl compound.

Look at the wall for the mechanism.

Question 55

What is Michael’s reaction and how is it used to alkylate a b-carbon?

Answer 55

Michael’s reaction is the reaction where the nucleophile in a conjugate addition reaction is an enolate ion.

Since the enolate ion is a weak base it will then add (via conjugate addition) to the b-carbon of a,b-unsaturated aldehydes and ketones, effectively alkylating the b-carbon. These enolate ions also add to the b-carbon of a,b-unsaturated esters and amides
because of the low reactivity of their carbonyl groups.

Question 56

What are the Enolate ions that work best in Michael reactions?

Answer 56

Enolate ions that work best in Michael reactions are formed from carbon
acids that are flanked by two electron-withdrawing groups—that is, enolate ions of b-diketones, b-diesters, b-keto esters, and b-keto nitriles.

Question 57

What is the mechanism of Michael’s reactions?

Answer 57

Look at the wall. It is with the enamine one.

Question 58

What is aldol addition?

Answer 58

An aldol addition is a reaction in which one molecule of a carbonyl compound—after a proton is removed from an a-carbon—reacts as a nucleophile and adds to the electrophilic carbonyl carbon of a second molecule of the carbonyl compound.

This reaction can occur between two molecules of an aldehyde or two molecules of a ketone. In each case, the reaction forms a new C¬C bond that connects the a-carbon of one molecule (the nucleophile)
and the carbon that was originally the carbonyl carbon of the other molecule (the electrophile). This is depicted in the digital notes.

Note: Since the addition reaction is reversible, good yields of the addition product are obtained only if it is removed from the solution as it is formed.

Question 59

How are good yields obtained in aldol addition (imppp)?

Answer 59

Since the aldol addition reaction is reversible, good yields of the addition product are obtained only if it is removed from the solution as it is formed.

Question 60

What are a B-hydroxyaldehyde and B-hydroxyketone?

Answer 60

They are products of the aldol addition:

• When the reactant is an aldehyde, the product is a B-hydroxyaldehyde, which is why the reaction is called an aldol addition (“ald” for aldehyde, “ol” for alcohol).
• When the reactant is a ketone, the
  product is a B-hydroxyketone.

Question 61

What is the mechanism for the aldol addition?

Answer 61

Look at the wall.

Question 62

What is a retro-aldol addition?

Answer 62

Because an aldol addition is reversible, when the product of an aldol addition is heated with hydroxide ion and water, the aldehyde or ketone that formed the
aldol addition product can be regenerated. This is known as the retro-aldol addition and is depicted on the wall.

Question 63

What are condensation reactions?

Answer 63

A condensation reaction is a reaction that combines two molecules while removing a small molecule

Question 64

What are the condensation reactions tackled in this course?

Answer 64

1. Aldol condenstation
2. Claisen condensation
3. Malonester synthesis
4. Acetoacetic ester synthesis

Question 65

What is the aldol condensation reaction?

Answer 65

it is the aldol addition reaction with the extra step of dehydrating the product.

aldol condensation forms an A,B-unsaturated aldehyde or an A, B-unsaturated ketone, called enones (“ene” for the double bond and “one” for the carbonyl group)

Question 66

Why is the dehydration of aldol addition products easier than the dehydration of normal alcohols?

Answer 66

The b-hydroxyaldehyde and b-hydroxyketone products of aldol addition reactions are easier to dehydrate than many other alcohols because the double bond formed when the compound is dehydrated is conjugated with a carbonyl group. Conjugation increases the stability of the product and, therefore, makes it easier to form

Question 67

What is something special about the dehydration step of aldol condensation?

Answer 67

Like any alcohol dehydration, aldol condensation takes place in acidic conditions. But, Unlike alcohols that can be dehydrated only under acidic conditions, b-hydroxyaldehydes and b-hydroxyketones can also be dehydrated under basic conditions.

Question 68

What is the E1cB reaction (elimination
unimolecular conjugate base)?

Answer 68

a two-step elimination reaction that forms a carbanion intermediate. E1cB reactions occur only when the carbanion can be stabilized by electron delocalization.

it is a third kind of elimination reaction in which the base-catalyzed dehydration proceeds by it.

Question 69

What is the mechanisms for the dehydration of b-hydroxyaldehyde and b-hydroxyketone under acidic and basic ( E1cB) conditions?

Answer 69

Look at the wall.

Question 70

In aldol condensations, is heat always required for dehydration?

Answer 70

no, Sometimes dehydration occurs under the conditions in which the aldol addition is carried out, without requiring additional heat. For example, in the reaction shown in digital notes, the aldol addition product
loses water as soon as it is formed because the new double bond is conjugated not only with the carbonyl group but also with the benzene ring. (Recall that the more stable the alkene, the easier it is formed.)

Question 71

What is crossed aldol addition and how many products are formed from it?

Answer 71

if two different carbonyl compounds are used in an aldol addition—known as a crossed aldol addition—four products can be formed because the reaction with hydroxide ion can form two different enolate ions (A- and B-) and each enolate ion can react with either of the two carbonyl compounds (A or B).

A reaction that forms four products clearly is NOT a synthetically useful reaction.

Question 72

How can we obtain primarily one product in the crossed aldol addition?

Answer 72

Case 1 - One Carbonyl Compound
Does Not Have A-Hydrogens: if one of the carbonyl compounds does not have any a-hydrogens it cannot form an enolate ion. That cuts the possible products from four to two. Then, if the carbonyl compound with a-hydrogens is added slowly to a solution of the carbonyl compound without a-hydrogens and a base, the chance that the carbonyl compound with a-hydrogens, after forming an enolate ion, will react with another molecule of carbonyl compound with a-hydrogens will be minimized because there is a much greater concentration of the carbonyl compound without a-hydrogens. Thus, the possible products are cut to essentially one. This crossed aldol condensation is sufficiently important to be given its own name—the Claisen-Schmidt condensation.

Case 2 - in the next flash card felt bad putting them into one

Question 73

Case 2

Answer 73

Case 2 - Both carbonyl compounds have A-Hydrogens: If both carbonyl compounds have a-hydrogens, primarily one aldol addition product can be formed if LDA is used to remove the a-hydrogen from the carbonyl carbon that is needed for the enolate ion. Because LDA is a strong base, all of the carbonyl compound is converted to an enolate ion, so none of that carbonyl compound is left for the enolate ion to react within an aldol addition. Therefore, an aldol addition does not occur until the second carbonyl compound is added to the reaction mixture. If the second carbonyl compound is added slowly, the chance that it will form an enolate ion and then react with another molecule of the same carbonyl compound will be minimized.

Question 74

What is the strategy to determine the starting materials needed for the synthesis of a compound formed by an aldol addition and aldol condensation?

Answer 74

This is shown in digital notes. It is nicer to show the whole page ykk. This is lowkey important.

Question 75

What is the Claisen condensation?

Answer 75

The Claisen condensation reaction is when two molecules of an ester undergo a condensation reaction. In this reaction, an alcohol is the small molecule removed. The product of a Claisen condensation is a B-keto ester

Connection note: The chemistry behind Claisen condensation is the same as the aldol addition, where one molecule of carbonyl compound is the nucleophile and a second molecule is the electrophile.

Question 76

What is the mechanism for the Claisen condensation?

Answer 76

Look at the wall

Question 77

What are the differences between Claisen and Aldol condensations?

Answer 77

1. the Claisen condensation is a nucleophilic acyl substitution reaction, whereas the aldol addition is a nucleophilic addition reaction. Since the carbon bonded to the negatively charged oxygen in an ester is also bonded to a group that can be eliminated, whereas the carbon bonded to the negatively charged oxygen in an aldehyde or a ketone is not bonded to such a group
2. In the last step, acid is used differently: in aldol, the acid (or sometimes base as we saw) is used to dehydrate the addition product. In Claisen, we will see in flashcard 79 ;)

Question 78

What are the two requirements for Claisen Condensation and why?

Answer 78

A successful Claisen condensation requires an ester with at least two α-hydrogens and a stoichiometric amount of base, rather than a catalytic amount.

This requirement arises because the Claisen condensation is reversible and initially favours the reactants, as they are typically more stable than the β-keto ester product. To drive the reaction to completion, a proton is removed from the β-keto ester intermediate.

This deprotonation occurs readily because the central α-carbon of the β-keto ester is flanked by two carbonyl groups, significantly increasing the acidity of its α-hydrogen compared to the α-hydrogen of the starting ester.

This is depicted in notes to make it easier to understand.

Question 79

Why is an acid added as a second step in Claisen condensation?

Answer 79

When the reaction is over, addition of acid to the reaction mixture reprotonates the b-keto ester anion and protonates the alkoxide ion, which prevents the reverse reaction from occurring. Lowkey neutralize.

Question 80

What is crossed Claisen condensation?

Answer 80

A crossed Claisen condensation is a condensation reaction between two different esters. Like a crossed aldol addition, a crossed Claisen condensation is a useful reaction only if it is carried out
under conditions that foster the formation of primarily one product. Otherwise, the reaction forms a mixture of products that are difficult to separate.

Question 81

How can we obtain primarily one product in the crossed Claisen condensation?

Answer 81

• Primarily one product will be formed from a crossed Claisen condensation if one of the esters has no a-hydrogens (and, therefore, cannot form an enolate ion) and the ester with a-hydrogens is added slowly to a solution of the ester without a-hydrogens and the alkoxide ion.
• If both esters have a-hydrogens, primarily one product can be formed if LDA is used to remove the a-hydrogen to form the ester enolate ion. The other ester is then added slowly to maximize the chance that it will react with the enolate ion and minimize the chance that it will form an enolate ion and react with another molecule of its parent ester.

Question 82

What are the other cross-condensation reactions?

Answer 82

A ketone can undergo a crossed condensation with an ester. The product is a 1,3-dicarbonyl compound. There 3 cases to consider:

1. A b-keto aldehyde is formed when a ketone condenses with a formate ester.
2. A b-keto ester is formed when a ketone condenses with diethyl carbonate.
3. A b-diketone is formed when a ketone condenses with a normal ester.

General note, we want one primary product so the two strategies discussed above (i.e adding slowly and LDA) is used here so be smart about it. This is all shown in notes.

Question 83

What is the Dieckmann condensation?

Answer 83

It is an intramolecular Claisen
condensation that occurs in 1,6-diesters (forms a 5-membered ring) and 1,7-diester (forms a 6-membered ring) upon the addition of a base.

The Dieckmann condensation, like the Claisen condensation, can be driven to completion by carrying out the reaction with enough base to remove a proton from the a-carbon of the b-keto ester product. When the reaction is over, acid is added to reprotonate the condensation product and neutralize any remaining base.

Question 84

What is the mechanism for the Dieckmann condensation?

Answer 84

It is basically the same as the Claisen condensation mechanism. The only difference is that the enolate ion and the carbonyl group undergoing nucleophilic acyl substitution are in the same molecule. Look at the wall.

Question 85

What are the possible molecules that can undergo intermolecular aldol additions?

Answer 85

Before I start, it is important to say ketones (most of the time) have multiple a-hydrogens so within a diketone compound there can be multiple enolate ions that form.

The molecules are:

• 1,4-diketone: can form one with a five-membered ring and one with a three-membered ring, but The greater stability of the five-membered ring draws the equilibria towards its formation where it ends up being the only product
• 1,6-diketone can potentially lead to either a seven- or a five-membered ring product. Again, the more stable product—the one with the five-membered ring—is the only product of the reaction.
• 1,5-Diketones and 1,7-diketones undergo intramolecular aldol additions to form six-membered ring b-hydroxyketones

Question 86

What is Robinsons annulation?

Answer 86

The Robinson annulation is a reaction that puts these two carbon–carbon bond-forming reactions (Michael and Aldol addition) together to form an a,b-unsaturated cyclic ketone. The Robinson annulation makes it possible to synthesize many complicated organic molecules.

Note that an annulation reaction is a ring-forming reaction.

Question 87

What are the steps for Robinsons annulation? repeated exam question!!

Answer 87

Look at the wall. Same paper as the Dickmann.

Question 88

How can we determine the starting materials needed for the synthesis of a 2-cyclohexenone by a Robinson annulation? Imp

Answer 88

Draw a line that bisects both the double bond and the bond between the b- and y-carbons on the other side of the molecule:

■ The b-carbon comes from the a,b-unsaturated carbonyl compound.

■ The y-carbon is the a-carbon of the enolate ion of the dicarbonyl compound that adds to the b-carbon in the Michael reaction.

■ The sp2 carbon of the bisected double bond closest to the g-carbon was a carbonyl carbon and the other sp2 carbon was an alkyl group.

This is depicted in digital notes

Question 89

How can carbonyl compounds lose CO2 (decarboxylation)? (long sorry)

Answer 89

Carboxylate ions do not lose CO2 due to the leaving group being a carbanion with localized electrons. Such carbanions
are very strong bases, which makes them very poor leaving groups.

If, however, the CO2 group is attached to a carbon that is adjacent to a carbonyl carbon, the CO2 group can be removed because the electrons left behind can be delocalized onto the carbonyl oxygen. Consequently, carboxylate ions with a carbonyl group at the 3-position lose CO2 when they are heated. The loss of CO2 from a molecule is called decarboxylation. (Shown in digital notes)

Decarboxylation is even easier under acidic conditions because the reaction is catalyzed by an intramolecular transfer of a proton from the carboxyl group to the carbonyl oxygen. The enol that is
formed immediately tautomerizes to a ketone. (Shown in digital notes)

note: We saw before that it is harder to remove a proton from an a-carbon if the electrons are delocalized onto the carbonyl group of an ester rather than onto the carbonyl group of a ketone. For the same reason, a higher temperature (~135 °C) is required to decarboxylate a b-dicarboxylic acid such as malonic acid than to decarboxylate a b-keto acid.

Question 90

What is malonic ester synthesis?

Answer 90

It is a condensation reaction that provides a way to synthesise carboxylic acid of any desired length by combining alkylation of an a-carbon and decarboxylation of a carboxylate ion with a carbonyl group at the 3-position.

The carboxyl group and the a-carbon of the carboxylic acid being synthesized come from malonic ester. The rest of the carboxylic acid comes from the alkyl halide used in the second step of the synthesis. Thus, a malonic ester synthesis forms a carboxylic acid with two more carbons than
the alkyl halide.

Question 91

What is the mechanism of the malonic ester synthesis?

Answer 91

Look at the wall MENTION THE MENOLIC POINT

Question 92

What is the strategy for Retrosynthetic Analysis of carboxylic acids formed from malonic ester synthesis?

Answer 92

What has already been mentioned:

We saw that when a carboxylic acid is synthesized by a malonic ester synthesis, the carboxyl group and the a-carbon come from a malonic ester. Any substituent attached to the a-carbon comes from
the alkyl halide used in the second step of the synthesis. If the a-carbon has two substituents, then two successive alkylations of the a-carbon will form the desired carboxylic acid.

There is a supporting image in the digital notes. But with your brain, you can work back to the starting material.

Question 93

What is acetoacetic ester synthesis?

Answer 93

Literally the same thing as malonic ester synthesis, with the difference in the starting material being acetoacetic ester, this leads to the formation of a methyl ketone of a desired length. The carbonyl group of the methyl ketone and the carbons on either side of it come from acetoacetic ester; the rest of the ketone comes from the alkyl halide used in the second step of
the synthesis.

IMP: An acetoacetic ester synthesis forms
a methyl ketone with three more
carbons than the alkyl halide.

Question 94

What is the mechanism for acetoacetic ester synthesis?

Answer 94

Look at the wall

Question 95

What is the strategy for Retrosynthetic Analysis of methyl ketones formed from acetoacetic ester synthesis?

Answer 95

What has already been mentioned:

When a methyl ketone is synthesized by an acetoacetic ester synthesis, the carbonyl carbon and the carbons on either side of it come from acetoacetic ester. Any substituent attached to the a-carbon comes from the alkyl halide used in the second step of the synthesis.

There is a supporting image in the digital notes. But with your brain, you can work back to the starting material.

Question 96

TO GET A GOOD GRADE DO THIS

Answer 96

idc how behind you are going through 17.20 it is so important!!!