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)

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Unacceptable topics:
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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.

 

[QofQuimica]

I just want to make sure I have this straight. For C-NMR we just see how many hydrogens are attached to that particular Carbon and add 1. For H-NMR we would look at the Carbon and see how many neighbours that carbon has and add one. Yes? So, for example, 1,1,2-tricholoropropane (first carbon has two Cl's, one H; second carbon has 1 Cl and 1 H; last Carbon has 3 H's) we would say that the first Carbon has 2 peaks because it's neighbour has 1 H add one, second has the 1 H from the first Carbon plus the 3 H's from last Carbon and add 1 for a total of 5 peaks and the third one has 2 peaks as well from the H's of the middle Carbon. Yes?

Thanks!

Assuming that the splitting due to carbons 1 and 3 were equivalent, yes.

 

[QofQuimica]

Wait, I thought for C-NMR they show you the carbons (odd # isotope, C 13 for example). Because in that graph, the benzene has 6 peaks, and benzene has 6 carbons. Carbonyl has 1 carbon. I hope I'm looking at this right 🙁

Right. The one farthest upfield is the methyl from the ester.

Also, how can we tell the number of carbons in a molecule if say every carbon has another carbon just like it (same chemical shift). So the graph would only show each TYPE of carbon, not how many of each type if a few are the same type (substituents and orientation)? Will there be more info given?

Yes. This happens if your molecule is symmetrical. So if, for example, you had methyl para-hydroxybenzoate instead of ortho like the one I posted, you would wind up only seeing four carbon peaks, because carbon 2 = carbon 6 and carbon 3 = carbon 5. In that case, you'd need something like the mass spec or the molecular formula to solve the structure. Alternatively, you would probably be able to do it with the integration ratio on the proton NMR. But you almost certainly couldn't do it with just the decoupled C-13 NMR.

 

[justmoi]

Hey! Does anyone have any tips on memorizing the ortho-para-meta directing groups for aromatic reactions? I've seen questions on these on practice tests and I understand they're pretty imp, but I'm having a hard time getting them all memorized without confusing them. Any mnemonics? Thanks!

 

[Pdiddy310]

Hey! Does anyone have any tips on memorizing the ortho-para-meta directing groups for aromatic reactions? I've seen questions on these on practice tests and I understand they're pretty imp, but I'm having a hard time getting them all memorized without confusing them. Any mnemonics? Thanks!

This is what worked for me:

(I just memorized the ortha/para deactivators and meta deactivators) - so far, its worked. 👍

1) ortho/para activating - everything else

2) ortho/para deactivating - Halogens

3) meta deactivating - NO2, SO3H, nitrile, N+R3, carbonyl compounds

 

[justmoi]

This is what worked for me:

(I just memorized the ortha/para deactivators and meta deactivators) - so far, its worked. 👍

1) ortho/para activating - everything else

2) ortho/para deactivating - Halogens

3) meta deactivating - NO2, SO3H, nitrile, N+R3, carbonyl compounds

thats very helpful, thanks a lot!

 

[osjx-82]

how come tert-butoxide can be used as a nucleophile in Williamson-Ether synthesis? I thought it was SN2, and SN2 is not favorable when there are steric hindrances...

does sterics just apply to the substrate, not the nucleophile?

 

[QofQuimica]

how come tert-butoxide can be used as a nucleophile in Williamson-Ether synthesis? I thought it was SN2, and SN2 is not favorable when there are steric hindrances...

does sterics just apply to the substrate, not the nucleophile?

If you want to form a t-butoxyether, you MUST use t-butoxide as the nucleophile. If you try to do it backward, you'll wind up getting an elimination instead. See the organic explanations post about the four mechanisms if you don't understand that point. It is true that the hindrance of the Nu can be a problem in Sn2 reactions, but the major issue with Sn2 is about the hindrance of the substrate more than the hindrance of the Nu.

 

[Foolins]

Is there a difference between the cis/trans and z/e designation for isomerism...or do they refer to the same thing and can be used interchangeably?

 

[QofQuimica]

Is there a difference between the cis/trans and z/e designation for isomerism...or do they refer to the same thing and can be used interchangeably?

It depends on how complicated the molecule is. If you're talking about something with two substituents (one on each carbon), the cis and the Z are the same thing, as are the trans and the E. If you're talking about something with more than two substituents, you should use the E/Z designation.

 

[legobikes]

too much stereochemistry! can anyone see anything wrong with all this:


enantiomers are mirror images and have the same phys/chem properties, but maybe different physiological properties.

diastereomers have different phys/chem/physiological properties.

meso compounds have chiral centers but their symmetry cancels out any optical activity.

anomers are cyclic carbohydrates that are diastereomers (differ at one chiral carbon) and are subject to flipping via mutarotation

epimers are aldoses (and other carbohydrates?) that differ only at chiral carbon (and are hence diastereomers)

D and L configurations are just based on the D and L glyceraldehydes (stupid conventions... argh) and so on diagrams where the carbonyl is at the top of the molecule they have an OH on the lowest left position.

D and L are independent of actual optical activity, where the real indication is +/- or d/l?

R and S are independent of actual optical activity.

 

[QofQuimica]

too much stereochemistry! can anyone see anything wrong with all this:


enantiomers are mirror images and have the same phys/chem properties, but maybe different physiological properties.

diastereomers have different phys/chem/physiological properties.

meso compounds have chiral centers but their symmetry cancels out any optical activity.

anomers are cyclic carbohydrates that are diastereomers (differ at one chiral carbon) and are subject to flipping via mutarotation

epimers are aldoses (and other carbohydrates?) that differ only at chiral carbon (and are hence diastereomers)

D and L configurations are just based on the D and L glyceraldehydes (stupid conventions... argh) and so on diagrams where the carbonyl is at the top of the molecule they have an OH on the lowest left position.

D and L are independent of actual optical activity, where the real indication is +/- or d/l?

R and S are independent of actual optical activity.

Looks mostly good to me, except that I'd add the following clarifications:

Diastereomers tend to have similar chemical properties; it's the difference in their physical properties (ex. solubility, affinity, etc.) that is significant.

Epimers are a subset of diastereomers that differ in stereochemistry at only a single chiral center when multiple chiral centers are present. Anomers are a subtype of epimers that are formed when a sugar cyclizes to form a new stereocenter with alpha and beta designations. (That's why anomers can mutarotate, as you said.) So, for example, glucose and galactose are epimers (at C-4), but they are not anomers. Alpha-D-glucopyranose and beta-D-glucopyranose are both anomers and epimers.

Only the L-isomer of a sugar has the OH group on the left at the highest-numbered (bottom) chiral carbon. If the OH group of that C is on the right, then the configuration is analogous to D-glyceraldehyde, and the designation is therefore D.

 

[legobikes]

Looks mostly good to me, except that I'd add the following clarifications:

Thank you!

 

[googlinggoogler]

Hi, I have a question about acidity.

One of the trends is that acidity increases as you go down a group because the greater the size of the atom, the longer the bond between the atom and H, and therefore, the more breakable it is.

However, can you explain to me why sp hybridized C-H bonds are more acidic than sp2 hybridized C-H bonds, is more acidic than sp3 hybridized C-H bonds? I initially explained to myself that sp has more s character than sp2, than sp3, and therefore the electrons are held closer to the carbon nucleus, which is why H+ can fall off so easily. But it seems to contradict the first trend (where acidity correlates to size) because a sp hybridized C-H bond would be the shortest of the three bonds, and acording to the first trend, the longer the bond between the atom and H, the more breakable it is.

 

[QofQuimica]

Hi, I have a question about acidity.

One of the trends is that acidity increases as you go down a group because the greater the size of the atom, the longer the bond between the atom and H, and therefore, the more breakable it is.

However, can you explain to me why sp hybridized C-H bonds are more acidic than sp2 hybridized C-H bonds, is more acidic than sp3 hybridized C-H bonds? I initially explained to myself that sp has more s character than sp2, than sp3, and therefore the electrons are held closer to the carbon nucleus, which is why H+ can fall off so easily. But it seems to contradict the first trend (where acidity correlates to size) because a sp hybridized C-H bond would be the shortest of the three bonds, and acording to the first trend, the longer the bond between the atom and H, the more breakable it is.

You can't use the bond length rule here because you are comparing like atoms. In other words, you are comparing C-H to C-H to C-H, not, say, H-Cl to H-Br to H-I. In this case, it is more helpful to think about it in terms of electronegativity instead of in terms of bond length, because every C atom is the same size as every other C atom, and the bond length isn't changing due to atom size. It's changing due to differences in electronegativity of the various hybrids. Basically, if the conjugate base being left behind after proton removal is more electronegative, that means it can better stabilize the negative charge it will hold once the proton is removed. So, since an sp-hybridized C is more electronegative than an sp2-hybridized C, it will be better able to stabilize the negative charge, and it will be a stronger acid. Same reasoning for why sp2-hybridized carbons are stronger acids than sp3-hybridized carbons. The differences are large: assuming no heteroatoms are present, an sp-hybridized carbon has a pKa of about 25, an sp2-hybridized carbon as a pKa of about 40, and an sp3-hybridized carbon has a pKa greater than 50. That is one H that ain't coming off. 😉

 

[googlinggoogler]

You can't use the bond length rule here because you are comparing like atoms. In other words, you are comparing C-H to C-H to C-H, not, say, H-Cl to H-Br to H-I. In this case, it is more helpful to think about it in terms of electronegativity instead of in terms of bond length, because every C atom is the same size as every other C atom, and the bond length isn't changing due to atom size. It's changing due to differences in electronegativity of the various hybrids. Basically, if the conjugate base being left behind after proton removal is more electronegative, that means it can better stabilize the negative charge it will hold once the proton is removed. So, since an sp-hybridized C is more electronegative than an sp2-hybridized C, it will be better able to stabilize the negative charge, and it will be a stronger acid. Same reasoning for why sp2-hybridized carbons are stronger acids than sp3-hybridized carbons. The differences are large: assuming no heteroatoms are present, an sp-hybridized carbon has a pKa of about 25, an sp2-hybridized carbon as a pKa of about 40, and an sp3-hybridized carbon has a pKa greater than 50. That is one H that ain't coming off. 😉

Wow, thanks for the reply 🙂

 

[natusss]

what does "stereogenic" mean?
Thanks 🙂

 

[4s4]

I think stereogenic just means a stereocentre, or a centre that is chiral (usually C with 4 different groups bonded to it).

I have a question: Why is a C bonded to OH more polar than a C double bonded to O? i.e. you are comparing polarities of cyclohexane with ketone vs. cyclohexanol

Thanks!

 

[Cloudcube]

Hey! Does anyone have any tips on memorizing the ortho-para-meta directing groups for aromatic reactions? I've seen questions on these on practice tests and I understand they're pretty imp, but I'm having a hard time getting them all memorized without confusing them. Any mnemonics? Thanks!

A prof taught me this: The meta directing groups carry a partial or full positive charge, while the ortho/para groups carry unshared pairs of electrons. The exception you have to watch out for is halogens because they don't follow the rule. It's worked for me so far.

 

[87138]

A prof taught me this: The meta directing groups carry a partial or full positive charge, while the ortho/para groups carry unshared pairs of electrons. The exception you have to watch out for is halogens because they don't follow the rule. It's worked for me so far.

A decent rule of thumb, but doesn't take into account the fact that alkyl groups are ortho-para activators. Make sure you remember that.

 

[QofQuimica]

what does "stereogenic" mean?
Thanks 🙂

Stereogenic atoms are ones where the interchange of two of the groups attached to them would invert the stereochemistry of the molecule. They are NOT necessarily chiral centers, although all chiral atoms are also stereogenic. For example, if you had cis-2-butene, the two carbons in the double bond are both stereogenic, but they are not chiral. They are stereogenic because if you exchange the methyl group (C1) attached to C2 with the H attached to C2, you will get the trans isomer. (The same thing happens for C3, which therefore is also stereogenic.) They are achiral because double bonds are flat, and therefore have an internal plane of symmetry.

 

[QofQuimica]

I think stereogenic just means a stereocentre, or a centre that is chiral (usually C with 4 different groups bonded to it).

I have a question: Why is a C bonded to OH more polar than a C double bonded to O? i.e. you are comparing polarities of cyclohexane with ketone vs. cyclohexanol

Thanks!

The size of a dipole depends on a lot of things, including the shape of the molecule and the bond length. But I would wager to say that in the example you've given, it would have to do with the fact that cyclohexanone is polar aprotic, while cyclohexanol is polar protic. (Read my explanations post about these terms if you're not familiar with them.)

 

[ssa915]

Regarding radical additions to alkenes, I know that Anti-markovnikov addition with HBr is a radical addition of the Br radical to the C-radical. But are all anti-markovnikov additions radical additions (like hydroboration, for example), or does this only apply to HBr addition?

 

[QofQuimica]

Regarding radical additions to alkenes, I know that Anti-markovnikov addition with HBr is a radical addition of the Br radical to the C-radical. But are all anti-markovnikov additions radical additions (like hydroboration, for example), or does this only apply to HBr addition?

Please do not start new threads in the subforum. Just post your question in the appropriate already existing thread. We check these threads daily.

In answer to your question, yes, you can have anti-Markovnikov additions without radicals. Hydroboration in particular does not go by a radical mechanism. Basically, any time you have a scenario where you are adding the heteroatom first before adding the H, you will get anti-Markovnikov addition. Instead of trying to memorize what reagents go Markovnikov versus anti-Markovnikov, try to understand the rationale for what's happening. The reason that these reactions have this regioselectivity is because they always want to form the more stable carbocation (or radical; remember that radical stability trends parallel carbocation stability trends). So whatever atom you add to the double bond first will always go on the less substituted carbon atom, leaving the carbocation (or radical) on the more substituted carbon because that forms the more stable intermediate. With radical addition of HBr, you're adding the Br first, so it goes to the less substituted atom. With BH3, you're adding the B first, and the same thing happens.

 

[ssa915]

Please do not start new threads in the subforum. Just post your question in the appropriate already existing thread. We check these threads daily.

In answer to your question, yes, you can have anti-Markovnikov additions without radicals. Hydroboration in particular does not go by a radical mechanism. Basically, any time you have a scenario where you are adding the heteroatom first before adding the H, you will get anti-Markovnikov addition. Instead of trying to memorize what reagents go Markovnikov versus anti-Markovnikov, try to understand the rationale for what's happening. The reason that these reactions have this regioselectivity is because they always want to form the more stable carbocation (or radical; remember that radical stability trends parallel carbocation stability trends). So whatever atom you add to the double bond first will always go on the less substituted carbon atom, leaving the carbocation (or radical) on the more substituted carbon because that forms the more stable intermediate. With radical addition of HBr, you're adding the Br first, so it goes to the less substituted atom. With BH3, you're adding the B first, and the same thing happens.

Sorry about that, I think I jumped ahead of myself when I posting..did it to fast.

I still dont understand what defines a radical reaction over forming a carbocation intermediate? Why doesn't the HBr/peroxide reaction proceed with a carbocation?

 

[QofQuimica]

Sorry about that, I think I jumped ahead of myself when I posting..did it to fast.

I still dont understand what defines a radical reaction over forming a carbocation intermediate? Why doesn't the HBr/peroxide reaction proceed with a carbocation?

You should assume that there is not a radical mechanism, unless you see one of two things: peroxides, which you've already seen form radicals; or light (often written as hv over the reaction arrow). Peroxides have an unstable O-O single bond, and that is why they tend to homolytically cleave (meaning one electron from the bond goes with one oxygen, and the other goes with the other oxygen, instead of both going with the same oxygen) into radicals. Light is very high energy, and so it is able to cause homolytic bond splitting that would not normally happen in the absence of light.

 

[supersash]

i have a last minute question about op/meta directors. i know most things are op, and i need to just remember which are meta (for some reason, this is the hardest thing EVER for me!)

is it safe to say carbonyl's are meta directors, or is there an exception to that? any help is appreciated.

sasha

 

[ssa915]

You should assume that there is not a radical mechanism, unless you see one of two things: peroxides, which you've already seen form radicals; or light (often written as hv over the reaction arrow). Peroxides have an unstable O-O single bond, and that is why they tend to homolytically cleave (meaning one electron from the bond goes with one oxygen, and the other goes with the other oxygen, instead of both going with the same oxygen) into radicals. Light is very high energy, and so it is able to cause homolytic bond splitting that would not normally happen in the absence of light.

So this is generally true for reactions with peroxide or light, but in hydroboration, this is an exception, since hydroboration, although it uses peroxide, doesnt use a radical mechanism, correct?

 

[Foolins]

I thought aromatics aren't on the MCAT. Why do we need to know anything about ortho meta directors?

 

[QofQuimica]

So this is generally true for reactions with peroxide or light, but in hydroboration, this is an exception, since hydroboration, although it uses peroxide, doesnt use a radical mechanism, correct?

Ah, ok, NOW I see why you're so confused! You're right; peroxide does get used with hydroboration, but not until AFTER the borane addition to the double bond is over. If you take a look at the reaction mechanism, you will see that the two compounds (borane and peroxide) are added in two separate steps. That's why this mechanism is different compared to the HBr reaction, where peroxides are added to the reaction WITH the HBr, and only in a catalytic amount.

 

[QofQuimica]

I thought aromatics aren't on the MCAT. Why do we need to know anything about ortho meta directors?

They're not directly tested, but they can be tested in other indirect ways, such as nomenclature, bonding/stability questions, and reactions. Alkenes can also appear in the guise of stereoisomerism (E vs. Z) or addition reactions. I assume that's why people are asking.

It's also possible that we have some DATers and PCATers in here, or even just some people who are taking organic now and want to ask questions.

 

[deleted106503]

These are probably the last questions I have. 🙂

For chirality, how do you determine if a carbon is chiral if the adjacent carbon is exactly like another.

i.e. say a carbon has a -H, -OH, -CH2CH3, and -CH2CH2CH3. So even though the last two groups are different, an ethyl and propyl, the carbons attached to the single carbon they share are both -CH2-. Is that "chiral" carbon really chiral because the four groups are different or is it achiral because two of the groups attach to the "chiral" carbon with the same exact -CH2- part?

For a chiral carbon in a ring, would it automatically be chiral if the ring was even numbered. i.e. say cyclohexane has a carbon with an -OH and an -H group (cyclohexanol). To that carbon, there are 5 others in the ring. So if you were to follow the ring from that carbon, you could pass 3 carbons on one direction and 2 in another until you get to the same point. Would the carbon be chrial because it's like an ethyl group and propyl group susbstituents, but just had a ring closure?

Now say it's a 7 membered carbon ring, leaving 6 other carbons in the ring besides the one with the hydroxyl group. If you do the same thing again by following along both side of the ring, you can split it into 2 3-Carbon chains, so two propyl groups. Since there would be two propyl groups (but just closed in a ring), would they be considered the same as on the carbon with the -OH making it achiral?

Any help would be appreciated. Thanks!

 

[QofQuimica]

These are probably the last questions I have. 🙂

For chirality, how do you determine if a carbon is chiral if the adjacent carbon is exactly like another.

i.e. say a carbon has a -H, -OH, -CH2CH3, and -CH2CH2CH3. So even though the last two groups are different, an ethyl and propyl, the carbons attached to the single carbon they share are both -CH2-. Is that "chiral" carbon really chiral because the four groups are different or is it achiral because two of the groups attach to the "chiral" carbon with the same exact -CH2- part?

For a chiral carbon in a ring, would it automatically be chiral if the ring was even numbered. i.e. say cyclohexane has a carbon with an -OH and an -H group (cyclohexanol). To that carbon, there are 5 others in the ring. So if you were to follow the ring from that carbon, you could pass 3 carbons on one direction and 2 in another until you get to the same point. Would the carbon be chrial because it's like an ethyl group and propyl group susbstituents, but just had a ring closure?

Now say it's a 7 membered carbon ring, leaving 6 other carbons in the ring besides the one with the hydroxyl group. If you do the same thing again by following along both side of the ring, you can split it into 2 3-Carbon chains, so two propyl groups. Since there would be two propyl groups (but just closed in a ring), would they be considered the same as on the carbon with the -OH making it achiral?

Any help would be appreciated. Thanks!

1) Yes, that carbon is chiral, because a propyl is not equivalent to an ethyl. Here's what you should do when you have two carbon substituents like that:

Since the first atom out (C in this case) is the same for both substituents, you must go out to the second atoms of each. For the propyl group, the first carbon is attached to a second carbon as well as two hydrogens: C, H, H. For the ethyl group, the same is true: carbon one is again attached to C, H, H. Therefore, you must go out to the third atoms. For the propyl, the second carbon is attached to one carbon plus two hydrogens, again giving: C, H, H. However, the second carbon of the ethyl is attached to three hydrogens, giving: H, H, H. You can cross out two H's from each (since they are identical), leaving you to compare C from the propyl with H from the ethyl. Since C has a higher atomic number than H, you would prioritize propyl before ethyl. Be careful here, because if you had a fluorine on your ethyl (CH2CH2F), then you would be comparing a C from propyl to a F from ethyl, and your higher priority group would then be ethyl.

In general: you should continue working your way outward from the chiral center until you find a difference in substituents. If you never do find a difference, the carbon is not chiral.

2) and 3) I'm sure you can now figure out from what I told you in part one that the answer to these questions is that chirality depends on whether the ring is symmetrical or not. 🙂

 

[deleted106503]

Thank you!!! Thanks for pointing out the CH2F beats out a CH2CH3!

2) and 3) I'm sure you can now figure out from what I told you in part one that the answer to these questions is that chirality depends on whether the ring is symmetrical or not. 🙂

So for cyclohexanol, the C with -OH will have 5 other ring carbons. If I do what you're saying, it would be like two ethyl groups, but both attached by a -CH2-. Would the MCAT ask me to determine if the chiral carbon is R or S? Wouldn't that depend on which chain (when ring broken) I put the CH2 on? Or maybe I got this all wrong...

 

[QofQuimica]

Thank you!!! Thanks for pointing out the CH2F beats out a CH2CH3!



So for cyclohexanol, the C with -OH will have 5 other ring carbons. If I do what you're saying, it would be like two ethyl groups, but both attached by a -CH2-. Would the MCAT ask me to determine if the chiral carbon is R or S? Wouldn't that depend on which chain (when ring broken) I put the CH2 on? Or maybe I got this all wrong...

Ok, take a look at this picture of 4-methylcyclohexanol. You might be tempted to say that carbons 1 and 4 are chiral, right? But actually they aren't, because the ring is identical from either carbon. Contrast that with 3-methylcyclohexanol, which is asymmetrical and chiral.

 

[christian15213]

Thank you!!! Thanks for pointing out the CH2F beats out a CH2CH3!



So for cyclohexanol, the C with -OH will have 5 other ring carbons. If I do what you're saying, it would be like two ethyl groups, but both attached by a -CH2-. Would the MCAT ask me to determine if the chiral carbon is R or S? Wouldn't that depend on which chain (when ring broken) I put the CH2 on? Or maybe I got this all wrong...

Yes you do

http://upload.wikimedia.org/wikipedia/en/thumb/7/71/Cyclohexanol.png/100px-Cyclohexanol.png

this is not chiral at all because there is not 4 steriogenic attachments to the carbon or any carbon... no chirality a or other wise...

listen, think of it like this... like the other gentlemen said... Take the initial C and start your problem like this C(mark here whatever is conected to the C in order) for the case of this molicule it would look like this

C(OH)
C(H)
C(CHHCHH(CH2)
C(CHHCHH(CH2) this is equal and there for is not a steriogenic carbon...

 

[deleted106503]

Alright, I got it now!!! Thanks! So if there's a plane of symmetry like in cyclohexanol and 4-methylcyclohexanol, it's the same as cutting it right down the middle. Which makes two equal sides so it's achiral. Why is it always the basics that screw me up?! haha, thanks again. 🙂

 

[Dave_D]

Ok, take a look at this picture of 4-methylcyclohexanol. You might be tempted to say that carbons 1 and 4 are chiral, right? But actually they aren't, because the ring is identical from either carbon. Contrast that with 3-methylcyclohexanol, which is asymmetrical and chiral.

Hi Q,

What might confuse some people on 4 methylcyclohexanol is that it does have 2 isomers, a cis isomer and a trans isomer.(Ok, so that caught me at first when you were explaining it.)

 

[QofQuimica]

Hi Q,

What might confuse some people on 4 methylcyclohexanol is that it does have 2 isomers, a cis isomer and a trans isomer.(Ok, so that caught me at first when you were explaining it.)

True. But cis/trans isomerism is not the same as stereoisomerism.

 

[christian15213]

True. But cis/trans isomerism is not the same as stereoisomerism.

Quimica I freaking love your ICON... I'm Diene... I do it all the time now... IT is so freaking NERDY I love it... LMAO..

Quimica is correct...

but to be honest I think my way of working it out is absolute best... The quimica explains it is vital to understanding but per the method of working it out my way sure cuts down on silly mistakes...

 

[wannaberockstar]

What is a cross product? Also, when judging which of two proposed mechanisms would produce a cross product how would we do that?

 

[christian15213]

What is a cross product? Also, when judging which of two proposed mechanisms would produce a cross product how would we do that?

Cross product of what? be specific... for example there is something called the crossed Adol reaction as well as the Crossed Claisen reaction... Which basically means that there isn't two identical molecules but rather one that fits some rule and another that doesn't... given some crossed result...

 

[QofQuimica]

Cross product of what? be specific... for example there is something called the crossed Adol reaction as well as the Crossed Claisen reaction... Which basically means that there isn't two identical molecules but rather one that fits some rule and another that doesn't... given some crossed result...

I thought maybe s/he meant a side product, but I'm not sure either. wannabe, the only cross product I know of is for vectors, not for organic reactions. 😉 Can you explain what you mean exactly? I don't think we really understand your question.

 

[deleted106503]

Cross product was mentioned in a practice exam passage. The passage is about the Claisen rearragement. One of the questions was something about which mechanism shows an absence of a cross-product. I guessed that the one where the substituent never broke off was whenn no "cross-products" were formed and it was correct. The other mechanism showed the substituent breaking off the ring, which then showed two separate compounds existing at once, and then reattaching. So I guessed that one was when a cross-product was formed since it could probably reattach anywhere it wanted. So was a cross product the part that breaks free off the original compound?

 

[wannaberockstar]

Yea, thanks WilliamsF1 - indeed that was the question I was referring to (I just wasn't sure if I was allowed to post it on here).
Also, what's the difference between bond energy and bond dissociation energy? Are they the same thing? More stable bonds would mean lower energy bonds with higher dissociation energies right?

Oh sorry...more thing...I'm not sure how to explain this clearly, but say you have an trans-1,2 dibromoethene molecule. In the proton NMR, would the hydrogens on either carbon be equivalent to each other?

 

[QofQuimica]

Yea, thanks WilliamsF1 - indeed that was the question I was referring to (I just wasn't sure if I was allowed to post it on here).
Also, what's the difference between bond energy and bond dissociation energy? Are they the same thing? More stable bonds would mean lower energy bonds with higher dissociation energies right?

Oh sorry...more thing...I'm not sure how to explain this clearly, but say you have an trans-1,2 dibromoethene molecule. In the proton NMR, would the hydrogens on either carbon be equivalent to each other?

Technically we're not supposed to post any specific AAMC questions on SDN, but it's ok to ask general questions. Just don't get too specific. Clear as mud, right? 🙂

1) Ok, so is your passage about a mixed Claisen condensation (where you preferentially react an enolized ketone with a non-enolizable aldehyde and get an unsaturated ketone)? I checked in one of my books, and that mixed Claisen condensation product can also be called a cross product. That is different, however, than the Claisen rearrangement, where you have a 3,3-sigmatropic rearrangement of an allyl vinyl ether to a carbonyl compound. Take a look at your passage, and tell me if you see an unsaturated ether undergoing a cyclic rearrangement to a carbonyl, or two carbonyl compounds being joined together to form an unsaturated ketone. I don't understand what you guys mean by things breaking off and reattaching.

Here is a picture of a Claisen rearrangement; is that what you're seeing:

2) Bond energy is the energy required to break a covalent bond homolytically (into neutral fragments). Bond energies are called bond dissociation energies when given for specific bonds, or average bond energies when summarized for a given type of bond over many kinds of compounds. A more stable bond will have a larger standard bond energy and therefore be more difficult to break.

3) If it's trans, then the molecule is symmetrical, so yes.

 

[wannaberockstar]

1) Ok, so is your passage about a mixed Claisen condensation (where you preferentially react an enolized ketone with a non-enolizable aldehyde and get an unsaturated ketone)? I checked in one of my books, and that mixed Claisen condensation product can also be called a cross product. That is different, however, than the Claisen rearrangement, where you have a 3,3-sigmatropic rearrangement of an allyl vinyl ether to a carbonyl compound. Take a look at your passage, and tell me if you see an unsaturated ether undergoing a cyclic rearrangement to a carbonyl, or two carbonyl compounds being joined together to form an unsaturated ketone. I don't understand what you guys mean by things breaking off and reattaching.

Here is a picture of a Claisen rearrangement; is that what you're seeing:

Yes, the passage was referring to a Claisen rearrangement; there were two proposed mechanisms for this rearrangement - one was the concerted rearrangement you illustrated above, and the other had the oxygen-C1 bond undergoing a heterolytic cleavage to form a carbocation and a phenolate anion. Then, the carbocation attacks the ortho-C of the ring to give a ketone that enolized to the product.
The question asked which mechanism would NOT produce a cross product. So yea, I think the second mechanism is a Claisen condensation where the condensation cross product is possible so that's why the 1st mechanism wouldn't produce the cross product.
Thanks for your help!

 

[christian15213]

Yes, the passage was referring to a Claisen rearrangement; there were two proposed mechanisms for this rearrangement - one was the concerted rearrangement you illustrated above, and the other had the oxygen-C1 bond undergoing a heterolytic cleavage to form a carbocation and a phenolate anion. Then, the carbocation attacks the ortho-C of the ring to give a ketone that enolized to the product.
The question asked which mechanism would NOT produce a cross product. So yea, I think the second mechanism is a Claisen condensation where the condensation cross product is possible so that's why the 1st mechanism wouldn't produce the cross product.
Thanks for your help!

what scares me is that you seem not to be understanding the difference between crossed and regular claisen... In order to be crossed one of the molecules isn't going to have a Beta acid hydrogen and the other is... The cleavage of the leaving group happens to the molecule that doesn't have the Beta acidic hydrogens...

Does that help at all?

 

[QofQuimica]

Yes, the passage was referring to a Claisen rearrangement; there were two proposed mechanisms for this rearrangement - one was the concerted rearrangement you illustrated above, and the other had the oxygen-C1 bond undergoing a heterolytic cleavage to form a carbocation and a phenolate anion. Then, the carbocation attacks the ortho-C of the ring to give a ketone that enolized to the product.
The question asked which mechanism would NOT produce a cross product. So yea, I think the second mechanism is a Claisen condensation where the condensation cross product is possible so that's why the 1st mechanism wouldn't produce the cross product.
Thanks for your help!

Ok, I think that I follow you. Basically, you are being presented with two alternative mechanisms for a Claisen rearrangement. The 3,3 sigmatropic rearrangement is the generally accepted mechanism, so that's why I was so confused. 😛

In that case then, yes, I would agree that the first mechanism (the 3,3 sigmatropic) would NOT form a cross product, because basically you are just rearranging the bonds in that mechanism, and it is concerted. In other words, you do not form any intermediates. Therefore, the only other thing that the reaction could do would be to reverse (i.e., reform the starting material). However, since the product carbonyl is more stable than the starting material allyl ether, this is not not likely to occur.

The alternative mechanism that they gave you is not concerted. Here, you are forming a reactive, high-potential intermediate (the carbocation), and since highly reactive compounds are not very selective, it is definitely possible that you could wind up with that carbocation reacting with some other electron pair besides the one you want it to react with. In that case, it would be reasonable to expect to see multiple products.

P.S. FYI, the second mechanism is NOT a Claisen condensation. See below for what the Claisen condensation looks like. This example is not mixed, and so no mixed product (or cross product) will form. To get a cross product from the Claisen condensation, you would need to have two different esters here, one of which is enolizable and the other of which is not.

 

[deleted106503]

Okay, I just went over my EK Orgo again today and I think they confused me with a chiral problem, geez!

Here's the struture they had (at least as close as I could find).

which they said was (-)-nicotine

I tried to find all the carbons where it would differ from (+)-nicotine. I figured that this would be where all the chiral ones are since the enantiomer rule in the book said all the absolute configurations of each chiral carbon must be the opposite (according to our text).

To me, it looks like Carbon 4 (the red arrow) is chiral and so is the one across with, Carbon 5 (black question mark).

Edit: I think I see where I messed up. C-5 doesn't have 4 substituents since it will always be double bonded to one of the Carbons next to it, making 3 substituents (achiral). Nevermind!!! 😀 I was using the technique from yesterday by assuming each Carbon flanking C-5 was identical meaning I had to go to the one connected to that one. Glad I realized this before Saturday! 🙂

 

[87138]

Okay, I just went over my EK Orgo again today and I think they confused me with a chiral problem, geez!

Here's the struture they had (at least as close as I could find).

which they said was (-)-nicotine

I tried to find all the carbons where it would differ from (+)-nicotine. I figured that this would be where all the chiral ones are since the enantiomer rule in the book said all the absolute configurations of each chiral carbon must be the opposite (according to our text).

To me, it looks like Carbon 4 (the red arrow) is chiral and so is the one across with, Carbon 5 (black question mark).

Edit: I think I see where I messed up. C-5 doesn't have 4 substituents since it will always be double bonded to one of the Carbons next to it, making 3 substituents (achiral). Nevermind!!! 😀 I was using the technique from yesterday by assuming each Carbon flanking C-5 was identical meaning I had to go to the one connected to that one. Glad I realized this before Saturday! 🙂

I used the same thought process on that very same question the first time around. Then it suddenly dawned on me that carbons in an aromatic ring can't be chiral.