Interesting O Chem Q

[wired202808]

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Which of the following carbonyl compounds is hydrolyzed the SLOWEST with an OH-?

a. CH-CO-Cl
b. CH-CO-OCH3
c. CH-CO-NH2
d. CH-CO-BR
e. CH-CO-C-CO-CH3

The answer says CH-CO-NH2
I thought it was CH-CO-OCH3, since O-CH3 is less activating then NH2, can someone please elaborate on the answer?

 

[PooyaH]

With this one, you should know amides are very un-reactive (due to resonance). Actually ester hydrolysis happens easily.

 

[wired202808]

With this one, you should know amides are very un-reactive (due to resonance). Actually ester hydrolysis happens easily.

but why is that slower? I thought that OH- being basic, would seek to find the one that is least basic aka more acidic hence more EWG. no?

 

[AlbinoPolarBear]

Which of the following carbonyl compounds is hydrolyzed the SLOWEST with an OH-?

a. CH-CO-Cl
b. CH-CO-OCH3
c. CH-CO-NH2
d. CH-CO-BR
e. CH-CO-C-CO-CH3

The answer says CH-CO-NH2
I thought it was CH-CO-OCH3, since O-CH3 is less activating then NH2, can someone please elaborate on the answer?

Look at the leaving group, and see how stable the conjugate base is.
The more stable the conjugate base, the easier it for hydrolysis to occur.

It would help if you memorized some pKa values:

At least know these:
HCl ~ HBr ~ -7
H2O ~ 16
NH3 ~ 33
CH4 ~ 48
H2 ~ 52

Substitution of the main atoms with carbon groups will usually slightly raise the pKa as it destabilizes the conjugated base (this won't work if substitution contributes resonance, eg. phenol, pka=10).

The questions asks for the slowest reaction, therefore, the least stable conjugate base. In this case, it would be the nitrogen containing compound.

 

[PooyaH]

^^^wheeeeeeeeeeeeeeee 👍

 

[zatoichi]

Look at the leaving group, and see how stable the conjugate base is.
The more stable the conjugate base, the easier it for hydrolysis to occur.

It would help if you memorized some pKa values:

At least know these:
HCl ~ HBr ~ -7
H2O ~ 16
NH3 ~ 33
CH4 ~ 48
H2 ~ 52

Substitution of the main atoms with carbon groups will usually slightly raise the pKa as it destabilizes the conjugated base (this won't work if substitution contributes resonance, eg. phenol, pka=10).

The questions asks for the slowest reaction, therefore, the least stable conjugate base. In this case, it would be the nitrogen containing compound.

Basicity and acidity are matters of thermodynamic/equilibrium stability. It has nothing to do with the kinetics of the reaction.

The reason why the amide undergoes saponification slowest is because of it's resonance stability which creates a larger PE differential between the transition state and amide, thus, leading to slower kinetics.

 

[AlbinoPolarBear]

Basicity and acidity are matters of thermodynamic/equilibrium stability. It has nothing to do with the kinetics of the reaction.

The reason why the amide undergoes saponification slowest is because of it's resonance stability which creates a larger PE differential between the transition state and amide, thus, leading to slower kinetics.

So from your reasoning, which hydrolysis reaction would you expect to react faster?

OH- with:
a) RCONH2
b) RCOCH3

 

[zatoichi]

So from your reasoning, which would you expect to react faster?

OH- with:
a) RCONH2
b) RCOCH3

The ester because the ester has lower resonance stability due to the greater electronegativity and able to stabilize the plus charge to a lesser extent than the nitrogen.

The lower stability increases its potential energy leading to a smaller activation energy.

You used thermodynamic trends to predict kinetics and that's what I disagree with, otherwise, the conclusion is the same. The op was asking for the reason behind it....

 

[AlbinoPolarBear]

The ester because the ester has lower resonance stability due to the greater electronegativity and able to stabilize the plus charge to a lesser extent than the nitrogen.

The lower stability increases its potential energy leading to a smaller activation energy.

You used thermodynamic trends to predict kinetics and that's what I disagree with, otherwise, the conclusion is the same. The op was asking for the reason behind it....

If you look at it carefully, it's not an ester, but the molecule is a ketone.

Use your reasoning behind it, and see if your reasoning still works with this example.

Amide or ketone? which one would react faster with hydrolysis. Will it kick out the NH2 (that resonates with the carbonyl) or the CH3 (which doesn't resonate)?

 

[zatoichi]

If you look at it carefully, it's not an ester, but the molecule is a ketone.

Use your reasoning behind it, and see if your reasoning still works with this example.

Amide or ketone? which one would react faster with hydrolysis. Will it kick out the NH2 (that resonates with the carbonyl) or the CH3 (which doesn't resonate)?

Oh, my bad...

Ketone is slower because the amide may have resonance stability, but also inductive effects contribute to greater potential energy relative to the ketone making it more susceptible to nucleophilic attack.

 

[AlbinoPolarBear]

Oh, my bad...

Ketone is slower because the amide may have resonance stability, but also inductive effects contribute to greater potential energy relative to the ketone making it more susceptible to nucleophilic attack.

I'm confused at what you're trying to get at.

At first you said that resonance of the N-C=O bond will slow the reaction down, but you threw that idea away.

Inductive means pulling electrons away, wouldn't Nitrogen donate bonds to the carbonyl carbon?

 

[spoog74]

I'm confused at what you're trying to get at.

At first you said that resonance of the N-C=O bond will slow the reaction down, but you threw that idea away.

Inductive means pulling electrons away, wouldn't Nitrogen donate bonds to the carbonyl carbon?

this is covered in chads videos. This is why i will ALWAYS recommend watching them. I knew the answer 5 seconds into reading it even BEFORE looking at the answer choices 🙂

 

[mh0000]

I'm confused at what you're trying to get at.

At first you said that resonance of the N-C=O bond will slow the reaction down, but you threw that idea away.

Inductive means pulling electrons away, wouldn't Nitrogen donate bonds to the carbonyl carbon?

Yes it would. It will reduce the electrophilicity of the carbonyl carbon and should reduce the energy of the reactant. This is what reduces the kinetic rate.
EDIT: To address albino's statement more directly: the nitrogen would inductively withdraw electrons from the carbonyl, but would donate electrons via resonance. The latter should have a larger effect.

 

[AlbinoPolarBear]

Yes it would. It will reduce the electrophilicity of the carbonyl carbon and should reduce the energy of the reactant. This is what reduces the kinetic rate.

What we're getting at right now is hydrolysis.

In hydrolysis, you need to KICK out the leaving group. The better the leaving group, the faster it would leave upon hydrolysis with OH.

He argued that the main reason for the slow hydrolysis of RCONH3 was due to the resonance ability of the NC=O plane, while I argued that the rate of the reaction was due to the leaving group stability.

From the answer choices, both of our reasonings worked.

But when I presented him with RCOCH3 vs RCONH2, his reasoning fails him.

 

[mh0000]

What we're getting at right now is hydrolysis.

In hydrolysis, you need to KICK out the leaving group. The better the leaving group, the faster it would leave upon hydrolysis with OH.

He argued that the main reason for the slow hydrolysis of RCONH3 was due to the resonance ability of the NC=O plane, while I argued that the rate of the reaction was due to the leaving group stability.

From the answer choices, both of your reasons worked.

But when I presented him with RCOCH3 vs RCONH2, his reasoning fails him.

If we're talking about the rate, then we have to discuss the activation energy. In this regard, you're both right: As I mentioned earlier, the amide will stabilize the the electrophile, reducing its potential energy. Now when we consider the transition complex, we have to consider the fact that an amide anion will be higher in energy than any other carb acid derivative leaving group. This will lead to a higher energy transition state. Both of these will contribute to a higher activation energy.

In regards to the ketone vs amide. I think the easiest way to determine the rate of hydrolysis would be to look at the electrophilicity of the carbonyl carbon. Here, I believe the amide would stabilize it more than the inductive effect of the methyl group for a ketone, making amide less reactive than a ketone.

Edit: "Ketone is slower because the amide may have resonance stability..."
Yes it does, but I disagree that this makes it MORE reactive as you seem to imply. It is the opposite. Is this what you are disagreeing with albino?

 

[AlbinoPolarBear]

If we're talking about the rate, then we have to discuss the activation energy. In this regard, you're both right: As I mentioned earlier, the amide will stabilize the the electrophile, reducing its potential energy. Now when we consider the transition complex, we have to consider the fact that an amide anion will be higher in energy than any other carb acid derivative leaving group. This will lead to a higher energy transition state. Both of these will contribute to a higher activation energy.

In regards to the ketone vs amide. I think the easiest way to determine the rate of hydrolysis would be to look at the electrophilicity of the carbonyl carbon. Here, I believe the amide would stabilize it more than the inductive effect of the methyl group for a ketone, making amide less reactive than a ketone.

Okay, for hydrolysis, it's a 2-step process.

The first step is the attack of the nucleophile (in this case OH) to the carbonyl, kicking one of the C=O bonds to the oxygen (creating a negatively charged oxygen). This step is reversible, as the lone pair can just kick out the nucleophile, and we're back to the beginning. In this step, you can argue for the polarity of the bond, and which groups activate the Carbonyl C the greatest. But ... it's reversible, so it does not carry much weight in the overall reaction as the second step...

The second step, is a lone pair on the negatively charged oxygen will regain the double bond, thereby kicking out the leaving group. This is the RATE DETERMINING STEP. This is why the LEAVING group is important in hydrolysis.

This was the point I was trying to get across. He's paying attention to the first step, while I was paying attention to the second step. I wanted for him to see it, but I guess he stopped replying.

To answer your question about the amide being less reactive than a ketone during hydrolysis, have you ever seen a methyl group leaving after being attacked by a hydroxyl ion? You won't because after nucleophilic attack with the hydroxyl group, when the negative charged oxygen wants to regain the double bond, it will kick out the hydroxyl group rather than the methyl group. Why? Because hydroxyl ions are MORE stable conjugated bases than the CH3-.

 

[rockclock]

Okay, for hydrolysis, it's a 2-step process.

The first step is the attack of the nucleophile (in this case OH) to the carbonyl, kicking one of the C=O bonds to the oxygen (creating a negatively charged oxygen). This step is reversible, as the lone pair can just kick out the nucleophile, and we're back to the beginning. In this step, you can argue for the polarity of the bond, and which groups activate the Carbonyl C the greatest. But ... it's reversible, so it does not carry much weight in the overall reaction as the second step...

The second step, is a lone pair on the negatively charged oxygen will regain the double bond, thereby kicking out the leaving group. This is the RATE DETERMINING STEP. This is why the LEAVING group is important in hydrolysis.

This was the point I was trying to get across. He's paying attention to the first step, while I was paying attention to the second step. I wanted for him to see it, but I guess he stopped replying.

To answer your question about the amide being less reactive than a ketone during hydrolysis, have you ever seen a methyl group leaving after being attacked by a hydroxyl ion? You won't because after nucleophilic attack with the hydroxyl group, when the negative charged oxygen wants to regain the double bond, it will kick out the hydroxyl group rather than the methyl group. Why? Because hydroxyl ions are MORE stable conjugated bases than the CH3-.

+1. Don't know what other people are arguing (haven't read the whole thread) but hard to go against any of this.

 

[mh0000]

Okay, for hydrolysis, it's a 2-step process.

The first step is the attack of the nucleophile (in this case OH) to the carbonyl, kicking one of the C=O bonds to the oxygen (creating a negatively charged oxygen). This step is reversible, as the lone pair can just kick out the nucleophile, and we're back to the beginning. In this step, you can argue for the polarity of the bond, and which groups activate the Carbonyl C the greatest. But ... it's reversible, so it does not carry much weight in the overall reaction as the second step...

The second step, is a lone pair on the negatively charged oxygen will regain the double bond, thereby kicking out the leaving group. This is the RATE DETERMINING STEP. This is why the LEAVING group is important in hydrolysis.

This was the point I was trying to get across. He's paying attention to the first step, while I was paying attention to the second step. I wanted for him to see it, but I guess he stopped replying.

Yes, I think we agree here.

To answer your question about the amide being less reactive than a ketone during hydrolysis, have you ever seen a methyl group leaving after being attacked by a hydroxyl ion? You won't because after nucleophilic attack with the hydroxyl group, when the negative charged oxygen wants to regain the double bond, it will kick out the hydroxyl group rather than the methyl group. Why? Because hydroxyl ions are MORE stable conjugated bases than the CH3-.

I think there is a degree of miscommunication on this. My statement was made already under the assumption that acyl substitution was not going to take place. This was what you were actually arguing for, which I implicitly agree with. So I was actually interpreting the question from a general electrophilicity of ketone vs amide. So either the focus of the question changed and I didn't adjust to it, or I was just off on what was being debated.

 

[needzmoar]

lol dang.. I guess i'm late to this chem party

 

[mh0000]

So, after thinking about this question a little more, I'm going to try to clarify some things:

Albino is correct in what he said. No acyl substitution will occur. If you don't know why, read his above posts. However, I think when asked about the hydrolysis of a ketone, assuming its talking about a possible acyl susbtitution would not be correct. I think the more appropriate assumption would be the addition of water across the carbonyl to form a gem diol (again, no acyl susbtitution will occur, like albino has stated). In this case, the first step of both addition of OH- to ketone and to an amide is the rate limiting step. THerefoe, it is the electrophilicity of the carbonyls that dictates the rate. So really, I think we are both correct, but just about different statements.

 

[rockclock]

So, after thinking about this question a little more, I'm going to try to clarify some things:

Albino is correct in what he said. No acyl substitution will occur. If you don't know why, read his above posts. However, I think when asked about the hydrolysis of a ketone, assuming its talking about a possible acyl susbtitution would not be correct. I think the more appropriate assumption would be the addition of water across the carbonyl to form a gem diol (again, no acyl susbtitution will occur, like albino has stated). In this case, the first step of both addition of OH- to ketone and to an amide is the rate limiting step. THerefoe, it is the electrophilicity of the carbonyls that dictates the rate. So really, I think we are both correct, but just about different statements.

+1. I think Albino's criterion is what you should be looking at if the question is specific to hydrolysis. Yours is appropriate for general reactivity.

 

[nateguy40]

Easy answer is:

a/d= have same reactivity
B/E= dont remember but thats unneccessary because
C= the least reactive

therefore C is slowest