Organic Chemistry Question Thread

[QofQuimica]

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

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

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

********

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.

---

Members don't see this ad.

 

[tncekm]

How can I systematically break down an NMR graph to identify a compound? Thanks again.

Sorry, but that requires a lot of explanation, and visual aids. I'd suggest venturing into the learning process on your own and then ask clarifying questions. You have access to a text book, right?

I found a great tutorial online long ago. I'll dig around for it.

 

[Foghorn]

How can I systematically break down an NMR graph to identify a compound? Thanks again.

If this is a serious question and you're not just messing with me cheesehead, I can come with some basics and post it on the thread in a couple of days.

 

[cheezer]

If this is a serious question and you're not just messing with me cheesehead, I can come with some basics and post it on the thread in a couple of days.

I think the question might be way to broad...BUT if you have time to post some basics for me that'd be cool. A few things to narrow it down:

1. I'm in Ochem 1
2. I'm covering Proton NMR right now.
3. Assume that I understand deshielding and the relationship spin-spin splitting has with the number of peaks on a graph (neighboring protons, n+1 etc.)

After that I don't know. I get to the point where I count the number of peaks, I identify the neigboring protons...and then I look at the ppm of the peak, compare it to some chart, and try to look for a group with the same ppm. That's when things start getting hazy and I waste a bunch of time. Anyways, is it just a matter of practice? Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)

 

[tncekm]

I think the question might be way to broad...BUT if you have time to post some basics for me that'd be cool. A few things to narrow it down:

1. I'm in Ochem 1
2. I'm covering Proton NMR right now.
3. Assume that I understand deshielding and the relationship spin-spin splitting has with the number of peaks on a graph (neighboring protons, n+1 etc.)

After that I don't know. I get to the point where I count the number of peaks, I identify the neigboring protons...and then I look at the ppm of the peak, compare it to some chart, and try to look for a group with the same ppm. That's when things start getting hazy and I waste a bunch of time. Anyways, is it just a matter of practice? Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)

Can you integrate the number of protons under a graph, etc? If you can, it seems you have sufficient knowledge and you just need to bust out some abstract thinking. It takes practice.

Maybe post an example and we can walk you through it (although its been over a year for me).

 

[cheezer]

Can you integrate the number of protons under a graph, etc? If you can, it seems you have sufficient knowledge and you just need to bust out some abstract thinking. It takes practice.

Maybe post an example and we can walk you through it (although its been over a year for me).

I see. Perhaps I haven't done enough practice problems.

 

[MundaneMD]

Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)

Err..I don't think so.


Isopropyl group = (CH3)2-CH-R

R may have it's own hydrogens, which would not give a septet.


There is also the possibility that you get a septet + doublet, but it is not an isopropyl group. Try this NMR:

CH(CH2OH)3

This would also give a septet and a doublet (along with a singlet from the OH groups). But it's not an isopropyl group.


I guess it's good to remember that a septet + doublet might be an isopropyl group, but there are always exceptions. Always double check to make sure that your assumption is right. I used tricks like that for starting points, but I always check to make sure that they made sense after I was done with the whole problem.

I still remember a few tricks:
triplet + quartet @ 1-4 ppm = ethyl group
singlet @ 5-6 ppm = OH group
multiplet @ 7 ppm = phenyl group

So yeah, always double check.

 

[fly1346]

can someone explain to me the reactivity rules for Aldehydes/ketones/carboxylic acids --> alcohols...furthermore the steric effects and the electric effects

thanks

 

[Foghorn]

I think the question might be way to broad...BUT if you have time to post some basics for me that'd be cool. A few things to narrow it down:

1. I'm in Ochem 1
2. I'm covering Proton NMR right now.
3. Assume that I understand deshielding and the relationship spin-spin splitting has with the number of peaks on a graph (neighboring protons, n+1 etc.)

After that I don't know. I get to the point where I count the number of peaks, I identify the neigboring protons...and then I look at the ppm of the peak, compare it to some chart, and try to look for a group with the same ppm. That's when things start getting hazy and I waste a bunch of time. Anyways, is it just a matter of practice? Also, are there any general assumptions I can make when I identify a certain # of neighboring protons (i.e. septet next to a doublet = isopropyl group)

Here's my handy dandy H-NMR problem solving guide.

Yeah, recognizing splitting patterns helps deduce what types of adjacent protons are present. The absence of proton signals in a spectrum also helps exclude out (some) functional groups.

Use the info available whether it's from H-NMR, C-NMR, IR or mass spec. They all complement each other & help confirm whether or not specific functional groups are absent/present in the molecule.

Here are the other spectral analysis:
Mass Spec
IR
C-NMR

 

[cheezer]

Here's my handy dandy H-NMR problem solving guide.

Yeah, recognizing splitting patterns helps deduce what types of adjacent protons are present. The absence of proton signals in a spectrum also helps exclude out (some) functional groups.

Use the info available whether it's from H-NMR, C-NMR, IR or mass spec. They all complement each other & help confirm whether or not specific functional groups are absent/present in the molecule.

you rule. did i ever tell you that you rule? well you do.

 

[jochi1543]

can someone explain to me the reactivity rules for Aldehydes/ketones/carboxylic acids --> alcohols...furthermore the steric effects and the electric effects

thanks

I'm not sure what the question is asking exactly. Do you want to know how reactive they are relative to alcohols or in terms of conversion to alcohols?

 

[StarTrack]

I have a question re: identifying the most stable isomer and exactly how do 'draw' it so that comparisons can be made.

For example.

Molecule A = cyclohexane with an isopropyl group on C1 projecting towards the viewer, and a methyl group projecting towards the viewer on C2.

Molecule B = exact same except that the C2-Methyl projects away from the viewer....


What are the rules in determining what is equitorial and what is axial when transforming them into chair conformations?

The wording of this question is a little confusing, but I'm going to try and take a crack at it. With regard to equitorial and axial, if I'm remembering the rules with stability and cyclohexanes correctly, cyclohexanes are always more stable when their constituent groups are in the equitorial position. That is, the methyl goup and isopropyl group are positioned parallel to the 'equator' of the chair conformation. I'm not really sure how this translates into projecting to the line of vision of the viewer. In this case, it really depends on how you're looking at the molecule. I suggest making a molecular model, if you have those little building blocks thingy's.
It may be easier if you think about this question in terms of steric hindrance rather than thinking about axial, equitorial conformations, and chair conformations. If both bulky groups are facing you, then the methyl and isoproply group are on the same side and they each experience mutual repelling forces from one another. If, however, one group faces you and the other projects away from you, then they are on opposite sides, and the repelant force that each group experiences is reduced. Hope this helps !

 

[Foghorn]

you rule. did i ever tell you that you rule? well you do.

I added the other spectra. See above. Also, I misnamed the actual molecule on the NMR guide example. It's supposed to be 3-pentanone NOT 3-propanone. I made the change on the pdf and posted a new link.

 

[Caesar]

deleted

 

[161927]

How do you know what the product looks like after an internal rearrangement that results in ring contraction or expansion? How can you systematically know where the carbocation ends up, and where the substituents go? Take this example below:

How did they know the carbocation went in the top right corner? I would have drawn the alkyl bromide on the top right corner and put the carbocation in the bottom right. And I really don't see where the 3-membered ring from the left side disappeared to. Can anyone demystify this for me?

 

[googlinggoogler]

Hi DCODY,
Here is my explanation, which might or might not help you (I hope it does!). I had a difficult time describing what happened in words and without drawings.

1. "And I really don't see where the 3-membered ring from the left side disappeared to."-->You see the triangle on the left (in the first step of the mechanism)? Where you see an arrow drawn, the electrons from that leg of the triangle attacked the bottown left corner of the triangle on the right(forming a bond there instead). Count the number of carbons now in the ring=4. Therefore, the triangle (3-membered ring) became a box (4-membered ring).

2. "How did they know the carbocation went in the top right corner? I would have drawn the alkyl bromide on the top right corner and put the carbocation in the bottom right"--> This is due to the attack of the electrons. Number the carbons in the 4 membered ring. Do you see that the electrons attacked carbon number 4 (if you numbered them clockwise)? Carbon number 4 is also where the alkyl bromide is located. That is why the alkyl bromide is on the bottom right (because that is where carbon number 4 is in the 4 membered ring after the attack).

Afterwards, there had to be a rearrangement because carbocations are more stable if they are tertiary than if they are secondary. I'm not sure why the proton is the one that did the rearrangement, but that is what I see more commonly than alkyl groups moving. Since the alkyl bromide did not move, the bottom right corner of the 4 membered ring would be a tertiery carbocation due to the rearrangement, and therefore that is where the carbocation is supposed to be located.

 

[Foghorn]

How do you know what the product looks like after an internal rearrangement that results in ring contraction or expansion?

In reality, it must be done empirically via experiments. (Organic) reactions are rarely 100% efficient, that's why purification steps are done in order to separate reaction components and products from each other. Among other things, the type of experiment(s) done can reveal what's occurring thermodynamically, kinetically and chemical structure(s) formed.

One of the driving forces for ring opening/rearrangement(s) is to relieve torsional and angle strain. The combination of these two is called ring strain and 3/4-membered rings are more "rigid" chemical structures relative to other larger cycloalkane molecules.

In the example, the cyclopropane analog relative to the cyclobutane analog has greater angle strain even though both are sp3 hybridized at all ring C atoms. If you recall, sp3 hybridization produces molecular orbitals with tetrahedral geometric angles of 109.5 degrees. It's preferable to maintain these angles in sigma bond formation rather than distorting, ~60 degrees (3-membered ring) or ~90 degrees (4-membered ring), because maximum overlap between orbitals is attained. Deviation from 109.5 degrees results in sub-optimal orbital overlap and a weakened sigma bond.

With regards to torsional strain, a 3-membered ring has greater torsional potential energy relative to a 4-membered ring. This energy results from the structure having one conformation. Viewing any two adjacent ring C atoms via Newman projection(s), cyclopropane's ring substituents have a completely eclipsed conformation similar to this. Keep in mind that another C atom is bonded to both C atoms in order to form a cyclopropane ring with internal angle ~60 degrees and all C atoms in the ring are coplanar. A "rigid" cyclopropane structure essentially "locks" substituents in a completely eclipsed conformation therefore there's little or no twisting/torsion along a C-C sigma bond to minimized repulsive forces between substituents unlike 4/5/6-membered rings where "puckering/folding" or (a) chair conformation(s) allows for partial or complete relief from angle and/or torsional strain.

In cyclobutane, ring C-atoms aren't all coplanar and structurally has more "wiggle" room since bond angles aren't as distorted relative to cyclopropane. As consequence, the internal bond angle between two adjacent sigma bonds in the ring "folds", allowing ring substituents to attain a slightly non-eclipsed conformation thus reducing torsional strain/potential energy. The "folding" comes at the energetic cost of having a bond angle slightly less than 90 degrees, but this is more than offset by a gain in overall stabilization energy from torsional relief.

For more in depth material on how to propose organic reaction mechanisms, the book authored by R.B. Grossman is an excellent reference.

 

[fly1346]

I'm not sure what the question is asking exactly. Do you want to know how reactive they are relative to alcohols or in terms of conversion to alcohols?

Yes relative to alcohols, and general reactions of them please
thanks

 

[BlackSails]

Why does benzene not undergo X2 halogenation? I understand the bonds are delocalized, but there are still pi orbital electrons ready for bonding, no?

 

[bozz]

What does being "soluble" in dilute acid/base mean?

Does a substance have to be a base to be soluble in dilute acid?

 

[BlackSails]

What does being "soluble" in dilute acid/base mean?

Soluble means that a decent amount of the compound can dissolve in the solvent. Diethyl ether is soluble in toluene, but not in water for example. Solubility is in general, related to how polar the molecule is.

Does a substance have to be a base to be soluble in dilute acid?

No.

 

[bozz]

The reason why I was asking was b/c I encountered a question asking us to choose whether a compound had a carboxylic acid or an amine... and the "condition" was that it was soluble in dilute acid

The answer was amine.. the explanation was that the amine was basic.

 

[chibii]

Soluble means that a decent amount of the compound can dissolve in the solvent. Diethyl ether is soluble in toluene, but not in water for example. Solubility is in general, related to how polar the molecule is.



No.

i dont think soo..=)

 

[BlackSails]

i dont think soo..=)

A neutral compound might need to be protonated to dissolve into some polar phase. Its done all the time in extractions.

 

[Richter915]

The reason why I was asking was b/c I encountered a question asking us to choose whether a compound had a carboxylic acid or an amine... and the "condition" was that it was soluble in dilute acid

The answer was amine.. the explanation was that the amine was basic.

the assumption i will make is that dilute acid means dilute acid in a polar solvent (like water). The carboxylic acid will remain neutral (as in, protonated or not deprotonated) which reduces it's solubility while the amine, being basic, will easily become protonated (and thus, charged) which makes it more soluble.

The question is poorly written if that is indeed the wording of it. For MCAT purposes...you can think of "like dissolves like" for solubility purposes (unless it's one of those big Ksp type questions, but those are usually in the Physical section). When they say "...because it's basic"...interpret that as "because it gets charged as a result of protonation". If the same question was asked but it was in a dilute base, the answer would be the carboxylic acid because it is an acid and when put in an alkaline solution, the acid becomes deprotonated and therefore, it becomes charged, and therefore, more soluble in a polar solvent.

 

[ashgore]

What role does heat play with acidification?

For example, in alcohol dehydration to alkene by addition of concentrated acid and heat.

 

[NickMB]

What role does heat play with acidification?

For example, in alcohol dehydration to alkene by addition of concentrated acid and heat.

heat...so beautiful. Oftentimes used for thermodynamic control where you would otherwise get a kinetic product. Heat makes atoms vibrate faster, thus increasing the rate at which reactants come in contact with each other. In thermodynamic control, the thermodynamic product is more stable, but the kinetic product is formed faster due to having a lower activation energy, thus heat is used to make sure that thermodynamic product formation is more favored.

 

[spoudaios]

1>. Can someone help me with disproportionation reaction? What is it and what are some examples?

2>. Regarding interconversion of: gauche/anti/eclipsed (conformational isomers) and chair/boat/twist (cyclohexanes), do they ALL exist in room temperature?


I am reading some posts and some of them relate to alkenes. I don't know when this starts, but alkene is no longer on the MCAT.

 

[SB100]

1>. Can someone help me with disproportionation reaction? What is it and what are some examples?

2>. Regarding interconversion of: gauche/anti/eclipsed (conformational isomers) and chair/boat/twist (cyclohexanes), do they ALL exist in room temperature?


I am reading some posts and some of them relate to alkenes. I don't know when this starts, but alkene is no longer on the MCAT.

A disproportion reaction is when not all of one reactant ion can be found 100% in one product. An example would be:

Cl2(g) + 6 OH-(aq) --> 5 Cl- + ClO3-(aq) + 3 H2O(l)

The chloride ion exists both in aqueous phase and as part of chlorate.

As for your second question, I'm not an expert on this but if I recall from organic chemistry all conformations exist in room temperature, just some more than others. The most common is obviously the chair because it is the least sterically hindered, with twist boat being more hindered than chair and regular boat conformation more hindered than both. For the same reason of steric hindrance, you find the conformation of atoms in a bond as anti, then gauche, then eclipse. It's all about which conformation has the lowest energy. Think lower energy to form = more stable.

 

[Richter915]

I think you're on the spot SB100. Here's an image that will help you visualize in case u ever forget though:

 

[BlackSails]

A disproportion reaction is when not all of one reactant ion can be found 100% in one product.

Another definition is that a molecule is disproportionated when part of it is reduced and the other part is oxidized.

 

[SB100]

Another definition is that a molecule is disproportionated when part of it is reduced and the other part is oxidized.

That too, although I never learned this in general chemistry and haven't come across it in my MCAT practice. I don't think this would be on the exam unless it was a passage.

 

[JurassicPat]

A have a question regarding acid catalysts:

During, for example, ester synthesis, when reacting the carboxylic acid and the alcohol, it is wise to react this under acidic conditions. This protonates the carbonyl oxygen, making the carbonyl carbon even more partial positive, and therefore more available to nucleophilic attack by the hydroxyl oxygen. My question:

How is a hydroxyl carbon more partial positive than a carbonyl carbon? I've always thought that a carbonyl group is more polar than a hydroxyl group. Thanks-

 

[Richter915]

A have a question regarding acid catalysts:

During, for example, ester synthesis, when reacting the carboxylic acid and the alcohol, it is wise to react this under acidic conditions. This protonates the carbonyl oxygen, making the carbonyl carbon even more partial positive, and therefore more available to nucleophilic attack by the hydroxyl oxygen. My question:

How is a hydroxyl carbon more partial positive than a carbonyl carbon? I've always thought that a carbonyl group is more polar than a hydroxyl group. Thanks-

one thing you must remember about carbonxyl groups is that the negative charge is "shared" between the oxygens which means that the electron density is also slightly spread onto the carbonyl carbon. By protonating the oxygen, there is no longer a negative charge which can distribute over the carbon which causes it to become more electropositive and therefore more nucleophilic.

I remember a few different reasons for making the conditions acidic, this being one of them. Other reasons i've heard was that the acid is used to stabilize the intermediate and to help form a good leaving group.

 

[JurassicPat]

...the negative charge is "shared" between the oxygens which means that the electron density is also slightly spread onto the carbonyl carbon...

Ahh, I do recall the dotted line spanning from oxygen to oxygen, going across the carbonyl carbon. Thanks.

 

[bozz]

Alcohol + HCl = Alkyl Halide

This can happen via a SN1 or SN2 reaction

According to an AAMC question, when we were given no additional information, the question basically gave a buncha compounds (primary, secondary, tertiary) and asked us which one was most readily prepared from an alcohol + hcl.

The answer was a tertiary compound (tert butyl chloride).... but couldn't it be a primary compound too (methyl chloride)? we don't know if a SN1 or SN2 reaction occured...

 

[NickMB]

Alcohol + HCl = Alkyl Halide

This can happen via a SN1 or SN2 reaction

According to an AAMC question, when we were given no additional information, the question basically gave a buncha compounds (primary, secondary, tertiary) and asked us which one was most readily prepared from an alcohol + hcl.

The answer was a tertiary compound (tert butyl chloride).... but couldn't it be a primary compound too (methyl chloride)? we don't know if a SN1 or SN2 reaction occured...

The universe gravitates towards stability. The question asks which is more readily prepared, not which ones can be prepared.

Alcohol relative reactivity order : 3o > 2o > 1o > methyl.

Preparation of a 3o alkyl halide is always faster than an a 1o alkyl halide because formation of a tertiary carbocation is far more favored. Also, recall the rates of SN1 vs. SN2:

SN1 rate = k [R-LG]
SN2 rate = k [Nu][R-LG]

Only the rate of SN2 depends on the nature of the nucleophiles, and better nucleophiles favor SN2. The reactivity trend of hydrogen halides is HI > HBr > HCl > HF.

Both the fact that a 3o alcohol is more reactive than a 1o alcohol and the fact that HCl is not as strong of a nucleophile to react faster with a 1o alcohol by SN2 than it reacts with a 3o alcohol by SN1 mean that a 3o alkyl halide is the one more readily formed by a reaction of a 3o alcohol with hydrogen chloride.

 

[161927]

Alcohol + HCl = Alkyl Halide

This can happen via a SN1 or SN2 reaction

According to an AAMC question, when we were given no additional information, the question basically gave a buncha compounds (primary, secondary, tertiary) and asked us which one was most readily prepared from an alcohol + hcl.

The answer was a tertiary compound (tert butyl chloride).... but couldn't it be a primary compound too (methyl chloride)? we don't know if a SN1 or SN2 reaction occured...

This reaction is almost exclusively Sn1, because -OH is a bad leaving group for Sn2 reactions. Reaction with acid converts OH to a better leaving group. It leaves, and a carbocation is formed. Because 3o carbocations are most stable (compared to primary and secondary), the tertiary substitution product is formed most readily.

 

[bozz]

Hey guys… I’ve got 3 questions for the organic chem. geniuses out there =)

1) A carboxylic acid is polar … and favors water…. Why is it that when you do an extraction, the carboxylic acid is in the ETHER solvent… in the first place? I understand that when it is DEPROTONATED, it will belong in water… but why is it in the ether? Or am I wrong?

2) I’m in OChem lab right now… and my professor said that a carboxylate anion is “IONIC” …. I thought the term ionic only referred to compounds with both metals and nonmetals… can it really be used for covalent compounds

3) What exactly does a “salt” mean… I thought it just meant something that disassociates in water into a nonmetal and a metal (NaCl) for example

Why is a carboxylate anion considered a salt… can anyone define a salt in an easy to understand manner

Thanks a lot!

 

[BerkReviewTeach]

1) A carboxylic acid is polar … and favors water…. Why is it that when you do an extraction, the carboxylic acid is in the ETHER solvent… in the first place? I understand that when it is DEPROTONATED, it will belong in water… but why is it in the ether? Or am I wrong?

The neutral carboxylic acid is amphopathic, meaning it can dissolve into both water and organic solvent. Because we typically use saltwater (brine solution) in this experiment, the solubility of the neutral organic acid in water is reduced. As a result, it shows a preference for the organic solvent. But the reality of extraction is that it depends on the partition coefficient (the ratio of how the solute distributes itself between the two solvents), so the organic acid is actually dissolving into both solvents, just not equally.

2) I'm in OChem lab right now… and my professor said that a carboxylate anion is "IONIC" …. I thought the term ionic only referred to compounds with both metals and nonmetals… can it really be used for covalent compounds

Ionic means it's an ion, be it a cation or anion. The carboxylate anion carries a negative charge, which makes it an ion. For every anion (negative charge), there must be a cation (positive charge). In organic chemistry, we just don't focus on it. But for every carboxylate anion, there is likely a sodium cation (or whatever counterion you used with the based added to deprotonate the carboxylate).

3) What exactly does a "salt" mean… I thought it just meant something that disassociates in water into a nonmetal and a metal (NaCl) for example

Why is a carboxylate anion considered a salt… can anyone define a salt in an easy to understand manner

That is exactly what a salt means: a compound that generates an anion and a cation. The carboxylate is half of the salt, with the unnamed cation being the second part of the salt.

The salt is NaO2CR

 

[bozz]

The neutral carboxylic acid is amphoteric, meaning it can dissolve into both water and organic solvent. Because we typically use saltwater (brine solution) in this experiment, the solubility of the neutral organic acid in water is reduced. As a result, it shows a preference for the organic solvent. But the reality of extraction is that it depends on the partition coefficient (the ratio of how the solute distributes itself between the two solvents), so the organic acid is actually dissolving into both solvents, just not equally.



Ionic means it's an ion, be it a cation or anion. The carboxylate anion carries a negative charge, which makes it an ion. For every anion (negative charge), there must be a cation (positive charge). In organic chemistry, we just don't focus on it. But for every carboxylate anion, there is likely a sodium cation (or whatever counterion you used with the based added to deprotonate the carboxylate).



That is exactly what a salt means: a compound that generates an anion and a cation. The carboxylate is half of the salt, with the unnamed cation being the second part of the salt.

The salt is NaO2CR

thanks a lot.. I really appreciate it

 

[nev]

.

 

[What up doc]

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?

 

[SketchLazy]

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?

Read this. Page 170

http://books.google.com/books?id=ZC...sig=FfIfhzc5qWXNC8KFe7LE9NM8p0c#PRA1-PA170,M1

 

[tncekm]

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?

That's for "just" an orbital. An S orbital is more stable than a P orbital, so an SP orbital is more stable than a P orbital because it has more S character. So as far as stability (via low energy) of "orbitals" go, S > SP Hybridized > P.

A HCCH molecule has its carbons in an SP state. BUT, there are a total of total of 4 orbitals, and SP only accounts for 2! The other two orbitals are P orbitals, which are high energy orbitals, and thus more reactive.

So, since an SP3 has 0 P orbitals, and an SP2 has 1 P orbital, and an SP has 2 P orbitals, in order of stability hybridized atoms we have: SP3> SP2 > SP.

Those extra P orbital, which want electrons really bad, are what cause alkenes and alkynes to try really hard to pull electrons, pulling them away from the -H, making those H more acidic.

Its been a year since O-Chem, so I hope I got this right, but I believe that's the case.

 

[BerkReviewTeach]

In class we learned that the more S character a hybridized orbital has, the lower energy this orbital is. Therefore, SP orbitals of C-H bonds in alkynes are lower E than SP2 orbitals of alkenes. What confuses me however is the fact that alkyne C--H are more acidic, and thus more reactive, than sp2 hybridized C--H orbitals of alkenes. I thought that reactivity corresponds to stability? Is there a concept I am missing here? Thanks?

A much easier way to think about this problem can be read on page 34 of the BR Organic Chemistry book. Basically, it states that we need to consider that the concept of acidity deals with bonds breaking in a heterolytic fashion (into ions). With an sp-orbital, the electrons are closer to the carbon nucleus than they are with the other hybird orbitals. Being that they are so close to carbon, the carbon holds on to both electrons more tightly than other hybrids (in essence, it's more electronegative). An sp-hybridized carbon can take the electron pair from the C–H bond and form a C:- and H+ more readily than longer hybrid orbitals.

 

[Engineer2MD]

Why are substances with double bonds UV active? Or has that already been discussed?

 

[tncekm]

Why are substances with double bonds UV active? Or has that already been discussed?

I'm not sure "why" is as important as knowing over what range of wavelengths that conjugated systems are photoactive--that would be the UV range, 200nm-400nm.

But, I'm not sure what level of detail you're looking for in an explanation as to "why", but it has to do with the fact that it just so happens that photons in the UV region have just the right of energy to promote an electron from a conjugated system into a higher energy state. Electrons and photons experience a 1:1 collision, and unless the photon has "just the right" amount of energy (it can be too little or too much), the electron won't be promoted to a higher energy level. If it does have the right amount of energy it will promote an electron to a higher energy level (after which time it will drop to a different energy level emitting a photon of light)

 

[ssh18]

Hi - I have a couple of quick orgo questions and I was hoping someone could help me out.

1. Nucleophiles can often attack unhybridized p-orbitals from either top or bottom side. Is there a way to know which side the nucleophile will attack from?

2. Acidic protons (protons attached to alpha carbons in carbonyl groups) are acidic because the lone pair that's left once alpha carbon is deprotonated can create resonance with the carbonyl. This provides stability..I understand that part but why does that mean that the proton is acidic (what is the relationship between being stable and acidic?)

3. Why is it that when an aldehyde or a ketone is protonated (carbonyl oxygen), the carbonyl carbon becomes more electrophilic than it was before?

Thanks so much!

 

[cheezer]

pyran with a carbocation para to the oxygen?

 

[biocmp]

1. No. Purely kinetics, if it is not being sterically hindered.

2. You first need to think conceptually about what an acid is. It is something capable of donating a H+. Now the strength of an acid is based on how likely it is to donate a H+. The more likely it donates, the stronger it is. So, if the molecule that donates a proton is able to assume a stable conformation, it is more likely to 'let go' of the proton. It's an energy thing. Everything wants to assume the lowest energy state possible.

3. edited, because it was wrong!