Organic Chemistry Question Thread

[QofQuimica]

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All users may post questions about MCAT, DAT, OAT, or PCAT organic chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what organic topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
-general, MCAT-level organic
-particular MCAT-level organic problems, whether your own or from study material
-what you need to know about organic for the MCAT
-how best to approach to MCAT organic passages
-how best to study MCAT organic
-how best to tackle the MCAT biological sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

********

If you really know your organic, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the Organic Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university organic chemistry TA teaching experience. In addition, I teach organic chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 15 on BS, 43 overall.

P.S. If you shorten "organic chemistry" to "orgo," not only will I not answer your questions, but during the BS section, your test form will backside attack you with a zillion strong nucleophiles (via the SN2 mechanism, of course).

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the BS section of the MCAT, and 36 overall.

 

[MSTPbound]

I had a basic question about bond conjugation.

First of all, conjugation is defined by book as alternating single and double bonds. Can a compound be conjugated if it has only one double bond between two single bonds? For some reason, I can't seem to accept that it is a conjugated species because whenever I think conjugated, I think the minimum is two double bonds.

As the above posters said, NO, conjugation cannot exist if if a compound has only one double bond between two single bonds. Conjugation is described as alternating single and double bonds, but requires at least two double bonds separated by a single bond.

Also, in terms of stability during an reaction, would something prefer a conjugated carbocation with the positive charge on a primary carbon or a nonconjugated carbocation with the positive charge on a secondary carbon?

For example:
CH2(+)-CH=CH-CH3 versus CH2=CH-CH(+)-CH3

Thanks in advance.

This question doesn't have a "simple" answer. It depends on the reaction conditions. If it's another elimination, i.e. another double bond is being formed, the carbocation on the secondary carbon is preferred as a target to re-establish a conjugated system. .

If a nucleophile is attacking the carbocation, a number of effects compete with each other, including carbocation stability through resonance, induction, and stability of the products.

Addition of HBr to 1,3-Butadiene is instructive. At very low temperature, the kinetic product is preferred, with substitution occuring at the secondary carbocation (so I suppose induction wins, in a sense). The result is a 1,2-addition, 3-bromo-1-butene.

At very high temperature, the 1,4 product, i.e. the thermodynamic product, dominates because even if kinetically, the 1,2-product is preferred, the activation energy for the reverse reaction is significantly lower, so the above reaction establishes an equilibrium, with the intermediate being regenerated. The 1,4-product, however is more stable because it has a more substituted double bond, so it is much more of an uphill battle for the intermediate to be regenerated here, so the thermodynamic product prevails: 1-bromo-2-butene.



Sorry if my previous post went off topic before - I didn't realize your question had not yet been answered. Hopefully this is helpful; if I'm off, I sure an O-Chem heavyweight will step in to correct/clarify.

-MSTPbound

 

[dochoov]

This is a primary allylic carbocation, CH2=CH2-CH2+ significantly more so than

this is +CH2-CH2=CH2-CH3

In fact, the latter isn't even considered its own species. It's a contributing resonance structure. The former is both.

 

[midn]

Thanks for the in-depth replies.

All the statements yall made are what I thought before, but my Princeton Review book said otherwise. I talked to my TPR organic teacher, and she agreed with the book despite me asking the same questions here about whether the species was actually conjugated.

The original problem asked for the product of the reaction of 1,3-butadiene reacted with HCl.
These are the steps that I thought it went through
Step 1: Hydrogen abstraction
CH2=CH-CH=CH2 + HCl -> CH2=CH-CH(+)-CH3 (I thought this was the most stable carbocation)
Book says it rearranges to this structure:
CH2(+)-CH=CH-CH3 and according to my TPR teacher, it is because this species is "conjugated" which is where my point of confusion was since I didn't think that's what constituted a conjugated species.

Actually, I just realized my mistake. This problem has nothing to do with conjugation, but rather forming the most subsituted alkene. Stupid! So I guess my question actually is: what is more favored: most sub alkene or secondary carbocation? I guess this goes back to kinetic versus thermodynamic product, but the problem makes no distinction (no reaction conditions given).

Friggin TPR teacher threw me off by saying that it was because the species was conjugated.

 

[BrokenGlass]

Thanks for the in-depth replies.

All the statements yall made are what I thought before, but my Princeton Review book said otherwise. I talked to my TPR organic teacher, and she agreed with the book despite me asking the same questions here about whether the species was actually conjugated.

The original problem asked for the product of the reaction of 1,3-butadiene reacted with HCl.
These are the steps that I thought it went through
Step 1: Hydrogen abstraction
CH2=CH-CH=CH2 + HCl -> CH2=CH-CH(+)-CH3 (I thought this was the most stable carbocation)
Book says it rearranges to this structure:
CH2(+)-CH=CH-CH3 and according to my TPR teacher, it is because this species is "conjugated" which is where my point of confusion was since I didn't think that's what constituted a conjugated species.

Actually, I just realized my mistake. This problem has nothing to do with conjugation, but rather forming the most subsituted alkene. Stupid! So I guess my question actually is: what is more favored: most sub alkene or secondary carbocation? I guess this goes back to kinetic versus thermodynamic product, but the problem makes no distinction (no reaction conditions given).

Friggin TPR teacher threw me off by saying that it was because the species was conjugated.

http://depts.washington.edu/chemcrs/bulkdisk/chem238A_win05/notes_10_09_11.pdf

Take a look at pages 4 and 5

 

[dochoov]

Isn't it true that we no longer need to know about alkenes for the MCAT? I've posted about this before. The AAMC study guide,

http://www.aamc.org/students/mcat/topics.pdf

emphasizes, no alkenes! But it seems to me that so many of the concepts in O chem are exemplified by alkene properties and reactions.

 

[BrokenGlass]

Isn't it true that we no longer need to know about alkenes for the MCAT? I've posted about this before. The AAMC study guide,

http://www.aamc.org/students/mcat/topics.pdf

emphasizes, no alkenes! But it seems to me that so many of the concepts in O chem are exemplified by alkene properties and reactions.

According to Examkrackers' O-Chem book (6th edition), alkenes are no longer covered on the MCAT. But like you said, it doesn't hurt to know this stuff.

 

[variablistic]

My recollection from organic is that the most substituted alkene is always favored. Granted, this is coming from Zaitsev's rule which only covers elimination rxns. Since the two compounds in question are rapidly interchanging, the preferred product is the one which thermodynamically is more stable, while I would expect you would have a significant contribution of the lesser product since its intermediate is more stable.

I would take the disclaimer don't quote me on this as well.

 

[ViolinCello]

I have seen 3 different orders for this mechanism, so can someone please clarify which one is correct?

a) ammonia or primary amine attacks the carbonyl carbon of ketone; the oxygen has a negative charge, and picks up 2 hydrogens (from where?) to make water(I understand b/c to be a good leaving group); water leaves, and the lone pair of electrons of the nitrogens forms the double bond with the carbon; hydrogen is abstracted(using what?) from the positively charged nitrogen of the mine to form end product
b) the oxygen in the ketone abstracts a hydrogen from a bronstead acid; then the primary amine attacks the carbonyl carbon; once attached, the OH then abstracts another H from the attached amine; the water leaves, imine is formed with + charges on the N; then H is abstracted to form final product(again using what?).
c) same as B, except the OH does not abstract an H from the amine when it is attached, but gets it from the solution(another bronstead acid?).

Thanks!

 

[BrokenGlass]

I have seen 3 different orders for this mechanism, so can someone please clarify which one is correct?

a) ammonia or primary amine attacks the carbonyl carbon of ketone; the oxygen has a negative charge, and picks up 2 hydrogens (from where?) to make water(I understand b/c to be a good leaving group); water leaves, and the lone pair of electrons of the nitrogens forms the double bond with the carbon; hydrogen is abstracted(using what?) from the positively charged nitrogen of the mine to form end product
b) the oxygen in the ketone abstracts a hydrogen from a bronstead acid; then the primary amine attacks the carbonyl carbon; once attached, the OH then abstracts another H from the attached amine; the water leaves, imine is formed with + charges on the N; then H is abstracted to form final product(again using what?).
c) same as B, except the OH does not abstract an H from the amine when it is attached, but gets it from the solution(another bronstead acid?).

Thanks!

Maybe this could be of some use:

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch17/ch17-3-3-1.html

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch17/ch17-3-3-3.html

 

[BrokenGlass]

My recollection from organic is that the most substituted alkene is always favored. Granted, this is coming from Zaitsev's rule which only covers elimination rxns. Since the two compounds in question are rapidly interchanging, the preferred product is the one which thermodynamically is more stable, while I would expect you would have a significant contribution of the lesser product since its intermediate is more stable.

I would take the disclaimer don't quote me on this as well.

It is not the case that most substituted alkene is always favored. Hofmann elimination is one obvious counterexample.

 

[cloosh]

how much do we have to know about condensation reactions and mechanisms in general, and specifically aldol condensations. i just get confused when i try to think of how one works. thanks.

 

[BrokenGlass]

how much do we have to know about condensation reactions and mechanisms in general, and specifically aldol condensations. i just get confused when i try to think of how one works. thanks.

Dehydration is just elimination of water (which results in formation of a new Pi bond). A hydroxy group (OH) and a hydrogen are eliminted from adjacent carbons and replaced by a Pi bond (for example via an E2 elimination mechanism). The result is loss of water. Dehydration is a subtype of condensation.

We need to know aldol condensation. Aldol condesation is when an aldehyde (or a ketone) reacts with another aldehyde (or ketone).

In the first step, a strong base rips off an alpha hydrogen from an aldehyde (or a ketone), forming an enolate ion. This newly-formed enolate ion acts as a nucleophile and attacks another aldehyde (or ketone), which serves as an electrophile.

A tetrahedral intermediate is formed (with a negative charge on what used to be the carbonyl oxygen of the electrohpile). This tetrahedral intemediate is an alkoxide ion. When tetrahedral intermediate gets protonated, an aldol is formed. This much is called aldol addition.

The reaction may not stop there. The next step maybe the loss of the hydroxy group along with the loss of another alpha hydrogen (i.e. dehydration/condenstation), resulting in an alpha-beta unsaturated ketone (or aldehyde).

 

[cloosh]

thanks a bunch brokenglass, it helped a lot.

could someone also clear up the difference between the Zaitsev and Hoffman rules? thanks in advance.

 

[BrokenGlass]

thanks a bunch brokenglass, it helped a lot.

could someone also clear up the difference between the Zaitsev and Hoffman rules? thanks in advance.

http://members.aol.com/logan20/zeitsev.html

 

[PalmettoGuy]

In the 2004 TPR Cracking the MCAT book there is the following question.

A researcher works with a pure sample of alkyl halide that rotates the plane of polarized light to the right. She subjects the substrate to a reaction in which the halide is replaced by an OH group according to an SN2 mechanism. Will she be justified in concluding that the product willl rotate the plane of polarized light to the left?

My answer:
"Yes, because the SN2 mechanism inverts configuration around chiral carbons."

Their answer:
"No, because substrate and product are not enantiomers."

Yeah, I got the fact that substrate and product are not enantiomers. However, earlier in the text reads

"If the original substrate is chiral, rotating the plane of polarized light to the right or left, the substitudted product is also chiral, but it rotates the plane of polarized light in the opposite direction. That is, SN2 substitution is accompanied by complete inversion of configuration."

So is the above statement true or not? No clear answer in my Organic Text either.

Thanks Folks!

 

[MSTPbound]

In the 2004 TPR Cracking the MCAT book there is the following question.

A researcher works with a pure sample of alkyl halide that rotates the plane of polarized light to the right. She subjects the substrate to a reaction in which the halide is replaced by an OH group according to an SN2 mechanism. Will she be justified in concluding that the product willl rotate the plane of polarized light to the left?

My answer:
"Yes, because the SN2 mechanism inverts configuration around chiral carbons."

Their answer:
"No, because substrate and product are not enantiomers."

Yeah, I got the fact that substrate and product are not enantiomers. However, earlier in the text reads

"If the original substrate is chiral, rotating the plane of polarized light to the right or left, the substitudted product is also chiral, but it rotates the plane of polarized light in the opposite direction. That is, SN2 substitution is accompanied by complete inversion of configuration."

So is the above statement true or not? No clear answer in my Organic Text either.

Thanks Folks!

The Orgo heavyweights will correct me if I'm wrong since I somehow forgot a TREMENDOUS amount of Orgo at the moment that I submitted my MCAT exam.

If the original substrate is chiral, the SN2 substitution WILL invert the configuration. Depending on priority of the nucleophile, this may or may not cause a change from R- to S- or vice versa.

That does NOT mean that light will necessarily rotate in a different direction. Rotation of light is an intrinsic physical property of the compound, so if you start with one compound (let's say, (S)-R-X), and your substitution results in a "DIFFERENT" compound (like (S)-R-OH, as opposed to the enantiomer of the original reactant which would be (R)-R-X), you can make no predictions about how rotation of polarized light will be affected. It depends on the physical properties of (S)-R-OH... may rotate light in the same, or the opposite direction as (S)-R-X.

If I'm correct, I hope this helps.

Good Luck,

-MSTPbound

 

[BrokenGlass]

In the 2004 TPR Cracking the MCAT book there is the following question.

A researcher works with a pure sample of alkyl halide that rotates the plane of polarized light to the right. She subjects the substrate to a reaction in which the halide is replaced by an OH group according to an SN2 mechanism. Will she be justified in concluding that the product willl rotate the plane of polarized light to the left?

My answer:
"Yes, because the SN2 mechanism inverts configuration around chiral carbons."

Their answer:
"No, because substrate and product are not enantiomers."

Yeah, I got the fact that substrate and product are not enantiomers. However, earlier in the text reads

"If the original substrate is chiral, rotating the plane of polarized light to the right or left, the substitudted product is also chiral, but it rotates the plane of polarized light in the opposite direction. That is, SN2 substitution is accompanied by complete inversion of configuration."

So is the above statement true or not? No clear answer in my Organic Text either.

Thanks Folks!

I think TPR's answer is correct but their explanation is shady.

ABSOLUTE configuration at a chiral center is either R or S. To assign R or S all we need is ONE molecule. In other words, we don't
need to compare the chiral center of the molecule in question to a chiral center of another molecule. We simply assign priorities to
the substituents attached to that chiral center to determine whether it's R or S. When we say configuration is R or S, we know
the exact stereochemistry at this chiral center.


RELATIVE configuration describes the position of substituents at a chirl center of one molecule relative to position of substituents
at a chiral center of another molecule without regard to the priority assignment needed for the determination of absolute
configuration. Inverstion of relative configuration is often described as being analogous to an umbrella being inverted in a strong
wind. Relative configuration is determined by comparing groups attached to two different chiral centers. When comparing two chiral
centers that have all identical substituents except for one, they have the same relative configuration if and only if the 3 identical groups occupy the same positions in space.

When you have an SN2 reaction, the addition of the new atom may alter the priorities around the chiral center, so while relative
configuration is inverted, the absolute configuration may or may not change because the priorities of the chiral carbon substituents
may or may not change as a result of losing one substituent (the leaving group) and adding another (the nucleophile).

So in an SN2 reaction it is possible for a chiral center to retain absolute configuration, but the relative configuration of the substituents at the chiral center is always inverted in an SN2 reaction.

The direction of rotation of plane-polarized light changes when absolute configuration changes. Since we don't know the absolute configuration before and after Sn2 took place (not enough information is given in the problem), we don't know what happens to direction of rotation of plane-polarized light.

Knowing that relative configuration changed tells us nothing about whether direction of rotation of plane-polarized light changed.

 

[shevie]

Hi
this is my first time posting on the orgo question board...hope this works-i was so hapy to come across this!!
With regard to acid catalyzed vs base catalyzed opening of epoxide rings, I remember when I took Orgo (back in 05), I never understood this and just came to terms with the fact that the mechanism would never click in my mind and just hoped for the best when it came to the exam

anyway-I STILL DONT GET IT and i read what it says in kaplan and i re-read what is explained in McMurry. Do I need to understand the mechanism or WHY Base catalyzed cleavage proceeds via "sn2 like" mechanism and acid catalyzed via "sn1 and some sn2" character? HELP!!!!!!!!!!!!! Can someone please clarify?????????

 

[Morpholino]

Dear Shevie,

Opening an epoxide in acid-catalyzed conditions is Sn1-like, because the epoxide begins to open by itself after the epoxide oxygen is protonated. The epoxide begins to open because the protonated oxygen is a good Sn1-type leaving group.

If the epoxide is asymmetric, then the epoxide will preferentially weaken along the C-O bond with the more substituted carbon. As the bond weakens, a partial positive charge builds up on the carbon and a partial negative charge builds up on the oxygen. Notice that the partial positive charge on the carbon resembles a carbocation, and more substituted carbocations are more stable (tertiary > secondary > primary). This explains why the epoxide preferentially begins to open along the C-O bond with the more substituted carbon.

In a base-catalyzed epoxide opening, the oxygen has no "incentive" to begin to leave, since an unprotonated oxygen is a bad leaving group. Thus, the reaction proceeds via Sn2.

Hope this helps!

 

[shevie]

thanks!!!!!!

 

[scotties123]

i recently took an organic chem 2 class and i know a good deal about these structures, but in both Ek and TPR, the nucleophilic attack process is BARELY covered. Should I review my class material about 1,2 additon and 1,4 addition, including reagants for both? or should i just assume this topic is barely covered on the MCAT and that's why it's barely covered in the books. Thanks guys.

 

[BrokenGlass]

i recently took an organic chem 2 class and i know a good deal about these structures, but in both Ek and TPR, the nucleophilic attack process is BARELY covered. Should I review my class material about 1,2 additon and 1,4 addition, including reagants for both? or should i just assume this topic is barely covered on the MCAT and that's why it's barely covered in the books. Thanks guys.

We need to know 1,2 additon and 1,4 addition for the MCAT.

 

[scotties123]

We need to know 1,2 additon and 1,4 addition for the MCAT.

i know that, but do we need to know which reagants cause which to occur? for ex: grignard's, LAH, Organolithiums all result in 1,2 addition.

theres a million more reagants, but im just not sure how many / which ones we need to know.

 

[MileyCyrus]

guys go on www.kaptest.com it will help you lots! with the mcats!🙄

 

[Crookshanks]

yep they still have a whole chapter dedicated to it,
i've noticed org II topics alot rather tha org I.

Does everyone agree with this? It was post # 31.

 

[rcd]

I don't understand alcohol acidity. EK 5th edition tells me that alcohols are acidic, but less acidic than water. Isn't anything that's less acidic than water basic?

EK also says that the order for acidity of alcohols goes from primary to secondary to tertiary. Is that right?

 

[killinsound]

I don't understand alcohol acidity. EK 5th edition tells me that alcohols are acidic, but less acidic than water. Isn't anything that's less acidic than water basic?

EK also says that the order for acidity of alcohols goes from primary to secondary to tertiary. Is that right?

1) it's all relative. you can say alcohol is a weaker acid than water, but also say alcohol is a stronger base than water.

2) yes that is correct. alkyl groups are electron donating, so you need to look at the stability of the conjugate base of the alcohol. because there is a negative charge on the oxygen of the conjugate base, and alkyl groups are electron DONATING (pushing electron density onto an already negatively charged oxygen makes it more reactive aka unstable), the least amount of alkyl groups an alcohol has, the more stable the conjugate base is.

 

[phospho]

okay, i know it's a stupid question, but it's bothering the crap out of me... would appreciate any help...

you're doing resonance structures, and you end up with a carbon that has 3 bonds with a positive formal charge... but that would be breaking the octet rule though! can someone tell me why it's still considered right? a good example would be the conjugated benzene ring when you're trying to eliminate the double bond between C2 and C3, moving it to between C1 and C2, leaving the C3 with only 3 bonds and a formal charge of +1. (Disregard the double bond between 6 and 1 and assume we movied it to an above functional group). I hope this makes sense. Please help. thanks! 🙂

 

[killinsound]

okay, i know it's a stupid question, but it's bothering the crap out of me... would appreciate any help...

you're doing resonance structures, and you end up with a carbon that has 3 bonds with a positive formal charge... but that would be breaking the octet rule though! can someone tell me why it's still considered right? a good example would be the conjugated benzene ring when you're trying to eliminate the double bond between C2 and C3, moving it to between C1 and C2, leaving the C3 with only 3 bonds and a formal charge of +1. (Disregard the double bond between 6 and 1 and assume we movied it to an above functional group). I hope this makes sense. Please help. thanks! 🙂

i think you're thinking of lewis structures. for resonance structures, almost anything can go (within reason).

However, resonance, like you know, is a combined state in which it exists in all of those separate states at the same time. The one contributing most of their state to the resonance structures are the ones that follow the most "rules" (ie: octect, negative charge on electronegative atom, etc.). so although it (being the +1 charge on carbon) is alot more unlikely than the regular triene structure of benzene, it probably still contributes somewhat.

 

[phospho]

i think you're thinking of lewis structures. for resonance structures, almost anything can go (within reason).

However, resonance, like you know, is a combined state in which it exists in all of those separate states at the same time. The one contributing most of their state to the resonance structures are the ones that follow the most "rules" (ie: octect, negative charge on electronegative atom, etc.). so although it (being the +1 charge on carbon) is alot more unlikely than the regular triene structure of benzene, it probably still contributes somewhat.

thank you for the response!
so basically it would be considered OK to have a carbon surrounded by only 3 single bonds instead of the usual 4 in a resonance structure, even if it contributes just a little bit to it...?

 

[killinsound]

it is OK but unstable

 

[BrokenGlass]

thank you for the response!
so basically it would be considered OK to have a carbon surrounded by only 3 single bonds instead of the usual 4 in a resonance structure, even if it contributes just a little bit to it...?

Octet rule applies to neutral elements only. If carbon has a formal positive charge of +1, it's not neutral and need not follow the octet rule. This has nothing to do with resonance.

 

[phospho]

killinsound and broken glass: thank you both... this helps so much😍

 

[rcd]

Octet rule applies to neutral elements only. If carbon has a formal positive charge of +1, it's not neutral and need not follow the octet rule. This has nothing to do with resonance.

Sorry, but you're mistaken. First, the octet rule applies to every atom, neutral or not (for instance, everything has an octet in R-COO-). There are violations, but these are restricted to weird things with Boron, Berryeleiumumum (number 4) and atoms with d orbitals (I'm looking at you, 3rd period and below) and are virtually unrelated to formal charges.

And second, this has everything to do with resonance because it has to do with stability.


Follow these rules, in this order of precedence for writing the most stable lewis structures and therefore the "major contributors:"

1. Use the proper # electrons and atoms.
2. Give everything an octet.
3. Make the formal charges favorable (get 'em at 0 if possible, or put the negative ones on the most electronegative atoms)

 

[phospho]

Sorry, but you're mistaken. First, the octet rule applies to every atom, neutral or not (for instance, everything has an octet in R-COO-). There are violations, but these are restricted to weird things with Boron, Berryeleiumumum (number 4) and atoms with d orbitals (I'm looking at you, 3rd period and below) and are virtually unrelated to formal charges.

And second, this has everything to do with resonance because it has to do with stability.


Follow these rules, in this order of precedence for writing the most stable lewis structures and therefore the "major contributors:"

1. Use the proper # electrons and atoms.
2. Give everything an octet.
3. Make the formal charges favorable (get 'em at 0 if possible, or put the negative ones on the most electronegative atoms)

lol, are we reading the same book?
ochem as a second language by David Klein?

 

[BrokenGlass]

Sorry, but you're mistaken. First, the octet rule applies to every atom, neutral or not (for instance, everything has an octet in R-COO-). There are violations, but these are restricted to weird things with Boron, Berryeleiumumum (number 4) and atoms with d orbitals (I'm looking at you, 3rd period and below) and are virtually unrelated to formal charges.

And second, this has everything to do with resonance because it has to do with stability.


Follow these rules, in this order of precedence for writing the most stable lewis structures and therefore the "major contributors:"

1. Use the proper # electrons and atoms.
2. Give everything an octet.
3. Make the formal charges favorable (get 'em at 0 if possible, or put the negative ones on the most electronegative atoms)

I am not mistaken. I have read the book your are referring to cover to cover a while back. Full octet and resonance are independent concepts. Full octet is desirable because it makes atoms more stable, whether or not reasonance is an issue. What you have listed are the rules for choosing the most stable resonance structure. Are you claiming that every time you have a carbocation you also have resonance?

A neutral carbon in a compound MUST have a formal charge of 0. A neutral carbon in a compound has a full octet.

A carbocation doesn't not have a full octet. A radical doesn't have a full octet. That is the reason why these intermediates are unstable and therefore very reactive. This has NOTHING
to do with resonance.

If you are still having trouble understanding this, consider a carbon bonded to 3 methyl groups.
This carbon has a formal positive charge of +1. There is no resonance involved here. NO RESONANCE.

 

[killinsound]

I am not mistaken. I have read the book your are referring to cover to cover a while back. Full octet and resonance are independent concepts. Full octet is desirable because it makes atoms more stable, whether or not reasonance is an issue.

A carbocation doesn't not have a full octet. A radical doesn't have a full octet. That is the reason why these intermediates are unstable and therefore very reactive. This has NOTHING
to do with resonance.

If you are still having trouble understanding this, consider a carbon bonded to 3 methyl groups.
This carbon has a formal positive charge of +1. There is no resonance involved here.

A neutral carbon in a compound MUST have a formal positive charge of 0.

i agree with broken glass...

let me try to clear some things up

it is true that full octet and resonance are separate concepts, but they are both intertwined.

resonance concept is the idea of a molecule being able to exist in different dot structures at the same time. they are inherently linked. however, the octet rules do not need apply when doing resonance structures. however, if you have a really weird resonance structure with a negative charge on a carbon and a positive charge on an oxygen, this resonance form, although very very weird, may only contribute .0000001%, but that's not saying that it does not exist.

ultimately, the resonance/octet concepts (like formal charges) are there to help us visualize what is happening w/o exactly existing, since it would be hard to talk about something you can't even picture.

 

[rcd]

phospho, we're not reading the same book 🙂, but the concepts are universal.

I am not mistaken. I have read the book your are referring to cover to cover a while back. Full octet and resonance are independent concepts. Full octet is desirable because it makes atoms more stable, whether or not reasonance is an issue. What you have listed are the rules for choosing the most stable resonance structure. Are you claiming that every time you have a carbocation you also have resonance?

A neutral carbon in a compound MUST have a formal charge of 0. A neutral carbon in a compound has a full octet.

A carbocation doesn't not have a full octet. A radical doesn't have a full octet. That is the reason why these intermediates are unstable and therefore very reactive. This has NOTHING
to do with resonance.

If you are still having trouble understanding this, consider a carbon bonded to 3 methyl groups.
This carbon has a formal positive charge of +1. There is no resonance involved here. NO RESONANCE.

That's not where I'm disagreeing with you.

Allow me to reiterate.

You said: "[the] Octet rule applies to neutral elements only."
This is never correct. Ever. The octet rule applies to non-neutral atoms just as well as neutral ones. For instance: [R-COO]-, [NH4]+. These guys aren't neutral, they have formal charges, and they all have octets.

Sure, you can violate the octet rule occasionally, but only when you have to for lack of electrons (or in other rare cases).

You're focusing on carbocations and not seeing the whole picture. Almost all other non-neutral atoms (C-, for instance) still have and seek octets.

 

[BrokenGlass]

phospho, we're not reading the same book 🙂, but the concepts are universal.



That's not where I'm disagreeing with you.

Allow me to reiterate.

You said: "[the] Octet rule applies to neutral elements only."
This is never correct. Ever. The octet rule applies to non-neutral atoms just as well as neutral ones. For instance: [R-COO]-, [NH4]+. These guys aren't neutral, they have formal charges, and they all have octets.

Sure, you can violate the octet rule occasionally, but only when you have to for lack of electrons (or in other rare cases).

You're focusing on carbocations and not seeing the whole picture. Almost all other non-neutral atoms (C-, for instance) still have and seek octets.

I am focusing on carbocations because the original question was on carbocations. What big picture are you talking about? Do you realize that octet rule applies to the MINORITY of elements in the periodic table (pretty much to a subset of the 2nd row of the periodic table)? So it's violated MORE than occasionally, certainly more often than YOU think it is. Elements in the 3rd and higher periods may have more than 8 valence electrons (e.g. PCl5). B and Be (both of whom are 2nd row elements) have less than 8 electrons in their valence shell when they form compounds (e.g. BF3). Your big picture is not as big as you think. Perhaps you need to get reacquainted with the periodic table?

My initial reply may not have been as precise as it could have been, but my point was that a carbocation is not expected to have a full octet simply because it's 2 electrons short of a full octet, which is reflected in its formal positive charge of 1. By contrast, a neutral carbon must always obey the octet rule in its compounds.

Finally, I was trying to make it clear that dragging resonance into this discussion only serves to confuse things because full octets are independent of resonance.

 

[rcd]

I am focusing on carbocations because the original question was on carbocations. What big picture are you talking about? Do you realize that octet rule applies to the MINORITY of elements in the periodic table (pretty much to a subset of the 2nd row of the periodic table)? So it's violated MORE than occasionally, certainly more often than YOU think it is. Elements in the 3rd and higher periods may have more than 8 valence electrons (e.g. PCl5). B and Be (both of whom are 2nd row elements) have less than 8 electrons in their valence shell when they form compounds (e.g. BF3). Your big picture is not as big as you think. Perhaps you need to get reacquainted with the periodic table?

My initial reply may not have been as precise as it could have been, but my point was that a carbocation is not expected to have a full octet simply because it's 2 electrons short of a full octet, which is reflected in its formal positive charge of 1. By contrast, a neutral carbon must always obey the octet rule in its compounds.

Finally, I was trying to make it clear that dragging resonance into this discussion only serves to confuse things because full octets are independent of resonance.

I didn't intend to insult you, and if I did I'm sorry for it. My goals continue to be to a) further understand relevant-MCAT topics and b) help others to do the same.

But more to my original point:
You said: "[the] Octet rule applies to neutral elements only."
That line is incorrect, in any interpretation. To be frank, I'm still wondering why you haven't agreed with me yet.

And actually, the octet rule does actually apply to most atoms. Sure, there are exceptions that I previously mentioned, but they are just that, exceptions. 3rd period elements and below occasionally break the rule o' eight. They don't do it normally.

 

[phospho]

I didn't intend to insult you, and if I did I'm sorry for it. My goals continue to be to a) further understand relevant-MCAT topics and b) help others to do the same.

But more to my original point:
You said: "[the] Octet rule applies to neutral elements only."
That line is incorrect, in any interpretation. To be frank, I'm still wondering why you haven't agreed with me yet.

And actually, the octet rule does actually apply to most atoms. Sure, there are exceptions that I previously mentioned, but they are just that, exceptions. 3rd period elements and below occasionally break the rule o' eight. They don't do it normally.

I hope neither of you wants to insult the other, because you've both proven to be very knowledgeable in organic chemistry. I can only speak for myself: Reading both of your replies to each other has taught me more chemistry than a week of studying 🙂

 

[BrokenGlass]

I didn't intend to insult you, and if I did I'm sorry for it. My goals continue to be to a) further understand relevant-MCAT topics and b) help others to do the same.

But more to my original point:
You said: "[the] Octet rule applies to neutral elements only."
That line is incorrect, in any interpretation. To be frank, I'm still wondering why you haven't agreed with me yet.

And actually, the octet rule does actually apply to most atoms. Sure, there are exceptions that I previously mentioned, but they are just that, exceptions. 3rd period elements and below occasionally break the rule o' eight. They don't do it normally.

Hmm, then we have the same goals.

Let me state one last time that it's NOT the case that the octet rule applies to every atom, neutral or not. A carbocation (arguably the most common intermediate in organic reactions) does not obey the octet rule. So your claim that "the octet rule applies to every atom, neutral or not" is incorrect under ANY interpretation. I am also wondering why you haven't agreed with me yet.

Exceptions to the octed rule are more numerous than you care to admit. For instance, transition metals seek to obtain 18 electrons in their valence shell, not 8. That doesn't seem rare to me, given the large number of transition elements. In addition, both you and me have stated numeroous other violations of the octet rule. I wasn't going to mention the transition elements, but you seem to be a big fan of big pictures.

 

[rcd]

Hmm, then we have the same goals.

Let me state one last time that it's NOT the case that the octet rule applies to every atom, neutral or not. A carbocation (arguably the most common intermediate in organic reactions) does not obey the octet rule. So your claim that "the octet rule applies to every atom, neutral or not" is incorrect under ANY interpretation. I am also wondering why you haven't agreed with me yet.

Exceptions to the octed rule are more numerous than you care to admit. For instance, transition metals seek to obtain 18 electrons in their valence shell, not 8. That doesn't seem rare to me, given the large number of transition elements. In addition, both you and me have stated numeroous other violations of the octet rule. I wasn't going to mention the transition elements, but you seem to be a big fan of big pictures.

Ok, you're cherry-picking at this point. The transition metals? I mean come on, they don't follow any traditional rules. And beyond that, they're the forgotten elements, only like 2 of them count for anything. Like Fe in heme. I think I saw Zn somewhere a couple weeks back. Maybe there's some more, but I mean come on.

In any event, you're putting words into my mouth. I didn't say the octet rule applies to every atom, I said it applies to most atoms. Sure, I slighted the transition elements, but only because everybody else does too.

You said the "octet rule does not apply to non-neutral atoms." That is a quote and not something I'm just saying you said. My whole[/i] point is that that quote is not true.


Ima bow out.

 

[BrokenGlass]

Ok, you're cherry-picking at this point. The transition metals? I mean come on, they don't follow any traditional rules. And beyond that, they're the forgotten elements, only like 2 of them count for anything. Like Fe in heme. I think I saw Zn somewhere a couple weeks back. Maybe there's some more, but I mean come on.

In any event, you're putting words into my mouth. I didn't say the octet rule applies to every atom, I said it applies to most atoms. Sure, I slighted the transition elements, but only because everybody else does too.

You said the "octet rule does not apply to non-neutral atoms." That is a quote and not something I'm just saying you said. My whole[/i] point is that that quote is not true.


Ima bow out.

You said "the octet rule applies to every atom, neutral or not", which is clearly false. These are your words, not the words I am putting into your mouth. Why are you having such a hard time accepting that? If you are looking for precision and perfection in other people's posts you need to first start my making sure your own posts are error free, which they are not.

 

[notsureyet]

Guys I gotta a question:

Can 3 different atoms around a Carbon atom be classified as R or S, or must it always be 4 ?

Btw,

How do you keep track of all the possible reations available for a mechanism, for example, If the reaction is bound to be a SN1 major product, how do you remember all the nuclephiles and solvents, making sense of it is easy, but I have trouble recalling all that information for writing out the mechanism.

 

[BrokenGlass]

Guys I gotta a question:

Can 3 different atoms around a Carbon atom be classified as R or S, or must it always be 4.

It must be 4. Only sp3 hybridized atoms can be chiral. If an sp3 hybridized atom has 4 different substituents, then we can assign an R or S absolute configuration to this chiral center.

Actually, there is a an exception to sp3 hybridization/chirality, but it almost never comes up.

 

[killinsound]

It must be 4. Only sp3 hybridized atoms can be chiral. sp3 hybridized atoms have 4 substituents. If these substituents are all different, then we can assign an R or S absolute configuration to this chiral center.

i'd also like to add that nitrogens, if bound to 4 R groups are not chiral.

 

[rcd]

What's the difference between nucleophilic addition and nucleophilic substitution?

 

[philios]

Hi Im going through the examkrackers Chemistry book, and came upon a study question I dont understand...(its #22 if that helps)
22)Which of the following best explains why sulfur can make more bonds than Oxygen?
a)Sulfur is more electronegative than oxygen
b)oxygen is more electronegative than sulfur
c)sulfur has 3d orbitals not available to oxygen
d)sulfur has fewer valence electrons

They say the answer is C, which confuses me..I thought sulfur was
[Ne]3s(^2)3p(^4) , which does not include 3d at all! Is it because sulfur has the potential to use the 3d orbital? Thanks!

EDIT* oops i thought this was the general chem thread

 

[rcd]

Is it because sulfur has the potential to use the 3d orbital?

Yes.