Organic Chemistry Question Thread

[QofQuimica]

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

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

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

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

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

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

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

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[QofQuimica]

hi Q of Q, can you please explain nucleophillicity and basicity, i feel like they can be confusing at sometimes. wht are their trends on the periodic table and how are they related in most orgo rxns (e.g. E1,Sn1,Sn2, and E2). like how elimination is favored by strong bases, while Sn2 is favored by strong nucleophiles over Sn1. Also, wht about electrophilicity and acidity is there a similar relationship? sorry if i mixed any concepts. ty in advance

and do you think we need to know reagents like LDA or pyridine?

I will be addressing this topic in more detail in an organic explanations thread in the future, but unfortunately it will not be ready before you August test-takers sit for the MCAT. Briefly, here is an explanation of these topics:

Basicity measures how well a substance can accept a proton (Bronsted-Lowry definition) or how readily it donates a lone pair (Lewis definition). Most MCAT questions will use the Bronsted-Lowry definition of a base. I have seen that EK's chemistry review notes book has a trend that relates basicity to the periodic table, but I generally gauge basicity by pKa. Since pKa is the negative log of Ka, the dissociation constant of an acid, substances with higher pKas are more basic. Strong bases favor E2 reactions over all other mechanisms, and E1 and Sn1 will never occur in a strongly basic medium. Sn2 is favored if the base is also a strong nucleophile and the alkyl halide is unhindered (primary).

Nucleophilicity is a measure of how polarizable an atom's or ion's electron cloud is. You can think of good nucleophiles as having relatively "sloppy" electron clouds. Nucleophilicity does not really follow basicity trends, except that if you are comparing two nucleophiles of the same element, the one with higher pKa will be more nucleophilic. (For example, both ethoxide and acetate are oxygen nucleophiles. Ethoxide has a much higher pKa than acetate, so it will be the better nucleophile. Note that you cannot do this comparison for two nucleophiles of different elements, such as acetate versus iodide. Acetate has a much higher pKa than iodide, but it is a much weaker nucleophile.) Good nucleophiles tend to be large and not too highly charged, or with the charge spread diffusely. They are not required for E1 or Sn1, but it is essential to have a good nucleophile for Sn2 reactions.

Acidity measures how readily a species will give up a proton (Bronsted-Lowry definition) and is also measurable by pKa. (The lower the pKa, the more acidic the species.) Strong acids have negative pKas. I have a previous post with the list of strong acids and bases. You should memorize this list; any acid not on it should be considered weak for MCAT purposes. Acids are related to electrophiles in that the H+ ion is itself an electrophile. If you use the Lewis acid-base definitions, you can think of protonation of a base as being a reaction between an electrophile (the proton) and a nucleophile (the base).

I would recommend that you know the two most common hindered strong bases, which are LDA and t-butoxide. If you see either of these bases on the MCAT, you should immediately consider the reaction as likely to go by the E2 mechanism.

Nutmeg has posted the link to a very nice table about the different reaction mechanisms with tips for distinguishing them.

 

[Scrub MD]

I know an impurity broadens the melting point but does it always lower it or only if the impurity has a lower MP.

 

[Nutmeg]

I know an impurity broadens the melting point but does it always lower it or only if the impurity has a lower MP.

Generally it will always lower the MP, because the differences in molecular size and shape disrupt the ability for an even crystal matrix to form. Impurities disrupt orderly packing.

Material scientists and engineers use a phase diagram to show the differences in melting points of alloys. A typical one looks like:

where you have a binary mixture of two components. While the intricacies of reading this chart are entirely beyond the scope of the MCAT, you can see here that the two verticle axes show the melting points of the pure components at the point where the "rabbit ears" touch the axis. In this diagram, pure beta has a higher melting point than pure alpha, but for both, you see that the melting range broadens and the melting temperature lowers as you move away from the pure component.

 

[Righty123]

I have two organic chemistry questions I was hoping someone would answer/explain to me:

1. What are the most stable configurations of 1,2- and 1,3-diisopropylcyclohexane, respectively?
a. trans trans
b. cis cis
c. trans, cis
d. cis trans

2. If a compound had an IR spectrum with four strong absorbances at 3460, 2950,1690 and 1050 cm^-1 then the compound must contain wihich of the following sets of bonds?
1. O-H, C-H, and C=0
2. O-H, C-H, and C=C
3. O-H, N-H, and C=C
4. O-H, C-O, and C-N

Any feedback is much appreciated. Thanks!

 

[mustangsally65]

I've got a couple questions from my EK 1001 Organic Chem book.

I never understood what a London Dispersion force is. Never got it from my undergrad chem courses, and the summary in EK doesn't do it for me.

Also, I had some questions in the book about which types of bonds are hardest to break, and it seems to me that the hardest to break are sigma bond, then triple bonds, then hydrogen bonds? I thought H-bonds were the strongest of them all? Where am I going wrong?

Thanks!

 

[QofQuimica]

1. What are the most stable configurations of 1,2- and 1,3-diisopropylcyclohexane, respectively?
a. trans trans
b. cis cis
c. trans, cis
d. cis trans

The best way to solve a problem like this is to draw out both structures to see which one is the most stable. In either case, it is best to have your substituents both lying equatorially rather than axially. So draw the two possible isomers (cis and trans) for each pairing of substituents, and see which one allows you to have both substituents equatorial. If you need more help, let me know. But try to solve it yourself first.

2. If a compound had an IR spectrum with four strong absorbances at 3460, 2950,1690 and 1050 cm^-1 then the compound must contain wihich of the following sets of bonds?
1. O-H, C-H, and C=0
2. O-H, C-H, and C=C
3. O-H, N-H, and C=C
4. O-H, C-O, and C-N

The 3460 tells you it must have an O-H or N-H, but that doesn't help you here. 1690, on the other hand, is a very helpful hint. Look in your organic book for what stretch comes at 1690. That stretch alone is sufficient to solve the problem.

 

[QofQuimica]

I never understood what a London Dispersion force is. Never got it from my undergrad chem courses, and the summary in EK doesn't do it for me.

I am working on writing up an explanation of intermolecular forces, but I haven't finished yet. See below.

Also, I had some questions in the book about which types of bonds are hardest to break, and it seems to me that the hardest to break are sigma bond, then triple bonds, then hydrogen bonds? I thought H-bonds were the strongest of them all? Where am I going wrong?

You are confusing intermolecular forces (hold molecules together in a liquid or solid) versus intramolecular forces (bonds that hold the atoms of a single molecule together). I have finished the explanation of intramolecular forces already, and I will try to get the intermolecular forces part done as soon as I can.

 

[mustangsally65]

Thanks a million, QofQ. Just wanted to show my appreciation to all the hard work you are putting in for us MCATers.

 

[MochaMD]

I'm tutoring Orgo right now and I can't seem to figure out something. I looked online but none of the websites do a good job explaining the reasoning. For heterocyclic compunds, how do you know when the lone pairs on say N are either part of the pi system or not when you're trying to determine if the compound is aromatic or not. The lone pairs on pyridine are not part of the pi system but the lone pairs on pyrrole are. Also with regards to that, I'm having trouble with determining the hybridization on the heteratoms like Nitrogen and Oxygen.

 

[QofQuimica]

I'm tutoring Orgo right now and I can't seem to figure out something. I looked online but none of the websites do a good job explaining the reasoning. For heterocyclic compunds, how do you know when the lone pairs on say N are either part of the pi system or not when you're trying to determine if the compound is aromatic or not. The lone pairs on pyridine are not part of the pi system but the lone pairs on pyrrole are. Also with regards to that, I'm having trouble with determining the hybridization on the heteratoms like Nitrogen and Oxygen.

If the compound requires the lone pair in order to obey Huckel's rule, then the lone pair will be part of the aromatic system. Pyridine has three double bonds, so it already has six pi electrons. Pyrrole, on the other hand, has only four pi bond electrons, so it needs the lone pair electrons to give it six. The heteroatoms must be sp2-hybridized if they are part of the aromatic system; an sp3-hybridized atom anywhere in the ring destroys aromaticity.

 

[SensesFail]

A few specific questions I came across in my Kaplan Lesson Book which my instructor didn't explain very well.


(From page 433, Kaplan Lesson Book, 2005)
1. Because of the peptide bond restriction to planar conformations, all of the following can be concluded about the atoms in the link EXCEPT:

A. the nitrogen lone pair has pie-overlap with the carbonyl pie-bond.
B. there is considerable positive-charge character on the nitrogen atom.
C. the nitrogen atom is sp3 hybridized.
D. there is considerable negative-charge character on the carbonyl carbon.

Correct answer is C. I chose B because I don't understand how nitrogen would have a partial positive charge. I thought it'd be withdrawing electron-density from the carbonyl carbon.


Next question.

(page 431, Kaplan Lesson Book, 2005)
2. The reason why a diazonium intermediate (R-N2+) is unstable is that:

A. the nitrogen-nitrogen bond is weak
B. it is a strong Lewis acid
C. it posesses a good leaving group
D. it is susceptible to attack by nucleophiles.

Correct answer is C. I chose D and thought it had to be either D or B, so I was way off.

Any help is much appreciated.

 

[Righty123]

The best way to solve a problem like this is to draw out both structures to see which one is the most stable. In either case, it is best to have your substituents both lying equatorially rather than axially. So draw the two possible isomers (cis and trans) for each pairing of substituents, and see which one allows you to have both substituents equatorial. If you need more help, let me know. But try to solve it yourself first.



The 3460 tells you it must have an O-H or N-H, but that doesn't help you here. 1690, on the other hand, is a very helpful hint. Look in your organic book for what stretch comes at 1690. That stretch alone is sufficient to solve the problem.

For the IR, I know that a C = O stretch is found around that range. But I guess the 1060 threw me off, b\c in Kaplan that value is used to indicate an ester bond, therefore, I thought the answer was D as a result.

And for the two conformations, I know that the sub's have to be equatorial. But, I always thought that it meant they had to be cis (same side). The answer says that it it is trans and cis, I don't understand how something that is trans can still be equatorial. I thought trans meant different sides of the molecule's plane?

 

[QofQuimica]

This question is concerning the Ochem Question
What Is the Effect of Hydrocarbon Branching on MP and BP?
You said as there's more hydrocarbon branching, MP and BP decreases. As I read your explanation it made sense. BUT, I'm also studying with Examkrackers and they're saying the opposite for MP. They say:

Increase branching = decrease BP , increase MP

Their explanation on the forum:
"The branching of an alkane gives it a more compact, three-dimensional structure, which packs more easily into solid structures; thus melting point tends to increase and boiling points decrease. Any decrease in surface area is not significant since tight van der Waals packing will apparently overcome it."

The effect of chain branching on alkane melting points is much harder to predict compared to its effect on alkane boiling points. In general, branching lowers van der Waals overlap between the two molecules. For example, n-pentane (mp = -130 C) has a higher mp than 2-methylbutane (mp = -160 C). However, in those cases where you have highly symmetrical isomers like 2,2,3,3-tetramethylbutane, the compounds will have abnormally high melting points because of their excellent packing properties. This is probably what the EK book is referring to in this context. To go back to my previous example, 2,2-dimethylpropane (mp = -17 C) has a much higher mp than either n-pentane or 2-methylbutane.

This information is probably beyond what you'd be expected to know about isomer properties for the MCAT, but I wanted to clarify this since a few people have asked about it.

 

[QofQuimica]

(From page 433, Kaplan Lesson Book, 2005)
1. Because of the peptide bond restriction to planar conformations, all of the following can be concluded about the atoms in the link EXCEPT:

A. the nitrogen lone pair has pie-overlap with the carbonyl pie-bond.
B. there is considerable positive-charge character on the nitrogen atom.
C. the nitrogen atom is sp3 hybridized.
D. there is considerable negative-charge character on the carbonyl carbon.

Correct answer is C. I chose B because I don't understand how nitrogen would have a partial positive charge. I thought it'd be withdrawing electron-density from the carbonyl carbon.

The carbonyl is more electron-withdrawing than the nitrogen is, so yes, the nitrogen does have a partial positive charge on it. In a tug-of-war between O and N, the more electronegative O will win. The reason why N cannot be sp3-hybridized is because its lone pair is actually resonating with the carbonyl double bond pi electrons; you have a delocalized pi system over all three atoms, and the entire amide portion of the molecule is flat (all three atoms are sp2-hybridized and in the same plane). There is a resonance structure you can draw for an amide having a double bond between N and C, with a positive charge on N and a negative charge on O; this rationalizes the partial positive on N from a Lewis structure perspective.

(page 431, Kaplan Lesson Book, 2005)
2. The reason why a diazonium intermediate (R-N2+) is unstable is that:

A. the nitrogen-nitrogen bond is weak
B. it is a strong Lewis acid
C. it posesses a good leaving group
D. it is susceptible to attack by nucleophiles.

Correct answer is C. I chose D and thought it had to be either D or B, so I was way off.

Nitrogen (N2) is an extremely stable molecule, owing to the triple bond between the two nitrogen atoms. The ability to form N2 in a reaction is a powerful force for driving a reaction to completion, not only because the product nitrogen is so stable, but also because it is a gas, which can escape from the flask, thereby removing a product and driving the reaction further to the right according to LeChatelier's principle. So, diazonium is a fantastic leaving group for these reasons. B is incorrect because N2 is not electron-deficient; recall that a Lewis acid is a lone pair acceptor, and typically on the MCAT you will only see the Lewis definition used for acids that lack a complete octet. D is incorrect because (-N2)+ is the leaving group, not the electrophile. The carbon to which the diazonium is bonded is the electrophile.

 

[QofQuimica]

For the IR, I know that a C = O stretch is found around that range. But I guess the 1060 threw me off, b\c in Kaplan that value is used to indicate an ester bond, therefore, I thought the answer was D as a result.

Good, you got it. You can rule out the two choices with alkenes; 1690 is probably too high to be a C=C stretch, and plus C=C isn't a strong absorbance. The C=O stretch is always strong and tends to come in the 1700-1800 range; you should look for it first thing when interpreting IR spectra. The 1060 stretch IS a little tricky, because it could be either a C-O or C-N stretch, so you're right that choice 4 is tough to rule out. However, having that carbonyl stretch show up along with a hydroxyl or amino stretch means that you have a carboxylic acid or amide, and so you'd expect to see a C-O or C-N bond in there anyway (1060 is actually too low for an acid C-O bond, so I'm guessing you might have an ether somewhere else in the molecule.) Since you have to pick the BEST answer choice, answer 1, which includes the carbonyl, is the right one.

And for the two conformations, I know that the sub's have to be equatorial. But, I always thought that it meant they had to be cis (same side). The answer says that it it is trans and cis, I don't understand how something that is trans can still be equatorial. I thought trans meant different sides of the molecule's plane?

Ok, the problem here is that you are unclear about how the axial and equatorial positions are oriented around the cyclohexane ring. Remember that a cyclohexane ring is not planar; in its most stable form, it is shaped like a chair. If you have a model kit, I suggest that you build a cyclohexane chair and take a look at the orientation of the equatorial positions. You will see that they alternate up and down around the ring in the opposite fashion that the axial positions do. (That is, if you have an up axial substituent on carbon 1, you will have a down equatorial substituent on carbon 1. Carbon 2 will then have a down axial substituent and an up equatorial substituent, carbon 3 will be like carbon 1, and so on.) So since the adjoining equatorial positions are oriented trans to one another (one up and one down), it is impossible for a 1,2-disubstituted cyclohexane to be diequatorial if the substituent configuration is cis.

 

[frankrizzo18]

The molecule phenol: For proton NMR, I would see a singlet corresponding to 3Hydrogens/A singlet corresponding to two hydrogens (because two H's are in the same environment)/A singlet corresponding to another two hydrogens (same arg as above)/Another singlet for the para proton. However, the rule for splitting refers to protons on adjacent carbons. Well, wouldn't the para proton neighbor a carbon with a proton on one adjacent side and a proton on the other adjacent side? Could there perhaps also be a multiplet that shows up in the spectrum? Thanks

Franky

 

[Learfan]

The molecule phenol: For proton NMR, I would see a singlet corresponding to 3Hydrogens/A singlet corresponding to two hydrogens (because two H's are in the same environment)/A singlet corresponding to another two hydrogens (same arg as above)/Another singlet for the para proton. However, the rule for splitting refers to protons on adjacent carbons. Well, wouldn't the para proton neighbor a carbon with a proton on one adjacent side and a proton on the other adjacent side? Could there perhaps also be a multiplet that shows up in the spectrum? Thanks

Franky

In a proton NMR of phenol, you probably will not see evidence of the OH proton due to exchange with the solvent. The place to look for confirmation of the phenolic proton is in the IR spectrum. If the proton spectrum is of insufficient resolution, you may just see a complex multiplet around a delta of 7.2 to 7.5. In a very high resolution spectrum you should see a clear pattern. After applying symetry considerations to the molecule and noting the mirror plane that bisexts the aromatic ring, it should be apparent that there are three unique forms of hydrogen on the ring. These are found ortho to the carbon with the OH (2H), meta to the carbon with the OH (2H) and para to the carbon with the OH (1H). The hydrogens on the carbon ortho to the one with the OH will each be split by one other hydrogen yielding a doublet. The hydrogens on the carbon meta to the one with the OH will be split twice by one hydrogen each yielding a doublet of doublets. Finally, the hydrogen on the carbon para to the one with the OH will be split once by two equivalent hydrogens affording a final doublet. Hope that helps.

 

[Righty123]

Just a couple of quick foundation questions:

1. How can larger atoms easily "shed" solvent molecules in protic solvents, while their large size also makes them more polarizable?

2. Also why are weak bases considered strong nucloephiles for Sn2 reactions?

3. Why do are larger halogens weaker bases, yet good leaving groups?

 

[Nitya2284]

I haev a question regarding unsaturation numbers

the general formula is (2C+2 - H)/2

In case of oxygen, halogens or nitrogen, what would you do to the general formula for each particular case?

 

[QofQuimica]

1. How can larger atoms easily "shed" solvent molecules in protic solvents, while their large size also makes them more polarizable?

Because polarizability isn't correlated with strength of the dipole interaction. Remember that dipoles are vectors (directional). Large, highly polarizable atoms have "sloppy" electron clouds, with diffuse spread of charge, and these types of ions actually tend to form weaker dipole interactions because the charges can move around.

2. Also why are weak bases considered strong nucloephiles for Sn2 reactions?

Some are and some aren't; you shouldn't make this generalization. For example, water or ammonia are both weak bases, and they are also poor nucleophiles. Just to confuse things more, some strong bases are also good nucleophiles, such as hydroxide or Grignard reagents.

3. Why do are larger halogens weaker bases, yet good leaving groups?

Weaker bases tend to be good leaving groups in general, because they often have to somehow stabilize a negative charge. In the case of the halides, this gets back to your first question about polarizability. Remember that the more that negative charge can be spread around, the more it will be stabilized. So highly polarizable atoms like iodide will stabilize a negative charge well, even though they are less electronegative than fluorine.

 

[QofQuimica]

I haev a question regarding unsaturation numbers

the general formula is (2C+2 - H)/2

In case of oxygen, halogens or nitrogen, what would you do to the general formula for each particular case?

Halogens should be counted like hydrogens, since they also make a single bond to C. You can ignore the oxygens because they have two bonds, and they won't affect the hydrogen count. Nitrogens, however, have three bonds, so you will have to add one more H to the numerator for each N. For example:

CH3CH2OCH3 (ethyl methyl ether) has 8 H's, just as you would predict from the 2N + 2 formula. (N = 3 here)

CH3CH2CH2Cl (1-chloropropane) has 7 H's plus 1 Cl, for a total of 8, as predicted.

(CH3)3N (trimethyl ether) has 9 H's [(2N + 2) + 1 for the nitrogen]

 

[QofQuimica]

For the IR question, is the answer choice # 1?The 1690 wavelenght indicates a carbonyl right? my book tells me that C=O shows up on IR b/t 1725-1750

That's true for aldehydes and esters. Ketones, carboxylic acids and amides show up below 1725; amides in particular come below 1700. Acid chlorides and acid anhydrides show up higher, around 1800. You won't be expected to know that kind of detail for the MCAT, but the overall range is anywhere from 1650-1800 for most carbonyls, with 1700-1750 being the most common range.

 

[challie2385]

What is a good way to remember which groups are electron-withdrawing and which are electron-donating? I see this question come up over and over again on practice MCATs, and still can't seem to get them straight. Thanks for your help!

 

[frany584]

I was doing a Kaplan subject test and the question asks what is the product of Br2 and 2-pentene. It states that the product would produce erythro enantiomers, but i thought to answer this you need to know the configuration of the alkene??

 

[QofQuimica]

What is a good way to remember which groups are electron-withdrawing and which are electron-donating? I see this question come up over and over again on practice MCATs, and still can't seem to get them straight. Thanks for your help!

In general, electron-withdrawing groups are highly electronegative (inductive withdrawers like -CF3) or contain polar multiple bonds (resonance withdrawers like -C=O or -NO2). Electron-donating groups have lone pairs that they can share (resonance donors like -NH2) or they can donate inductively (like -CH3). The trickiest ones to remember are the halogens, which behave somewhat differently than most other atoms with lone pairs (they tend to be overall withdrawers). Try to determine which of those categories a chemical group falls into, and it should help you figure out whether it is a donor or a withdrawer.

 

[QofQuimica]

I was doing a Kaplan subject test and the question asks what is the product of Br2 and 2-pentene. It states that the product would produce erythro enantiomers, but i thought to answer this you need to know the configuration of the alkene??

They are talking about the Br atoms, not the alkyl chains, and that is a confusing way of explaining it IMHO. Since the two Br atoms go on trans to one another, one is pointing down instead of up, and you'll have to rotate half of the product molecule around so that you can view it as a Fisher projection. (Remember that the horizontal substituents should all be pointing up out of the page in a Fisher projection.) When you do that, the two Br atoms will end up on the same side, which is indeed the erythro product. (The opposite product is called threo and would have the two Br's on opposite sides of the Fisher projection.)

 

[frany584]

Thanks for the explanation, their explanation was confusing me!

 

[QofQuimica]

Thanks for the explanation, their explanation was confusing me!

It's understandable. Usually people use erythro and threo to describe Fisher projections showing the stereochemistry of carbohydrates, not of bromination products.

 

[challie2385]

In general, electron-withdrawing groups are highly electronegative (inductive withdrawers like -CF3) or contain polar multiple bonds (resonance withdrawers like -C=O or -NO2). Electron-donating groups have lone pairs that they can share (resonance donors like -NH2) or they can donate inductively (like -CH3). The trickiest ones to remember are the halogens, which behave somewhat differently than most other atoms with lone pairs (they tend to be overall withdrawers). Try to determine which of those categories a chemical group falls into, and it should help you figure out whether it is a donor or a withdrawer.

Thanks Q, for the explanation. What about the constituent -OCH3 on, say, a benzene molecule? At first I would assume this is electron-withdrawing because the O group is electronegative, but it also has lone pairs that make it electron-donating. What property should be prioritized in a case like this?

 

[QofQuimica]

Thanks Q, for the explanation. What about the constituent -OCH3 on, say, a benzene molecule? At first I would assume this is electron-withdrawing because the O group is electronegative, but it also has lone pairs that make it electron-donating. What property should be prioritized in a case like this?

You're correct that you have two opposing forces to take into consideration here. The inductive effect (through sigma bonds) is withdrawal due to oxygen's greater electronegativity. But the resonance effect (through pi bonds) is donation via the lone pair on oxygen. So what wins? For oxygen, the resonance donation does, and you end up with -OCH3 being a donor group. The same is true for nitrogen, with the one important exception being a protonated aniline (-NR3+), which is inductively withdrawing only (no lone pairs left to donate). But for halogens, the tide starts to turn: that is why the halogens are all slightly withdrawing overall, but still direct o/p. They just aren't as good of resonance donors as N and O are.

 

[juiceman311]

Question about carbocation stability;

So the tertiary carbocation is the most stable, now if you have 3 R groups off of that center C, those are all electron withdrawing groups correct? Doesn't that make it less reactive? I don't know if I'm comparing apples and oranges by comparing an alkane's stability to a benzene, but I would think some degree of similarity should exist. Basically, WHY is the tertiary carbocation MORE reactive in Sn1 reactions? Is it because the carbon has high elecrophilic character and thus will reaction with a nucleophile easier? ... I may have just answered my own question, I guess you can't compare apples and oranges, the last thing I said must be correct...but I'll wait for some affirmation from a higher authority.

Thanks

 

[el.harpo]

Hey QofQ, Learfan and Nutmeg,

I was wondering if any of you guys could clarify the concept of the unique sets for H' NMR. How do you look at a molecule and decide how many peaks it will have? The book I'm using says to use mirror planes and rotational symmetry but this doesn't do anything for me. Thanks.

 

[QofQuimica]

Question about carbocation stability;

So the tertiary carbocation is the most stable, now if you have 3 R groups off of that center C, those are all electron withdrawing groups correct?

No. Alkyl groups are inductive electron donors, not ewgs. That is why they stabilize carbocations, which are electron deficient.

Basically, WHY is the tertiary carbocation MORE reactive in Sn1 reactions?

It isn't. Tertiary carbocations are more stable and therefore LESS reactive than secondary or primary carbocations. But the problem is that secondary carbocations are unstable to the point that it is difficult to form them long enough for them to be able to react, and primary carbocations are even worse. Tertiary carbocations are still relatively reactive, but they are stabilized enough that they exist long enough for a reaction to occur.

 

[QofQuimica]

Hey QofQ, Learfan and Nutmeg,

I was wondering if any of you guys could clarify the concept of the unique sets for H' NMR. How do you look at a molecule and decide how many peaks it will have? The book I'm using says to use mirror planes and rotational symmetry but this doesn't do anything for me. Thanks.

In general, each chemically unique proton will have its own peak. So you should count the unique chemical groups when you look at the molecule. I think your book is trying to tell you that if you have a molecule that is symmetrical, the two sides of it are not chemically or magnetically unique, so they will be represented by a single peak. For example, neopentane

C(CH3)4

would only have one peak on proton NMR, since all four methyl groups are chemically and magnetically equivalent.

 

[N1DERL&]

Hi Q!! *wave*

Could you explain the Benedict's test for me please? For some reason, I just can't get mind around it!

Also, do we need to memorize the monosaccharides and their linkages to: lactose, sucrose, maltose, cellobiose?

Thank you!!

 

[QofQuimica]

Hi Q!! *wave*

Could you explain the Benedict's test for me please? For some reason, I just can't get mind around it!

Also, do we need to memorize the monosaccharides and their linkages to: lactose, sucrose, maltose, cellobiose?

Thank you!!

Well, well, well, look who the panda dragged in. I haven't seen you in ages. Are you taking the test next week?

Benedict's test is a redox reaction that is used to test for reducing sugars. The Benedict reagent has copper (II) in it, which is reduced to copper (I) oxide while the sugar is oxidized up to the acid. Only sugars with aldehydes in them (like glucose) or with ketones that can tautomerize to aldehydes (like fructose) will undergo the reaction. Here's an illustration; the Cu(II) ion is blue, but the Cu2O ppt. is dark red:

I would not bother to memorize the monosaccharides' structures.

 

[juiceman311]

No. Alkyl groups are inductive electron donors, not ewgs. That is why they stabilize carbocations, which are electron deficient.



It isn't. Tertiary carbocations are more stable and therefore LESS reactive than secondary or primary carbocations. But the problem is that secondary carbocations are unstable to the point that it is difficult to form them long enough for them to be able to react, and primary carbocations are even worse. Tertiary carbocations are still relatively reactive, but they are stabilized enough that they exist long enough for a reaction to occur.

Thank you so much!! Yeah, I realized about 30 min after I posted I said the wrong thing about the R group being donating...Again thanks for the clarification, makes perfect sense.

 

[freesolo]

really simple question ive been missing free points on this, how many stereocenters do compounds have? I know that a stereocenter is a carbon connected to four diff groups, but sometimes i see a carbon in a ring connected to two CH2 groups, and than a Cl and maybe a CH3, after the CH2 groups there are C with other groups attatched to them. But i thought we only look at what the stereocenter is directly connected to. I think this is a confusing question, if you get what im asking please help lol.

 

[hippocampus]

what is the difference between a nucleophilic addition and an electrophilic addition?

 

[QofQuimica]

really simple question ive been missing free points on this, how many stereocenters do compounds have? I know that a stereocenter is a carbon connected to four diff groups, but sometimes i see a carbon in a ring connected to two CH2 groups, and than a Cl and maybe a CH3, after the CH2 groups there are C with other groups attatched to them. But i thought we only look at what the stereocenter is directly connected to. I think this is a confusing question, if you get what im asking please help lol.

I think what you are asking is how far you have to go out before you can stop looking for differences between two groups attached to an asymmetric carbon that join to form a ring, and the answer is that you have to go out all the way to the other end of the ring. So if you have a ring, and there is one carbon in it with two CH2 groups, a Cl, and a methyl attached to it, that carbon will be achiral if the rest of the ring is symmetrical (i.e., 1-chloro-1-methylcyclohexane is not chiral). It will be chiral (a stereocenter) if there is a substituent at the 2 or 3 position on the ring (i.e., 1-chloro-3-methylcyclohexane is chiral-there are actually TWO stereocenters), but again achiral if the substituent is at the 4 position (i.e., 1-chloro-4-methylcyclohexane is NOT chiral, because it would be symmetrical.) If you have trouble picturing this, draw out the molecules and you will see what I mean about the symmetry.

 

[QofQuimica]

what is the difference between a nucleophilic addition and an electrophilic addition?

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.

 

[frankrizzo18]

Which molecule is more polar? butane or pentane? I thought it should be pentane because the end carbons will be inductively donating in a northward direction (or southward) together!!

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.

 

[QofQuimica]

Which molecule is more polar? butane or pentane? I thought it should be pentane because the end carbons will be inductively donating in a northward direction (or southward) together!!

Is this a practice MCAT question??? I wouldn't call either one of them polar. Pentane will have greater van der Waals forces because it's larger. Maybe that's what you meant?

 

[frankrizzo18]

I believe the four answer choices were:
A) Pentane, due to inductive donation
B) Butane, due to inductive donation
C) Pentane, due to better sigma overlap
D) Butane, due to better sigma overlap

It was just there when I took it, and I was like oh, Pentane would have a higher rf value because it is more polar, but that did not really help me!!

Is this a practice MCAT question??? I wouldn't call either one of them polar. Pentane will have greater van der Waals forces because it's larger. Maybe that's what you meant?

 

[QofQuimica]

I believe the four answer choices were:
A) Pentane, due to inductive donation
B) Butane, due to inductive donation
C) Pentane, due to better sigma overlap
D) Butane, due to better sigma overlap

It was just there when I took it, and I was like oh, Pentane would have a higher rf value because it is more polar, but that did not really help me!!

Hmm. I'm not really sure what to make of this question. Maybe it is trying to get you to look at the extra carbon as an alkyl substituent? I.e., the fourth carbon of butane is R-CH3, while the fourth carbon of pentane is R-CH2-R'? I'm guessing this because of the two answer choices that talk about inductive donation; the pentane would have greater inductive donation at carbon 4 than the butane would at carbon 4. I don't think that there would be a signficant difference in the orbital overlap between the two, so I'm inclined to rule out choices C and D. Is there a solution given?

On a sort of related note, more polar substances have LOWER Rf values on TLC or CC, unless you are doing reverse phase. But RP chromatography is definitely beyond the scope of the MCAT.

 

[GCT]

pentane and butane are not polar, as you may have learned in the classroom they are considered non-polar. Although the higher molecular weight nonpolar compounds may dissolve in water to an appreciable extent.; one may find it interesting though to consider the relative induced dipoles for butane and pentane, one may wish to consider the time range, distance between dipoles, although boiling point and intermolecular attraction measures may not be so direclty relevant.

 

[frankrizzo18]

I can not reasonably get the solution. If I say any more I may get in trouble. I may have remembered the question incorrectly but when I saw it, I remeber picking pentane and inductive donation because the two end carbons would be inductively donating in a direction northward or southward! But, I remembered the question because It seemed like a brainteaser. (For the rf value I said it backwards, sorry)

Hmm. I'm not really sure what to make of this question. Maybe it is trying to get you to look at the extra carbon as an alkyl substituent? I.e., the fourth carbon of butane is R-CH3, while the fourth carbon of pentane is R-CH2-R'? I'm guessing this because of the two answer choices that talk about inductive donation; the pentane would have greater inductive donation at carbon 4 than the butane would at carbon 4. I don't think that there would be a signficant difference in the orbital overlap between the two, so I'm inclined to rule out choices C and D. Is there a solution given?

On a sort of related note, more polar substances have LOWER Rf values on TLC or CC, unless you are doing reverse phase. But RP chromatography is definitely beyond the scope of the MCAT.

 

[medworm]

I seem to recall that certain reagents act as Nucleophile, but not sure which ones. I'd like to compile a short list so I don't have to waste time pondering later while taking the test. Help? Thanks.

Good luck everyone!

 

[QofQuimica]

I can not reasonably get the solution. If I say any more I may get in trouble. I may have remembered the question incorrectly but when I saw it, I remeber picking pentane and inductive donation because the two end carbons would be inductively donating in a direction northward or southward! But, I remembered the question because It seemed like a brainteaser. (For the rf value I said it backwards, sorry)

It *is* definitely an odd question. I'm not sure how much I've been able to help you, but if it's any consolation, I can't imagine that you'd run into something like this on the MCAT. They'd be much more likely to ask you to compare pentane with butanol rather than butane in terms of polarity. They might ask you to do something like compare the bp of pentane versus butane though.

 

[QofQuimica]

I seem to recall that certain reagents act as Nucleophile, but not sure which ones. I'd like to compile a short list so I don't have to waste time pondering later while taking the test. Help? Thanks.

I would advise you against this strategy, as there are literally infinite different nucleophiles out there. Rather, you are better off if you understand what a nucleophile is, so that you can identify them in any reaction that may be thrown your way. In general, nucleophiles are electron-rich species that contain lone pairs (i.e., Lewis bases) or nonpolar multiple bonds (alkenes, alkynes, and arenes) that they can donate to the electrophile, which is electron-deficient.

Here is a brief explanation of nucleophilicity that I posted previously:

Nucleophilicity is a measure of how polarizable an atom's or ion's electron cloud is. You can think of good nucleophiles as having relatively "sloppy" electron clouds. Nucleophilicity does not really follow basicity trends, except that if you are comparing two nucleophiles of the same element, the one with higher pKa will be more nucleophilic. (For example, both ethoxide and acetate are oxygen nucleophiles. Ethoxide has a much higher pKa than acetate, so it will be the better nucleophile. Note that you cannot do this comparison for two nucleophiles of different elements, such as acetate versus iodide. Acetate has a much higher pKa than iodide, but it is a much weaker nucleophile.) Good nucleophiles tend to be large and not too highly charged, or with the charge spread diffusely. They are not required for E1 or Sn1, but it is essential to have a good nucleophile for Sn2 reactions.