Organic Chemistry Question Thread

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.

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.

Kussemek said:

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?

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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.

Shrike said:

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?

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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.

Shrike said:

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

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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.

Abe said:

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?

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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.

EMT2ER-DOC said:

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?? ?

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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.

EMT2ER-DOC said:

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.

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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.

Abe said:

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?

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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.

gujuDoc said:

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?????

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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).

Santosh said:

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.

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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

Santosh said:

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!

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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.

Jezzielin said:

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!

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Yes, we will add this to the list.

Jezzielin said:

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

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

Nutmeg said:

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.

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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.

Andrew99 said:

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.

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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.

Kussemek said:

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

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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.