Organic Chemistry Question Thread

[QofQuimica]

---

Members don't see this ad.

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.

 

[PRamos]

As far as I know, there is not any difference between a quartet of triplets or the reverse. It's just going to depend on which neighbors you consider first. I don't think that EK is smoking anything, but they are definitely simplifying things. This is something you have to be careful about on the MCAT; if you over-think things, you can actually get questions wrong.

The standard description lists the splitting with a bigger J-value first. If the triplet had a J-value of 3.0 Hz and the quartet had a J-value of 4.0 Hz, then it would look like four triplets (ignoring the overlapping) in a ratio of 1(1:2:1) : 3(1:2:1) : 3(1:2:1) : 1(1:2:1). Hence the name quartet of a triplet. The problem lies in the resolution of the NMR (strength of the B field). If the J-values are harmonics or if the resolution is too low, then the peaks all sort of blend together into a multiplet. I would assume that the MCAT would describe the the Hs on C-2 of 1-halopropane as a sextet, because the difference in J-values would be minimal enough to get overlapping. A q of t seems like overkill, although the answer choices would have the ultimate say in what to call it. A q of t and t of q each have twelve signals. A q of t would show peaks in a 1:2:1:3:6:3:3:6:3:1:2:1 ratio. A t of q would show peaks in a 1:3:3:1:2:6:6:2:1:3:3:1 ratio. That definitely seems like overkill.

BTW, your explanations are excellent. I wish I would have known about this section the first time I studied.

 

[QofQuimica]

I'm thinking hso4 is a crummy(IE weak) base since it's the conjugate base of a strong acid.(Of course assuming I haven't completely forgotten all of my gen chem II) So I'd expect E1 with possible re-arrangement. (Q is probably going to come by now and let me know how wrong I am 😀 )

No, you're right. Bisulfate will not be a good Nu nor a strong base. It will protonate the alcohol to make it into water, which is a good LG. The water molecule will fall off, leaving behind a secondary carbocation, which can sometimes rearrange as you said. At that point, you can either lose a neighboring proton to form a double bond (E1) or if there is something nucleophilic in the mixture, you can get the Sn1. Typically, Sn1 and E1 co-occur, because formation of the carbocation is the RDS for both mechanisms. The one thing you can do to favor E1 is to heat the reaction. Or, you can just not have any nucleophile besides water and bisulfate available, in which case Sn1 will not occur. (Bisulfate won't readily act as a Nu, and if water acts as a Nu, you'll just end up with the same alcohol you started with in the first place.)

P.S. Yes, kevin, structural isomers are the same thing as constitutional isomers.

 

[md2011]

Can you tell me the signal range for some typical structures on C13NMR spectrum? for -OH, carbonyl, carbon carbon double bonds, and anything else that you think important. I cannot believe AAMC test 8 tested the C13NMR, sigh.

Second question, for cis-1-tert-butyl-4-cholrocyclohexane, what is the most stable conformation? The answer said it's larger tert-butyl substituent in the equatorial, and smaller chloro on the axial. Why not both on equatorial orientation?

Thank you so much

 

[Saluki]

Double post...

 

[Saluki]

Here's a link I think might help with the C13 shifts:

http://www.wfu.edu/~ylwong/nmr/c13/

or maybe this one is better

http://www.mnstate.edu/jasperse/Chem355/C-13 NMR Summary.doc.pdf

I should have kept my Kaplan books

I'll let Q answer the other question and correct me if the link above isn't for the right thing 😛

 

[Dave_D]

Second question, for cis-1-tert-butyl-4-cholrocyclohexane, what is the most stable conformation? The answer said it's larger tert-butyl substituent in the equatorial, and smaller chloro on the axial. Why not both on equatorial orientation?

Thank you so much

The quick answer is because if both were equatorial it'd be trans. What you need to do is draw out the chair conformation like is pictured here
chair conformation

Look at the carbon in the upper left, lets call that 1. Therefore carbon 4 is the one in the lower right. Ok, got that? There's a couple of ways to think about this but really cis means both substituents are on the same side of the ring.(Alternatively it means both in the same direction, either both up or both down.) So lets imagine that tert-butyl is equitorial on the upper left. That corresponds to the "slightly down pointing" H. Now look at the the carbon in the lower right which is carbon 4. Which one points down? It's the axial one. Q, got any better way of explaining this one?

 

[md2011]

The quick answer is because if both were equatorial it'd be trans. What you need to do is draw out the chair conformation like is pictured here
chair conformation

Look at the carbon in the upper left, lets call that 1. Therefore carbon 4 is the one in the lower right. Ok, got that? There's a couple of ways to think about this but really cis means both substituents are on the same side of the ring.(Alternatively it means both in the same direction, either both up or both down.) So lets imagine that tert-butyl is equitorial on the upper left. That corresponds to the "slightly down pointing" H. Now look at the the carbon in the lower right which is carbon 4. Which one points down? It's the axial one. Q, got any better way of explaining this one?

I understand that. Thank you so much. You guys are wonderful!

 

[kevin86]

hey by the way, with nmr H and C13, does the height of the peaks and well lines suggest intensity, like the more of a particular similar proton or carbon there is, the higher it is?

Also those reactions involving a proton added to a say ketone to make the carbon more electrophilic and like the SN1 and E1 reaction when an OH group is made into water that leaves, are those protons catalytic and gets regenerated?

 

[QofQuimica]

hey by the way, with nmr H and C13, does the height of the peaks and well lines suggest intensity, like the more of a particular similar proton or carbon there is, the higher it is?

Also those reactions involving a proton added to a say ketone to make the carbon more electrophilic and like the SN1 and E1 reaction when an OH group is made into water that leaves, are those protons catalytic and gets regenerated?

It's not the height that matters; it's the area under the curve. That's what the integral is.

Yes, they often are. We may be able to come up with an example where they aren't, but I can't think of one off the top of my head right now.

 

[md2011]

Here is a question asks for what type of intermolecular interaction can C2H5OH undergo with water that C4H9OCH3 can not. The answer said to be hydrogen bonding. I thought ether can hydrogen bond with water. So I was wrong? Can you confirm, and explain why ether cannot hydrogen bond with water? Thanks

 

[md2011]

QofQ, for your hand rule for determing the R,S of a chiral center, the highest priority atom does not have to start from the index finger, right?

 

[Dave_D]

Here is a question asks for what type of intermolecular interaction can C2H5OH undergo with water that C4H9OCH3 can not. The answer said to be hydrogen bonding. I thought ether can hydrogen bond with water. So I was wrong? Can you confirm, and explain why ether cannot hydrogen bond with water? Thanks

I'd like to know this as well. I checked my orgo book(Solomons for what it's worth.) and it states ethers can't h-bond with themselves but can form h-bonds with water.(Paraphrasing what is stated on p497 of the 8th edition.) Any ideas Q?

 

[QofQuimica]

Here is a question asks for what type of intermolecular interaction can C2H5OH undergo with water that C4H9OCH3 can not. The answer said to be hydrogen bonding. I thought ether can hydrogen bond with water. So I was wrong? Can you confirm, and explain why ether cannot hydrogen bond with water? Thanks

Ethers are hydrogen bond acceptors, but they cannot donate. Depending on the exact wording of the question, you should probably classify them as having dipole-dipole interactions. I would say you are probably not technically wrong, since ethers ARE H-bond acceptors, but that's not what the question was looking for. Whenever you get a question like this, focus on what is DIFFERENT between the two molecules. And most importantly, don't overthink these things. You know that there is something different between the intermolecular interactions of an ether versus those of an alcohol, and that difference is the answer to the question. It's that simple.

 

[QofQuimica]

QofQ, for your hand rule for determing the R,S of a chiral center, the highest priority atom does not have to start from the index finger, right?

If you're using MY method, which apparently is only taught by my college professor and no one else in the whole world ( ), you should align your THUMBS with the FOURTH priority substituent. Then you curl the fingers of each hand, and the hand whose fingers curl in the correct order is the correct orientation. (It's R for your right hand, S for your left.) If this method is difficult for you to visualize, don't use it. Use one of the others that you can do easily.

 

[jsong812]

Hey Q,

Generally speaking, are acid-catalyzed reactions continuously being regenerated throughout the reactions whereas base-catalyzed reactions are not? Or do they depend on the type of reaction? thanks!

 

[QofQuimica]

Hey Q,

Generally speaking, are acid-catalyzed reactions continuously being regenerated throughout the reactions whereas base-catalyzed reactions are not? Or do they depend on the type of reaction? thanks!

In general, a catalyst is a species that gets regenerated. You can recognize it easily because the catalyst will be used in less than a stoichiometric amount. Either acids or bases can be catalysts or can be consumed, depending on the reaction.

 

[kevin86]

1. I have a question about the oxidation states of the carbon in organic compounds. How do you know if the anomeric carbon in glucose has a +2 and +3 in the carboxylate form. And also whats the redox rule for organic compound involving how many electronegative atoms they are connected to.

2. are anomers diasteromers or conformers.

 

[QofQuimica]

1. I have a question about the oxidation states of the carbon in organic compounds. How do you know if the anomeric carbon in glucose has a +2 and +3 in the carboxylate form. And also whats the redox rule for organic compound involving how many electronegative atoms they are connected to.

2. are anomers diasteromers or conformers.

1) Assuming you have a carbon (R group) attached to the other end of the carboxylic acid, you will get one valence electron from the R group, and the two oxygens will steal the other valence electrons. So a carboxylic acid carbon should have an oxidation state of +3. If it's formic acid, though, the carbon will receive both electrons from the C-H bond, and then it will have an oxidation state of +2.

In general, you should pretend like all of the bonds between unlike atoms are ionic. Any atom less electronegative than C (like H) gives both bond electrons to C. Any atom more electronegative than C (like O or N) take both bond electrons away from C. Another carbon will have the same electronegativity, so each carbon receives one electron. Once you figure out how many electrons carbon "gets," subtract that number from 4, which is carbon's group number. A carbon atom attached to lots of electronegative atoms will have a positive oxidation state, while a saturated C will have a negative one.

2. Anomers are not interchangeable without breaking a bond. Therefore, they are stereoisomers, not conformers.

 

[eli2k]

Hi,

I have a question from the 9R biological sciences section. If you are doing a distillation, and the bp of a compound was 160 C, and you're keeping the flask at 90 C, what's the chance that your compound is going to end up distilling, since there's such a large temperature difference already?

Thanks.

 

[QofQuimica]

Hi,

I have a question from the 9R biological sciences section. If you are doing a distillation, and the bp of a compound was 160 C, and you're keeping the flask at 90 C, what's the chance that your compound is going to end up distilling, since there's such a large temperature difference already?

Thanks.

If the distillation is being done under vacuum, you will probably be able to boil the liquid because lowering pressure lowers bp. If it's being done at atmospheric pressure, it won't boil, but you will still get some amount of vapor pressure. (Imagine heating water up to 60 degrees. It's not boiling, but you still do see SOME steam, right?) I have to say that this is not an efficient way to distill your compound, however. Did they say that you can distill a compound with a bp of 160 at 90 degrees at atmospheric pressure???

 

[md2011]

If the distillation is being done under vacuum, you will probably be able to boil the liquid because lowering pressure lowers bp. If it's being done at atmospheric pressure, it won't boil, but you will still get some amount of vapor pressure. (Imagine heating water up to 60 degrees. It's not boiling, but you still do see SOME steam, right?) I have to say that this is not an efficient way to distill your compound, however. Did they say that you can distill a compound with a bp of 160 at 90 degrees at atmospheric pressure???

They are saying (the answer choice) the purpose of control the temperature at 90 degree is to keep the liquid of 160 bp from being distilled (the other liquid to be distilled has bp of 85 degree). I think what eli2k might be wondering is why worrying about distillation of the 160bp with such big temperature gap.

 

[QofQuimica]

They are saying (the answer choice) the purpose of control the temperature at 90 degree is to keep the liquid of 160 bp from being distilled (the other liquid to be distilled has bp of 85 degree). I think what eli2k might be wondering is why worrying about distillation of the 160bp with such big temperature gap.

Ok, I follow you now. There are two liquids with widely divergent BPs that they are trying to separate by distillation. In that case, yes, the separation will go better if you boil below one compound's BP. Remember that when you distill, what you are trying to do is change the composition of the vapor pressure relative to the composition of the liquid. For example, if I start with a 50-50 mix of compounds A and B, and A boils at 85 C while B boils at 160, I want to have a high proportion of A in the vapor phase (higher than 50%), but only a little B (less than 50%). Conversely, I want most of my B to be in the liquid phase, and not my A. If I heat the liquid to 160, both liquids will boil. But if I keep the temperature between the two BPs, then A will boil but not B, and there will be much more A in the vapor in comparison to B. Thus, I can only separate A from B if I can enrich my vapor in A with respect to B. (If I can't do this, I get what's called an azeotrope, which cannot be separated by distillation because the vapor and liquid phases each have the same proportions of A and B.)

 

[md2011]

Ok, I follow you now. There are two liquids with widely divergent BPs that they are trying to separate by distillation. In that case, yes, the separation will go better if you boil below one compound's BP. Remember that when you distill, what you are trying to do is change the composition of the vapor pressure relative to the composition of the liquid. For example, if I start with a 50-50 mix of compounds A and B, and A boils at 85 C while B boils at 160, I want to have a high proportion of A in the vapor phase (higher than 50%), but only a little B (less than 50%). Conversely, I want most of my B to be in the liquid phase, and not my A. If I heat the liquid to 160, both liquids will boil. But if I keep the temperature between the two BPs, then A will boil but not B, and there will be much more A in the vapor in comparison to B. Thus, I can only separate A from B if I can enrich my vapor in A with respect to B. (If I can't do this, I get what's called an azeotrope, which cannot be separated by distillation because the vapor and liquid phases each have the same proportions of A and B.)

Does that mean, between the low bp (for A) and high bp (for B), the lower the temperature, the higher proportion of A in vapor?

 

[eli2k]

Ok, I follow you now. There are two liquids with widely divergent BPs that they are trying to separate by distillation. In that case, yes, the separation will go better if you boil below one compound's BP. Remember that when you distill, what you are trying to do is change the composition of the vapor pressure relative to the composition of the liquid. For example, if I start with a 50-50 mix of compounds A and B, and A boils at 85 C while B boils at 160, I want to have a high proportion of A in the vapor phase (higher than 50%), but only a little B (less than 50%). Conversely, I want most of my B to be in the liquid phase, and not my A. If I heat the liquid to 160, both liquids will boil. But if I keep the temperature between the two BPs, then A will boil but not B, and there will be much more A in the vapor in comparison to B. Thus, I can only separate A from B if I can enrich my vapor in A with respect to B. (If I can't do this, I get what's called an azeotrope, which cannot be separated by distillation because the vapor and liquid phases each have the same proportions of A and B.)

Thanks. Okay, that makes sense then; you want to keep the temperature just above the boiling point of whatever target compound you are trying to distill in order to minimize the amount of the other compound, which boils at a much higher temperature, from distilling due to vapor pressure that you mentioned. They mentioned two boiling points, 83 C and 161 C, 90 makes sense.

 

[QofQuimica]

Does that mean, between the low bp (for A) and high bp (for B), the lower the temperature, the higher proportion of A in vapor?

Yes. As you keep raising the temperature above A's bp and closer to B's, you will keep increasing the proportion of vapor pressure for B, which will worsen your separation.

 

[WestTexasRambler]

Ms. Q,

QUick Q on the H NMR shift of alkenes vs alkynes. From what I read: the S character differential between alkenes and alkynes results in the alkene being shifted further downfield. This even though, it seems to have less S character as it is sp2 hybridized 33% S character vs 50% S character for the alkyne. This also seems counterintuitive since the alkyne or more acidic, which seems like it would indicate that there is more deshielding going on thus the increase in acidity. Save me from myself.

Thanks,


Charlie

 

[QofQuimica]

Ms. Q,

QUick Q on the H NMR shift of alkenes vs alkynes. From what I read: the S character differential between alkenes and alkynes results in the alkene being shifted further downfield. This even though, it seems to have less S character as it is sp2 hybridized 33% S character vs 50% S character for the alkyne. This also seems counterintuitive since the alkyne or more acidic, which seems like it would indicate that there is more deshielding going on thus the increase in acidity. Save me from myself.

Thanks,


Charlie

The good news is that this is definitely beyond the scope of the MCAT. 😉 You are right that an sp-orbital has more s-character than an sp2-orbital does. In general, you would predict that to cause greater deshielding as you said. However, alkyne protons appear further upfield versus alkene protons due to the orientation of their electron clouds. In other words, the triple bond electrons are shielding the alkyne protons from the magnetic field.

 

[WestTexasRambler]

The good news is that this is definitely beyond the scope of the MCAT. 😉 You are right that an sp-orbital has more s-character than an sp2-orbital does. In general, you would predict that to cause greater deshielding as you said. However, alkyne protons appear further upfield versus alkene protons due to the orientation of their electron clouds. In other words, the triple bond electrons are shielding the alkyne protons from the magnetic field.

Makes sense. I could not find anything definitive in any of the TPR text. Thank you very much. 👍

Take care,


Charlie

 

[j-med]

For an acid-catalyzed esterification reaction between methylbutanol and acetic acid.. (i.e. making methylbutyl acetate).. First, the resulting mixture was washed with 5% K2CO3 for a few times, after all CO2 are released, aqueous layer removed, then NaCl was added to wash the organic layer. The final organic layer was then dried with CaCl2.

Related to that passage,

WHY would CaCl2 not be a good drying agent for drying a solution of n-butanol dissolved in ether?


Thanks!

 

[QofQuimica]

For an acid-catalyzed esterification reaction between methylbutanol and acetic acid.. (i.e. making methylbutyl acetate).. First, the resulting mixture was washed with 5% K2CO3 for a few times, after all CO2 are released, aqueous layer removed, then NaCl was added to wash the organic layer. The final organic layer was then dried with CaCl2.

Why was saturated NaCl used intead of water to wash the organic layer?

And why would CaCl2 not be a good drying agent for drying a solution of n-butanol dissolved in ether?

Finally, what are ways to improve the yield of an esterification reaction?

Thanks!

These sound like HW questions. We will help people with HW questions as long as they don't abuse this, and as long as they make an attempt to answer the questions themselves first. Try to look the answers up and/or figure them out on your own first. The third one in particular is definitely in your lecture textbook. If you're still having trouble, tell us what you've got, and we'll point you in the right direction.

 

[j-med]

Hey QofQuimica,

You caught me before I could change it! I was just editing my post and saw your reply when I came back to this page!.. okay.. so I think I found the answer to the first and the third... but I'm still not sure about the second--- CaCl2-- why is it not a good drying reagent for n-butanol dissolved in ether?


As for the first, I said we alter the equilibrium by LeChatelier's principle? Though.. the experiment already used excess acid and also was heating under reflux/ azeotropic distillation, so I can't think of much else.. I suggested that we use better drying agent.. such as some molecular sieve that may allow absorption of water in higher temperature better than normal drying reagent, or use supercritical fluids to better control phase behavior and to allow very high presssure to be employed.. how does that sound?

Yeah.. I figured out the NaCl after looking more closely at the mechanisms...
But if you still have suggestion for that remaining question, please let me know. Thanks!

 

[QofQuimica]

Hey QofQuimica,

You caught me before I could change it! I was just editing my post and saw your reply when I came back to this page!.. okay.. so I think I found the answer to the first and the third... but I'm still not sure about the second--- CaCl2-- why is it not a good drying reagent for n-butanol dissolved in ether?


As for the first, I said we alter the equilibrium by LeChatelier's principle? Though.. the experiment already used excess acid and also was heating under reflux/ azeotropic distillation, so I can't think of much else.. I suggested that we use better drying agent.. such as some molecular sieve that may allow absorption of water in higher temperature better than normal drying reagent, or use supercritical fluids to better control phase behavior and to allow very high presssure to be employed.. how does that sound?

Yeah.. I figured out the NaCl after looking more closely at the mechanisms...
But if you still have suggestion for that remaining question, please let me know. Thanks!

I knew you could do it. 🙂 You're definitely on the right track.

For the first question, you're essentially trying to draw water out of the reaction mixture by using the brine solution. The water in the organic layer will be attracted to the heavily concentrated salt solution. I'm not sure what your procedure is, but it's probably better to think of it in terms of osmosis rather than Le Chatelier's Principle, because you probably aren't washing with brine to drive the reaction forward. (In other words, you usually wash the reaction AFTER it's over, not during!)

Calcium chloride is generally used to dry hydrocarbons (solvents). It's not a great drying agent to use with alcohols because it will form adducts with them. Carbonyl compounds can also react with some of the impurities present in calcium chloride. (You almost certainly have some calcium hydroxide in there.) So since you are using an alcohol as a reagent and forming an ester product, you don't want to use a drying agent that reacts with hydroxyl groups and carbonyls!

For the third question, here's where you want to manipulate Le Chatelier's Principle. Generally you do one of two things to improve ester formation (or both): one is to use excess alcohol, and the other is to remove the water as it is formed. Often, the alcohol can be used as the solvent and is therefore present in huge excess. But even if you use a different solvent, your limiting reagent is often the acid and you use excess alcohol. In your case, since you're using acetic acid, which is a liquid and readily available, you have used that in excess instead. But it's the same idea.

 

[colorado_doc_2B]

I would appreciate any help you can give me with this hydrogen NMR question. "A compound with the formula C7H16O4 gives three NMR signals: a triplet at delta = 1.9 that integrates to 2, a singlet at delta = 3.3 that integrates to 12, and a triplet at 4.4 that integrates to 2."

So, we know that there are three different "types" of hydrogens. We have a CH2 group that's adjacent to two other hydrogens, 4 CH3 groups that are adjacent to no other hydrogens, and a CH2 group that is adjacent to 2 other hydrogens. I've spent hours trying to draw a compound that meets those criteria, but nothing fits. Any ideas? 😕

I almost forgot to mention, the unsaturation number is zero, so there are no rings or double bonds.

 

[colorado_doc_2B]

Never mind--I just figured it out! Sometimes just writing out your question is enough to make the answer seem obvious. 🙂

 

[j-med]

Thank you so much for your explanation, Q! You're so good at this!! 👍 🙂 Thanks!

 

[QofQuimica]

Never mind--I just figured it out! Sometimes just writing out your question is enough to make the answer seem obvious. 🙂

Glad I could help. 😉

 

[j-med]

Hi Dr. Q,

I must say I need help again.. I'm stuck calculating this:..

How do you calculate the yield of a polymer?

I mean.. from the results I have.. I'm pretty sure I have way more number of moles for the resulting polymer, than the starting materials'...

😕 😕 any hint would be great!


In this case, (to simplify the numbers, I will use just use an approx ratio),
I have 3 mole of hexane diamine
1 mole of sebacoyl dichloride
but somehow I got in the resulting product (nylon 6.10), that
I have like 10 moles!?!
What happened?
I'm pretty sure my data are what it is.. just that I'm missing some concepts to tie this together.
I tried either by mass or by moles.. and it's still definitely bigger..
Please help!
😕

Thanks!

p.s. For the molecular weight of Nylon, do I just add the MW of the two reactants together, or do I subtract out 2 times of HCl?

 

[QofQuimica]

Hi Dr. Q,

I must say I need help again.. I'm stuck calculating this:..

How do you calculate the yield of a polymer?

I mean.. from the results I have.. I'm pretty sure I have way more number of moles for the resulting polymer, than the starting materials'...

😕 😕 any hint would be great!


In this case, (to simplify the numbers, I will use just use an approx ratio),
I have 3 mole of hexane diamine
1 mole of sebacoyl dichloride
but somehow I got in the resulting product (nylon 6.10), that
I have like 10 moles!?!
What happened?
I'm pretty sure my data are what it is.. just that I'm missing some concepts to tie this together.
I tried either by mass or by moles.. and it's still definitely bigger..
Please help!
😕

Thanks!

p.s. For the molecular weight of Nylon, do I just add the MW of the two reactants together, or do I subtract out 2 times of HCl?

Sorry it took me so long to answer; I'm out of town right now. 😳

I don't know too much about polymer chemistry, but I do know that you definitely can't have more moles of product that you have of SM. 😉 So there's definitely something wrong with your calculation. I would guess that one of your monomers is being used in excess. If that's the case, then the other should be the limiting reagent, and if you got 100% yield, that's the maximum number of moles you can have. As for getting the MW, you will probably need to use an average MW. Polymers usually form in a range of sizes. So I would suggest that you figure out what the mean MW is; hopefully the polymer formed in a normal distribution of sizes. Sorry if this doesn't help you much; it's definitely outside my area of expertise.

For anyone else reading this, you do NOT need to know about polymer chemistry for the MCAT. 🙂

 

[j-med]

Sorry it took me so long to answer; I'm out of town right now. 😳
..............
(abbreviated)

Thanks Q! You pointed out a good point, after double checking with people, turns out it is just impurities-- in this case, water, since nylon is very absorbant. And the molecular weight I am supposed to use one functional unit for it... so I guess even if it's almost 300% yield.. it's gonna be fine!.. ;-p

hope you had a good out of town weekend! 🙂

 

[curioso]

I spent about an hour on the following problem, and came up with some answer. But I cant decide if it's correct or not. It's about conformations and potential energy vs. angle of rotation graphs. Here we go:

The anti conformation of 1,2-dichloroethane, Cl-CH2-CH2-Cl, is 4.81 kj/mol more stable than the gauche conformation. The two energy barriers (measured relative to the energy of the gauche conformation) for carbon-carbon bond rotation are 21.5 kj/mol and 38.9 kj/mol.

Sketch a graph of potential energy vs. angle of rotation about the carbon-carbon bond. Show the energy differences on your graph and label each minimum and maximum with the appropriate conformation of 1,2-dichloroethane.


My answer:
I believe the 38.9 kj/mol energy barrier has to be the eclipsed conformation (where, on a Newman Projection, the Cls on both carbons are directly next to each other). So that's where we start off. Because the two energy barrier measurements are measured relative to the gauche conformaion (which is at 60 degrees, where both Cls are 60 degrees to each other), the potential energy diagram will drop to 0. At 120 degrees (where Cls are 120 degrees to each other), the potential energy will rise to 21.5 (again, because this measurement was made relative to the gauche conformation.) At the anti point, however, (180 degrees), the graph will drop -4.81 because that is the stability difference between anti and gauche conformation. Then at 240, the graph will again rise to 21.5 (since both Cls are 120 degrees from each other), drop to 0 at 300 (another gauche point), then rise to 38.9 at 360.

Now, this makes sense to me. However, I looked at the sample problem in the early chapter and in their problem, the two energy barriers (which in that problem are 18.8 and 15.1) subtract to give the difference in stability, which is 3.72. Yet in my answer, 38.9 and 21.5 do NOT subtract to give the stability difference (which, for the above problem is 4.81).

So, what am I doing wrong?

 

[QofQuimica]

I spent about an hour on the following problem, and came up with some answer. But I cant decide if it's correct or not. It's about conformations and potential energy vs. angle of rotation graphs. Here we go:

The anti conformation of 1,2-dichloroethane, Cl-CH2-CH2-Cl, is 4.81 kj/mol more stable than the gauche conformation. The two energy barriers (measured relative to the energy of the gauche conformation) for carbon-carbon bond rotation are 21.5 kj/mol and 38.9 kj/mol.

Sketch a graph of potential energy vs. angle of rotation about the carbon-carbon bond. Show the energy differences on your graph and label each minimum and maximum with the appropriate conformation of 1,2-dichloroethane.


My answer:
I believe the 38.9 kj/mol energy barrier has to be the eclipsed conformation (where, on a Newman Projection, the Cls on both carbons are directly next to each other). So that's where we start off. Because the two energy barrier measurements are measured relative to the gauche conformaion (which is at 60 degrees, where both Cls are 60 degrees to each other), the potential energy diagram will drop to 0. At 120 degrees (where Cls are 120 degrees to each other), the potential energy will rise to 21.5 (again, because this measurement was made relative to the gauche conformation.) At the anti point, however, (180 degrees), the graph will drop -4.81 because that is the stability difference between anti and gauche conformation. Then at 240, the graph will again rise to 21.5 (since both Cls are 120 degrees from each other), drop to 0 at 300 (another gauche point), then rise to 38.9 at 360.

Now, this makes sense to me. However, I looked at the sample problem in the early chapter and in their problem, the two energy barriers (which in that problem are 18.8 and 15.1) subtract to give the difference in stability, which is 3.72. Yet in my answer, 38.9 and 21.5 do NOT subtract to give the stability difference (which, for the above problem is 4.81).

So, what am I doing wrong?

Ok, so the question here is whether the difference in energy between the two eclipsed positions (full and half) must be the same as the difference in energy between the two staggered positions (anti and gauche). In the sample problem, it apparently was coincidentally. But I'm not sure why that would always have to be true. It doesn't look like it is for your dicholoroethane molecule. Evidently the two chlorines have a greater repulsion in the total eclipsed position than whatever other molecule they showed you in your book.

 

[curioso]

I just assumed it would be true since it worked out that way for the practice problem.

Thanks for the response though.

 

[87138]

Stupid question, but if you have to cyclohexanes that are simply just interconverted chairs of each other (say two different chair forms of trans-1,3-dimethycycolhexane), what is the specific named relationship between the two. Conformational isomers? Or is it more accurate to say they're the "same compound?"

 

[QofQuimica]

Stupid question, but if you have to cyclohexanes that are simply just interconverted chairs of each other (say two different chair forms of trans-1,3-dimethycycolhexane), what is the specific named relationship between the two. Conformational isomers? Or is it more accurate to say they're the "same compound?"

Yes, they are conformational isomers, and yes, they are technically the same molecule. Pick the appropriate answer based on context. 🙂

 

[osjx-82]

1) what difference do polar protic and polar aprotic solvents make for various reactions?

2) can someone explain the following descriptions in Kaplan books? For PROTIC solvents, larger atoms tend to be better nucleophiles as they can shed their solvent molecules and are more polarizable. (HUH??!?!)
for APROTIC solvents, nucleophiles are "naked" and nucleophile strength is related to basicity. (HUH!??????????????????)

 

[xylem29]

Hey, I haven't found a question posted about this yet so I'm going to ask it - For the MCAT, do we have to re-learn/re-memorize the reaction mechanisms for each rxn, reagents & solvents etc, for all the functional groups that we learned in Orgo class? Do we also have to rememorize how each functional group can be prepared etc? Do we have to relearn synthesis or retrosynthesis again? B/c then this would require that we know all the functional groups and the reactions for them.

 

[QofQuimica]

1) what difference do polar protic and polar aprotic solvents make for various reactions?

2) can someone explain the following descriptions in Kaplan books? For PROTIC solvents, larger atoms tend to be better nucleophiles as they can shed their solvent molecules and are more polarizable. (HUH??!?!)
for APROTIC solvents, nucleophiles are "naked" and nucleophile strength is related to basicity. (HUH!??????????????????)

1) Polar protic solvents stabilize carbocations, so they help for Sn1. Polar aprotic solvents destabilize nucleophiles, so they help for Sn2. I wrote two explanations posts about this in the gen chem explanations thread.

2) See my post about basicity vs. nucleophilicity in the explanations thread. I don't have time to get into a whole thorough explanation about the mechanisms now, but I'll write a post about it when I get a chance. You might want to go back and review the chapters in your organic textbook about the substitution mechanisms. In general, though, you should correlate nucleophilicity with basicity only when the atoms are the same. (ex. an alcoxide is more basic than an acetate, and they are both oxygen-containing species, so you can conclude that the alcoxide is also more nucleophilic than the acetate).

 

[QofQuimica]

Hey, I haven't found a question posted about this yet so I'm going to ask it - For the MCAT, do we have to re-learn/re-memorize the reaction mechanisms for each rxn, reagents & solvents etc, for all the functional groups that we learned in Orgo class? Do we also have to rememorize how each functional group can be prepared etc? Do we have to relearn synthesis or retrosynthesis again? B/c then this would require that we know all the functional groups and the reactions for them.

I'm not a huge advocate of memorization in general, so my answer is biased that way here. 😉 It's a lot better if you learn and understand the mechanisms instead, because a lot of MCAT questions like to test you on familiar material but in new scenarios that you haven't seen before. (There were organic reactions that *I* hadn't seen on the test before, and I have my PhD in organic.) I would say that you should be familiar with all of the major functional groups and understand how to convert them from one to another. (Ex. how do you get an acid from a primary alcohol?) You will not need to do any synthesis or retrosynthesis problems on the MCAT.

 

[xylem29]

I'm not a huge advocate of memorization in general, so my answer is biased that way here. 😉 It's a lot better if you learn and understand the mechanisms instead, because a lot of MCAT questions like to test you on familiar material but in new scenarios that you haven't seen before. (There were organic reactions that *I* hadn't seen on the test before, and I have my PhD in organic.) I would say that you should be familiar with all of the major functional groups and understand how to convert them from one to another. (Ex. how do you get an acid from a primary alcohol?) You will not need to do any synthesis or retrosynthesis problems on the MCAT.

Thank god for the synthesis being out! Okay, cool - when I took orgo, memorizing each reaction for each functional group and their respective mechanisms wasn't pure memorization anyway, eventually - you start to understand the trends behind them, but okay, so that basically means i have to relearn my orgo stuff. Damn this is will be a pain.

 

[priam18]

I don't know if this has been answered already..if it has, please show the link..

How exactly can one tell what makes a strong acid strong, and what makes a weak acid weak?

Obviously one answer is to look at a pkA table (strong acid has a pkA greater than that of hydronium, weak acid has one weaker than hydronium), but I want to understand exactly what determines acid strength. I know that acid strength increases as you go to the right and down on a periodic table (due to greater number of electrons and electrons not being held so tightly to the nucleic center as they are farther away from the nucleus)...but what about a problem like this:

Which is a stronger acid? Sodium Carbonate Na2CO3 or Sodium Acetate CH3-C-O-Na (there's also a double-bond O on the central C)

In a problem like the above, it's not a simple case of looking at the periodic table to determine which has greater acidity (or is it?) I believe acidity is partially based on how 'available' electrons are for removal from an atom, but how exactly does one determine that availability?