Organic Chemistry Question Thread

[QofQuimica]

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

Acceptable topics:
-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.

 

[kingbrew]

If t-butyl bromide is added to Mg in anh. ether, what would be the impurity formed?

 

[spicedmanna]

If t-butyl bromide is added to Mg in anh. ether, what would be the impurity formed?

The principle side-reaction in the prepartion of alkyl magnesium halides is called "Wurtz Coupling", in which you get R-R as the side-product, R- being the original alkyl group of the alkyl halide used in the preparation. You can think of this reaction sort of like, "R-Mg-X + R-X ---> R-R + Mg(X)2".


Here is a good review of Grignard reactions; it's where I got my answer (pg. 7):

http://www.joe-harrity.staff.shef.ac.uk/meetings/GrignardReagentsReviewMeeting.ppt

 

[travelbug73]

Why do higher temperatures favor E1 over SN1? Can somebody please explain that in terms of reaction energy diagrams?

Thanks

 

[spicedmanna]

Why do higher temperatures favor E1 over SN1? Can somebody please explain that in terms of reaction energy diagrams?

Thanks

Here's my crude guess:

Let's look at a simple solvolysis reaction.

First, note that per given reaction where both mechanisms are possible, both E1 and SN1 reactions share a common carbocation intermediate, which is generated through of the dissociation of the leaving group. This rate determining first step (highest activation energy) is generally referred to as the "ionization" step.

R-LG <---> R+ + LG- (1)

Second, note that the remaining steps for the substitution reaction (addition) are reversible and have lower activation energies than the first step.

R+ + OH2 <---> R(OH2)+ (2)

R(OH2)+ + OH2 <----> ROH + H3O+ (3)

In elimination, the solvent or base extracts the beta-hydrogen from the carbocation in (1), resulting in the formation of a stable double bond. I think that this step probably has a higher activation energy than the addition step of the substitution reaction. The resulting alkene product is generally more stable (i.e., lower energy), however, than the substitution product, and the remaining steps are generally not as readily reversible because of the higher activation energy required to break the double bond (614 kj/mol). The energy required to break the C-O bond is 360 kj/mol, in comparison.

If you increase the temperature, you place the reaction into thermodynamic control; the more stable product of a reaction is favored. Increasing temperature allows the products of a reaction to better overcome the activation energies for the reverse steps. The substitution product will reverse more readily to the common carbocation intermediate than the more stable elimination product (if it occurs). Futhermore, the increased temperature allows the second step of the elimination reaction (breaking of the C-H bond) to occur more readily than at room temperature to generate the more stable alkene. Thus, increasing temperature favors the elimination reaction in two ways.

 

[Absurdist]

Does E2 happen faster at a bromine attached to a primary carbon or at a bromine attached to a tertiary carbon?

Also, when doing R,S notation and the smallest priority group attached to the chiral center is not sticking in or out, am i supposed to treat it like it is sticking in?

 

[QofQuimica]

Does E2 happen faster at a bromine attached to a primary carbon or at a bromine attached to a tertiary carbon?

Also, when doing R,S notation and the smallest priority group attached to the chiral center is not sticking in or out, am i supposed to treat it like it is sticking in?

First question: E2 doesn't care too much about sterics of the substrate, as long as the molecules can get into an antiperiplanar geometry. That being said, tertiary compounds often react faster just because of the relief of steric strain when you go from the hindered sp3 carbon to the planar sp2 carbon in the double bond.

Second question: No. You will either have to rotate the molecule in your mind, or else swap two of the substituents. If you do the latter, don't forget to switch the designation, because every time you make a swap, you change the configuration.

 

[magilla]

My question is: Without learning the structures by heart, how could i
distinguish the structures of the following common names?
t-butyl,
neopentyl,
isopropyl,
isobutyl,
sec-butyl

Thanks.

 

[Absurdist]

Thanks Q.

I also have one other question:

What is the product when you treat 1,1-dibromopentane with t-butoxide?

I know t-butoxide is the base for E2 reactions, so i said the product is 1-pentene, but I don't know if that's right because that C1 is still missing a hyrdogen.

 

[novawildcat]

My question is: Without learning the structures by heart, how could i
distinguish the structures of the following common names?
t-butyl,
neopentyl,
isopropyl,
isobutyl,
sec-butyl

Thanks.

memorize what the prefixes mean. neo, t (aka tert), iso, sec all indicate specific structures. every organic chemistry student should have memorized all the names for n-hydrocarbons for 1-10 carbons long. dont memorize all them, but just break them down by prefix and suffix.

Thanks Q.

I also have one other question:

What is the product when you treat 1,1-dibromopentane with t-butoxide?

I know t-butoxide is the base for E2 reactions, so i said the product is 1-pentene, but I don't know if that's right because that C1 is still missing a hyrdogen.

hint: 1-pentene is wrong, you need to change an e somewhere in the name to a y.

 

[QofQuimica]

My question is: Without learning the structures by heart, how could i
distinguish the structures of the following common names?
t-butyl,
neopentyl,
isopropyl,
isobutyl,
sec-butyl

Thanks.

Yeah, I'd agree with nova that the best thing to do is to learn what the prefixes and suffixes all mean. If you know what tertiary, iso, and sec are, and you also know the hydrocarbon chains, you can figure out the structures based on that.

 

[QofQuimica]

Thanks Q.

I also have one other question:

What is the product when you treat 1,1-dibromopentane with t-butoxide?

I know t-butoxide is the base for E2 reactions, so i said the product is 1-pentene, but I don't know if that's right because that C1 is still missing a hyrdogen.

You're halfway there. Go through the reaction step by step. The first thing that happens is that you lose one HBr due to E2, right? That gives you a pentene as you said. But you didn't start with just one Br; you started with two. So what pentene intermediate do you have left after losing that HBr, and what happens to it? That's the hint that nova is trying to give you. 😉

P.S. I'm not sure that t-butoxide is a strong enough base to do what you're trying to do here; the reaction might actually stop after one step.

 

[commuter9]

I was a little confused about electrophoresis. In electrophoresis is the anode positive or negative? (I thought it was negative because in a galvanic cell it is negative, but I'm guessing electrophoresis is like a electrolytic cell so the anode is positive?)

 

[QofQuimica]

I was a little confused about electrophoresis. In electrophoresis is the anode positive or negative? (I thought it was negative because in a galvanic cell it is negative, but I'm guessing electrophoresis is like a electrolytic cell so the anode is positive?)

You're guessing right. 🙂 Don't forget, you have to put energy into an electrophoresis apparatus to get it to work. So, that means it's an electrolytic cell, not a galvanic cell.

 

[commuter9]

You're guessing right. 🙂 Don't forget, you have to put energy into an electrophoresis apparatus to get it to work. So, that means it's an electrolytic cell, not a galvanic cell.

thanks Q. wishing i was a chem master like you...🙂

 

[midn]

Is there any free software online that will let me draw 3D molecules and manipulate them (rotate it, etc.). A plus would be if I could draw them in Fischer diagrams. I'm really bad at stereoisomerism and this would really help me to visualize it.

Thanks.

 

[spicedmanna]

Is there any free software online that will let me draw 3D molecules and manipulate them (rotate it, etc.). A plus would be if I could draw them in Fischer diagrams. I'm really bad at stereoisomerism and this would really help me to visualize it.

Thanks.

Yes. Try, ACD/ChemSketch. It's freeware.

www.acdlabs.com

Good luck!

 

[midn]

Thanks a bunch. I actually figured out how to manipulate the Fischer diagrams on paper in regards to chirality, but this will help me for other formats.

 

[midn]

I've got another general question. Where can I find a list of functional groups and their relative inductive effects. My O-Chem book doesn't have a concise list and I've lost my intuition on inductive effects.

 

[voirlesetoiles]

My organic manual says that an alcohol behaves as an electrophile if a C-O bond is broken and behaves as a nucleophile if the O-H bond is broken. (the molecule shown is in the form R-O-H2 with the oxygen having a +1 charge)

I do not understand this at all. I don't understand the idea of an alcohol being electrophilic or nucleophilic. The oxygen gains the electrons regardless of if C-O or O-H is broken. I don't understand how it is an electrophile if the C-O bond is broken and a nucleophile when the O-H bond is broken.

Any help on this would be much appreciated. Thanks.

 

[estairella]

My organic manual says that an alcohol behaves as an electrophile if a C-O bond is broken and behaves as a nucleophile if the O-H bond is broken. (the molecule shown is in the form R-O-H2 with the oxygen having a +1 charge)

I do not understand this at all. I don't understand the idea of an alcohol being electrophilic or nucleophilic. The oxygen gains the electrons regardless of if C-O or O-H is broken. I don't understand how it is an electrophile if the C-O bond is broken and a nucleophile when the O-H bond is broken.

Any help on this would be much appreciated. Thanks.

It's just a general guideline...

ROH2+ -> R+ & H2O, R+ is electrophile (carbon is electrophilic)
ROH2+ -> ROH & H+, ROH is nucleophile (oxygen is nucleophilic)

This helps you see what is happening in a reaction. For example, if an alcohol is protonated, and then the O attacks another molecule, and solution is acidified (H+ ions), then you know nucleophilic attack has occurred. If, on the other hand, an alcohol is protonated and then water is released, you know carbon has acted as an electrophile... e.g.

ROH2+ & Br- -> RBr + H2O

 

[5moreminutes]

The reaction of Compound IV with HBr primarily differs from that of Compound II in that it proceeds through which of the following intermediates?
A secondary carbocation
A vinylic anion
A vinylic cation
A tertiary carbocation
This is a kaplan question that I am getting a little confused on
I know that B and D are definitely wrong and C is right, but I dont know why A is wrong, because I thought that when we add HBr to an alkene it adds the H to the less substituted and then the carbocation that is formed is secondary and so choice A would be true too

please help! thanks 🙂

 

[tamtam513]

I was wondering if you could go over what makes a certain chair confirmation stable...? Thanks!

 

[Anastasis]

I was wondering if you could go over what makes a certain chair confirmation stable...? Thanks!

The mantra of organic chemistry and stereochemistry:
Steric hindrance.

If you have a cyclohexane ring substituted at the C1 & C4 in the cis conformation then there will be one equatorial and one axial in either chair conformation so it will interconvert between the two chairs...
but if it is in the trans conformation then they will either be both axial or both equatorial... in THAT situation the conformation with both subsitutents in the equatorial postition is energetically favored because if they were both in the axial position there would be greater steric interactions.

Maybe it would help if you could get your hands on some models and build the ring. Then you would be able to see the idea behind what I'm describing.

Hope that answered your question.

 

[midn]

The reaction of Compound IV with HBr primarily differs from that of Compound II in that it proceeds through which of the following intermediates?
A secondary carbocation
A vinylic anion
A vinylic cation
A tertiary carbocation
This is a kaplan question that I am getting a little confused on
I know that B and D are definitely wrong and C is right, but I dont know why A is wrong, because I thought that when we add HBr to an alkene it adds the H to the less substituted and then the carbocation that is formed is secondary and so choice A would be true too

please help! thanks 🙂

The transition structure is cyclic with Br binding two carbons. The next nucleophile attacks "below" (anti) with respect to the first Br and adds with regular Markovnikov regioselectivity.

If you drew the mechanism through a regular SN1 with the positive charge residing on one carbon to form a carbocation, you'll still get the same answer. I have no idea why C is right since I do not have the compounds in front of me.

 

[BrokenGlass]

Why do the O-H bonds of alcohols have higher absorption frequency (measured in wavenumbers) than the O-H bonds of caboxylic acids in IR Spectroscopy?

 

[spicedmanna]

Why do the O-H bonds of alcohols have higher absorption frequency (measured in wavenumbers) than the O-H bonds of caboxylic acids in IR Spectroscopy?

This post should lead you in the right general direction to answer the question:

http://forums.studentdoctor.net/showpost.php?p=4253801

Also, in addition to the factors introduced in the aforementioned post, consider that the hydrogen in the hydroxyl group of a carboxylic acid is considerabley more acidic than the hydrogen of the hydroxyl group of an alcohol. Part of the reason for this increase in acidity is due to resonance stabilization and delocalization of the negative charge on the conjugate base of the carboxylic acid; also, the presence of two electron withdrawing groups (the two oxygens) pull electron density away from the hydrogen atom. In comparison, the conjugate base of the alcohol has a fairly localized negative charge. In general, an acid is stronger, has a higher Ka value(tendency to lose a hydrogen ion), if the conjugate base of that acid can be stabilized by resonance. Thus, it can be said that the hydrogen atom on the carboxylic acid is more "weakly" bonded to the oxygen than the hydrogen atom on the hydroxyl group of the alcohol. I bet this effects the oscillation tendency of the -OH stretch on the carboxylic acid.

Yet another factor to consider is that carboxylic acids can dimerize. You can have -OH--O=C-, times two, for a pair of molecules, right? This phenomenon, I would guess, further effects the oscillation.

These are just my best guesses.

 

[BrokenGlass]

Thank you spicedmanna. That really helps!

 

[BrokenGlass]

O-H alcohol is about 3400 cm-1 (given)
C-H alkene is about 3000 cm-1 (given)
N-H amine (need to determine)
Relative bond polarities determine relative absorption intensities.
The greater the change in dipole moment, the more intense the absorption.
Changes in dipole moments depend on respective electronegativity values.
But since we are interested in absorption frequencies, not absorption
intensities, relative EN values of C,N,O are not the answer.
Stronger bonds and lighter atoms give rise to higher absorption frequencies.
N is lighter than O but heavier than C. so based on atomic mass alone, N-H
should absorb at a higher frequency than O-H and at a lower frequency than C-H.
N is smaller than C but larger than O. So N-H is shorter than C-H, but longer
than O-H. shorter bonds are stronger than longer bonds. So based on bond
strength alone, N-H should absorb at a higher frequency than C-H and at a lower
frequency than O-H.
So looks like we have two competing factors here. How
do we know that bond strength effect outcompetes the atomic mass effect here?
Without checking any tables, how do we know tha N-H absorbs somewhere between
o-H and N-H? That's what the answer says.

 

[spicedmanna]

O-H alcohol is about 3400 cm-1 (given)
C-H alkene is about 3000 cm-1 (given)
N-H amine (need to determine)
Relative bond polarities determine relative absorption intensities.
The greater the change in dipole moment, the more intense the absorption.
Changes in dipole moments depend on respective electronegativity values.
But since we are interested in absorption frequencies, not absorption
intensities, relative EN values of C,N,O are not the answer.
Stronger bonds and lighter atoms give rise to higher absorption frequencies.
N is lighter than O but heavier than C. so based on atomic mass alone, N-H
should absorb at a higher frequency than O-H and at a lower frequency than C-H.
N is smaller than C but larger than O. So N-H is shorter than C-H, but longer
than O-H. shorter bonds are stronger than longer bonds. So based on bond
strength alone, N-H should absorb at a higher frequency than C-H and at a lower
frequency than O-H.
So looks like we have two competing factors here. How
do we know that bond strength effect outcompetes the atomic mass effect here?
Without checking any tables, how do we know tha N-H absorbs somewhere between
o-H and N-H? That's what the answer says.

Hmm. I don't really know, but here's my best hair-brained guess. First of all, remember there are different kinds of vibrations: stretches, bends, wags, etc. From the look of your numbers, it appears you are examining stretches. That's most like a spring-mass system. I would think that for X-H stretches, the bond length/strength has a much greater impact than other factors that might have an influence. We are assuming here naturally that X >>>> H with respect to atomic mass, and we are essentially looking at the oscillation of H; presumably X (in your case, X=N, C, or O) is also attached to the rest of the organic molecule (RX-H in your case is, -C=C-H, -C-O-H, or -C-N-H). The frequency of the stretch between R and X is less than the frequency of the stretch between X and H. Here's the way I am currently thinking about it: if RX is like a boulder and H is like a tiny pebble, and they are attached together by a spring with a spring constant that is proportional to the bond strength, then H will oscillate faster or slower depending mostly on the spring constant. The stiffer your spring, the higher the frequency of oscillation (w=sqrt(k/m)), where m is the mass of H, which is, of course, the same among the three moeities.

Something interesting to note: primary amines can have asymmetric N-H stretches, thus leading to two peaks.

 

[BrokenGlass]

Acidity is affected by 3 factors: bond strength, bond polarity, and the conjugate base stability. We know that carboxylic acids are more acidic
than alcohols. So "more acidic" could be the result of weaker bonds,
more polar bonds, more stable conjugate bases, or some combination of
these factors that outcompete the remaining factors. So I was thinking about
this some more and it looks like we concluded "weaker bond" from "more
acidic." Bond polarity is not an issue since the bond is O-H in both cases.
Conjuagate base of a carboxylic acid is more stable than the conjugate base
of an alcohol, which is why carboxylic acids are more acidic than alcohols. If
I follow this line of reasoning, then it's not immediately apparent that bond
strength is the difference between carboxylic acids and alcohols (which is what we care about for IR Spec)....It seems that the difference in acidity
is due to differences in conjugate base stability, so is the conclusion "more acidic" means "weaker bond" still valid? Thank you!

 

[BrokenGlass]

I wrote earlier that "based on atomic mass alone, N-H
should absorb at a higher frequency than O-H and at a lower frequency than C-H." I think that would be impossilbe since there are no numbers larger than 3400 and at the same time less than 3000. Is this reasonable? Thank you.

 

[spicedmanna]

It seems that the difference in acidity
is due to differences in conjugate base stability, so is the conclusion "more acidic" means "weaker bond" still valid? Thank you!

Well, I think you might be oversimplifying it a little; things are rarely that cut and dry in the world of organic chemistry. I am quite hesitant to make any hard rules, but I have seen it modelled that way. As you mentioned before, however, it is actually a combination of several factors that make carboxylic acid more acidic than alcohols.

It's true that resonance stabilization plays a highly significant role, and it is also true, I think, that the O-H bond in a carboxylic acid is weaker than the O-H bond in the alcohol. The two aren't mutually exclusive principles. We know that the two predominate factors in acidity here in this situation are resonance and induction, both of which can be related directly to bond strength. Remember that carboxylic acid has TWO electron withdrawing groups, the carbonyl oxygen and the hydroxyl oxygen, whereas the unsubstituted alcohol has only one, the hydroxyl oxygen. Furthermore, you have significant delocalization of pi electrons across the carbonyl group and the hydroxyl group of the carboxylic acid, where this doesn't happen in the alcohols. Not only do you have a strong inductive effect at play here, you have a strong resonance component as well, where electron density is being sucked away from the hydrogen. These factors impart significant ionic character to the acidic hydrogen. One can theorize, from these observations, that the O-H bond is weaker in the carboxylic acid as a result. This theory is supported visually by looking at an electrostatic potential (electron density) map of a carboxylic acid, which clearly shows the lack of electron density in the region of the acidic hydrogen, and by it's apparent acidity measured by hydrogen ion concentration. Naturally, if the bond is weaker, it will increase the tendency for the hydrogen to dissociate from the moeity, thereby, using the bronsted-lowry definition, increasing acidity.

In terms of IR stretching behavior of the O-H, you really need to look at the influence of hydrogen bonding and dimerization that takes place between molecules of carboxylic acid, as well. IR is part science and part artistry.

Hope this helps, as a starting point at least.

 

[spicedmanna]

I wrote earlier that "based on atomic mass alone, N-H should absorb at a higher frequency than O-H and at a lower frequency than C-H." I think that would be impossilbe since there are no numbers larger than 3400 and at the same time less than 3000. Is this reasonable? Thank you.

Well, I'm not sure what you are getting at. There are some groups that absorb higher than 3400, a non-hydrogen-bonded OH, for instance, and definitely many groups that absorb lower than 3000, carbonyl stretches for instance. Indeed, some groups can even span a large range, for instance, the OH group of the carboxylic acid, which can absorb broadly from 2500 to 3300, due to dimerization and hydrogen bonding.

A more plausible reason for the appearance of the NH stretch band (amine) between =CH and -O-H is probably because bond strength is the more important factor, since we are essentially looking at the oscillation of the H relative to the rest of the molecule. This is my thought, anyway. I'm too far away from organic chemistry to remember the physics behind it. Suffice it to say that this level of knowledge isn't necessary for the MCAT.

 

[BrokenGlass]

Well, I'm not sure what you are getting at. There are some groups that absorb higher than 3400, a non-hydrogen-bonded OH, for instance, and definitely many groups that absorb lower than 3000, carbonyl stretches for instance. Indeed, some groups can even span a large range, for instance, the OH group of the carboxylic acid, which can absorb broadly from 2500 to 3300, due to dimerization and hydrogen bonding.

A more plausible reason for the appearance of the NH stretch band (amine) between =CH and -O-H is probably because bond strength is the more important factor, since we are essentially looking at the oscillation of the H relative to the rest of the molecule. This is my thought, anyway. I'm too far away from organic chemistry to remember the physics behind it. Suffice it to say that this level of knowledge isn't necessary for the MCAT.

I think what I was trying to say is that since absorbance values for O-H and
=C-H are given, we can use these numbers to help determine the absorbance
value for N-H. And since it's impossilbe to have a number greater than 3400 but less than 3000, we could eliminate one of the competing factors.

 

[strive01]

hello,

I am a freshman right now and i was thinking of taking orgranic chem I and II over the summer and then taking physics I and biochem I (1st semester soph yr) and physics II and anatomy & phys (2nd semester) then have my sophmore summer to practice MCATs and maybe do some summer research during the day and come back home and do MCATs at night and take it in august before my junior year.

I was wondering if that is a good gameplan? I am a bit scared of the MCATS because I am not a good standardized test taker and want the time over the summer when classes arent there to really prac the MCATs. I also EMTing and in other clubs during the school year so it would be hard fitting in MCATs over the semester.

If taking orgo chem during my freshman summer is a bad idea, what else can i do?

thank u for the advice

 

[Anastasis]

hello,

I am a freshman right now and i was thinking of taking orgranic chem I and II over the summer and then taking physics I and biochem I (1st semester soph yr) and physics II and anatomy & phys (2nd semester) then have my sophmore summer to practice MCATs and maybe do some summer research during the day and come back home and do MCATs at night and take it in august before my junior year.

I was wondering if that is a good gameplan? I am a bit scared of the MCATS because I am not a good standardized test taker and want the time over the summer when classes arent there to really prac the MCATs. I also EMTing and in other clubs during the school year so it would be hard fitting in MCATs over the semester.

If taking orgo chem during my freshman summer is a bad idea, what else can i do?

thank u for the advice

It depends. Will those be your only summer classes? If so, maybe it's doable. I took them during the school year so I don't have any personal experience but from what I've read on SDN summer orgo can be difficult. (Vast amount of information anyway so cramming it into 1/2 the time makes it seem like even more).

If I was in your position I would take physics over the summer and take Organic during the school year. But that's just a preference of mine. If you're good at Chemistry and think the shortened schedule shouldn't affect you negatively then I don't think there is any problem to your schedule.

Does your school require both Orgo I & II or just orgo I before Biochem?

 

[5moreminutes]

Why is this, can someone please explain as to why alkanes are electron donating or direct me to the relevant post

thanks a bunch!

 

[spicedmanna]

Why is this, can someone please explain as to why alkanes are electron donating or direct me to the relevant post

thanks a bunch!

When people think of the electron-donating effects of alkyl groups, they often attribute it to a phenomenon known as hyperconjugation, or the specific overlap of a sigma bond with a p-orbital, or donation of sigma bond electrons into a pi-system.

Let's first examine a simpler system, like a primary carbocation (say, CH3CH2+, for example). Normally, when the carbocation is formed, you have an sp2-hybridized carbon with an empty p-orbital. Well, remember that there is relatively free rotation about the methyl group. When one of the three C-H sigma bonds comes into alignment with the empty p-orbital, overlapping occurs, and the sharing of the electron density from the electron-rich sigma bond to the electron-poor, empty p-orbital. This helps stabilize the system.

Now, let's take hyperconjugation to an aromatic system, like benzene. It's essentially the same as above. Consider toluene (benzene substituted with a methyl group). Remember that benzene is aromatic and has a pi-system. The carbons on the ring are sp2-hybridized and there is significant overlap of the p-orbitals and pi-electrons. Well, consider that there is rotation about the methyl group and when one of the C-H sigma bonds of methyl group come into close alignment with the p-orbitals, the electron density in the sigma bonds can overlap with the pi-system, resulting in the donation of electron density into the aromatic ring. Thus, alkyl groups function as both o,p-activators and electron-donating groups (EDG) in electrophilic aromatic substitution. Naturally, hyperconjugation doesn't have as significant an effect as, say, the donation of an available lone pair of electrons in the nitrogen of aniline into the pi-system, but it is significant enough to lend additional stabilization in EAS.

The above explain the effect of alkyl substitution in EAS and in the stabilization of carbocation systems quite well. However, we should also take the time to note more generically, that the second often-cited reason for electron-donating effects of alkyl groups falls into the category of inductive effects. These effects are transmitted through the sigma system (sigma bonds) and are quite distance-dependant (effects decrease with distance). Essentially, inductive effects work by the shifting of electron density in response to differences in electronegativity between two entities in a molecule. The greater the difference in electronegativity, the bigger the polarization of electron density in the molecule. There are plenty of electrons in the sigma system of alkyl groups that can be polarized and "transmitted" when inductive effects come into play.

Hope these thoughts provide some insight.

 

[endofevangelion]

There was this one question on the Kaplan Qbank that really confused me (and made me wonder whether I really understood NMR 🙁 )

Imagine Cyclopentanone, and then adding an oxygen atom in between the carbonyl carbon and one of its alpha carbons (making it a 6 ring structure)

Now the former structure should have 2 triplets right? (due to symmetry?)

But with the oxygen added to the structure, shouldnt it throw off the symmetry and produce additional peaks?

 

[5moreminutes]

When people think of the electron-donating effects of alkyl groups, they often attribute it to a phenomenon known as hyperconjugation, or the specific overlap of a sigma bond with a p-orbital, or donation of sigma bond electrons into a pi-system.

Let's first examine a simpler system, like a primary carbocation (say, CH3CH2+, for example). Normally, when the carbocation is formed, you have an sp2-hybridized carbon with an empty p-orbital. Well, remember that there is relatively free rotation about the methyl group. When one of the three C-H sigma bonds comes into alignment with the empty p-orbital, overlapping occurs, and the sharing of the electron density from the electron-rich sigma bond to the electron-poor, empty p-orbital. This helps stabilize the system.

Now, let's take hyperconjugation to an aromatic system, like benzene. It's essentially the same as above. Consider toluene (benzene substituted with a methyl group). Remember that benzene is aromatic and has a pi-system. The carbons on the ring are sp2-hybridized and there is significant overlap of the p-orbitals and pi-electrons. Well, consider that there is rotation about the methyl group and when one of the C-H sigma bonds of methyl group come into close alignment with the p-orbitals, the electron density in the sigma bonds can overlap with the pi-system, resulting in the donation of electron density into the aromatic ring. Thus, alkyl groups function as both o,p-activators and electron-donating groups (EDG) in electrophilic aromatic substitution. Naturally, hyperconjugation doesn't have as significant an effect as, say, the donation of an available lone pair of electrons in the nitrogen of aniline into the pi-system, but it is significant enough to lend additional stabilization in EAS.

The above explain the effect of alkyl substitution in EAS and in the stabilization of carbocation systems quite well. However, we should also take the time to note more generically, that the second often-cited reason for electron-donating effects of alkyl groups falls into the category of inductive effects. These effects are transmitted through the sigma system (sigma bonds) and are quite distance-dependant (effects decrease with distance). Essentially, inductive effects work by the shifting of electron density in response to differences in electronegativity between two entities in a molecule. The greater the difference in electronegativity, the bigger the polarization of electron density in the molecule. There are plenty of electrons in the sigma system of alkyl groups that can be polarized and "transmitted" when inductive effects come into play.

Hope these thoughts provide some insight.

THankyou very much 🙂 that cleared up the fuzziness

 

[Jasonskizzle]

I was wondering if someone could explain what product I would get and by what mechanism I would use for this reaction.

1-chloro-1-methylcyclohexane + ethanol

I just started learning Sn1, Sn2, E1,E2 reactions and I know the ethanol is 2 carbons and an OH, I was just wondering how I was going to go about doing this problem.

 

[spicedmanna]

I was wondering if someone could explain what product I would get and by what mechanism I would use for this reaction.

1-chloro-1-methylcyclohexane + ethanol

I just started learning Sn1, Sn2, E1,E2 reactions and I know the ethanol is 2 carbons and an OH, I was just wondering how I was going to go about doing this problem.

Gosh, it's been awhile... I've been straining the depths of my mind remembering this stuff, and I used to be pretty good at it! 😳

Okay, let's look at this reaction. In these types of reactions, you always have an electrophile and a nucleophile. The first step is to identify which is which. So, it looks like your ethanol reagent is the nucleophilic species. It is a weak nucleophile and also a weak base, but it is definitely polar. Now, we know that the 1-chloro-1-methylcyclohexane must be the electrophilic species in the reaction. It is a tertiary cycloalkyl halide, with a good leaving group. It is also important to determine the solvent in the reaction. Since there are no other reagents present, it looks like the solvent is also ethanol. Sterics can also be a factor, so it would usually be wise to consider this as well.

Next, we need to determine which mechanism is the most favored by the reagents and conditions. First, is our reaction going to be unimolecular or bimolecular in terms of rate? I think this reaction is most likely going to be unimolecular.

Recall that the unimolecular reactions in this case are SN1 and E1. These types of reactions involve the formation of a carbocation intermediate; thus, we are looking for conditions that favor the formation of a carbocation in our reaction. Clearly a more substituted alkyl halide (electrophile) can result in a more substituted carbocation; we know that the more substituted your carbocation is, the more stable it will be, and the more readily it is to form. Our alkyl halide is a tertiary cycloalkyl halide and can form a tertiary carbocation, so this fact is consistant with our hypothesis that we are dealing with a unimolecular reaction. Furthermore, a polar solvent, which is present in our reaction, helps to stabilize carbocation intermediates and help pull the leaving group off of the alkyl halide; this increases the rate of the reaction. We also know that our nucleophile is weak and is weakly basic; this further favors the hypothesis that we are dealing with a unimolecular reaction (if our nucleophile had been strongly basic, we would have considered E2). Lastly, it is important to have a good leaving group on the alkyl halide in order for the unimolecular reactions to be favored; this is certainly the case in our reaction.

There will be competition between SN1 and E1. You are likely to get a mixture of products. Both occur through a similar mechanism. First, the leaving group dissociates from the electrophile (that's why it's important to have a good leaving group), forming a carbocation intermediate, which is solvated and stabilized by the polar solvent present in the reaction mixture. This is the rate-determining, or slow, step of the reaction. In the second, or fast, step of the reaction, the final product is formed, either by eliminating a hydrogen to form a double bond, or by substitution.

So with our reaction, the first step would be the dissociation of the chloride from the 1-chloro-1-methylcyclohexane, forming the resultant carbocation intermediate, 1-methylcyclohexylium ion. It is stabilized by the ethanol solvent. Next, you have two possible, competing pathways. In the substitution reaction pathway, the nucleophilic region of the ethanol (the oxygen) attacks the electrophilic carbon of the 1-methylcyclohexylium ion (the carbon with the positive charge, of course) and forms the substitution product, 1-ethoxy-1-methylcyclohexane. Due to the planar nature of the carbocation intermediate, you would get addition of the nucleophile above and below the plane of the ring, usually resulting in the formation of enantiomeric products. In this case, however, the product is not chiral, so you get only one substitution product, 1-ethoxy-1-methylcyclohexane. Since the product is a cyclohexane, I believe the predominate conformer would be the chair form, in which the ethoxy group is equitorial, in order to minimize strain. For the elimination reaction, the oxygen of the ethanol pulls a hydrogen off from the carbon that is adjacent to the carbon with the positive charge of the carbocation (there are two of them in this case, so you get two different products), forming the resultant elmination products, 1-methylcyclohexene and methylenecyclohexane. Elimination follows Zaitev's rule; the more highly substituted alkene product predominates. In this case, the more highly substituted alkene would be the 1-methylcyclohexene product.

I hope this gives you a starting point. Good luck!

 

[wishuponastar]

does anybody have an easy way to remember the axial and equatorial relationships in cis and trans disubstituted cyclohexanes...the 1,2 cis disubstituted axial equatoiral or equatorial axial stuff? a mnemonic or catchy phrase or something interesting please! thanks.

 

[Nevadanteater]

There was this one question on the Kaplan Qbank that really confused me (and made me wonder whether I really understood NMR 🙁 )

Imagine Cyclopentanone, and then adding an oxygen atom in between the carbonyl carbon and one of its alpha carbons (making it a 6 ring structure)

Now the former structure should have 2 triplets right? (due to symmetry?)

But with the oxygen added to the structure, shouldnt it throw off the symmetry and produce additional peaks?

the 6 ring structure:

-O-CH2-CH2-CH2-CH2-Carbonyl- (in a ring)

right?

if that is the strucutre you'll get 4 peaks, two triplets and two multiplets (the multiplets will probably overlap on the chemical shift)

perhaps you have the reaction wrong? or did it give you that structure directly?

 

[amestramgram]

shouldn't you get 4 different sets of peaks for the six membered ring? Since there is no symmetry with it.

does anybody have an easy way to remember the axial and equatorial relationships in cis and trans disubstituted cyclohexanes...the 1,2 cis disubstituted axial equatoiral or equatorial axial stuff? a mnemonic or catchy phrase or something interesting please! thanks.

if you draw a 6 membered ring well in the chair form, axial should be straight up and down - and equatorial should be everything else 🙂

if your additional groups appear to be pointing towards each other, they probably have steric hindrance.

 

[amestramgram]

sorry to double post

 

[wedogogirl]

.

 

[amestramgram]

certainly. material from that class will be on the exam.

 

[Chunkeroni]

if you are planning on taking the MCAT in april you might be ok...i was not quite finished with ochem2 when i took the MCAT last april and do not think it negatively affected me..just make sure to study things that you may have not gotten to and ask for help!

 

[dochoov]

There was this one question on the Kaplan Qbank that really confused me (and made me wonder whether I really understood NMR 🙁 )

Imagine Cyclopentanone, and then adding an oxygen atom in between the carbonyl carbon and one of its alpha carbons (making it a 6 ring structure)

Now the former structure should have 2 triplets right? (due to symmetry?)

But with the oxygen added to the structure, shouldnt it throw off the symmetry and produce additional peaks?

Yes, I think the former structure should have two triplets of equal integration. I think that adding an oxygen in that ring (making it a lactone) would make it no longer symmetrical. What was the answer according to Kaplan?