Getting concepts confused in Gen Chem vs. Ochem

[bozz]

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for example.. consider the inductive effect

in gen chem, we learned that the more Oxygens an acid has, the more acidic it is.. b/c it is an electron withdrawer and makes the charge on the cental atom "more positive"

in ochem, if we're looking at an aldehyde vs. an ester, for example, the oxygen on the ester is considered to be electron "donating" and makes the carbonyl group "less positive"

can anyone clear this confusion.. any general trends/rules as to whether something is electron donating/withdrawing...
do I just have to memorize that an ester is electron donating.. isn't it better to understand why?

 

[TangoNovember]

http://myweb.unomaha.edu/~dstack/2250/Overheads/EWG_EDG.PDF

hope this helps.

good luck!

 

[brianmartin]

This is going to be pretty low yield considering how little o-chem is on the MCAT. I wouldn't worry about things like this. If you know enough to ask the question you are asking, then you know more than enough for the MCAT.

I'm not really sure what your question is though...are you comparing esters and carboxylic acids?

 

[bozz]

Well I was comparing inorganic "oxyacids" such as

Perchloric acid (HClO4) with organic stuff like an ester for example. I guess in general more Oxygen atoms = more acidic. In perchloric acid, Oxygen's electronegativity causes it to act as an electron withdrawing...

However, in O Chem, we are told to memorize that -OH, and OCH3 groups are electron donating.

I was wondering why this discrepancy exists... or if there is an easy way to remember if something is electron withdrawing/donating without memorization.... in order to reconcile the confusion between ochem and g chem

 

[georgia_md]

for example.. consider the inductive effect

in gen chem, we learned that the more Oxygens an acid has, the more acidic it is.. b/c it is an electron withdrawer and makes the charge on the cental atom "more positive"

in ochem, if we're looking at an aldehyde vs. an ester, for example, the oxygen on the ester is considered to be electron "donating" and makes the carbonyl group "less positive"

can anyone clear this confusion.. any general trends/rules as to whether something is electron donating/withdrawing...
do I just have to memorize that an ester is electron donating.. isn't it better to understand why?

Yes, well, technically, it depends on which atom is more electronegative. The more electronegative an atom is , the more "affinity" it has for electrons. The general trend on the periodic table is that electronegativity increases from left to right, from bottom to top.

But generally, in Gen Chem, the Lewis definition of acid is used. A Lewis acid is a substance that accepts an electron pair. And a Lewis base is the substance that donates the electron pair.

In organic chemistry, the bronsted-lowry definition is usually used. This says that an acid is a substance that donates a proton(H+). And the base is the substance that accepts the proton.



Though they are both polar, there are partial charges on both aldehydes and esters, that stabilizes both these substances. It really depends on what they are reacted with, solvents, reactants, and reaction conditions, to know which substance(s) will be donating or accepting protons/electrons if at all a reaction takes place.

 

[georgia_md]

Well I was comparing inorganic "oxyacids" such as

Perchloric acid (HClO4) with organic stuff like an ester for example. I guess in general more Oxygen atoms = more acidic. In perchloric acid, Oxygen's electronegativity causes it to act as an electron withdrawing...

However, in O Chem, we are told to memorize that -OH, and OCH3 groups are electron donating.

I was wondering why this discrepancy exists... or if there is an easy way to remember if something is electron withdrawing/donating without memorization.... in order to reconcile the confusion between ochem and g chem

In O Chem, hydoxide(-OH) and methoxide(-OCH3) are considered as nucleophiles. Nucleophiles are electron rich substances, that donate electrons to electron poor substances. These electron poor substances are called electrophiles(philia means love, so this is like, they love electrons).

The definitions of nucleophiles and electrophiles are similar to those of lewis acids and lewis bases, but the difference is, that nucleophiles and electrophiles are primarily used only for organic substances.

 

[georgia_md]

In Organic Chem, the general trend is, if that a substance is electron rich, then it will donate an electron pair. The substance that is electron poor, will accept that pair of electrons.

Organic chem is quite different from General Chem. You're going to learn all new set of rules.

I coasted pretty easily through gen chem and did well. But I actually had to study hard when I got into Organic chemistry. There are reasons and trends behind every reaction, and if you recognize the trends, you will do fine. Good luck

 

[bozz]

haha the opposite happened for me... I cruised thru O Chem.. so learning O Chem rules wasn't so bad... it was all new, didn't have to unlearn anything

But it's been a while since I've taken ochem... and from MCAT gen chem prep, I feel like a gen chem master.... except I've forgotten everything ochem related 😀

Thanks for the help

 

[Vihsadas]

This is going to be pretty low yield considering how little o-chem is on the MCAT. I wouldn't worry about things like this. If you know enough to ask the question you are asking, then you know more than enough for the MCAT.

I'm not really sure what your question is though...are you comparing esters and carboxylic acids?

Hold on a second now...It's true that most reports are that there is less organic on the exam. Perhaps I had a freak exam, but there were 4 organic passages, and a number of discretes on my exam.

 

[Foghorn]

However, in O Chem, we are told to memorize that -OH, and OCH3 groups are electron donating

....when the C it's attached to carries a full/partial (+) charge during the transition state of a reaction. When this happens, O can donate electron density through sigma bond(s) in order to stabilize the intermediate. As a consequence, transition state energy is lowered relative to an analog molecule undergoing the same rxn but without an electron donating substituent.

 

[georgia_md]

....when the C it's attached to carries a full/partial (+) charge during the transition state of a reaction. When this happens, O can donate electron density through sigma bond(s) in order to stabilize the intermediate. As a consequence, transition state energy is lowered relative to an analog molecule undergoing the same rxn but without an electron donating substituent.

OH- is a poor leaving group and usually a good nucleophile. That means during the transition state, energy is increased, NOT decresed.

 

[Vihsadas]

You don't know what you're talking about. LOL . OH- is a poor leaving group and usually a good nucleophile. That means during the transition state, energy is increased, NOT decresed.

Yes, the transition state energy is increased in a reaction. But what if you compare two different reactions, one with an e- donating group, and one without? Essentially all he's said is that when you stabilize the intermediate you lower the activation energy, which is true. He said: "relative to an analog molecule undergoing the same rxn but without an electron donating substituent."

PS: Humility is your friend. If someone else says something incorrect, just correct them, and let it be. There's nothing to be gained by using smilies, or other prose in an attempt to condescend another student.

 

[georgia_md]

Yes, the transition state energy is increased in a reaction. But what if you compare two different reactions, one with an e- donating group, and one without? Essentially all he's said is that when you stabilize the intermediate you lower the activation energy, which is true. He said: "relative to an analog molecule undergoing the same rxn but without an electron donating substituent."

PS: Humility is your friend. If someone else says something incorrect, just correct them, and let it be. There's nothing to be gained by using smilies, or other prose in an attempt to condescend another student.

Uh, actually, no. Activation energy is usually at the rate determining steps, it's highest peaks are the transition states. The intermediate comes after the transition state. The energy lowers at the intermediate because bonds were either made or broken in the transition states.


You're right, I didn't mean to insult anybody. I know people who try to pass "complex" answers, just so they can look more intelligent when they are actually not. But I apologize if I offended any one. Thanks.

 

[Vihsadas]

Uh, actually, no. Activation energy is usually at the rate determining steps, it's highest peaks are the transition states. The intermediate comes after the transition state. The energy lowers at the intermediate because bonds were either made or broken in the transition states.


You're right, I didn't mean to insult anybody. I know people who try to pass "complex" answers, just so they can look more intelligent when they are actually not. But I apologize if I offended any one. Thanks.

That's true. I misread intermediate with transition state. An intermediate should have lower energy because it's essentially an "intermediate product", not a transition state. If we were talking about stabilizing the partial positive charge of a carbonyl carbon in a transition state, then the energy of the transition state would be lower, and activation energy would be decreased.

 

[Foghorn]

OH- is a poor leaving group and usually a good nucleophile. That means during the transition state, energy is increased, NOT decresed.

And somehow you manage to assume that I indicated OH- or a methoxy substituent was acting as a LG? Try to read for comprehension next time...m'kay

 

[BerkReviewTeach]

for example.. consider the inductive effect

in gen chem, we learned that the more Oxygens an acid has, the more acidic it is.. b/c it is an electron withdrawer and makes the charge on the cental atom "more positive"

in ochem, if we're looking at an aldehyde vs. an ester, for example, the oxygen on the ester is considered to be electron "donating" and makes the carbonyl group "less positive"

can anyone clear this confusion.. any general trends/rules as to whether something is electron donating/withdrawing...
do I just have to memorize that an ester is electron donating.. isn't it better to understand why?

If you consider the oxyacid example you presented, it has to do with the nature of the additional oxygen atoms. The addition of oxygens in an oxyacid manifest themselves as =O bonds, which are electron-withdrawing. This is pretty much the exact same thing in organic chemistry as general chemistry. The basic rule to follow is that resonance beats the inductive effect. When you increase the oxygen count of an oxyacid (consider H2SO4 versus H2SO3), you are actually increasing the number of electron withdrawing groups attached to the central atom and thereby decreasing the electron density on the protic hydrogens through resonance. This makes them more acidic, but it's by resonance rather than the inductive effect.

When comparing an ester to an aldehyde (or more so a ketone), the additional oxygen is in the -O- form (and not =O form) . This means that the oxygen has a lone pair of electrons that it can donate through resonance (as opposed to a pi-bond that can break and pull electons away). The alkoxy group donates electron density to the carbonyl group, and in doing so, makes it electron richer. This means that the carbonyl carbon will carry less of a partial positive charge and therefore be less electrophilic. As a result, the aldehyde carbonyl group is more reactive than an ester carbonyl group in reactions involving attack of the carbonyl. For instance, NaBH4 will react with an aldehyde, but not an ester, because the ester is not electrophilic enough. To reduce an ester, you need a stronger reagent like LiAlH4. You may also recall that when treating an ester with a Grignard reagent, it will first substitute an alkyl group for the alkoxy group to generate a ketone, which then readily undergoes addition of a second alkyl group, because the ketone is more reactive than the ester with a Grignard nucleophile.

The thing is that you need to determine whether the additional O increases the pi-bonds to O or whether it serves as a lone pair source. And this entails resonance rather than the inductive effect.

 

[BerkReviewTeach]

OH- and OR- are considered to be electron donating

...when the C it's attached to carries a full/partial (+) charge during the transition state of a reaction. When this happens, O can donate electron density through sigma bond(s) in order to stabilize the intermediate. As a consequence, transition state energy is lowered relative to an analog molecule undergoing the same rxn but without an electron donating substituent.

Right on! You are absolutely correct that an OR-group versus an analog of let's say an R-group is a better electron donor and thereby stabilizes the intermediate (and likewise the transition states) more so than the R-group would. This would lead to a lower activation energy for the OR-transition state than the R-transition state, exactly as you have stated. Contrary to the poster who has opted to insinuate that you were trying "to pass 'complex' answers, just so they can look more intelligent when they are actually not", your answer is factually right on the mark and helpful. I appreciate that you toss in your comments.

Uh, actually, no. Activation energy is usually at the rate determining steps, it's highest peaks are the transition states. The intermediate comes after the transition state. The energy lowers at the intermediate because bonds were either made or broken in the transition states.

Vihsadas makes a great point that you might want to tone down the attitude a bit, especially when you are insulting people for trying to seem intelligent and then you post generalizations that aren't altogether sound.

First, there is one rate-determining step in a reaction (the highest transition), and not rate-determining stepS. There can certainly be more than one transition state in a reaction mechanism, but only one can be the highest. If you are thinking of a reaction pathway, then there can be local maxima surrounding different relative high-energy points, but technically speaking, a reaction mechanism has but one highest point.

Second, there are separate transition states that both preceed and follow an intermediate, not just preceed as you stated. If there were not a climb on both sides of the intermediate on the energy diagram, it would never form. And as for the lowering of the energy from transition state to intermediate, the bond formation and breakage associated with the transition state may or may not play a significant role, depending on the mechanism of the reaction and the environment. Quite often, the reduced energy level of an intermediate from a transition state is the result of solvation and/or rehybridization. Consider the carbocation intermediate. The transition state involves a bond breaking, which results in a rehybridization from sp3 to sp2 and the coordination of solvent to the cation as it proceeds to an intermediate planar carbocation.

This is generally a friendly place where people try to help one another, and that is exactly what Foghorn was doing before being ambushed. I for one love that Foghorn shares his/her insights.

 

[meme me]

Perhaps you folks could ditch the egos for a while? A simple question was asked, no need to turn it into an O-chem knowledge competition. I'm sure you've all confused the poor MCAT studier who asked the question more than anything.

 

[lumanyika]

If you consider the oxyacid example you presented, it has to do with the nature of the additional oxygen atoms. The addition of oxygens in an oxyacid manifest themselves as =O bonds, which are electron-withdrawing. This is pretty much the exact same thing in organic chemistry as general chemistry. The basic rule to follow is that resonance beats the inductive effect. When you increase the oxygen count of an oxyacid (consider H2SO4 versus H2SO3), you are actually increasing the number of electron withdrawing groups attached to the central atom and thereby decreasing the electron density on the protic hydrogens through resonance. This makes them more acidic, but it's by resonance rather than the inductive effect.

When comparing an ester to an aldehyde (or more so a ketone), the additional oxygen is in the -O- form (and not =O form) . This means that the oxygen has a lone pair of electrons that it can donate through resonance (as opposed to a pi-bond that can break and pull electons away). The alkoxy group donates electron density to the carbonyl group, and in doing so, makes it electron richer. This means that the carbonyl carbon will carry less of a partial positive charge and therefore be less electrophilic. As a result, the aldehyde carbonyl group is more reactive than an ester carbonyl group in reactions involving attack of the carbonyl. For instance, NaBH4 will react with an aldehyde, but not an ester, because the ester is not electrophilic enough. To reduce an ester, you need a stronger reagent like LiAlH4. You may also recall that when treating an ester with a Grignard reagent, it will first substitute an alkyl group for the alkoxy group to generate a ketone, which then readily undergoes addition of a second alkyl group, because the ketone is more reactive than the ester with a Grignard nucleophile.

The thing is that you need to determine whether the additional O increases the pi-bonds to O or whether it serves as a lone pair source. And this entails resonance rather than the inductive effect.

there's the magic word i was waiting for!! RESONANCE 👍


thanks berkreviewteach.

 

[bozz]

Right on! You are absolutely correct that an OR-group versus an analog of let's say an R-group is a better electron donor and thereby stabilizes the intermediate (and likewise the transition states) more so than the R-group would. This would lead to a lower activation energy for the OR-transition state than the R-transition state, exactly as you have stated. Contrary to the poster who has opted to insinuate that you were trying "to pass 'complex' answers, just so they can look more intelligent when they are actually not", your answer is factually right on the mark and helpful. I appreciate that you toss in your comments.



Vihsadas makes a great point that you might want to tone down the attitude a bit, especially when you are insulting people for trying to seem intelligent and then you post generalizations that aren't altogether sound.

First, there is one rate-determining step in a reaction (the highest transition), and not rate-determining stepS. There can certainly be more than one transition state in a reaction mechanism, but only one can be the highest. If you are thinking of a reaction pathway, then there can be local maxima surrounding different relative high-energy points, but technically speaking, a reaction mechanism has but one highest point.

Second, there are separate transition states that both preceed and follow an intermediate, not just preceed as you stated. If there were not a climb on both sides of the intermediate on the energy diagram, it would never form. And as for the lowering of the energy from transition state to intermediate, the bond formation and breakage associated with the transition state may or may not play a significant role, depending on the mechanism of the reaction and the environment. Quite often, the reduced energy level of an intermediate from a transition state is the result of solvation and/or rehybridization. Consider the carbocation intermediate. The transition state involves a bond breaking, which results in a rehybridization from sp3 to sp2 and the coordination of solvent to the cation as it proceeds to an intermediate planar carbocation.

This is generally a friendly place where people try to help one another, and that is exactly what Foghorn was doing before being ambushed. I for one love that Foghorn shares his/her insights.

ooohhhhh.... thanks big time... helped out a lot!
btw, I went through your chem books.. and they were great... I'm so much more confident now... I never understood titrations or acid/base gen much much at all. It's weird b/c the actual "text" part isn't the most helpful. It's the few 6-7 examples in each chapter that do the trick for me. 👍

I'm hoping for a titration passage on the MCAT.. for me to own 😀
Until then, I havta work on my ochem