[Haaaaaaanh]

For this reaction http://i.imgur.com/Y2Y1c.png?1?5611

We have to show how it occurs and it has to be subsitution because we haven't "learned" elimination yet. By this I mean it is a practice question for a test coming up where elimination will not be tested. I have found an answer but it relies on the fact that the carbon that is bonded to the oxygen in the products that is also bonded methyl group, I assumed it has an unwritten hydrogen. I got both products with the assumption that the H3PO4 would remove the hydroxide group and a hydrogen would shift from the benzene ring because it is better to have a + charge on that ring so it can stablize. I got the product but I am not 100% sure it is correct again because there is a possibility that the carbon bonded to the oxygen and methyl simply does not have a hydrogen. This is why I am wondering if this reaction is possible if the product is supposed to have a carbon bonded only to a ch3 and oxygen with no double bonds and basically not fulfil its octet.

[PhDoc]

It's very important to forget "hydroxide" in Organic Chemistry when dealing with carbon compounds. Hydroxide (from sodium or potassium hydroxide) was pounded into you in General Chemistry.

When you see the OH group bound to carbon, it's an alcohol.

With respect to your problem, why is the H3PO4 present in the equation? There's a certain mechanism at play here. Identify the most nucleophilic center, and also identify the most electrophilic center. What happens when the nucleophile goes after the electrophile? Another electrophile is generated. What happens with this new electrophile?

Please read the article, 95% of Organic Chemistry in 5 Minutes.

http://lennoxtutoring.com/2012/05/12/95-of-organic-chemistry-in-5-minutes/

You'll realize that mostly everything you're learning in Organic Chemistry has to do with a nucleophile going after an electrophile. The **trick** is in learning what's the most nucleophilic center, and what's the most electrophilic center.

Please also review your acid pKa data. Knowledge of pKa will help you ace the substitution / elimination exam.

Best of luck to you!

[Haaaaaaanh]

Ok sorry, I need to review my functional groups which is my weakness. However I was taught that if a nucleophile is too "bulky" then it will not proceed. Is this correct? Or is it that it will be a weak nucleophile but a nucleophile nonetheless? I recently learned substitution so what I do is locate a leaving group which I find easy because usually it is a halogen or something obvious. However in this I can't seem to find one. I was thinking it was solvolysis, which is SN1 reaction and I had a vague idea that the solvent interacts with the nucleophile after the attack but I am not sure what the leaving group is. Thank you for the article I will read it right now.

[PhDoc]

Bulky nucleophiles can react with methyl and primary alkyl halides, in addition to analogous compounds containing good leaving groups. The common bulky nucleophiles in undergraduate organic chemistry doing this are tert-butoxide, tert-butylamine and diisopropylamine. Amide bases, i.e. NH2(-), RNH(-) or R2N(-), are far more basic than nucleophilic; hence they're horrible nucleophiles.

Something is a very good nucleophile if its conjugate acid is </= 10 (oxygen resonators don't count); something is a very good leaving group if its conjugate acid is < 0. This is why knowledge of pKa is so EXCEPTIONALLY important.

Knowledge of pKa in addition to the equation,

Nu(-)  +  E-X  ----->  Nu-E   +   X(-)

will get you through 95% of organic chemistry. It's that simple.

[Haaaaaaanh]

OK thank you very much now I understand how to use pKa. So I tried it again and I am not sure if this is correct but I wrote that the alcohol group on the Carbon that is also attached to a deterium gets protonated by the H3PO4. This makes h2O and h2Po4 and the H2O gets attacked by the nucleophilic OH (other alcohol) which makes an oxonium ion. The extra H+ then gets taken by the H2PO4 and that is one of the products? Is that correct because that is what my friend told me but I am not sure if it is correct or not and I was wondering if maybe this reaction has a hydride shift?

[PhDoc]

Which carbon will form the most stable carbocation intermediate upon protonation and departure of the alcohol group? Why is this carbocation stable?

Take note that you're generating diastereomers. What does this tell you about the mechanism of the reaction?

There is no hydride shift. What is the thermodynamic "payoff" for a hydride shift?

You're making progress. Solving this one will be a good learning experience. Please keep going. Once you've solved the problem, you can tutor your friend.

 

[Haaaaaaanh]

Ok thanks I really want to solve this problem as I feel it will bring me confidence because our proffesor said no one has solved this problem before as a student in her class and I have until friday to solve it.

Ok So the most stable carbocation would definitely be the C-Ch3 carbon because of the methyl group and its electron donating properties right? I am pretty sure about this. I see that to generate a diasteromer there has to be an SN2 reaction otherwise there can't be an inversion because to my knowledge an SN1 reaction is racemized because it can attacked from either side but I could be wrong.

Ok so since it is a diasteromer does that mean that a molecule has to be rearranged since an inversion would cause an enatiomer. This is where I am stumped like is this correct that after protonation of the alcohol group the H3Po4-> H2PO4- which is charged and more nucleophilic? That way it can attack and form the C-O bond on one of the carbons but I don't understand how the rest of the molecule (h2Po3) leaves and I think since this is solvolysis that the solvent of H3PO4 must react with it somehow but that is where I am stumped so can you please steer me in the right direction or help me see something I may be overlooking. Thanks for the help by the way!

[NL-bucky]

Ok thanks I really want to solve this problem as I feel it will bring me confidence because our proffesor said no one has solved this problem before as a student in her class and I have until friday to solve it.

Ok So the most stable carbocation would definitely be the C-Ch3 carbon because of the methyl group and its electron donating properties right? I am pretty sure about this. Correct!I see that to generate a diasteromer there has to be an SN2 reaction otherwise there can't be an inversion because to my knowledge an SN1 reaction is racemized because it can attacked from either side but I could be wrong. For the top carbon you get both diastereomers --> i.e. racemization, SN1 as you established before. For the bottom one, you say there is inversion. Are you sure?

Ok so since it is a diasteromer does that mean that a molecule has to be rearranged since an inversion would cause an enatiomer. This is where I am stumped like is this correct that after protonation of the alcohol group the H3Po4-> H2PO4- which is charged and more nucleophilic? That way it can attack and form the C-O bond on one of the carbons but I don't understand how the rest of the molecule (h2Po3) leaves and I think since this is solvolysis that the solvent of H3PO4 must react with it somehow but that is where I am stumped so can you please steer me in the right direction or help me see something I may be overlooking. Thanks for the help by the way! The "lysis" in solvolysis means a bond is broken, "solvolysis" means as much as "a bond is broken under influence of the solvent". That is what you established in the first paragraph, wasn't it? Now, identify the strongest nucleophile.

[Haaaaaaanh]

Isn't there an inversion because the Deterium was at the front now it is at the back and the hydrogen is at the back now it is at the front? I am unsure what a better nucleophile is would it be the OH or the H2PO4-?

[PhDoc]

What happens to the stereogenicity of a center that becomes a carbocation? Once the carbocation is formed, is there any stereofacial preference for attack by an incoming nucleophile, i.e. is there any preference for the "top" over the "bottom" of the sp2 hybridized center?

What happens when you rotate carbon-carbon bonds?

Is anything really happening to the chiral center on the bottom? Determine the R/S configuration before and after reaction. Is it the same?

What do your answers thus far to my questions tell you about the mechanism of this reaction?

You're really getting close to a solution. Keep on going... the learning is in the struggling to master the concepts.

[Haaaaaaanh]

Ok so wouldn't the nucleophile attack from the back or bottom where the hydrogen is because the CH3 would like block it off I guess? Or would it attack from where the h2o is leaving I am unsure because our prof mentioned how the leaving group sometimes blocks off the nucleophile which leads to a more likelihood of a bottom attack. Oh so I guess the prof just wanted to trick us by doing that, so if it is still R/S that means that an inversion did not occur unless for example a Br is replaced by something of lesser priority? Your questions make me feel that this is completely an SN1 reaction based on there is no backside attack which means that the nucleophile must be weak. SO would OH- be the nucleophile?

[PhDoc]

The pKa of the alcohol functionality is approximately 16. Protonation leads to a functionality with a pKa of approximately -1.5. This is an excellent leaving group, so it leaves as water with zero charge. The result is formation of a stable secondary benzylic carbocation on the top of your molecule.

What can happen with bond rotations in both the top and bottom of the molecule to generate both diastereomers observed? What is the nucleophile leading to ring closure? What is the most abundant basic species in solution that will abstract the "final proton" leading to a neutral species?

[Haaaaaaanh]

OK so I understand the OH becoming the leaving group after being protonated into water.

The bond rotations on the top and bottom hmm... the only way I know it can change is by a backside and become inverted I don't know anything else sorry I am not that smart at this stuff so maybe the concept of newman projections and how anti formations are wanted  so the bond could rotate?

I still can't figure out the nucleophile either I am guessing it is H2PO4- after donating its hydrogen but even then it is still an acid. The most abundant basic species would it be the other OH on the other carbon?

[PhDoc]

To understand what's happening in this reaction, it's important to realize there is no true "backside attack" in SN1 reactions. The backside attack to which you refer is endemic to the SN2 reaction where the HOMO of incoming nucleophile interacts with the LUMO (sigma antibonding orbital) of the electrophile. That's not happening here.

The top group is the one forming the most stable carbocation because it's benzylic and secondary. This is more stable than the benzylic and primary carbocation that would form on the bottom. Most importantly, this is what explains the outcome of the reaction.

Once that top carbocation is formed, it's sp2 hybridized and trigonal planar. Keep in mind it can rotate to place the methyl group in the front or the back.

You need to rotate bonds on the bottom carbon to place oxygen in the correct trajectory for bonding with the empty p-orbital, which undergoes concomitant rehybridization to an sp3 orbital.

The last step in the process in removal (HPO4- [pKa ~7] / H2O [pKa 15.7]) of the H+ from the protonated ether. The most basic species in solution is the alcohol [pKa ~16]. Since H2O is the second most basic species in solution (delta pKa = 0.3), it may also be involved. The proton might be abstracted by another secondary benzylic alcohol, thereby setting it up for the reaction.

It's unlikely the HPO4- will be the base in the deprotonation step (delta pKa = 9).

This is the answer to your question. The answer has very little value. What's important is that you understand the logic behind the answer.

Your success in Organic Chemistry depends upon you ability to apply the information learned in a logical manner. What was important here was (1) identification of the mechanism based upon the results; (2) deduction that pKa data as well as carbocation stability support the results; (3) identification of the nucleophile(s); and (4) identification of the penultimate electrophile.

The mechanistic Organic Chemistry will continue to build. Even though this exam is challenging, down the road you'll consider it easy in comparison to that on the horizon. The learning is in the struggle.

Good luck!

[Haaaaaaanh]

Thank you very much for helping me I just want to clarify so basically the top Carbon's alcohol is protonated  I understand that and the H3PO4 becomes a H2PO4- so afterwards does the carbon at the bottom's alcohol also get protonated by the H2PO4- turning it into HPO42-? Or does the H2PO4- bond with the top carbocation and does not protonate the bottom alcohol giving us C-O-POOOH? Sorry don't know how to phrase it basically the O that lost its hydrogen for protonation of alcohol bonds with the carbon since it is the most negative? Then does the alcohol group at the bottom replace the O that bonded with the carbon leaving us with C-O and a carbocation on the bottom which bonds with the O to give us the product?