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.

Short answer: Branching will decrease mp and bp. Condensing and freezing happen for alkanes because of dispersion forces caused by temporary, induced dipoles, which are the only intermolecular forces holding nonpolar molecules like alkanes together in the liquid and solid states. Branched alkanes have smaller dispersion forces compared to straight-chain alkanes of the same MW, so they will be harder to liquify or freeze.

Long answer: Let's imagine that we are cooling down a gaseous alkane like hexane that is straight-chained, versus another of the same MW that is branched, like 2,3-dimethylbutane. Remember that gases don't have any intermolecular interactions, at least not if they're ideal. As we lower the temperature, the molecules stop moving as much, and they begin to have intermolecular interactions that are due to induced, temporary dipoles.

What is an induced, temporary dipole? Well, the electrons in a bond or in an atom are not stationary. They are constantly in motion. Sometimes, by sheer chance, they are not equally distributed on the atom or in the bond, and this inequality of charge forms a dipole. It is temporary because soon thereafter, as the electrons continue to move, they distribute more equally again, and so the dipole goes away. But in the meantime, that short-lived dipole has affected its neighbors, and it has caused similar dipoles to form in them as well. This is why these dipoles are induced. So you can imagine a bunch of molecules, held together by small dipoles that are constantly forming and unforming. But there are always some dipoles present at any given time, and that is what holds the liquid or solid together.

Ok, so now we need to consider what kinds of molecules will have the strongest induced dipoles. The strength of the induced dipoles is directly proportional to the surface areas of the molecules that are coming into contact. This is intuitive, because if contact can be made over a greater area, there is a greater chance that electrons will distribute unequally at some point over that surface, causing the temporary dipole and inducing dipoles in the neighboring molecule. You may know that the shape with the smallest surface area-to-volume ratio is a sphere. So molecules that are more spherical (ie, highly branched) do not have very much surface area relative to molecules that are long and extended (straight chains). That is why branched molecules have weaker dispersion forces versus straight-chains. Since they have weaker dispersion forces, the branched molecules will tend to want to stay in the gaseous phase longer, and you'll have to cool them further to force them to condense into a liquid. This means that branched compounds have a lower bp (condense at a lower temperature) versus straight chains.

As we continue to cool, the molecules continue to move closer and closer together, and their interactions continue to increase. Eventually, we reach a point where they begin to crystallize, or at least form an amorphous solid. So we need to consider how well the molecules pack together at this point. Branched compounds are like little spheres, or like porcupines. It's hard to get them to pack well, and this means that you will have to cool them to a lower temperature to freeze them (lower mp) compared with straight chains, which can stack up nicely, more like a cord of firewood. So the mp of a branched compound will be lower than that of a straight chain, assuming that they have the same MW.

There is an exception to this melting point trend for highly branched, symmetrical isomers (http://forums.studentdoctor.net/showpost.php?p=2815590&postcount=64).

That is often true, although if your substrate (alkyl halide) is primary, you will probably get mostly SN2 product, unless your base is bulky. Two things should immediately tip you off that a reaction goes by E2: one is a strong, bulky base like t-butoxide or LDA (lithium diisopropylamine), and the other is the heat symbol, which looks like a triangle (it's the greek letter delta). Heat tends to favor eliminations over substitutions in general. If the substrate is secondary and you mix it with a strong base, you may still get some SN2 product, but E2 will probably predominate. And if it's tertiary, E2 is the only possibility.

Question:Which of the following is one of the products of a reaction between propyl-magnesium bromide and ethyne?

A. 1-pentyne
B. propane
C. 1-pentene
D. propene

You are trying to mix two electron-rich species here (a Grignard with an alkyne) so you cannot do any kind of nucleophilic attack on the alkyne using the Grignard. We spend a lot of time in organic chemistry focusing on how Grignard reagents are such good nucleophiles, which they are. However, they are also very powerful bases; the pKa of propyl magnesium bromide is estimated to be about 50. (In comparison, the pKa of NaOH, a common strong inorganic base, is about 16.)The protons of acetylene (the common name for ethyne; no one ever calls it ethyne!) are fairly acidic for an organic compound; they have a pKa of about 25. Remember that each time you go up or down a pKa unit, that is a decrease or increase of 10 TIMES in acidity. So, since the difference in pKa between this Grignard and acetylene is ~25 pKa units, we are talking about an acidity difference of 10 ^ 25, and you are gonna get one heck of an acid-base reaction when you mix the two. The Grignard gets protonated, forming propane, and the acetylene gets deprotonated, forming the anion. The correct answer is B (propane).

This is how I learned to do it as an undergrad, although if you find it confusing, there's no need for you to change your current method. If you want to use this technique, what you do is align your thumbs (of both hands) in the same direction of the fourth priority group (which isn't always H, incidentally) and then curl your fingers. If the priorities go in the direction of the fingers on your right hand, the configuration is R. If they go in the direction of the fingers on your left hand, the configuration is S. The benefit of this method is that you don't have to worry about swapping groups if your low-priority group is sticking out at you; you merely need to point your thumbs outward, and the technique still works just as easily as if it were pointing away.

This mnemonic comes to us courtesy of Turkeyman.

So you have uncle Ben(Benzene), and he's a lazy leech.
If you lend him money to go party(electron donating), he'll love you, and want you sit across or next to him at the dinner table (ortho/para).
If you don't lend him any money, he'll hate you and want at least 1 person between you and him (meta).

One exception is that Uncle Ben respected Uncle Hal (halides) even if he didnt give Ben any electrons, because of his character. So, in effect, he would always let Uncle Hal sit across or next to him at the dinner table.

Question: in gas chromatography, does the area under the curve give the amount of the substance?

It gives the relative proportions. So, for example, you could conclude based on the peak areas that you have 60% of Compound 1 and 40% of Compound 2, but you could not conclude that you have exactly 6 mg or 6 g or 6 kg of Compound 1 from GC (unless you have also injected a known amount of a GC standard, but that is more complex than you need to know for the MCAT). Although, you can rule out the larger amounts I gave just because it would be impossible to inject that much onto the column.

Question: why does a compound with a higher boiling point have a longer retention time?

GC separates compounds based on two main characteristics: boiling point, and to a lesser extent, polarity. The columns used for GC are a bit different than the ones used for liquid chromatography; the stationary phase in GC is actually a viscous liquid, and the columns are typically not packed, but rather are "open tubular". That means there is an open space in the center for the gas (your mobile phase) to go through. The column is very long (30 m....that's METERS....is a common length) and it is coiled and placed into an oven. The temperature is raised over time, typically from RT to some other moderate temperature. (You can't raise it too much, or you'll destroy the column and/or decompose your compound.) After injection, all of the compounds are vaporized before entering the column. Compounds that are low-boiling remain vaporized and travel rapidly through the column. Compounds that are high-boiling adsorb to the stationary phase more readily, and therefore come off later. If the two compounds have the same boiling point, they might still be separated if they have different affinities for the stationary phase (different polarities).

In general, it is best to think of chromatography as an equilibrium. Both the stationary and mobile phases are "competing" for the compounds. Some compounds prefer the stationary phase, and others prefer the mobile phase, but it is not a winner-takes-all scenario. So, in GC, once the compound vaporizes, it is carried a ways by the gas, then it adsorbs to the stationary phase, unsticks and travels a bit further, sticks again, etc., etc., all down the length of the column.

You should know the common common names like benzene, acetyl, and formyl. Knowing the butyl isomers is probably helpful also. But you do not need to memorize extensive lists of common names. BTW, these common names are not "out of date"; IUPAC grandfathered many of them in, so it is actually acceptable to use either name (ex. acetone versus propanone) under IUPAC rules.

Question: When I do an extraction, how do I know which layer is on the top and which one is on the bottom?

Figuring out which layer is which in a separatory funnel is often confusing for students. Basically, the denser layer is always the one on the bottom. This is true regardless of whether the denser layer is aqueous or organic. Many organic solvents, such as hexanes, ethyl acetate, and diethyl ether, are less dense than water. These solvents will form the top layer when in a separatory funnel with an aqueous solution. But there are some organic solvents, like dichloromethane and chloroform, that are denser than water. These solvents will form the bottom layer when mixed with an aqueous solution in a separatory funnel.

The density of pure water is 1 g/cm3. (You should memorize this for the MCAT as well as your organic lab!) Knowing this, you can determine which layer will be on top if you look up the extraction solvent's density before you go to class. Alternatively, if you are in the middle of an extraction and you just aren't sure, you can add a drop of water to your sep funnel, and see into which layer it dissolves.

My best recommendation for organic lab students is this: When in doubt, never throw away any of your layers until you are absolutely certain that you won't need them any more.

When distinguising between enantiomers and diastereomers can I assume that if there are two chiral centers that the molecule in question is diastereomers and likewise if there is one than they are enantiomers?

When distinguising between enantiomers and diastereomers can I assume that if there are two chiral centers that the molecule in question is diastereomers and likewise if there is one than they are enantiomers?

No. A molecule with two chiral centers has one enantiomer and also two diastereomers, assuming that none of the four isomers is a meso compound. So if you have two or more chiral centers in your molecule, and you are asked whether it and one of its isomers are enantiomers or diastereomers, what you must do is look at the absolute configuration of every stereocenter. A pair of enantiomers will be inverted at EVERY stereocenter. If even one stereocenter has its configuration retained, then they are a pair of diastereomers.

For example, if your first molecule is (R,R), then its enantiomer will be (S,S) and its two diastereomers will be (R,S) and (S,R). Note that only the (S,S) isomer has both stereocenters inverted from the original (R,R). The (R,S) and (S,R) isomers are diastereomers of the (R,R) isomer because they each have one stereocenter that is inverted from the original, but another that retains the R configuration.

What percent, approximately, of MCAT problems asking for a Lewis structure as an answer, can be solved by doing nothing but eliminating answer choices that have the wrong charge, and choosing a radical if one's called for while eliminating radicals if they're not called for? What about elminating structures that are entirely unlike what's called for, e.g., we're asked for an intermediate structure but the answer choice adds an extra carbon or something. In other words, how often can I get the problem right while knowing essentially nothing?

How many problems does an average MCAT have that ask for Lewis structures?

What NMR resonances do I absolutely have to know? What other experimental values do I absolutely have to know?

You should know the common common names like benzene, acetyl, and formyl... But you do not need to memorize extensive lists of common names.
I had some trouble on one of my MCATs because I didn't know what acetate was. (Yeah, not a common problem among premeds, I know.) Could you eventually give a complete list of the molecules and functional groups that I have to know for the test?

I had some trouble on one of my MCATs because I didn't know what acetate was. (Yeah, not a common problem among premeds, I know.) Could you eventually give a complete list of the molecules and functional groups that I have to know for the test?

Yes, we will do that. I would like to set up an organic topics and explanations thread eventually, as we already have for gen chem and physics.

An acetyl group is merely CH3C=O (methyl attached to a carbonyl). Acetate, then, would have another O attached to the other side of the carbonyl (CH3COO-). It's the conjugate base of acetic acid (CH3COOH). Formyl is very similar to acetyl; it's HC=O (a proton attached to a carbonyl). These two groups are nearly always called by their common names, and many other common names contain them in both organic and biochem (ex. acetyl Co A), so it's smart to learn them.

What percent, approximately, of MCAT problems asking for a Lewis structure as an answer, can be solved by doing nothing but eliminating answer choices that have the wrong charge, and choosing a radical if one's called for while eliminating radicals if they're not called for? What about elminating structures that are entirely unlike what's called for, e.g., we're asked for an intermediate structure but the answer choice adds an extra carbon or something. In other words, how often can I get the problem right while knowing essentially nothing?

How many problems does an average MCAT have that ask for Lewis structures?

In my experience, Knowledge of lewis structures will be either required or helpful in solving two to three problems on the test. You will need to know enough to actually work the problem, not just ensure you are taking a reasonable guess.

What NMR resonances do I absolutely have to know? What other experimental values do I absolutely have to know?

You should know broad ranges rather than memorizing specific numbers, because many groups can move up or downfield depending on what else is present in the molecule. I would recommend knowing the following NMR shift ranges:


aromatic protons (attached to a benzene ring): ~7-8 ppm
vinyl protons (attached to a double bond): ~5.5-6.5 ppm
protons on a C attached to an electronegative atom like O, N, or a halogen: ~3-4 ppm
protons on a C alpha to a carbonyl: ~2-3 ppm
most other aliphatic protons: ~0.5-2 ppm
aldehyde protons (proton directly attached to the carbonyl): ~9-10 ppm
carboxylic acid protons (attached to the O in the COOH): ~10-12 ppm


You should also know these common IR stretches:

-OH or -NH stretch: broad, around 3300 wavenumbers
carbonyl C=O stretch: sharp peak around 1675-1750 wavenumbers
aromatic and double bond C=C stretch: around 1400-1650 wavenumbers


Most others would be given to you in the passage. I will be posting more info about HNMR and how to interpret it in a later post.

What NMR resonances do I absolutely have to know? What other experimental values do I absolutely have to know?

You will need to know the shifts for protons on an aliphatic backbone, aromatic ring, alpha to a ketone, alpha to an ester, on the carbon or alpha to the carbon with a halogen, on an acetylenic functional group, on an amine, on an aldehyde and on or alpha to a double bond. In addition, I would understand the splitting patterns associated with nearby protons on ajacent carbon atoms both aliphatic and aromatic. Know the principal of Pascals numerical triangle for split intensities. I would also know the carbon shifts for aliphatic carbons, ketonic carbons, aromatic carbons, ester carbons, carbons with an alcohol group, olefinic carbons, carboxylic acids, acetylenic carbons and nitriles. Understand the spliting patterns in a coupled vs. a proton decoupled carbon spectrum. Finally, I would know the positions on the IR chart which are strongly suggestive if not diagnostic of several functional groups such as amine, alcohol, nitrile, acid, ester, ketone, double bond, triple bond and aliphatic hydrogen. Hit the charts until it becomes second nature to look at the spectrum and immediately grasp what functional groups are likely to be present. In addition, I would have an understanding of the chromophores that are likely to interact with UV light.

i am having a problem determining absolute configurations in rings, but i get the idea for the most part when not in rings. anyway, i am looking at a pentose ring with two deutrium's coming out of the two bottom verticies, one D each. I am having trouble assigning priority because it keeps coming out as R for me, but according to this book its S. Any help would be appreciated.

i was looking under topic for organic and i found no alkenes or alkynes i knew of alkynes no longer being on there no alkenes either? So i can skip a whole chapter in kaplans book about elimination, hydroboration ozonolysis, potassium permanganate ect?

i was looking under topic for organic and i found no alkenes or alkynes i knew of alkynes no longer being on there no alkenes either? So i can skip a whole chapter in kaplans book about elimination, hydroboration ozonolysis, potassium permanganate ect?

That does not sound like good idea. You need to know both the electrophillic addition mechanisms, the elimination machanisms and trends in attacking species for double and triple bonds as well as their hydrogen and carbon spectral characteristics. This is all basic ochem. It may very well be tested. Not knowing these things would be to your detriment.

:(

i cant find them under aamc topics maybe im blind lol

Learfan, are you sure? I thought AAMC had announced that alkenes and alkynes would no longer be tested.

No, this isn't my field, and I may have it wrong, but I'm pretty sure this is what TPR teaches us.

Learfan, are you sure? I thought AAMC had announced that alkenes and alkynes would no longer be tested.

No, this isn't my field, and I may have it wrong, but I'm pretty sure this is what TPR teaches us.

Sorry, I may be an outdated fossil. I thought that such concepts were so basic to the practice of organic chemistry that there was no means to understand a significant portion of the science without comprehending the reaction chemistry and spectroscopy of alkenes. I can understand why triple bonds are no longer tested but alkene chemistry??????? As of 2004, Kaplan was still covering the material in their large review book.

Obviously AAMC is the real authority so their word is law. Guess I am getting old and out of date. Someone come visit me in the old chemists home. Speak up sonny, I cant hear very well anymore you know.

yep they still have a whole chapter dedicated to it,
i've noticed org II topics alot rather tha org I.

i was looking under topic for organic and i found no alkenes or alkynes i knew of alkynes no longer being on there no alkenes either? So i can skip a whole chapter in kaplans book about elimination, hydroboration ozonolysis, potassium permanganate ect?

Although the AAMC claims not to test alkenes on the MCAT, this is not really true. For example, you need to know about all of the following topics:


E1 and E2 reactions (produce alkene products)
addition reactions (start with an alkene reactant)
geometric isomerism (cis and trans or E and Z diastereomers around a double bond)
fatty acids (kinked cis bonds produce oils while straight trans bonds produce solid fats)


I'm sure that you guys can come up with some more ideas besides these. But the main point is that alkenes are an essential part of organic chemistry, and they DO show up, one way or another, on many AAMC topics that you need to know for the test. So I'd recommend that you know the basics about them.

I have found an easy way to remember which way is the R configuration and which was is the S configuration.

When you draw the letter R, when you draw the loop of the top of the R you move in the CLOCKWISE direction, therefor R=clockwise. When you draw the S, the top of the letter loops down pointing in the COUNTERCLOCKWISE direction, therefor S=counterclockwise.

Any comments?? :thumbup: :thumbdown ?

I have found an easy way to remember which way is the R configuration and which was is the S configuration.

When you draw the letter R, when you draw the loop of the top of the R you move in the CLOCKWISE direction, therefor R=clockwise. When you draw the S, the top of the letter loops down pointing in the COUNTERCLOCKWISE direction, therefor S=counterclockwise.

Any comments?? :thumbup: :thumbdown ?

Works perfectly, as long as you add the caveat that the fourth priority substituent must be pointing away from you. If it's pointing out toward you, then the designations will be reversed.

I have another trick. If you heard this one, sorry.

Because the reaction conditions are the same except for one factor how do you decide if the reaction is Sn1 or E1?

if you add hEat!! heat drives the reaction to E1.

I have another trick. If you heard this one, sorry.

Because the reaction conditions are the same except for one factor how do you decide if the reaction is Sn1 or E1?

if you add hEat!! heat drives the reaction to E1.

Yes, heat does generally favor eliminations over substitutions. Keep in mind, though, that you usually will get a mix of SN1 and E1 products rather than clean reactions of just one or the other. This occurs because these two mechanisms have the exact same rate-determining step (carbocation formation), and your solvent can typically act as both a base and a nucleophile.

alright im getting overwhelmed with all these reactions some of which we didnt even talk about! Should i just memorize that this reagent does this and this does this? Like PCC ive never even heard of. Will the mcat tell you what the reagent does?

alright im getting overwhelmed with all these reactions some of which we didnt even talk about! Should i just memorize that this reagent does this and this does this? Like PCC ive never even heard of. Will the mcat tell you what the reagent does?

You don't need to memorize many reagents for the MCAT, but PCC is one that I'd suggest that you do learn. It stands for pyridinium chloro chromate, and it's a mild oxidizing agent that will convert primary alcohols to aldehydes, and secondary alcohols to ketones. PCC is special because it stops at the aldehyde; most oxidizing agents, like KMnO4, Jones reagent (CrO3/H+) and dichromate (Cr2O7) will oxidize primary alcohols all the way up to the carboxylic acid. I suggest that you memorize all four of these oxidizing agents.

You should also learn two reductive reagents: LAH (lithium aluminum hydride) and NaBH4. LAH is a strong reducing agent, and it will reduce any carbonyl (usually to the alcohol, except for amides, which are reduced to amines). NaBH4 is a milder reducing agent that will reduce aldehydes and ketones but not esters, amides, or carboxylic acids.

I would say that if you know those half dozen oxidation and reduction reagents, you should be fine for most MCAT questions.

Q,

I have a few questions for a friend of mine.........

1. How does the valence configuration of an atom, in terms of s and p orbitals, relate to its hybridization in terms of s and p character?????

How does the valence configuration of an atom, in terms of s and p orbitals, relate to its hybridization in terms of s and p character?????

I think your friend is confusing two concepts: the order of orbital filling in a lone atom (Aufbau principle) and the process of combining p and s orbitals to make hybrid orbitals in an atom that is bonded to other atoms in a molecule. There is not a direct relationship between the number of valence electrons in an atom and orbital hybridization; for example, a carbon atom can be sp, sp2, OR sp3 hybridized, but it still has four valence electrons in each case, right?

One thing that students should be careful about as far as hybridization is concerned is that elements in row 2 (carbon, nitrogen, oxygen, and fluorine) must NEVER exceed their octets. That is, they should never have more than four bonds. Elements in row 3 and lower do on occasion exceed their octets, and they can do this by bringing in d-orbitals when they hybridize. Thus, phosphorus (but never nitrogen) can use one of its 3d orbitals to make five sp3d hybrids (leading to five substituents and trigonal bipyramidal molecular geometry), and sulfur (but never oxygen) can use two 3d orbitals to make six sp3d2 hybrids (leading to six bonds and octahedral geometry).

pi bonds are highly electronegative; since they cant rip of a neighbouring electron they still 'suck' some negative charge...this makes the alkhene more stable. The more substituted the double bond the more stable the alkhene since the doulbe bond is inductively electron withdrawing....uhhh...WHAT DOES ALL THAT MEAN?...i know its quite elementary; but i'm a bit unclear about 'stability' itself...i dont have a complete idea...someone help! ...also, if someone could explain the trends in carbocation,alkhyl groups (and any other such trends relating to how sub'd a carbon can be) (ie. tertiary>secondary>primary)..THANKS in ADVANCE!

Is there any of making the memorization of chemical shift ranges /IR absorption peaks EASIER? Is there some kind of rationale that could be applied? thanks.

Is there any of making the memorization of chemical shift ranges /IR absorption peaks EASIER? Is there some kind of rationale that could be applied? thanks.

There are not very many NMR shifts or IR absorptions that you need to know for the MCAT. Basically, for NMR, the protons that are having electron density withdrawn from them (they are connected to electronegative atoms or electron withdrawing groups) will tend to appear further downfield (larger ppms). For IR, the peaks correspond to the energies of the stretching and bending of bonds; if you only know the carbonyl and -OH/-NH stretches, you should be fine (though I suggest knowing C=C bonds as well). For my complete list of suggested NMR and IR absorbances to know, see this post: http://forums.studentdoctor.net/showpost.php?p=2721706&postcount=17

:) Can anyone do a E1/SN1 and E2/SN2 breakdown summary - it is always so hard to get this info together. I get so confused and would love to be able to have this all in one post. Thanks in advance!

pi bonds are highly electronegative; since they cant rip of a neighbouring electron they still 'suck' some negative charge...this makes the alkhene more stable. The more substituted the double bond the more stable the alkhene since the doulbe bond is inductively electron withdrawing....uhhh...WHAT DOES ALL THAT MEAN?...i know its quite elementary; but i'm a bit unclear about 'stability' itself...i dont have a complete idea...someone help! ...also, if someone could explain the trends in carbocation,alkhyl groups (and any other such trends relating to how sub'd a carbon can be) (ie. tertiary>secondary>primary)..THANKS in ADVANCE!

I don't exactly understand your first sentence, particularly the part about ripping and sucking electrons....I think you mean that sp2-hybridized carbons are more electronegative than sp3-hybridized carbons? This is true because of the sp2-hybrids' greater s-character. The more s-character a hybridized atom's orbitals have, the more electronegative that atom will be. This is because having greater s-character means that the electrons spend more time close to the nucleus on average, and electrons want to be close to the nucleus b/c it is positively charged. Thus, an sp-hybridized carbon is more electronegative than an sp2-carbon, and an sp2-hybridized carbon is more electronegative than an sp3-carbon.

To answer the second part of your question, you need to understand that alkyl groups are electron donating groups. We can't draw resonance structures very easily to show how alkyl groups donate electron density like we can for pi donors, though, because alkyl groups are sigma donors. This means that they donate inductively, through a sigma bond (i.e., a bond dipole, with the positive end on the alkyl group and the negative end pointing to the double bond or carbocation.) The reason that this happens is that both double bond carbons and carbocation carbons are sp2-hybridized, but the alkyl group carbons are sp3-hybridized. So, since the sp2-hybrids are more electronegative than the sp3-hybrids, they can pull the bond electron density disproportionately toward themselves, in effect forcing the sp3-hybridized alkyl substituents to donate electron density to them. This inductive donation stabilizes the sp2-hybridized atoms because it helps offset or neutralize their relatively large electronegativity.

Hope this helps.

:) Can anyone do a E1/SN1 and E2/SN2 breakdown summary - it is always so hard to get this info together. I get so confused and would love to be able to have this all in one post. Thanks in advance!

Yes, we will add this to the list.

Does anyone have a quick trick to know which acids are strong and which are weak? Same for bases??

Does anyone have a quick trick to know which acids are strong and which are weak? Same for bases??

Yes. http://forums.studentdoctor.net/showpost.php?p=2711972&postcount=6

What is an azeotropic mixture and where is this concept likely to come up on the MCAT? A few examples of such mixtures would be helpful as well.

Much appreciated

What is an azeotropic mixture and where is this concept likely to come up on the MCAT? A few examples of such mixtures would be helpful as well.

Much appreciated
An azeotropic mixture is one where both constituents evaporate at the same rate--hence, it cannot be distilled beyond the azeotropic point.

A common example is ~95% ethyl alcohol in 5% water. You cannot distill ethanol to be pure ethanol, because once it gets to the 95% purity, further distilation makes vapor that has the same composition as the liquid.

I don't expect it will come up on the MCAT, but it is fair game. The main thing you need to understand is that the vapor and liquid have the same component percentages (whereas if you boil wine, the vapor has more ethanol than the liquid).

An azeotropic mixture is one where both constituents evaporate at the same rate--hence, it cannot be distilled beyond the azeotropic point.

A common example is ~95% ethyl alcohol in 5% water. You cannot distill ethanol to be pure ethanol, because once it gets to the 95% purity, further distilation makes vapor that has the same composition as the liquid.


You actually CAN separate an azeotrope further, but it requires adding something that can break the azeotrope. To use Nutmeg's example, it is possible to distill 100% ethanol, but you have to add benzene to the 95% ethanol in order to do it. Life lesson: don't drink 200 proof ethanol; it still has traces of benzene in it. :eek:

You actually CAN separate an azeotrope further, but it requires adding something that can break the azeotrope. To use Nutmeg's example, it is possible to distill 100% ethanol, but you have to add benzene to the 95% ethanol in order to do it. Life lesson: don't drink 200 proof ethanol; it still has traces of benzene in it. :eek:
Well, if we're going to get technical, once you add the entraining agent, the solution is no longer an azeotrope.
:p ;) :D

I gotta stop this line of conversation. I'm getting flashbacks of trying to compile my ASPEN file to distill pure ethanol for a fuel additive as part of my final project for Process Design. It was twice as much work to get from 95% to 100% as it was to get from 10% beer to 95%. :scared: :scared: :scared:

http://www.pen.k12.va.us/Div/Winchester/jhhs/math/humor/comics/computer/coderage.jpg

But is there someway or somewhere to get a list of major reactions? Just general formulas of things like


Esterification, EAS, Sn1, Sn2, E1, E2, Alcohols, Hydrogenation, Hydroboration, Hydration, etc.

Also, what are some key reactants that will trigger you to realize what type of reaction it is? Examples might include PCC or LAH, heat, acid, base, O3...

Thanks for any help.

am i correct in assuming, that based on the topic guide from mcat, that we dont need to be able to identify aromatic cmpds?

But is there someway or somewhere to get a list of major reactions? Just general formulas of things like


Esterification, EAS, Sn1, Sn2, E1, E2, Alcohols, Hydrogenation, Hydroboration, Hydration, etc.

Also, what are some key reactants that will trigger you to realize what type of reaction it is? Examples might include PCC or LAH, heat, acid, base, O3...

Thanks for any help.

We plan to add a topics in organic chemistry thread in the future. For now, however, I suggest that you visit Dr. Alfa Diallo's mcatpearls site (http://www.mcatpearls.com).

am i correct in assuming, that based on the topic guide from mcat, that we dont need to be able to identify aromatic cmpds?

Do you mean whether you need to understand what aromaticity is? I would suggest that you do learn the four rules for identifying an aromatic compound:

1. it's cyclic
2. it's conjugated (p-orbital on each atom in the ring)
3. it's planar
4. it obeys Huckel's rule (4n + 2 electrons) where n is an integer greater than or equal to zero.

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?

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 (http://forums.studentdoctor.net/showpost.php?p=2773903&postcount=5). 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 (http://ludwig.chem.selu.edu/sarah/CHEM266/SN1vsSN2vsE1vsE2.html) with tips for distinguishing them.

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

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:

http://www.answers.com/main/content/wp/en/thumb/4/4a/300px-Binary_phase_diagram.PNG

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.

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!

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!

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.

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 (http://forums.studentdoctor.net/showpost.php?p=2756252&postcount=4) already, and I will try to get the intermolecular forces part done as soon as I can.

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

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.

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.

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.

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?

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.

(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.

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. :thumbup: 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.

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

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.

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?

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?

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.

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]

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.

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!

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??

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.

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.)

Thanks for the explanation, their explanation was confusing me! :p

Thanks for the explanation, their explanation was confusing me! :p

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

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?

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.

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

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.

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.

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.

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!! :p

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!! :p

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:

http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA5/STILLS64/TRAM18/GENERAL/1001123.JPG


I would not bother to memorize the monosaccharides' structures.

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.

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.

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

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.

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.

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.

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. :p Pentane will have greater van der Waals forces because it's larger. Maybe that's what you meant?

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. :p Pentane will have greater van der Waals forces because it's larger. Maybe that's what you meant?

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.

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.

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.

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!

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.

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.

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.

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 (http://http://forums.studentdoctor.net/showpost.php?p=2753020&postcount=32), but luckily there aren't many of them.

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!

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.

Bumping the organic thread

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?

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.

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. :p

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.

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.

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?

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!

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.

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.

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 :love: ) 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

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 :love: ) 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. :thumbup:

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?

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. :thumbup:

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.

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!

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. :o 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'.

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.

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.

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.

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.

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.

bump

bumping this thread

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,

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.

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?

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.

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!

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.

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 f