Using previous posts you should be able to suggest the synthesis. It takes two or three steps.
Solution:
Mandelic acid is an antibacterial which can be used orally. It is also used in skin cream to prevent acne and wrinkles.
Mandelic acid
Mandelic acid can be made from benzaldehyde.
Using previous posts you should be able to suggest the synthesis. It takes two or three steps.
Solution:
Using previous posts you should be able to suggest the synthesis. It takes two or three steps.
Solution:
Hydrolysis of nitriles
Nitriles can be viewed as primary amide which has lost a molecule of water:
Reacting a primary amide with a dehydrating reagent is one way nitriles can be made.
Likewise, nitriles can be made into primary amides by acid-catalysed hydrolysis. Mechanism:
Reacting a primary amide with a dehydrating reagent is one way nitriles can be made.
Likewise, nitriles can be made into primary amides by acid-catalysed hydrolysis. Mechanism:
Of course we also know that if the conditions are vigorous enough, the amide will hydrolyse into a carboxylic acid and ammonia. So overall a long acid reflux can turn a C≡N into a COOH.
Hydrolysing amides
There are two ways of doing this. The first one is protonating the C=O to make it more electrophillic.
The second is brute force by OH-:
The second is brute force by OH-:
Both methods require vigorous conditions - high acid/base concentration, high heat, and long reaction times. Secondary and tertiary amides hydrolyse even more slowly.
A secret weapon to force the hydrolysis of a tertiary amide is to use a very strong base such as Potassium tert-butoxide. This deprotonates the tetrahedral intermediate:
It may be difficult to make an amine anion leave, but something has to, and O2- or a carbanion are much harder to push off.
Using an acid to aid attack on C=O
This has two uses:
1. It can protonate an electrophile, making it more open to attack.
2. It can protonate a leaving group - turning a poor leaving group into a good one
Both of these effects are shown in the most common example of acid catalysts - the formation of an ester:
Alternatively the OR in the tetrahedral intermediate could be protonated instead - which would lead back to the reactant. We drive the formation of product by removing water using distillation or a drying agent.
We can also use an excess of one of the reactants. If we want to fully esterificate an expensive carboxylic acid, we can use an excess of alcohol, and vice versa.
To reverse the reaction, we can use an excess of water.
The same principles can be used to convert the ester of one alcohol into the ester of another. For example, the reaction below can be driven to the right by distilling off methanol:
Another example of transesterifaction is in the commercial production of PET (polyethylene terephthalate) which is the polymer used in plastic drink bottles. It is the ester of terephalic acid and ethylene glycol.
It is produced commercially by transesterifying dimethyl terephthalate, distilling off methanol to drive the reaction.
I don't know why this process is cheaper than making it directly.
1. It can protonate an electrophile, making it more open to attack.
2. It can protonate a leaving group - turning a poor leaving group into a good one
Both of these effects are shown in the most common example of acid catalysts - the formation of an ester:
Alternatively the OR in the tetrahedral intermediate could be protonated instead - which would lead back to the reactant. We drive the formation of product by removing water using distillation or a drying agent.
We can also use an excess of one of the reactants. If we want to fully esterificate an expensive carboxylic acid, we can use an excess of alcohol, and vice versa.
To reverse the reaction, we can use an excess of water.
The same principles can be used to convert the ester of one alcohol into the ester of another. For example, the reaction below can be driven to the right by distilling off methanol:
Another example of transesterifaction is in the commercial production of PET (polyethylene terephthalate) which is the polymer used in plastic drink bottles. It is the ester of terephalic acid and ethylene glycol.
It is produced commercially by transesterifying dimethyl terephthalate, distilling off methanol to drive the reaction.
I don't know why this process is cheaper than making it directly.
Amines do not attack carboxylic acids
This is because the acid will quickly lose its proton:
Almost no nucleophile will attack a carboxylate anion. Organolithiums are one exception.
One way of forcing this reaction to take place is to use high heat - which dehydrates the salt:
I don't know the mechanism. It is also seen as a bad way of making amides - but I don't why.
And unlike alcohols, the reaction cannot be catalyzed by protonating the C=O, since any acid will preferentially produce an ammonium ion - which is obviously a poor nucleophile.
Almost no nucleophile will attack a carboxylate anion. Organolithiums are one exception.
One way of forcing this reaction to take place is to use high heat - which dehydrates the salt:
I don't know the mechanism. It is also seen as a bad way of making amides - but I don't why.
And unlike alcohols, the reaction cannot be catalyzed by protonating the C=O, since any acid will preferentially produce an ammonium ion - which is obviously a poor nucleophile.
Predicting the substitution reactions of carboxylic acid derivatives
Amides can be formed from esters by an attacking amine, but esters cannot be form from amides by attack with alcohol.
Both cases would form a tetrahedral intermediate, but the RO- would always be preferentially kicked off:
There are other factors than leaving group ability which decide whether a substitution takes place, such as how well the intermediate forms. The amide is so bad an electrophile that the immediate is very unlikely to form.
So for a sucessful attack we require:
1. A good enough electrophile
2. A good enough nucleophile
3. A leaving group on the defending molecule which is better at leaving than the attacking nucleophile.
Leaving group ability is a good guide to how good a nucleophile something is - good nucleophiles are bad leaving groups. Hence a good nucleophile will have a high pKaH. Another way to think about it is that something which wants to form a bond to hydrogen will be likely to want to form a bond to carbon.
For example, below are four carboxylic acid derivatives with different leaving groups, each attacked by H2O. Try to relate the reaction conditions to the table above:
The difficulty of attacking an amide is due to n -> pi* overlap. This can be depicted as resonance:
Both cases would form a tetrahedral intermediate, but the RO- would always be preferentially kicked off:
There are other factors than leaving group ability which decide whether a substitution takes place, such as how well the intermediate forms. The amide is so bad an electrophile that the immediate is very unlikely to form.
So for a sucessful attack we require:
1. A good enough electrophile
2. A good enough nucleophile
3. A leaving group on the defending molecule which is better at leaving than the attacking nucleophile.
Leaving group ability is a good guide to how good a nucleophile something is - good nucleophiles are bad leaving groups. Hence a good nucleophile will have a high pKaH. Another way to think about it is that something which wants to form a bond to hydrogen will be likely to want to form a bond to carbon.
A reaction particularly suited to substitution will have more relaxed reaction conditions. Eg. Will react faster, at a lower temperature, and without needing a catalyst.
For example, below are four carboxylic acid derivatives with different leaving groups, each attacked by H2O. Try to relate the reaction conditions to the table above:
The difficulty of attacking an amide is due to n -> pi* overlap. This can be depicted as resonance:
Or spacial overlap of the filled n and empty π* orbital:
Or as an MO diagram:
Now we can see why an amide is such a bad nucleophile - because resonance raises the LUMO of the molecule.
This MO approach also gives another explanation for why an amide is a worst base than an amine - a lower lone pair energy makes it less favorable to donate them into a proton's empty 1s orbital
Amines attack acyl chlorides to give amides
The mechanism is very similar to attack by alcohol:
Notice that two equivalents of ammonia are required.
The attack can be done by other amines, not just ammonia:
While ammonia is easy to add in excess, protonating amines can be wasteful if the amine is rare or expensive. A base such as NaOH could deprotonate the amines. But OH- is itself a nucleophile. The solution is to use two separate phases:
This two-phase strategy is called the Schotten-Bauman synthesis.
Notice that two equivalents of ammonia are required.
The attack can be done by other amines, not just ammonia:
While ammonia is easy to add in excess, protonating amines can be wasteful if the amine is rare or expensive. A base such as NaOH could deprotonate the amines. But OH- is itself a nucleophile. The solution is to use two separate phases:
This two-phase strategy is called the Schotten-Bauman synthesis.
pKaH is a guide to leaving group ability
pKaH refers to the pKa of a leaving group's conjugate acid. The lower the value, the better a leaving group it is:
The leaving group ability of an anion depends on how stable it is at carrying negative charge. While pKa is the negative of a molecule's ability to dissociate into a proton and an anion. So this relation should also make sense intuitively.
Remember this is only a guide. For example, it does not factor in the stability of the carbonyl it creates by leaving.
The above table explains which group is pushed out in the attack below:
The leaving group ability of an anion depends on how stable it is at carrying negative charge. While pKa is the negative of a molecule's ability to dissociate into a proton and an anion. So this relation should also make sense intuitively.
Remember this is only a guide. For example, it does not factor in the stability of the carbonyl it creates by leaving.
The above table explains which group is pushed out in the attack below:
Nucleophilic catalysts
This process is usually excluded from reaction mechanisms.
There are many compounds which act as both good nucleophiles and good leaving groups. These can produce an intermediate which makes the molecule more reactive with a weaker nucleophile. These are called nucleophillic catalysts. Below is an example with pyridine helping an ROH attack a C=O:
Pyridine is useful above because it acts as a solvent, a weak base, and a catalyst.
Notice pyridine will end up protonated instead of the chloride ion. So at least two equivalents of pyridine are required.
There are many compounds which act as both good nucleophiles and good leaving groups. These can produce an intermediate which makes the molecule more reactive with a weaker nucleophile. These are called nucleophillic catalysts. Below is an example with pyridine helping an ROH attack a C=O:
Pyridine is useful above because it acts as a solvent, a weak base, and a catalyst.
Notice pyridine will end up protonated instead of the chloride ion. So at least two equivalents of pyridine are required.
Alcohol attacks acid chlorides and acid anhydrides to give esters
The difference to attacking an aldehyde is that we now have a better leaving group than RO-:
Same with attacking acid anhydrides:
Same with attacking acid anhydrides:
N,N,3,5-tetramethylaniline NMR
It should be obvious that 7.24 is the CDCl3 solvent and 6.38 are the aromatic protons. The two other peaks are from the two different methyl environments.
Both nitrogen and benzene pulls in electrons inductively. But the nitrogen methyls are the most deshielded. The trick is to consider resonance:
[7]-para-cyclophane NMR
This gives a great view of aromatic anisotropy:
A cyclophane consists of one aromatic unit and an aliphatic chain - which forms a bridge between two non-adjacent positions.
A cyclophane consists of one aromatic unit and an aliphatic chain - which forms a bridge between two non-adjacent positions.
Effect of copper salts on conjugate addition
An organocopper can be made by transmetallating a grignard reagent. This also acts as a nucleophilillic carbanion:
Copper is less electropositive, so it produces a less-charged (or "softer") carbanion. And we know soft nucleophiles are a good way to encourage conjugate addition instead of direct attack. Example:
Another method to produce an organocopper is reacting it with two equivalents of alkyl lithium:
Which also has a soft carbanion. These are unstable and must be used immediately, this is why they are kept at low temperatures.
Reacting these happens much better by adding Me3SiCl:
Copper is less electropositive, so it produces a less-charged (or "softer") carbanion. And we know soft nucleophiles are a good way to encourage conjugate addition instead of direct attack. Example:
Another method to produce an organocopper is reacting it with two equivalents of alkyl lithium:
Which also has a soft carbanion. These are unstable and must be used immediately, this is why they are kept at low temperatures.
Reacting these happens much better by adding Me3SiCl:
Though its mechanism is not fully understood.
Conjugate addition: how to tell which atom gets attacked
1. Kinetic and thermodynamic products
Reactions are faster at high temperature. So for the above reaction with a high heat and a long reaction time, we will get mostly the thermodynamic product, because systems tend towards the most stable state if given enough time. At low temperature and a short reaction time, we will get mostly the kinetic product, because it is created faster.
Why is the cyanohydrin formed faster? It is because the carbonyl carbon is more delta-positive, so electrostatic attraction will encourage it to attack here.
Why the nitrile overall more stable? Because it has an oxygen pi bond rather than a carbon pi bond. The oxygen pi bond has a lower energy.
Addition to the carbonyl isn't always reversible, in the above case this would mean the cyanohydrin would dominate the products, even with a high reaction time and high temperature. In other cases, the reactivity of the C=O will effect the rate it is attacked, hence the final ratio of the products:
2. Reactivity of C=O
Reactions are faster at high temperature. So for the above reaction with a high heat and a long reaction time, we will get mostly the thermodynamic product, because systems tend towards the most stable state if given enough time. At low temperature and a short reaction time, we will get mostly the kinetic product, because it is created faster.
Why is the cyanohydrin formed faster? It is because the carbonyl carbon is more delta-positive, so electrostatic attraction will encourage it to attack here.
Why the nitrile overall more stable? Because it has an oxygen pi bond rather than a carbon pi bond. The oxygen pi bond has a lower energy.
Addition to the carbonyl isn't always reversible, in the above case this would mean the cyanohydrin would dominate the products, even with a high reaction time and high temperature. In other cases, the reactivity of the C=O will effect the rate it is attacked, hence the final ratio of the products:
2. Reactivity of C=O
Example:
Both reactions above are irreversible, but you can see that a conjugate addition can happen faster than a direct attack on C=O - if a relatively unreactive type of C=O is used.
3. Steric hindrance
This should be obvious:
4. Hard and soft nucleophiles
Successful attack is governed by two types of interactions, electrostatic attraction and orbital overlap. Most attacks use a mixture of both. The dominant type of interaction depends on the species involved.
Small electronegative atoms with high charge density tend to interact mostly under electrostatic interaction. These are called hard nucleophiles. Eg. Oxygen, fluorine and chlorine atoms.
Larger less electronegative atoms, with less charge density (from more diffuse orbitals) tend to interact mostly via orbital overlap. These are called soft nucleophiles. Eg. Sulfur, phosphorous, bromine and iodine atoms.
Nitrogen is usually hard, but tends to soften as alkyl groups are added to it.
Electrophiles can be called hard or soft under the same description. Eg. H+ is a very hard electrophile while Br2 is a soft one. Soft electrophiles tend to react with soft nucleophiles, and hard electrophiles tend to react with hard nucleophiles.
This explains why alkenes react with bromine. They are both soft, and the interactions are purely from orbital overlap rather than charges. Water, being a hard nucleophile, does not react with bromine.
See if you can tell how all this applies to conjugated carbonyls:
Soft nucleophiles such as R-SH will tend to attack the larger lower-charged lobe of the β-carbon. Hard nucleophiles such as carbanions will tend to attack the higher-charged carbonyl carbon.
Conjugate addition double attack
Sometimes both places will be attacked by nucleophiles, such as this example with borohydride:
This doesn't always happen. The nucleophile might only be suitable to attacking one center. Or one of the electrophillic centers might be too unreactive to react. For example, ester carbonyls are not as reactive as aldehydes or ketones, so the molecule below is only attacked once:
This doesn't always happen. The nucleophile might only be suitable to attacking one center. Or one of the electrophillic centers might be too unreactive to react. For example, ester carbonyls are not as reactive as aldehydes or ketones, so the molecule below is only attacked once:
OH conjugate addition
This needs an acid or base catalyst.
Base
Alcohols are not very good nucleophiles, but alkoxide anions are great ones. So we can use a base to deprotonate some of the alcohols.
It can also happen intramolecularly:
Intramolecular reactions are usually much faster than intermolecular ones, this is why the above reaction can use MeO- as a base without having it compete by acting as a nucleophile.
Acid
Alcohols are not very good nucleophiles, but a C=OH+ is very easy to attack. So we can make the reaction go by protonating the C=O:
Base
Alcohols are not very good nucleophiles, but alkoxide anions are great ones. So we can use a base to deprotonate some of the alcohols.
It can also happen intramolecularly:
Intramolecular reactions are usually much faster than intermolecular ones, this is why the above reaction can use MeO- as a base without having it compete by acting as a nucleophile.
Acid
Alcohols are not very good nucleophiles, but a C=OH+ is very easy to attack. So we can make the reaction go by protonating the C=O:
Amine conjugate addition 3
The product of an amine conjugate addition is still an amine, which can attack another molecule if it has a proton to lose:
This can even happen in the same molecule:
Notice that the hydroxylamine always attacks using the lone pair on nitrogen, not oxygen. Nitrogen is a better nucleophile. This is obvious from experimental data, but can also be viewed from a MO perspective - nitrogen is more electropositive so its lone pair is higher in energy.
Notice that the hydroxylamine always attacks using the lone pair on nitrogen, not oxygen. Nitrogen is a better nucleophile. This is obvious from experimental data, but can also be viewed from a MO perspective - nitrogen is more electropositive so its lone pair is higher in energy.
Amine conjugate addition 2
Amines can also attack conjugated bonds of acids or esters:
The reaction requires a high temperature, can you tell why?
It is because ammonia is basic, and will first deprotonate the molecule:
Negatively charged molecules are worst electrophiles, even if the charge is not centered on the point of attack. So heat is required to make the attack happen at a reasonable rate.
The reaction requires a high temperature, can you tell why?
It is because ammonia is basic, and will first deprotonate the molecule:
Negatively charged molecules are worst electrophiles, even if the charge is not centered on the point of attack. So heat is required to make the attack happen at a reasonable rate.
Conjugate addition using amines
The proton-transfer step can be catalysed by a base, for higher yields:
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