Organic Chemistry Question Thread

[QofQuimica]

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

Acceptable topics:
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Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
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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.

 

[medworm]

Good point -- I was trying to have a framework for determining what a reagent does. Of course, there are the obvious oxidation agents or protonators, but I just don't recall what the more obscure ones like PCC does.

I'm keeping my fingers crossed this Saturday.

 

[QofQuimica]

Good point -- I was trying to have a framework for determining what a reagent does. Of course, there are the obvious oxidation agents or protonators, but I just don't recall what the more obscure ones like PCC does.

Unfortunately, you will have to memorize the common reducing and oxidizing agents, but luckily there aren't many of them.

 

[challie2385]

Hi Q, can you tell me if I have this concept correct? Cis and trans on a cyclohexane ring refers to whether adjacent constituents are axial/axial or equatorial/equatorial and axial/equatorial, respectively, right? Or is there something else to it? Thanks!

 

[QofQuimica]

Hi Q, can you tell me if I have this concept correct? Cis and trans on a cyclohexane ring refers to whether adjacent constituents are axial/axial or equatorial/equatorial and axial/equatorial, respectively, right? Or is there something else to it? Thanks!

The cis/trans axial/equatorial relationship varies depending on the number of carbons between the two substituents. That is, 1,2-diaxial substituents will be trans, but 1,3-diaxial substituents will be cis. Trans and cis mean that the substituents are on opposite sides or the same side of the ring, respectively, regardless of whether they are axial or equatorial.

 

[QofQuimica]

Bumping the organic thread

 

[Twitch]

So when Organic I started we were told not to memorize. Is there a fine line (if any) between what to memorize and what not to memorize as it relates to reactions and mechanisms?

We're using the McMurry book and are on the chapter covering alkenes. Even the book says to memorize the reactions (though it doesn't say anything about the mechanism).

So what's the deal - memorize or not?

Take for instance the halohydrin formation. At the point in the mechanism where water acts as a nucleophile and attacks the C; well why couldn't the Br- hanging out there have attacked it instead. Sure you wouldn't get a product with an alcohol group if that happened. But it seems to me that if you're just given a reactant and told to draw the mechanism and figure out the product - well, there seems to be several ways to skin that cat.

Perhaps it's a blend of the two extremes - (i) memorize the basic reactions and mechanisms (ii) At the same time understand the reaction and the associated mechanism (as to WHY it happens this way) so if the same type of scenario is presented in say a larger molecule you can work it out.

If memorize is the answer what should I record on my notecard? The reactions, reagents, products AND the mechanism?

 

[Nutmeg]

Perhaps it's a blend of the two extremes - (i) memorize the basic reactions and mechanisms (ii) At the same time understand the reaction and the associated mechanism (as to WHY it happens this way) so if the same type of scenario is presented in say a larger molecule you can work it out.

If memorize is the answer what should I record on my notecard? The reactions, reagents, products AND the mechanism?

yes, indeed. You ultimately have to memorize because different professors/texts will have different ways of accomplishing a single objective, but you need to understand mechanisms and trends. As time goes along, you're understanding will get progressively better, and you might eventually rely on memorization less and less. But for now, memorize as needed. Memorization will ultimately help you understand in this matter. And some of the reagents are quirky and bizarre--you have no real choice but to memorize a lot of them.

 

[QofQuimica]

Take for instance the halohydrin formation. At the point in the mechanism where water acts as a nucleophile and attacks the C; well why couldn't the Br- hanging out there have attacked it instead. Sure you wouldn't get a product with an alcohol group if that happened. But it seems to me that if you're just given a reactant and told to draw the mechanism and figure out the product - well, there seems to be several ways to skin that cat.

You're right: you could form the dihalide as a side product. However, because there is so much more water in the rxn mixture compared to Br-, the halohydrin would be predicted to be the major product. In general, very few organic reactions are completely clean. You almost always get side reactions and minor products that must then be separated out. Of course, in a sophomore level textbook, they are mostly going to focus on the major products so as not to confuse the students. But rest assured, in the lab, those minor products do form. 😛

Perhaps it's a blend of the two extremes - (i) memorize the basic reactions and mechanisms (ii) At the same time understand the reaction and the associated mechanism (as to WHY it happens this way) so if the same type of scenario is presented in say a larger molecule you can work it out.

If memorize is the answer what should I record on my notecard? The reactions, reagents, products AND the mechanism?

I would say that you should memorize the *reagents* for various reactions, but you should *not* memorize the reactions themselves, nor the mechanisms. You should also memorize the functional groups and the IUPAC nomenclature rules. However, you really need to actually understand mechanisms in order to do well in your organic class and also on the MCAT, which is why I advise against memorizing them. Instead, when you are faced with a new reaction in your textbook, try to work out all of the possible pathways, and then pick the likeliest one based on the reaction conditions. As Nutmeg said, you will get better at doing this with practice, so you also should work all of the practice problems in your book. McMurray isn't a bad book; I have a copy of it myself, though I used Ege as an undergrad.

 

[Twitch]

I would say that you should memorize the *reagents* for various reactions, but you should *not* memorize the reactions themselves, nor the mechanisms.

Thanks Nutmeg & QofQuimica. As usual you guys are here to guide us in the right path.

Could you elaborate further on the suggestion of memorizing reagents. Were you referring contextually or generically? I"m glad you clarified not memorizing reactions/mechanisms. I had a bad feeling about doing that.

 

[QofQuimica]

Thanks Nutmeg & QofQuimica. As usual you guys are here to guide us in the right path.

Could you elaborate further on the suggestion of memorizing reagents. Were you referring contextually or generically? I"m glad you clarified not memorizing reactions/mechanisms. I had a bad feeling about doing that.

I'd have to say that memorizing contextually is preferable. For example, you need to memorize that the reagents used to make a halohydrin from an alkene are molecular bromine and water. You should see molecular bromine with an alkene, and like Pavlov's dog, associate it with addition across the double bond. Memorization of reagents allows you to answer "predict the product" questions efficiently. Such questions just test your knowledge of reaction conditions, and they are not conceptual in nature. On the other hand, if you're asked to explain why the bromine and hydroxyl groups end up trans to each other, that is a mechanistic question. You should *not* memorize the fact that the stereochemistry of a bromine addition to a double bond is trans, because then you won't be able to explain *why* that happens. Rather, you should work out the mechanism during your study time so that you intuitively understand that the two substituents *must* end up trans any time you add a nucleophile to a bromonium ion via Sn2. Does that help?

 

[Twitch]

I'd have to say that memorizing contextually is preferable. For example, you need to memorize that the reagents used to make a halohydrin from an alkene are molecular bromine and water. You should see molecular bromine with an alkene, and like Pavlov's dog, associate it with addition across the double bond. Memorization of reagents allows you to answer "predict the product" questions efficiently. Such questions just test your knowledge of reaction conditions, and they are not conceptual in nature. On the other hand, if you're asked to explain why the bromine and hydroxyl groups end up trans to each other, that is a mechanistic question. You should *not* memorize the fact that the stereochemistry of a bromine addition to a double bond is trans, because then you won't be able to explain *why* that happens. Rather, you should work out the mechanism during your study time so that you intuitively understand that the two substituents *must* end up trans any time you add a nucleophile to a bromonium ion via Sn2. Does that help?

It does. Thanks!

 

[Twitch]

Anyone have any tips or a link for Fischer Projections? Flattening the stereocenter's substituents seems easy however it seems there are several ways (directions?) to flatten and not all of them are right.

 

[QofQuimica]

Anyone have any tips or a link for Fischer Projections? Flattening the stereocenter's substituents seems easy however it seems there are several ways (directions?) to flatten and not all of them are right.

I know. It takes some practice before you can do this easily without struggling. If you haven't already, get a hold of a model kit, and use that to help you visualize. With some practice, you will be able to visualize the 2D to 3D spatial relationship in your head, and you'll be able to draw the projections correctly.

 

[Twitch]

I know. It takes some practice before you can do this easily without struggling. If you haven't already, get a hold of a model kit, and use that to help you visualize. With some practice, you will be able to visualize the 2D to 3D spatial relationship in your head, and you'll be able to draw the projections correctly.

I think I figured it out. The key is getting a hold of the two substituents that are coming out at you (as if to hug you 😍 ) those two go on the horizontal. Once I have this drawn, the other two go top and bottom. FWIW, I also found a nifty site that lets you move molecules around and provides a quick quiz: http://www.saintmarys.edu/~pbays/Stereochemistry.html

 

[QofQuimica]

I think I figured it out. The key is getting a hold of the two substituents that are coming out at you (as if to hug you 😍 ) those two go on the horizontal. Once I have this drawn, the other two go top and bottom. FWIW, I also found a nifty site that lets you move molecules around and provides a quick quiz: http://www.saintmarys.edu/~pbays/Stereochemistry.html

Ok, good, I'm glad you figured it out. Practice on a few more examples, and I'm sure you'll be fine come test day. 👍

 

[Twitch]

Q1) The pKa value of an acetylenic proton is approximately 25, which of the following compounds is/are not sufficently basic to remove this proton? (pKa of conjugate acid are given in parenthesis):
(a) BuLi (60)
(b) LiN(i-Pr)2 (40)
(c) NaOH (16)
(d) NaHCO3 (6)

The answer is (a) & (b)? Could someone explain why (or explain the question 🙂 ? I understand that pKa ~ pH so the higher the pKa the less acidic. From LeChatlier's (sp?) principle the eq lies to the side of the weaker acid.
UPDATE: I think I got it. the answer is c & d (hence the cause for confusion above). a&b WILL deprotonate!

Q2) If you have 3 chiral centers, how many diastereomers can you have?

For this, 2^n = 8 total stereoisomers:

A A'
B B'
C C'
D D'

where the primes are enantiomers. If you leave out the first row then you're left with 6 diastereomers??? Could someone explain? So, can one say that in general you have (2^(n-1) - 1)*2 diastereomers?

Q3) In CIP priorties, is CO2H > CN, since O beats N?

 

[QofQuimica]

Q1) The pKa value of an acetylenic proton is approximately 25, which of the following compounds is/are not sufficently basic to remove this proton? (pKa of conjugate acid are given in parenthesis):
(a) BuLi (60)
(b) LiN(i-Pr)2 (40)
(c) NaOH (16)
(d) NaHCO3 (6)

The answer is (a) & (b)? Could someone explain why (or explain the question 🙂 ? I understand that pKa ~ pH so the higher the pKa the less acidic. From LeChatlier's (sp?) principle the eq lies to the side of the weaker acid.
UPDATE: I think I got it. the answer is c & d (hence the cause for confusion above). a&b WILL deprotonate!

You got it. 👍

Q2) If you have 3 chiral centers, how many diastereomers can you have?

For this, 2^n = 8 total stereoisomers:

A A'
B B'
C C'
D D'

where the primes are enantiomers. If you leave out the first row then you're left with 6 diastereomers??? Could someone explain? So, can one say that in general you have (2^(n-1) - 1)*2 diastereomers?

Four stereocenters will yield a maximum of eight stereoisomers. You are right that each individual isomer could have up to six diastereomers. But, there could actually be fewer stereoisomers if some of them are meso compounds.

Q3) In CIP priorties, is CO2H > CN, since O beats N?

Yes. One bond to O beats a zillion bonds to N.

 

[Twitch]

Four stereocenters will yield a maximum of eight stereoisomers. You are right that each individual isomer could have up to six diastereomers. But, there could actually be fewer stereoisomers if some of them are meso compounds.

Thanks for the verification. Wouldn't four stereocenters yield 2^4 = 16 stereoisomers? Note there were 3 stereocenters in the question. BTW, why did we discard the first row? Is there another way to calc # of diastereomers from # of chiral centers? Help!

 

[QofQuimica]

Thanks for the verification. Note there were 3 stereocenters in the question. BTW, why did we discard the first row? Is there another way to calc # of diastereomers from # of chiral centers?

Oops, sorry. 😳 Long day. I meant three. We didn't "discard" the first row. If the question is simply, what is the max. number of stereoisomers, it's eight total. But each one of your molecules only has a max. of six diastereomers to it. For example, if A is the isomer we are considering, then A' is its one enantiomer, and all of the others are diastereomers to both A and A'.

 

[PRamos]

Three stereocenters will yield a maximum of eight stereoisomers. You are right that each individual isomer could have up to six diastereomers. But, there could actually be fewer stereoisomers if some of them are meso compounds.

Good point that you should consider the possibility of a structure being meso. You should do this whenever there is an even number of chiral centers. As a point of interest in this example, with an odd number of stereogenic centers, it is not possible to have a meso compound. Meso compounds have an even number of chiral centers, because they must be paired about the mirror plane slicing through the middle of the molecule.

RRR SSS
RRS SSR
RSR SRS
SRR RSS

Any one of the eight structures you choose is an enantiomer of the structure on the same line and a diastereomer of the other six structures. Enantiomers, because they are mirror images, vary at all of the stereogenic centers.

 

[QofQuimica]

Good point that you should consider the possibility of a structure being meso. You should do this whenever there is an even number of chiral centers. As a point of interest in this example, with an odd number of stereogenic centers, it is not possible to have a meso compound.

No, that's not true. All that is necessary to have a meso compound is for the molecule to be SYMMETRICAL. You are correct that you can't have a meso compound if you have a single stereocenter, but you CAN have a meso compound if you have three stereocenters, or five, or any other odd number besides one. Consider a molecule having COOH, then three stereocenters with H on the left, OH on the right, and a second COOH. That would be an example of a meso compound with three stereocenters.

 

[PRamos]

No, that's not true. All that is necessary to have a meso compound is for the molecule to be SYMMETRICAL. You are correct that you can't have a meso compound if you have a single stereocenter, but you CAN have a meso compound if you have three stereocenters, or five, or any other odd number besides one. Consider a molecule having COOH, then three stereocenters with H on the left, OH on the right, and a second COOH. That would be an example of a meso compound with three stereocenters.

In all due respect, you are incorrect here. You are making the classic mistake of considering the Fisher projection of a aldopentose after it has been treated with nitric acid. You get the aldaric acid you describe above, but the central carbon (carbon three of the original aldopentose) is no longer a stereogenic center following the nitric acid oxidation. The central carbon now has two equivalent groups attached to it (two CH(OH)-COOH groups), so it is no longer chiral. Hence, the aldaric acid formed has only two stereogenic centers. So while the original aldopentose had three sites of chirality, the meso product has only two.

The even number rule can be found in Streitweiser, Heathcock, and Kosower, on page 136 if you are interested.

BTW, if you have any lingering doubts about your example, try assigning either R or S to carbon-3. You will discover upon prioritizing that the central carbon cannot be assigned an R or an S. Hence, the oxidative conversion takes a 2R,3R,4R aldopentose (D-ribose in specific) and converts it into a 2R,4S diacid product.

 

[QofQuimica]

In all due respect, you are incorrect here. You are making the classic mistake of considering the Fisher projection of a aldopentose after it has been treated with nitric acid. You get the aldaric acid you describe above, but the central carbon (carbon three of the original aldopentose) is no longer a stereogenic center following the nitric acid oxidation. The central carbon now has two equivalent groups attached to it (two CH(OH)-COOH groups), so it is no longer chiral. Hence, the aldaric acid formed has only two stereogenic centers. So while the original aldopentose had three sites of chirality, the meso product has only two.

You're right, I hadn't considered that the middle chiral center would no longer be chiral, so there are still technically an even number of chiral centers in an odd-carbon meso compound. Thank you for pointing this out.

 

[PRamos]

You're right, I hadn't considered that the middle chiral center would no longer be chiral, so there are still technically an even number of chiral centers in an odd-carbon meso compound. Thank you for pointing this out.

Let's just say I learned that the hard way on a midterm about three years ago. I think almost the entire class missed that question.

 

[QofQuimica]

bump

 

[QofQuimica]

bumping this thread

 

[bravotwozero]

Hi,

I need to figure out how to do the following synthesis problem. Please help!
(see question 1)
http://www.erin.utoronto.ca/~w3chm243/Lab4questions.doc

 

[QofQuimica]

Hi,

I need to figure out how to do the following synthesis problem. Please help!
(see question 1)
http://www.erin.utoronto.ca/~w3chm243/Lab4questions.doc

Hi bravotwozero,

We don't do people's homework for them, but here's a hint. The trick to solving synthesis problems like this is to start with the product and work your way backward to the starting material. That is called retrosynthesis, and it works much better than trying to go forward from the starting material. Think about what you could have made your product from, always keeping in mind that you want to eventually make your way back to the SM. You may need to try a few times if you get "stuck" due to going down some wrong pathways.

Try it yourself first, and let us know if you have a specific question that we can help you with.

 

[bravotwozero]

Fair enough.

I was trying to do a retrosynthesis. This is what i'm unsure of:

if you do a Friedel Crafts acylation, isn't the carbon with the acyl group attached to it going to directly join the ring? In this case, I want to attach an alkyl chain to the benzene ring, while having the acyl group on the adjacent Carbon. (click on the link mentioned previously to get an idea of what i'm talking about). Can this be done using Friedel Krafts acylation?

 

[QofQuimica]

Fair enough.

I was trying to do a retrosynthesis. This is what i'm unsure of:

if you do a Friedel Crafts acylation, isn't the carbon with the acyl group attached to it going to directly join the ring? In this case, I want to attach an alkyl chain to the benzene ring, while having the acyl group on the adjacent Carbon. (click on the link mentioned previously to get an idea of what i'm talking about). Can this be done using Friedel Krafts acylation?

Nope. So you're going to have to come up with another method. But it's a good thought though.

 

[jp104]

Question:

Say I have a molecule of 3-Chlorotoulene. Is it also correct to name this thing:

1-methyl-3-Chlorobenzene, or is it 3-chloro-Methylbenzene?

IUPAC confuses me sometimes!

Thanks!

 

[QofQuimica]

Question:

Say I have a molecule of 3-Chlorotoulene. Is it also correct to name this thing:

1-methyl-3-Chlorobenzene, or is it 3-chloro-Methylbenzene?

IUPAC confuses me sometimes!

Thanks!

Nomenclature of aromatics can be difficult because there are so many common names. It is actually correct to call that compound 3-chlorotoluene (or m-chlorotoluene) under IUPAC rules. Any time that a substituent (like a methyl) is present that gives the benzene ring a new base name, it is assumed to be on carbon one, and the new parent name is used.

 

[jp104]

Nomenclature of aromatics can be difficult because there are so many common names. It is actually correct to call that compound 3-chlorotoluene (or m-chlorotoluene) under IUPAC rules. Any time that a substituent (like a methyl) is present that gives the benzene ring a new base name, it is assumed to be on carbon one, and the new parent name is used.

Thank you! Xylene is another one that confuses me. It's a 1,2-Dimethylbenzene (if I am describing it correctly). But, I think Xylene is a common name?

But, thanks for your explanation!!

 

[QofQuimica]

Thank you! Xylene is another one that confuses me. It's a 1,2-Dimethylbenzene (if I am describing it correctly). But, I think Xylene is a common name?

But, thanks for your explanation!!

Yes, xylene is the common name for a benzene ring having two methyl groups on it. Xylene with no modifier is actually ambiguous; what you just described is o-xylene. You can also have m-xylene (1,3-dimethylbenzene) or p-xylene (1,4-dimethylbenzene). I hope I'm not making your confusion worse. 😛

You don't need to know too many common names of aromatics for the MCAT. But I would recommend learning at least the common monosubstituted ones, including toluene (methylbenzene), phenol (hydroxybenzene), and aniline (aminobenzene). You don't need to worry about the disubstituted ones like xylene, cresol, etc.

 

[smoky]

ethyl 4-aminobenzoate or ethyl4-iodobenzoate?

c-n bond is more polar than c-i bond...but what about the whole molecule? I am not sure how to figure this out...

Thanks!

 

[QofQuimica]

ethyl 4-aminobenzoate or ethyl4-iodobenzoate?

c-n bond is more polar than c-i bond...but what about the whole molecule? I am not sure how to figure this out...

Thanks!

You need to consider the type of intermolecular interactions that each molecule can have. If you don't know what I mean, take a look at my post about intermolecular interactions in the Organic Explanations thread.

 

[smoky]

Thank you.

 

[NontradICUdoc]

to determine if a molecule is aromatic it has to satisfy Huckels Rule of 4n+1

How do you determine n?

 

[QofQuimica]

Sorry. Yes 4n+2. Up until now I have been counting the c=c and multiplying by 2 and then adding any extra electrons. I am not sure if I am doing this correctly. Please explain it to me.

n is any integer of zero or greater. Most common examples that you know of aromatic compounds are going to have an n=1, like the example jota gave. You should count the pi electrons and the lone pairs inside the ring, not the atoms themselves. It sounds like you are doing it correctly by counting the bonds and lone pairs.

 

[NontradICUdoc]

When hydrolyzing an anhydride, do you always get 2 carboxylic acids?

Also, whenever I see LAH, I should always think reduction and loss of an oxygen, correct?

During Transesterification, does the alcohol only add 1 carbon to the original ester? Does that mean that if the alcohol is methanol, the entire molecule is now part of the ester?

 

[QofQuimica]

When hydrolyzing an anhydride, do you always get 2 carboxylic acids?

yes

Also, whenever I see LAH, I should always think reduction and loss of an oxygen, correct?

yes

During Transesterification, does the alcohol only add 1 carbon to the original ester? Does that mean that if the alcohol is methanol, the entire molecule is now part of the ester?

I am not sure if I understand your question, but the entire alcohol becomes part of the ester except for the hydroxyl group's proton. So yes, methanol would add 1 carbon to the ester. But ethanol would add 2 carbons, propanol would add 3, and so on.

 

[premedrocky]

Perspective. In all addition reactions, you are adding something across a double (or triple) bond. In nucleophilic additions, you have a polar multiple bond, like a carbonyl, that serves as the electrophile, so we say we are adding the nucleophile to it. In electrophilic additions, you have a nonpolar multiple bond, like an alkene, that serves as the nucleophile, so we say we are adding the electrophile to it.

How does this reasoning work with substitution reactions (ie. electrophilic vs. nucleophilic substitution). I'm having trouble with this concept for some reason. I understand the addition reactions better now thanks to this explanation, but I'm still shaky about the substitution reactions.

Thanks!

 

[QofQuimica]

How does this reasoning work with substitution reactions (ie. electrophilic vs. nucleophilic substitution). I'm having trouble with this concept for some reason. I understand the addition reactions better now thanks to this explanation, but I'm still shaky about the substitution reactions.

Thanks!

Again, you want to always identify the nucleophile and the electrophile. You must never pair together two nucleophiles or two electrophiles. Nucleophilic substitutions occur when nucleophiles attacks electrophiles that have polar bonds, such as an alkyl halide. These electrophiles should not be sp2-hybridized. In other words, a vinyl halide (where the halogen is attached directly to a double bond) is NOT an electrophile for nucleophilic substitution reactions because you cannot form an SN2 transition state, and vinyl carbocations are usually not stable.

Electrophilic substitutions that you should be familiar with occur with aromatic compounds. Here, you have the aromatic ring serving as the nucleophile and attacking a Lewis acid, which is the electrophile. You must have a Lewis acid or some other electron-deficient species because aromatic rings are not very powerful nucleophiles. Remember that increased stability means decreased reactivity; this rule will be true in general.

Hope this helps.

 

[suckermc]

heres a question. they give you two aromatic looking amines and ask you what they are (aromatic, primary, secondary, etc) the rings have an nh2 group attached at the first position. why would it be considered primary and not aromatic?

 

[st.exupery]

Hi... Can't figure this out...

NMR Signals for: 2-Chloropentane

Answer says: 5 peaks (c1 through c-5 respectively): 3H doublet, 1H multiplet, 2H quintet, 2H sextet and 3H triplet.

I get all except the 2H quintet. Wouldn't c-3 be a quartet if 2H is a sextet? Is it a quintet since it would seperately split by c-2 and c-4 (1+1) + (2+1)? If so, wouldn't that then mean that c-4 should be split separately as well into a septet? Thanks

 

[QofQuimica]

heres a question. they give you two aromatic looking amines and ask you what they are (aromatic, primary, secondary, etc) the rings have an nh2 group attached at the first position. why would it be considered primary and not aromatic?

If it's an aniline, I would call that both primary and aromatic. If it's a pyridine ring, it's aromatic but not primary. Are you only allowed to pick one choice? What an odd question. 😛

 

[QofQuimica]

Hi... Can't figure this out...

NMR Signals for: 2-Chloropentane

Answer says: 5 peaks (c1 through c-5 respectively): 3H doublet, 1H multiplet, 2H quintet, 2H sextet and 3H triplet.

I get all except the 2H quintet. Wouldn't c-3 be a quartet if 2H is a sextet? Is it a quintet since it would seperately split by c-2 and c-4 (1+1) + (2+1)? If so, wouldn't that then mean that c-4 should be split separately as well into a septet? Thanks

Hmm, it looks like they are just asking you to count the neighbors and assume they are the same. If so, then C-3 should be a quartet like you said because there are three total neighboring Hs. On a real NMR, you would probably see a multiplet for C-2, C-3, and C-4, and you would probably also have trouble with the peaks overlapping with one another where it would be difficult to resolve the separate shifts. I'm guessing since they picked "sextet" for C-4 that they probably wanted "quartet" as the answer for C-3. Ask around and/or look online to see if anyone else has noticed the same error in your book.

 

[suckermc]

If it's an aniline, I would call that both primary and aromatic. If it's a pyridine ring, it's aromatic but not primary. Are you only allowed to pick one choice? What an odd question. 😛

only one choice, and the answer was primary. i thought it was aromatic ..maybe because the benzylamine is not aromatic, so both are primary?

 

[QofQuimica]

only one choice, and the answer was primary. i thought it was aromatic ..maybe because the benzylamine is not aromatic, so both are primary?

Ah, ok, that does change things. If the question was something like this:

Which of the following classifications applies to both aniline and benzylamine?

then yes, they are both primary amines. But only the aniline is an aromatic amine.

 

[suckermc]

Ah, ok, that does change things. If the question was something like this:

Which of the following classifications applies to both aniline and benzylamine?

then yes, they are both primary amines. But only the aniline is an aromatic amine.

when i grow up, i wanna be just like you dr q. thanks.