Organic Chemistry Question Thread

[QofQuimica]

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

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

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

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

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

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

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

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[Calheesh]

To answer question number 3:

The reason why the cabonyl carbon becomes more electrophilic (more prone to nucleophilic attack) is because you now have an oxygen with a positive charge (+) that is going to be even more electron withdrawing thus decreasing the electron density around the carbonyl carbon. You can also look at it in terms of resonance. When carbonyl oxygen gets portonated, you can see that you can delocalize the pi electrons onto the oxygen and you get a positive charge on the carbon in the other contributor.

 

[biocmp]

good answer for #3. That sounds much better!

 

[DoctorRx1986]

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!

1. In many cases, the nucleophile will tend to attack from the site that is less sterically hindered (cluttered by large groups such as a tert-butyl substituent). The reason for this is based on kinetics...it is faster to attack a group that is minimally hindered and has only protons attached than it is to attack a carbocation, for instance, that has two bulky groups adjacent to it. In the case of a carbonyl group such as an aldehyde or ketone, both of which contain sp2 hybridized carbons, the nucleophile can theoretically attack the electrophilic carbonyl from both sides and therefore has a 50% of attacking one side and 50% the other. This would result in the formation of a pair of enantiomers. In this particular case, it comes down to equal probability of attacking the sp2 carbonyl center.

2. Resonance and acidity are related because the former stabilizes the negative charge throughout the molecule, thereby giving the molecule the "tendency" to donate its proton with more ease and incur a negative charge. In other words, by losing a proton and incurring a negative charge, the molecule is still very stable because this charge can be evenly dispersed throughout several atoms. This is why the molecule is acidic.

3. Remember that oxygen has an electronegativity of 3.5 and as a result, it is quite electronegative (up there with fluorine). When protonated, it is highly positively charged and now a second resonance structure can be drawn for the carbonly, in which the double bond of the carbonyl donates its electrons to the charged oxygen...this leaves an even more positive charge on the carbon (it is now beyond being delta positive) and thus more electrophilic, susceptible to nucleophilic attack.

Hope this helps.

 

[unsung]

pyran with a carbocation para to the oxygen?

Aromatic, I believe. One of the lone pairs on the oxygen would participate in the conjugated pi system, while the other would sit it out. The cation on the C para to the oxygen ensures that every atom in the ring participates in the pi electron delocalization. The total of 6 pi electrons means it's 4N + 2, by Huckel's Rule, making it an aromatic compound.

 

[unsung]

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

I think it has to do with the fact that the pi to pi* transition occurs at UV wavelengths, whereas it takes more energy to make other transitions such as n to pi*, which would have to occur at even shorter wavelengths, that aren't registered by the UV equipment. Someone correct me if I'm wrong.

 

[tncekm]

Aromatic, I believe. One of the lone pairs on the oxygen would participate in the conjugated pi system, while the other would sit it out. The cation on the C para to the oxygen ensures that every atom in the ring participates in the pi electron delocalization. The total of 6 pi electrons means it's 4N + 2, by Huckel's Rule, making it an aromatic compound.

Ditto.

Tertiary carbocation = sp2, therefore, has p orbital available, so the system is conjugated and huckels is satisfied..

 

[cheezer]

Ditto.

Tertiary carbocation = sp2, therefore, has p orbital available, so the system is conjugated and huckels is satisfied..

ugh, i missed that on my quiz. what the hell was i thinking

 

[tncekm]

ugh, i missed that on my quiz. what the hell was i thinking

Its easy to miss... but you won't next time

 

[tncekm]

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?

Yes, flip a coin jk

 

[AtheGre]

why does that mean that the proton is acidic (what is the relationship between being stable and acidic?)
!

You might want to think of this one in terms of equilibrium and acid/base chemistry. When something is acidic, it is in part because it has a stable conjugate base (via resonance in our case with the alpha carbon) and thus the ratio of conj. base to protonated acid (the Keq) gives it a pKa value that is relatively acidic compared to neighboring carbons that are farther away from the carbonyl group.

 

[tncekm]

You might want to think of this one in terms of equilibrium and acid/base chemistry. When something is acidic, it is in part because it has a stable conjugate base (via resonance in our case with the alpha carbon) and thus the ratio of conj. base to protonated acid (the Keq) gives it a pKa value that is relatively acidic compared to neighboring carbons that are farther away from the carbonyl group.

Yes, like you said its conjugate base is stable because of the resonance, which delocalizes electrons. When a carbanion can have its lone pair delocalized that immediately increases its stability considerably because carbon doesn't like to have a lone pair at all.

 

[blastula]

Why do saturated fats have higher heat of combustion than unsaturated fats?

This is from EK Organic question 76 lecture 4. They just say that saturated fats have the highest energy, twice of carbs and proteins.

 

[tncekm]

Why do saturated fats have higher heat of combustion than unsaturated fats?

This is from EK Organic question 76 lecture 4. They just say that saturated fats have the highest energy, twice of carbs and proteins.

This is a guess... couldn't find any info.

CH4 + 2O2 -> CO2 + 2H20 + heat

I'm guessing that when a hydrocarbon is unsaturated it limits how much combustion can take place because of the lesser number of Hydrogens available to reduce O2 in the redox reaction.

e.g. 2-propene = CH2=CH-CH3, C3H6 vs CH3-CH2-CH3, C3H8.


If we simplify it to just H2 and O2, it makes more sense:
6H (or 3H2) + excess O2 -> 3H2O + 3 Heat units
8H (or 4H2) + excess O2 -> 4H20 + 4 Heat units

 

[ssh18]

Thanks so much! that is really helpful!

I have another question about nucleophilic addition reactions. When an aldehyde or a ketone reacts with a primary amine in an acidic solution, it produces an imine. What if the primary amine structure is such that there are two primary amines on both sides and a carbonyl group in the middle. Which N would act as the nucleophile? The one that's closer to the carbonyl or farther away? and what's the reason behind it? Thanks!!

 

[tncekm]

Thanks so much! that is really helpful!

I have another question about nucleophilic addition reactions. When an aldehyde or a ketone reacts with a primary amine in an acidic solution, it produces an imine. What if the primary amine structure is such that there are two primary amines on both sides and a carbonyl group in the middle. Which N would act as the nucleophile? The one that's closer to the carbonyl or farther away? and what's the reason behind it? Thanks!!

Sorry, I'm not quite understanding the question. From what I gather you want to know what would happen if:

HN-C(O)-NH reacted with R-C(O)R (R=C, H), and which HN-R would act as the Nu, but its symmetrical?

If you mean HN-N-C(O)H then there is still only one primary amine.

If you had HN-C(O)-CH2-N*H then the NH labeled (*) would be more likely to react because it would have more e- density and would be a stronger nucleophile. I don't know if it would actually react though.

 

[portugal]

I need to see what this mechanism looks like because i have never seen one like this before its all new i just started organic chem at usyd

1. The chemical action of 48%HBr on trans-1,2-dimethyloxirane results in a stereoselective reaction via protonation of the oxirane oxygen followed by ring opeing via SN2 mechanism

i need some one to show me what this mechanism would look like because i cant write it out thank you,

ps. also what would the diff be in the stereochemical outcome if it was to be an SN1 mechanism

 

[aliDO]

can somebody please tell me how 4-chloro-1-cylcohexene has a chiral carbon? Examkrackers says carbon 4 is chiral but doesn't it have two CH2 groups attached to it? I must be having a crazy moment or something.

 

[iA-MD2013]

can somebody please tell me how 4-chloro-1-cylcohexene has a chiral carbon? Examkrackers says carbon 4 is chiral but doesn't it have two CH2 groups attached to it? I must be having a crazy moment or something.

it is chiral. you're right...next to c4, there are 2 ch2 groups. but, when determining chirality, you look at the next substituent. on one side we have CH2-CH=CH-R while the other side has CH2-CH2-CH=CH-R. so, there are 4 different substituents attached to the carbon, making it chiral.
hope that helps

 

[SketchLazy]

I need to see what this mechanism looks like because i have never seen one like this before its all new i just started organic chem at usyd

1. The chemical action of 48%HBr on trans-1,2-dimethyloxirane results in a stereoselective reaction via protonation of the oxirane oxygen followed by ring opeing via SN2 mechanism

i need some one to show me what this mechanism would look like because i cant write it out thank you,

ps. also what would the diff be in the stereochemical outcome if it was to be an SN1 mechanism

Here's the mechanism for SN2:

For SN1, you would have a carbocation intermediate, so the reaction would not be stereoselective and you would have both R and S configurations.

 

[BlackSails]

Here's the mechanism for SN2:

For SN1, you would have a carbocation intermediate, so the reaction would not be stereoselective and you would have both R and S configurations.

Also note that the sterochemistry is different when you open the ring in base. Also, you can do the Sn2 even with tertiary substrates here, because of the dipole.








Question: How the hell is it possible to remember all of the carbonyl reactions?

 

[iA-MD2013]

Also note that the sterochemistry is different when you open the ring in base. Also, you can do the Sn2 even with tertiary substrates here, because of the dipole.








Question: How the hell is it possible to remember all of the carbonyl reactions?

theyre all of the same format...makes it really easy to remember. write them all down on one piece of paper. youll find that you will memorize them much faster than you expected.

 

[travelbug73]

Between these two which is a stronger nucleophile?

H3CO- and Cl-

I'm inclined to pick Cl-. I thought CN- and the halide anions made better nucleophiles. The answer is H3CO- because it is a stronger base. I did not think a stronger base necessarily made it a better nucleophile, what am I missing here?
Though the passage is about SN1 vs SN2, there is no mention of mechanism in this question. Regardless, I thought Cl- would be better for SN2 and neither would be good for SN1 since SN1 would require a weak Nu and a weak base

Thank you in advance.

 

[RSAgator]

Stronger bases generally make better nucleophiles unless they are sterically hindered (preventing a nucleophillic attack). If something is a stronger base, it essentially means that it more strongly seeks to form a bond in order to become more stable. This is why you have nucleophillic attack. The nucleophile, an electron donor, seeks to become more stable by forming a bond with the electrophile, an electron acceptor. Another way to think of it is in terms of the leaving group. Cl- is a great leaving group because it is very stable in solution.

Also consider HCl. HCl is a strong acid because it has a strong tendency to disassociate into H+ and Cl-. This is because the Cl- ion is very stable in solution. The conjugate of a strong acid is a weak base, and we know a weak base has less tendency to form a new bond (which is why H+ and Cl- don't spontaneously reform HCl). This is also why a strong base such as OH- has a high tendency to act as a nucleophile and is a poor leaving group. A lot of organic chemistry can be explained by stabilities, and when you consider how strong a nucleophile something is you can also draw conclusions about how weak a leaving group it is.

 

[prettyslick]

What is the difference between carbonyl group and acetyl group? Both have C=O double bond. Whats the difference between the two?

 

[tncekm]

An acetyl group is different from a carbonyl group, although it does contain a carbonyl group.

C=O is a carbonyl, as you know.

So, something with R-CH₂-(C=O)-R is a ketone with a carbonyl group.

Now, something with R-CH[(C=O)-CH&#8323]-R would have an acetyl group. The acetyl group is like having an acetone hanging off the edge of a larger parent carbon chain.

http://www.crscientific.com/diagram%20-%20Ac%20groups1a.jpg

 

[prettyslick]

So Acetyl group is:

And Carbonyl group is simply C=O double bond.

so, in other words, an acetyl group HAS a carbonyl group in it??

 

[iA-MD2013]

So Acetyl group is:

And Carbonyl group is simply C=O double bond.

so, in other words, an acetyl group HAS a carbonyl group in it??

yep, exactly.

 

[Kaustikos]

Couple questions for you guys.

1) For absolute configuration of an organic compound, can you determine it from a line/non-3d arrangement? If so, how do you go about doing it? Do you use the fischer method and assume horizontal is sticking out and vertical is sticking in the background? I get easily tripped up on this and would appreciate an explanation.

2) This is more of a confirmation than question, but the only way for a carbon to be chiral other than having 4 different substituents is to have 3 or less substituents with a double bond/triple bond, correct?

Thanks.

 

[tncekm]

From a fischer or hawthorn projection you can find the absolute configuration, but IMHO those would be classified as "3D" just as the dashes and wedges would. You cold also look at boat/chair/etc conformations. However, other than I can't think of any others.

But really, you need to understand whether or not the depiction you're looking at give you 3D information, even if it isn't in the wedge-dash form.

 

[Kaustikos]

This is about Michael Additions and Aldo Condensation.

In Aldol Condensation, we get the usual product that, at high temps, can further condense into water and a form of an aldehyde with a double bond alpha to the carbonyl. They state that aldol condensation primarily occurs with similar aldehydes/ketones (acetaldehyde with acetaldehyde).
Remember that product.


In Enol formation, one can make an enolate anion with a strong base (like the one we form in aldol condensation), and that the enolate can attack alkene groups alpha/beta to carbonyl carbons and form an entirely different product without the double bond.

So my question; would an aldehyde/ketone form an enolate and then attack the final product of a aldol condensation and make an entirely different product? Or is there something restricting the aldehyde (acetaldehyde in this case, since reactants are the same) from doing that and just going into aldol condensation instead without forming the michael addition?

 

[Kaustikos]

Here's the mechanism for SN2:

For SN1, you would have a carbocation intermediate, so the reaction would not be stereoselective and you would have both R and S configurations.

That's actually SN1, nto SN2. That's an acid-based cleavage of an oxirane which causes a carbocation intermediate to form on the most stable carbon. The bromide anion would then attack that cation and form the product.

For an SN2, you would start with the bromide anion attacking the least sterically hindered carbon and then have the OH form from the resulting H in solution.

 

[tncekm]

This is about Michael Additions and Aldo Condensation.

In Aldol Condensation, we get the usual product that, at high temps, can further condense into water and a form of an aldehyde with a double bond alpha to the carbonyl. They state that aldol condensation primarily occurs with similar aldehydes/ketones (acetaldehyde with acetaldehyde).
Remember that product.


In Enol formation, one can make an enolate anion with a strong base (like the one we form in aldol condensation), and that the enolate can attack alkene groups alpha/beta to carbonyl carbons and form an entirely different product without the double bond.

So my question; would an aldehyde/ketone form an enolate and then attack the final product of a aldol condensation and make an entirely different product? Or is there something restricting the aldehyde (acetaldehyde in this case, since reactants are the same) from doing that and just going into aldol condensation instead without forming the michael addition?

After the aldol has formed and condensed it is conjugated, so I don't see why you wouldn't get michael addition products if you were to react react it with something like HBr. And, there doesn't seem to be anything that would further stop you from reacting a ketone or aldehyde (after the enolate is formed) with your product, either, in another aldol addition, condensation.

 

[iA-MD2013]

This is about Michael Additions and Aldo Condensation.

In Aldol Condensation, we get the usual product that, at high temps, can further condense into water and a form of an aldehyde with a double bond alpha to the carbonyl. They state that aldol condensation primarily occurs with similar aldehydes/ketones (acetaldehyde with acetaldehyde).
Remember that product.


In Enol formation, one can make an enolate anion with a strong base (like the one we form in aldol condensation), and that the enolate can attack alkene groups alpha/beta to carbonyl carbons and form an entirely different product without the double bond.

So my question; would an aldehyde/ketone form an enolate and then attack the final product of a aldol condensation and make an entirely different product? Or is there something restricting the aldehyde (acetaldehyde in this case, since reactants are the same) from doing that and just going into aldol condensation instead without forming the michael addition?

You're right...it can happen. This is the reason why these reactions are so messy. It is a mixture of thermodynamics and kinetics that allows one to determine what are the best reaction conditions to forming the desired product (in this case, the michael addition product). It may require much higher temperatures for the side reaction to occur...therefore, running the reaction at 0 deg may allow for a minimal amount of side product. It must be determined experimentally.

 

[Kaustikos]

Thanks guys

 

[What up doc]

why is pyridine aromatic? are all rings with n's in them aromatic?

 

[Kaustikos]

why is pyridine aromatic? are all rings with n's in them aromatic?

That wouldn't always be the case, I think. Pyridine and Pyrrole are the only ones I know of off the top of my head. Pyridine because it doesn't donate its electrons to the group, thus making the 4n+2 pi rule fit for the aromatic and pyrrole DOES donate, making the rule fit for it as well. I wouldn't be surprised if there was a ring structure with an N that didn't fit this and wasn't aromatic.

 

[tncekm]

Exactly.

Nitrogen's lone pair of electrons can contribute to the resonance forms, so its electrons are included in 4n + 2 rule.

That means there are 6 conjugated, contributing electrons. So, 4n + 2 = 6 --> n = 1, so its aromatic.

 

[Kaustikos]

Exactly.

Nitrogen's lone pair of electrons can contribute to the resonance forms, so its electrons are included in 4n + 2 rule.

That means there are 6 conjugated, contributing electrons. So, 4n + 2 = 6 --> n = 1, so its aromatic.

Just for clarification; that above is pyrrole, not pyridine.

 

[tncekm]

Yeah, my bad I had to google pyrrole to get the image, and I still wrote the wrong name

 

[IronMan2009]

Hey just wondering if there is a good rule of thumb to have when you have withdrawing groups and donating groups on the same compound. Take for example and alkoxide ion. It is a stronger base than water but you have the carbonyl group on the end of the molecule which is electron withdrawing which is 2 carbons from the basic oxygen. You also have the electron donating alky group which is directly attached to the carbon with the basic oxygen on it.

How do you know that the carbonyl group wouldn't withdraw enough of the charge as compared to the alkyl group donating it? Is this solely by the distance? If so is that how we can say its a stronger base than water?

 

[Kaustikos]

Hey just wondering if there is a good rule of thumb to have when you have withdrawing groups and donating groups on the same compound. Take for example and alkoxide ion. It is a stronger base than water but you have the carbonyl group on the end of the molecule which is electron withdrawing which is 2 carbons from the basic oxygen. You also have the electron donating alky group which is directly attached to the carbon with the basic oxygen on it.

How do you know that the carbonyl group wouldn't withdraw enough of the charge as compared to the alkyl group donating it? Is this solely by the distance? If so is that how we can say its a stronger base than water?

That's a good question. Generally, the carbonyl group is more withdrawing than the alkyl's donating group. The distance is what factors into the basicity of the oxygen in that how far away that withdrawing group is. The more alkyl groups, the more basic the oxygen becomes.
I also tend to look at it from the perspective of activating/deactivating for aromatics which goes by the rule that if the oxygen/nitrogen is directly connected to the aromatic, it can donate the electrons to the pi orbitals and if the oxygen is one carbon away, it tends to take the electron. So, if you have the O=CH2-CH2-O (example), then the withdrawing capabilities of the carbonyl group isn't so great because of how far the oxygen is from the oxygen.

To answer your second question, I don't know. It seems to me that the alkyl group's electron-donating capabilities are what make the alkoxides more basic than water.

 

[IronMan2009]

Thanks for the tip.

One more question, as far as equilibrium goes with organic reactions, the reaction tends to form the product that has the most reactive base as a leaving group so it won't leave and react in an aqueous solution? This would make the overall compound with the highly reactive basic leaving group less reactive in general?

So comparing acyl chloride R-C-O-Cl with an amide R-C-O-NR2, the amide is less likely to be reactive in a nucleophilic substitution because if the O-NR2 bound is broken then NR2 anion is a much stronger base than Cl anion.

And I guess to further up on that thought, NR2 is a strong electron donating group so the dipole that is created around the electrophillic carbonyl carbon would be less positive because the nitrogen's electrons would donate into that dipole and cancel the positive out?

Thanks for the help, April 18th MCAT'er here, gotta get these last trends ironed out.

 

[161927]

Thanks for the tip.

One more question, as far as equilibrium goes with organic reactions, the reaction tends to form the product that has the most reactive base as a leaving group so it won't leave and react in an aqueous solution? This would make the overall compound with the highly reactive basic leaving group less reactive in general?

So comparing acyl chloride R-C-O-Cl with an amide R-C-O-NR2, the amide is less likely to be reactive in a nucleophilic substitution because if the O-NR2 bound is broken then NR2 anion is a much stronger base than Cl anion.

And I guess to further up on that thought, NR2 is a strong electron donating group so the dipole that is created around the electrophillic carbonyl carbon would be less positive because the nitrogen's electrons would donate into that dipole and cancel the positive out?

Thanks for the help, April 18th MCAT'er here, gotta get these last trends ironed out.

Yes the chloride withdraws electron density from the carbonyl carbon, making it more electrophilic and vulnerable to nucleophilic attack. When comparing Cl- vs NR2- as leaving groups, Cl is better because it the charge is less polarized. It can be spread over the large electronegative element, whereas the nitrogen is more basic in its anionic form. These are two reasons why R-CO-Cl is more susceptible to nucleophilic attack than R-CO-NR2.

The third, most important, reason is that the amine lone pair can resonate with the carbonyl pi bond. Disruption of the resonance stabilization is not favored, and this is why nucleophilic attack on the carbonyl or protonation of the amine may not be the preferred reaction. This is the same reason for why a carboxylic acid is unlikely to undergo attack by a nucleophile.

 

[wannabedocta]

Say we have aldehyde and a ketone. How would we determine which would become the nucleophile and electrophile? I'm assuming the molecule with the more acidic alpha hydrogen would be the nucleophile. What is more acidic, ketone or aldehyde??

 

[Kaustikos]

Say we have aldehyde and a ketone. How would we determine which would become the nucleophile and electrophile? I'm assuming the molecule with the more acidic alpha hydrogen would be the nucleophile. What is more acidic, ketone or aldehyde??

Well, first off, aldehydes and ketones don't tend to react with each other, but instead with their own kinds. So there wouldn't be any adol condensations for these two, I don't think.

In terms of nucleo/electrophilicity; I would have to guess that the H- bonded to the carbonyl carbon is what determines how nucleophilie/electrophilic the compounds are. The fact that the carbonyl compound for the ketone is more substituted probably has something to factor into the acidity of the alpha hydrogen. My guess is that since an aldehyde has no alkyl group (electron donating and which would thus decreas the acidity) makes it a better nucleophile than a ketone. A ketone, on the other hand, has an alkyl group to which donates electrons to decrease the nucleophilic capabilities of the ketone. But that's just guessing.

I'm almost positive that like reacts with like for aldehydes and ketones. So if you put the two together, you would have aldol condensation of likes instead of nucleophilic attack by one over the other.

 

[iA-MD2013]

Well, first off, aldehydes and ketones don't tend to react with each other, but instead with their own kinds. So there wouldn't be any adol condensations for these two, I don't think.

Well...that's not exactly true. You can have mixed aldol condensations, but it's not emphasized as much in ochem because we're more concerned with an aldol condensation occurring between the same molecule in one container. They're both acidic and the differences in their acidity is negligible, which is why you will get 4 different products (even more because the products can also undergo aldol condensation with the SM)...but, in general, an aldehyde is more acidic than an ketone. This is because the ketone has methyl groups to stabilize the positive charge on the carbonyl. The aldehyde has more of a positive charge on its carbonyl...which makes a negative charge on the alpha carbon more stable.

 

[Kaustikos]

Well...that's not exactly true. You can have mixed aldol condensations, but it's not emphasized as much in ochem because we're more concerned with an aldol condensation occurring between the same molecule in one container. They're both acidic and the differences in their acidity is negligible, which is why you will get 4 different products (even more because the products can also undergo aldol condensation with the SM)...but, in general, an aldehyde is more acidic than an ketone. This is because the ketone has methyl groups to stabilize the positive charge on the carbonyl. The aldehyde has more of a positive charge on its carbonyl...which makes a negative charge on the alpha carbon more stable.

Sorry, I worded tha incorrectly. I didn't mean to say that they never. And again, like he said also, the alkyl group on the ketone is what affects the acidity.

 

[wannabedocta]

Ok, great thanks. I realized where I went wrong. For an aldol condensation with an aldehyde and a ketone, I deprotonated the lone H on the aldehyde, rather than the alpha hydrogen bound to the R group of the aldehyde so my resultant molecule came out really funky.

 

[wannabedocta]

Why is that an alkene reacts with HBr to form the anti-markovnikov product in the presence of peroxide yet if it reacts with HI or HCl it forms the markovnikov product instead?

 

[iA-MD2013]

Why is that an alkene reacts with HBr to form the anti-markovnikov product in the presence of peroxide yet if it reacts with HI or HCl it forms the markovnikov product instead?

who cares why...just know it
It's actually because the bond dissociation energies of the homolytic cleavages makes it more favorable to form the antimarkovnikov product. I think it's important to know peroxide indicates that a radical reaction is going on...I don't think anything else is important.