Organic Chemistry Question Thread

[QofQuimica]

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

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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.

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[MundaneMD]

During oxymercuration, an intermediate is the mercurinium ion (Hg+(OAc) bonded to both Cs across the double bond). Then, water attacks the more highly substituted C (which can hold more + charge) from the backside, breaking the ring and leading to an organomercurial alcohol (following deprotonation of water).

This mechanism seems to greatly resemble Sn2 in that both utilize backside attack on an electrophilic C.

My question is:

Why can water attack a tertiary C (bonded to Hg as part of the mercurinium ion) in oxymercuration, whereas in SN2, the nucleophile CANNOT attack a tertiary C due to steric hinderance ? Isn't there steric hinderance also in the nucleophilic attack of mercurinium ion?

I hope that question made sense. Thanks in advance!

Consider the fact that when the Hg atom is bridged, it has three resonance forms. They are the bridged form, along with two minor contributing forms.

Looking at the minor form with the partial positive charge on the substituted carbon, it is essentially a carbocation. Obviously, it's not a "traditional" carbocation, because the double bridged form is the dominant form.

Still, at the instance where the ring opens up to leave the partial positive charge on the highly substituted carbon, consider the geometry of the molecule. A carbon with only three bonds, and no lone pairs --> somwhere between tetrahedral, and trigonal planar (which would have decreased steric hindrance). So the water molecule attacks at this instance, but anti to the mercury (which would still provide steric hindrance).

Let me pre-empt any question you might have about why the same process does not occur on the other bridged carbon. The contribution of that resonance form is negligible, so it does not have a positive charge.

In an SN2 reaction, the nucleophile has to attack first, but it can't due to the hindrance. In this reaction, the nucleophile has to wait for the bridge to open up, so that the carbon will have a partial positive charge. This is actually closer to the behavior of an SN1 reaction.

 

[BlackSails]

Consider the fact that when the Hg atom is bridged, it has three resonance forms. They are the bridged form, along with two minor contributing forms.

Looking at the minor form with the partial positive charge on the substituted carbon, it is essentially a carbocation. Obviously, it's not a "traditional" carbocation, because the double bridged form is the dominant form.

Still, at the instance where the ring opens up to leave the partial positive charge on the highly substituted carbon, consider the geometry of the molecule.

Molecules dont flip between resonance forms though, they are in an averaged form of all of them. My professor explained it like this:
Dr. Jekyll is sometimes Dr. Jekyll and sometimes Mr. Hyde. This is an equilbrium between two forms.

Frankenstein is half man and half monster, but he is this all the time, he does not alternate between man and monster. He is a resonance form.

 

[unsung]

Molecules dont flip between resonance forms though, they are in an averaged form of all of them. My professor explained it like this:
Dr. Jekyll is sometimes Dr. Jekyll and sometimes Mr. Hyde. This is an equilbrium between two forms.

Frankenstein is half man and half monster, but he is this all the time, he does not alternate between man and monster. He is a resonance form.

Haha.. my professor explained it by showing us a picture of Counselor Troi in the episode where she was altered to be a Romulan... and asking us-- "So, anyone know who this is?"

(The idea being that Counselor Troi is half Betazoid, half human... but she doesn't change between being a Betazed and a human- she's half and half all the time.)

I love it.

 

[MundaneMD]

First of all, you guys have fun orgo teachers.

Second, you're right. The resonance structures don't flip from one to another. But the partial charges still exist on the carbons, as if the molecule did flip from one to another.

And one thing I found in my orgo book...they say that computer modeling shows that the bond between the more substituted carbon and the mercury is longer than the other carbon-mercury bond, which also makes the substituted carbon easier to attack.

 

[unsung]

Thanks for the replies... and, I have yet another question. This one has been bugging me!

I get that absolute configuration is R/S assignment for naming chiral centers. But the concept of "relative configuration" (D/L, not to be confused with "d/l = +/-")
is endlessly confusing to me... and a question on an EK exam clearly showed me that I did not grasp the concept .

Not to repeat the question here, but basically the problem told us a reaction proceeds with "retention of configuration", and we're told the product is (S). Then we're asked to pick the compound that could be a reactant.

I thought, oh, I just have to look for the (S) reactant... but no, that was TOTALLY wrong. The answer was actually the reactant with (R) configuration. In the answer key, the explanation was something like "retention of configuration does not mean absolute configuration is retained; it means there's no inversion."

So basically, we were supposed to draw the (S) product, then get the structure of the reactant from it (which actually ends up being (R) ), and match it to the answer choice.

Can anyone shed some light on the topic? What does "retention of configuration" mean if it doesn't refer to retention of absolute configuration ???

I've tried researched, googled, etc. to no avail.

 

[poserboarder]

When you need to determine organic products for organic reactions (Sn1, Sn2, E1, E2) when you aren't given which reaction is taking place, how can you tell if the compound written above the yield arrow is going to be a nucleophile that will attack the substrate resulting in a substitution reaction or if it is a base that will pull off a Hydrogen resulting in an elimination reaction.

This is my second time around in Organic, doing better this time, but I screwed up the second test, because I couldn't tell if it was a nucleophile or a base. Our professor gave us a chart to determine what type of reaction it was, but it seemed to always come down to know if it was a nucleophile or a base.

Check out the chart:

 

[omar586]

how can u tell how many a compound can have stereoisomers??

 

[unsung]

When you need to determine organic products for organic reactions (Sn1, Sn2, E1, E2) when you aren't given which reaction is taking place, how can you tell if the compound written above the yield arrow is going to be a nucleophile that will attack the substrate resulting in a substitution reaction or if it is a base that will pull off a Hydrogen resulting in an elimination reaction.

This is my second time around in Organic, doing better this time, but I screwed up the second test, because I couldn't tell if it was a nucleophile or a base. Our professor gave us a chart to determine what type of reaction it was, but it seemed to always come down to know if it was a nucleophile or a base.

Check out the chart:

Substitution and elimination almost always go hand in hand. Say you have methoxide CH3O- , it's both a strong nucleophile and a strong base. This makes it likely to participate in Sn2 (as a nucleophile) and in E2 (as a base).

Otoh, methanol CH3OH is a comparative weaker nucleophile/base. This makes it likely to participate in Sn1 (as a nucleophile) and in E1 (as a base).

So, okay, the part of determining the order of reaction is easy... but how do you determine whether substitution or elimination predominates?

1) look at the substrate

If the substrate is really hindered (for ex: halide is bonded to a tertiary carbon, or worse yet neopentyl group), Sn2 basically is NOT going to happen.

If there's too much steric hinderance, the nucleophile can't get close enough to the C to participate in substitution. So instead, it's going to abstract a H from the neighboring C instead, allowing elimination to predominate over substitution.

So to state that rule of thumb more clearly: if the reaction involves a *strong* nucleophile/base, you've narrowed down your possible reactions to Sn2/E2. Next, if the substrate is hindered, Sn2 becomes unlikely, so the reaction is going to be E2.

(Note: hinderance of the substrate doesn't affect Sn1/E1 in the same way, as a tertiary carbocation is readily attacked by the nucleophile in Sn1, because it is a charged species, despite the fact that it is tertiary.

This actually has to do with the last question I asked, which I later spoke to my professor about, and that was his explanation for why steric hinderance matters for Sn2, but not for Sn1 or oxymercuration. Basically, steric hinderance is a huge consideration for Sn2, because the tertiary C is a neutral species. Whereas for Sn1, a planar tertiary carbocation carries a positive charge, which is enough to get the nucleophile to overcome steric hinderance and substitute.)


2. look at the nucleophile/base

Is it big and bulky? (like t-butoxide, or triethylamine) Or is it something small like methoxide? Something big and bulky is going to favor elimination- again, because of steric considerations. That is, something big and bulky is more likely to act as a base than as a nucleophile.

So that's also something helpful to keep in mind for synthesis problems, if you want to make the elimination product, while minimizing substitution products: use a bulky base.

Hope that helps.

 

[poserboarder]

That does help, but my problem is I always look at what's written above the arrow and treat everything as a nucleophile only (which you're saying is okay anyway) and refer to that chart.

But to my understanding, not all strong nucleophiles can also be strong bases and vice versa. Same for good nuc/bases, or poor nuc/bases.

This is how I determine if the nucleophile is strong:

weak = no negative charge present
moderate = negative charge on a halogen or resonance stabilized
strong = negative charge present, but not on a halogen

Is it okay to do this, or is there going to be a point where I'm going to actually have to sit down and determine if that's a base or a nucleophile right off the bat?

 

[Chemist0157]

That's a good start, but, as unsung said, you have to pay attention to the "bulkiness" of the nucleophile/base.

One more thing to watch out for with these reactions is temperature. High temperature favors elimination while room temperature (and below) favor substitution.

 

[poserboarder]

That's a good start, but, as unsung said, you have to pay attention to the "bulkiness" of the nucleophile/base.

I get what you're saying, but that shouldn't matter right? Look at the chart, Strong nucleophiles pair w/ strong bases, good pairs w/ good, and poor pairs w/ poor.

One more thing to watch out for with these reactions is temperature. High temperature favors elimination while room temperature (and below) favor substitution.

I read that in an o-chem help book, but our professor didn't teach us that or throw any temps on the test. Also he completely dropped the use of polar aprotic solvents (strong indicators of Sn2) this year, so determining reaction was difficult.

 

[Chemist0157]

It's not always that way though. Ethoxide is a good nucleophile but a weak base, while t-butyoxide is a bad nucleophile but a strong base. A compound won't always be good a nucleophile AND a good base.

 

[unsung]

I get what you're saying, but that shouldn't matter right? Look at the chart, Strong nucleophiles pair w/ strong bases, good pairs w/ good, and poor pairs w/ poor.

Okay, I don't know about that chart... but do you want to post some problems that you're working on? It's much easier to talk in terms of specifics than in the abstract.

Anyway, I thought of a really crappy metaphor . Say you're really really hungry. And there's a delicious cake in the next room through a door. There's also a ham sandwich in the same room as you. If you're really hungry, you might grab both the ham sandwich and the cake. (Or if you're not really very hungry at all, you'll probably just sit there and not grab either.)

Fine. But let's say there's three henchmen guarding the door to the next room... you might be really really hungry- but it's unlikely you're getting through that guarded door to get to the cake. So you'll probably grab the ham sandwich.

Alternatively, say the door isn't guarded, BUT, you're really really obese. So you can't fit through that damn door. So- same result, even though you're really hungry, you're going to have to grab that ham sandwich, instead of the cake.

So, whether or not you eat the cake (act as a nucleophile- Sn2) or eat the ham sandwich (act as a base- E2) has to do with two things: 1) how hungry are you? (really hungry = 2nd order reaction, not so hungry = 1st order reaction) and 2) what do you have access to? (elimination vs. substitution)

You've got part 1) figured out-- how to determine whether a species is potentially "strong" or "weak" (i.e. hungry or not hungry).

BUT just because you're really hungry (potential strong nucleophile/base) doesn't mean you're going to end up grabbing either the cake or the sandwich (*acting* as a nucleophile or base)... perhaps it's hard to get near the cake because of the "henchman" (steric hinderance)... or perhaps you're too fat to get through the door (bulky nucleophile = bad nucleophile).

To determine whether substitution actually occurs or whether elimination actually occurs, it's necessary to consider the circumstances limiting (or promoting) "access". It's not enough to just look at the species itself...

 

[poserboarder]

To determine whether substitution actually occurs or whether elimination actually occurs, it's necessary to consider the circumstances limiting (or promoting) "access". It's not enough to just look at the species itself...

If this is the case, then this might just solve my problems, because I'm putting too much thinking into this.

How about let me ask this:

When you're given an organic reaction and you must predict the product, if you just look at what's written above the yield arrow (which will be the nucleophile or the base)... if you cross EVERYTHING else out and just look above the yield arrow... Is there a way to tell if that molecule will be a nucleophile or a base right off the bat?

Also this might be helpful to bring up as well...

So I always thought this was the way to determine strength of the nucleophile:

poor = no negative charge present

good = negative charge on a halogen or a resonance stabilized structure (lone pair, single bond, double bond)

strong = negative charge is present, but not on a halogen

So let me ask this:

If you're looking at the nucleophile (or base) written above the yield arrow and don't take any other part of the problem into account, what are the rules you use to determine the strength of it? Are my rules accurate or do I need to amend them in some way? I know for sure there have been times when my definition of a strong nucleophile have failed me. I should get this straight.

Oh and thanks for your analogy, haha that was great!

And I will try to post some example problems that are giving me trouble, but maybe after I get an answer to my above couple of questions, because I might be all right after that.

Thanks kindly.

 

[SLagraize]

if you are doing gas chromatography and get overlapping peaks, how would you adjust the following variables to resolve the overlap:
1. sample volume
2. column length
3. column temperature
4. injection rate
5. mobile phase flow rate
i don't understand what would cause the peaks to overlap beside the fact that the gc cannot distinguish them properly. thanks for your help.

 

[SBK]

Can someone tell me about orbitals and unpaired electrons?

For example, does a lone pair of electrons occupy an orbital? I.e., Is NH3 (with a lone pair attached to N) sp2 or sp3 hybridized?

What about a radical? Is CH3 (radical) sp3 or sp2?

 

[poserboarder]

Can someone tell me about orbitals and unpaired electrons?

sp3 with 0 lone pairs = tetrahedral
sp3 with 1 lone pair = trigonal pyramidal
sp3 with 2 lone pairs = bent
sp2 with 0 lone pairs = trigonal planar
sp2 with 1 lone pair = bent
sp with 0 lone pairs = linear

For example, does a lone pair of electrons occupy an orbital?

Yes. Lone pairs and bonds occupy orbitals.

Is NH3 (with a lone pair attached to N) sp2 or sp3 hybridized?

In Ammonia (NH3), the nitrogen atom is sp3 hybridized... meaning all four orbitals are arranged in a tetrahedral structure as you would initially suspect... BUT only three of the orbitals in this arrangement are responsible for bonds, so if you look at how the atoms are connected, you don't see a tetrahedron... INSTEAD you see a trigonal pyramidal arrangement.

What about a radical? Is CH3 (radical) sp3 or sp2?

A radical is just a single electron, not bonded to anything. Basically it's just a lone pair (2 electrons not bonded to anything) minus one of the electrons.



Hope all that helps you some. By the way to anyone viewing this thread, I'm still desperately seeking an answer to my question about three posts up from this one. Thanks kindly.

 

[unsung]

When you're given an organic reaction and you must predict the product, if you just look at what's written above the yield arrow (which will be the nucleophile or the base)... if you cross EVERYTHING else out and just look above the yield arrow... Is there a way to tell if that molecule will be a nucleophile or a base right off the bat?

I don't think there's a foolproof way right off the bat. Here's what I usually do: I do what you do, which is first, look above at the yield arrow. And from that, what I can determine is "strong" or "weak".

If it's strong, I write preliminarily Sn2/E2. If it's weak, I write preliminarily Sn1/E1. Then I look at the other factors, and I eliminate possibilities that are impossible. (Ex: tertiary C means Sn2 not going to happen, so only E2 is possible).

Although, SOMETIMES, you can get some clues about whether it's going to act as a nucleophile or base JUST from the species itself. And that's when it's something really bulky, like t-butoxide. So yes, if you see t-butoxide, you know first of all, that it's strong. And secondly, you can sort of know that it's more likely to eliminate than substitute, because of it's "bulk". But, this is still not a guarantee, because if you have a really easily approachable substrate (say methyl chloride), I don't think it's impossible for a bulky nucleophile to approach and substitute (please correct me if I'm wrong). Although elimination is still going to be more easy than substitution. Again though, oftentimes, both occur.

Also this might be helpful to bring up as well...

So I always thought this was the way to determine strength of the nucleophile:

poor = no negative charge present

good = negative charge on a halogen or a resonance stabilized structure (lone pair, single bond, double bond)

strong = negative charge is present, but not on a halogen

So let me ask this:

If you're looking at the nucleophile (or base) written above the yield arrow and don't take any other part of the problem into account, what are the rules you use to determine the strength of it? Are my rules accurate or do I need to amend them in some way? I know for sure there have been times when my definition of a strong nucleophile have failed me. I should get this straight.

Okay... so, your rules look good to me. I don't know if this helps any, but basically I just do this:

If what's above the yield arrow is basically a solvent (i.e. methanol CH3OH, or ethanol or whatever), I call that "weak", and write preliminarily: Sn1/E1.

If it's the conjugate base of the solvent (i.e. methoxide CH3O-, or CH3CH2O-, etc.), I call that "strong", and write preliminarily Sn2/E2.

That's pretty much it. It gets trickier if the problem asks to compare between two reagents and determine which one is *better* at Sn1, or Sn2. For ex, all else being equal, is the reagent with the N as the nucleophile the stronger nucleophile, or the reagent that has O? O is more electronegative than N, so that means it holds on to its electrons more tightly, so it's a weaker nucleophile, and N is the stronger nucleophile.

So... it really depends on the type of problem, what my strategy is. It usually involves thinking about how stable the species is. More electronegative = more stable = weaker nucleophile. etc. etc.

 

[poserboarder]

Okay... so, your rules look good to me.

Something isn't right with my rules and I think that's what screwing me up. -CN and -N3 are both considered good nucleophiles, but according to my definition they would be strong (negative charge on something other than a halogen).

How about:

strong = any -O or -N... exception being -N3.

Suggestions??

I'll post some of the nucelophiles we are given when I get a chance to give you a better idea what I'm working with, because usually what he gives us are not solvents or conjugate bases of solvents. What type of solvents are you speaking of anyway? The only solvents that come to mind at the moment are polar aprotic (DMF, DME, etc..) and polar protic and these are written under the arrow.

 

[fly1346]

Hi would someone mind helping me with 1H NMR. I tried reading the post in the MCAT forum, but thats very basic... Im lost when it comes to actual problems when you have to figure out the structure of the compound when your given the NMR data, plays alot of mind games.

Can someone please explain the upfield, downfield, N+1, different H's etc

THANKS~!!!

 

[MundaneMD]

1H NMR

1H NMR is used to identify the positions of hydrogen atoms within a given molecule.

Rule for getting peaks:
2(I) + 1

This means that if there are atoms of hydrogen surrounding a specific hydrogen atom, and they have a spin of (I) (all hydrogens have a spin of 1/2), the signal of that specific hydrogen will contain 2(I) + 1 peaks.

Example:
Ethanol (CH3CH2OH)

In the molecule of ethanol, there are three hydrogens attached to the carbon on the left (A), and two hydrogens attached on the other carbon (B). There is also a hydrogen on the oxygen which is attached to (B).

(A) has three equivalent hydrogens. (Let me know if you don't understand this statement). In a 1H NMR, each of the hydrogens on (A) will have see that there are two neighboring hydrogens on (B). The (A) hydrogens will not see the other (A) hydrogens, because they are all equivalent. Hydrogens can see other hydrogens that are up to three bond lengths away. (There is a bond between the three hydrogens to (A), a bond between (A) and (B), and a bond between (B) and the other hydrogens. Three bonds.

Using the 2(I) + 1 rule, the (A) hydrogens see two neighboring hydrogens, each with a spin of 1/2.
2(2 hydrogens on (B))x(1/2) + 1 = 3. Triplet.

Repeat this rule for the (B) hydrogens:
2(3 hydrogens on (A)) x (1/2) + 1 = 4. Quartet.

Before we do the OH group, let me tell you (very simplisticly) that the oxygen acts as a barrier for the OH hydrogen from seeing the other hydrogens. So if you see a hydrogen connected to an oxygen, it will not be able to "communicate" with the other hydrogens. It will see 0 neighboring hydrogens as a result.

One more time for the hydrogen on the OH group:
2(0 hydrogens) x (1/2) + 1 = 1. Singlet.

Thus, there will be a quartet due to three hydrogens attached to (A), a triplet from the two hydrogens attached to (B), and a singlet for the hydrogen attached to the oxygen.

Now. Upfield/Downfield.
Oxygen has a lot of electon density (look for lone pairs, and double/pi bonds). Electron-rich atoms will shift connected hydrogens downfield (to the left of the 0ppm starting point). Thus, you can expect that the hydrogen singlet on the OH group will be more downfield than the other hydrogen signals.

On the other hand, things with less electron density will not be as downfield as the oxygen. An example would be the transition metals, which are very upfield.

Take a look at the chart in your orgo text book (every book has one) for the upfield/downfield shifts of different groups. You'll notice trends such as the COOH group being very downfield (think about it, a double bonded oxygen, AND a OH group. ahhhh! ).

The key to NMR problems is being able to see different hydrogen environments. In this problem, there are three hydrogen environments, which produce three different signals. If a molecule is symmetric, (propane), there will be one signal from the middle carbon, and one (not two) signals from the terminal carbons. Why? The two terminal carbons are perfectly symmetrical, so the NMR only see's one type of hydrogen environment.

Hope that helps. BTW, you're right, NMR seems like a game. Once you get good at it, I would go as far as to say it's pretty fun. NMR was always my favorite part of orgo.

 

[poserboarder]

I finally developed a pretty solid list of different types nucleophiles and found a common trend so that you can determine the strength of any given nucleophile.

I'll upload it here when I get a chance, it should definitely help anyone struggling when predicting organic products when substitution and elimination are happening at the same time and you are trying to figure out which one predominates.

 

[poserboarder]

Okay here it is... for students struggling with substitution/elimination reactions, this may solve your problems and frustrations.

If you're asking yourself questions like:

-How can I determine the order of this reaction?
-How can I determine if this reaction is substitution or elimination?
-How can I predict the organic product(s) of this sub./elim. reaction?
-What is the strength of this nucleophile?
-How do I know if this molecule is acting as a nucleophile for substitution or a base for elimination?

The key to predicting if sub. or elim. will predominate so you can predict the outcome, is knowing two things. The substrate (very easy to determine). And the strength of the nucleophile.

For me, my professor did not hand out a list of nucleophiles to memorize and at times the book was difficult to follow, so I was very lost. I have been where you are at right now and after seeing my professor numerous times, discussing with friends, and doing a lot of my own studying/practicing... I organized some very helpful tools that will help you succeed with these reactions.

Here is a chart. You match up your substrate and strength of your nucleophile to determine what type of reaction will predominate.



Here are the rules for determining nucleophile strength (I came up with myself after looking at trends and had them confirmed by my professor). Also on this page are tips for differentitating between sub. and elim. and some example nucleophiles drawn out in kekule (lewis dot) form. Just a quick note, the lewis dot structures are not drawn to match molecular geometry.

Forgot to throw temp. rules on this page. 100 degrees C or higher will favor elim. products. Room Temp. (18-23 degrees C) or lower will favor sub. products.



Finally here is just a page on some things you might want to know about sub. reactions. Like what a polar aprotic solvent looks like, leaving group (LG) information, and what a pseudohalogen is and what they look like.



I hope this all helps you as it did me. I was able to go back to my old test and correctly do all the problems I missed. I understand this stuff like never before and thanks to my better understanding, I am actually able to help out a couple friends in my class (and you guys here at sdn) that need some clarity.

If you're trying to print these off and are having problems, get into print preview, try landscape view and choosing the shrink image to fit option.

If you have any questions at all about my scans (i.e. how do i use this chart) or any of the info written on the pages, I will be glad to help. Also if you just have any random question about sub./elim. I will be happy to help you the best I can.

 

[phospho]

Okay here it is...

you freakin' rock...I can't begin to tell you how helpful that was...thank you so much...

 

[Foghorn]

1H NMR

1H NMR is used to identify the positions of hydrogen atoms within a given molecule.

Rule for getting peaks: 2(I) + 1

What if spin-spin splitting isn't 1st order, how do you figure that out?

 

[MundaneMD]

What if spin-spin splitting isn't 1st order, how do you figure that out?

You figure that out by buying a NMR instrument with a higher resolution.

Seriously though, I have never encountered any questions about higher orders of splitting in my Orgo class, my MCAT studies, or even my upper-level chemistry classes. This is because only the older NMR instruments had this problem. Higher resolution NMR instruments have eliminated this problem.

Have you come across a problem about different orders of NMR spectra? Or does your question stem from intellectual curiosity?

 

[cheezer]

foghorn is the mack daddy of chemistry. he was probably asking out of amusement.

 

[prmdbeach17]

Is there a list somewehere of concepts/rxns we don't need to know for the MCAT? I was scanning this thread and saw that alkenes, alkynes & hydroborations are things that won't be tested...but can anyone make/does anyone have a comprehensive list? Thanks!

 

[prmdbeach17]

And I wanted to ask if it's actually true that we don't need to know anything about alkynes (eg, H2/Lindlar's, H2/surface metal) or alkenes (eg, conversions to alcohols or halohydrins)? It seems like it would be focusing mostly on Organic 2 and skipping a lot of organic 1...

 

[161927]

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.

 

[Foghorn]

You figure that out by buying a NMR instrument with a higher resolution.

Seriously though, I have never encountered any questions about higher orders of splitting in my Orgo class, my MCAT studies, or even my upper-level chemistry classes. This is because only the older NMR instruments had this problem. Higher resolution NMR instruments have eliminated this problem.

Have you come across a problem about different orders of NMR spectra? Or does your question stem from intellectual curiosity?

I'm messing with you dude. cheezer is right.

You don't need a higher resolution NMR apparatus though at the cost of signal peaks not being well-defined. Using different (magnetic) pulse sequences works and also at different magnetization angles. For example, Nuclear Overhauser Effects (NOEs) can help figure out stereochemistry at a specific chiral center. These techniques are taught in upper div/grad O-Chem classes and you'll never see them on the MCAT.

Pascal's (n +1) rule, aka 1st order splitting, taught in basic O-Chem are good for H-atoms separated by not more than 3-bonds i.e., H-atoms are attached on C-atoms adjacent to each other. Therefore the maximum number of peaks observed from this formula is 7. A general rule of thumb is when more than 7 NMR-peaks are observed for a particular proton, it's an indicator of higher ordered splitting.

foghorn is the mack daddy of chemistry. he was probably asking out of amusement.

It's the Lounge effect
Nah that would be QofQuimica. She has a Ph.D. in Chem. I've only taken one graduate course in NMR theory and that's because I couldn't find another class to fill my schedule last year. And stupid me didn't realize how much work was involved. I was the only undergrad in there so no curve for me

 

[jochi1543]

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.

Hmmmm, I'm pretty sure those 2 are the same compound. I'm too lazy to build a model, but imagine flipping it. All that matters is that the OH is trans to the hydrogen. Also, either way the OH carbon is numbered 1, and the next one would be #2, so no matter how you flip it, it should be the same.

 

[Foghorn]

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.

You've got to keep in mind Markovnikov's rule for this type of addition reaction plus the fact that the mechanism of the reaction produces two different types of 3-membered ring mercurinium ion intermediate. There's no preference which intermediate forms and both occur with equal probability; the reaction produces a racemic mixture.

Hmmmm, I'm pretty sure those 2 are the same compound. I'm too lazy to build a model, but imagine flipping it. All that matters is that the OH is trans to the hydrogen. Also, either way the OH carbon is numbered 1, and the next one would be #2, so no matter how you flip it, it should be the same.

The alcohol molecules are non-superposable mirror images of each other with a plane of symmetry "cutting" though all the C-atoms. A more accurate figure, for either enantiomer, is drawing 2 chair conformations that are fused together to form the bicyclic molecule.

 

[MundaneMD]

I'm messing with you dude. cheezer is right.

Bah.

I knew something was fishy. You're usually the one answering questions, not asking them.

 

[ask1andy]

the reaction produces a racemic mixture.

You are correct sir!

 

[premd]

Which is correct? If you have an ester with an ethyl group attached to the oxygen attom and a group XYZ on the other side, should it be named:


A. ethyl XYZ carboxylate

OR

B. XYZ ethanoate?


Kaplan says that A is correct, but according to my textbook, a carboxylate is used to refer to the conjugate base of a carboxylic acid and not an ester.

 

[kronickm]

Which is correct? If you have an ester with an ethyl group attached to the oxygen attom and a group XYZ on the other side, should it be named:


A. ethyl XYZ carboxylate

OR

B. XYZ ethanoate?


Kaplan says that A is correct, but according to my textbook, a carboxylate is used to refer to the conjugate base of a carboxylic acid and not an ester.

Both wrong. Group attached to oxygen-yl, carbon chain on side of carbonyl-oate.

should be ethyl XYZoate

 

[unsung]

In an organomercu... rxn with an alkene, what is the stereochemistry, inverted right? SO if I have a dicyclohexane (two linked cyclohexanes) and a H at both vertices, and a double bond extending from a vertex to an adjacent ring carbon, how do I know if the H is sticking out or is behind? How do I predict the stereochemical outcome?Hmm I'll draw a picture.

Wait a minute, unless I'm not thinking clearly, I believe oxymercuration-demercuration gives the Markovnikov product (i.e. the more highly substituted alcohol). So, the OH should be attached to the tertiary C, not the secondary carbon.

In terms of the stereochemistry of the outcome, in THIS case, you'll only get one product. But in the picture of the products that you drew (which would be the products for hydroboration-oxidation), the two alcohols are enantiomers. Although in this case, oxymercuration-demercuration would give only 1 product, in general, sometimes it gives 1 product, and sometimes it gives 2 products (pair of enantiomers). Although, I'm finding that with my professor at least, he doesn't require us to give both enantiomers for the product (just one of the pair)... but, ymmv.

 

[161927]

Wait a minute, unless I'm not thinking clearly, I believe oxymercuration-demercuration gives the Markovnikov product (i.e. the more highly substituted alcohol). So, the OH should be attached to the tertiary C, not the secondary carbon.

In terms of the stereochemistry of the outcome, in THIS case, you'll only get one product. But in the picture of the products that you drew (which would be the products for hydroboration-oxidation), the two alcohols are enantiomers. Although in this case, oxymercuration-demercuration would give only 1 product, in general, sometimes it gives 1 product, and sometimes it gives 2 products (pair of enantiomers). Although, I'm finding that with my professor at least, he doesn't require us to give both enantiomers for the product (just one of the pair)... but, ymmv.

Yeah you just caught my error! The OH would attach at the linkage point rather than the secondary carbon.

 

[Enthalpy430]

edit: Answered my own question.

 

[cheezer]

edit: Answered my own question.

hydride is preferred, if a hydride shift isn't possible, then methyl.

and based on your nomenclature the carbon in the 2nd position has one methyl group and a hydrogen attached to it prior to any rearrangement occurring


*edit I want my two minutes back.

 

[Enthalpy430]

hydride is preferred, if a hydride shift isn't possible, then methyl.

and based on your nomenclature the carbon in the 2nd position has one methyl group and a hydrogen attached to it prior to any rearrangement occurring


*edit I want my two minutes back.

In 2-methylpentan-3-ol, the 2 carbon has two methyl groups attached to it and the hydrogen. For whatever reason, I was thinking of a scenario in which moving either a methyl or a hydride would form a tertiary carbocation.

But in the molecule I provided, moving the methyl off the 2 carbon would create a secondary carbocation at the 2 carbon. So, I deleted it since it was erroneous to begin with.

But thanks for answering my question about which shift would be preferred.

 

[cheezer]

In 2-methylpentan-3-ol, the 2 carbon has two methyl groups attached to it and the hydrogen. For whatever reason, I was thinking of a scenario in which moving either a methyl or a hydride would form a tertiary carbocation.

I hope Foghorn catches this, but I'll spout off what I think: 2-methylpentan-3-ol has one methyl and one hydrogen attached to the second carbon, otherwise the nomenclature would be 2,2 -dimethylpentan-3-ol and the 2nd carbon would not have a hydrogen. What you're proposing is five bonds to the second carbon and that is impossible. In 2,2 -dimethylpentan-3-ol the only way to form a tertiary carbocation is through the methyl shift, however we're dealing with the above structure, correct?

But in the molecule I provided, moving the methyl off the 2 carbon would create a secondary carbocation at the 2 carbon. So, I deleted it since it was erroneous to begin with.

But thanks for answering my question about which shift would be preferred.

Yeah. Hydride first, methyl second.

 

[MundaneMD]

I hope Foghorn catches this, but I'll spout off what I think: 2-methylpentan-3-ol has one methyl and one hydrogen attached to the second carbon, otherwise the nomenclature would be 2,2 -dimethylpentan-3-ol and the 2nd carbon would not have a hydrogen. What you're proposing is five bonds to the second carbon and that is impossible. In 2,2 -dimethylpentan-3-ol the only way to form a tertiary carbocation is through the methyl shift, however we're dealing with the above structure, correct?

Eh.

2-methylopentan-3-ol is the same as saying 1,1-dimethylbutan-2-ol (but calling it this does not follow IUPAC rules).

So it does, in fact, have two methyl groups attached, although one of them is part of the pentane chain. Carbon #2 in the pentane chain does have two methyl groups, technically.

 

[cheezer]

Eh.

2-methylopentan-3-ol is the same as saying 1,1-dimethylbutan-2-ol (but calling it this does not follow IUPAC rules).

So it does, in fact, have two methyl groups attached, although one of them is part of the pentane chain. Carbon #2 in the pentane chain does have two methyl groups, technically.

oh, i see what he was getting at...duh!

 

[unsung]

Thanks for the replies... and, I have yet another question. This one has been bugging me!

I get that absolute configuration is R/S assignment for naming chiral centers. But the concept of "relative configuration" (D/L, not to be confused with "d/l = +/-")
is endlessly confusing to me... and a question on an EK exam clearly showed me that I did not grasp the concept .

Not to repeat the question here, but basically the problem told us a reaction proceeds with "retention of configuration", and we're told the product is (S). Then we're asked to pick the compound that could be a reactant.

I thought, oh, I just have to look for the (S) reactant... but no, that was TOTALLY wrong. The answer was actually the reactant with (R) configuration. In the answer key, the explanation was something like "retention of configuration does not mean absolute configuration is retained; it means there's no inversion."

So basically, we were supposed to draw the (S) product, then get the structure of the reactant from it (which actually ends up being (R) ), and match it to the answer choice.

Can anyone shed some light on the topic? What does "retention of configuration" mean if it doesn't refer to retention of absolute configuration ???

I've tried researched, googled, etc. to no avail.

Okay geniuses... surely somebody can shed some light on my question? Please?

 

[MundaneMD]

Okay geniuses... surely somebody can shed some light on my question? Please?

Can you post the exam question that you are refering to (if it's allowed on SDN)?

In the meantime, this might help:

(Note: I stole the following information from http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch07configurations.html)

In this diagram, they say that the absolute configuration for the parent molecule is known.

After the reaction with the TsCl, only the relative configuration is known.

In this case, since the parent molecule was R, and the product is R, it is known as retention of the configuration (with respect to the relative configuration).

The image goes on to show that since the middle molecule reacts with KCN, the configuration is inverted, so the configuration is not retained.

Absolute is +/-, and R/S
Relative is only R/S

 

[SBK]

I have a question re: identifying the most stable isomer and exactly how do 'draw' it so that comparisons can be made.

For example.

Molecule A = cyclohexane with an isopropyl group on C1 projecting towards the viewer, and a methyl group projecting towards the viewer on C2.

Molecule B = exact same except that the C2-Methyl projects away from the viewer....


What are the rules in determining what is equitorial and what is axial when transforming them into chair conformations?

 

[Foghorn]

I have a question re: identifying the most stable isomer and exactly how do 'draw' it so that comparisons can be made.

For example.

Molecule A = cyclohexane with an isopropyl group on C1 projecting towards the viewer, and a methyl group projecting towards the viewer on C2.

Molecule B = exact same except that the C2-Methyl projects away from the viewer....


What are the rules in determining what is equitorial and what is axial when transforming them into chair conformations?

Learn the differences, how to manipulate and what type of information you can get from chair conformations, Newman projections and Fisher projections. If you can do that, you can answer your question(s).

 

[cheezer]

How can I systematically break down an NMR graph to identify a compound? Thanks again.