Enable contrast version

Tutor profile: Zachary S.

Online
Zachary S.
Ph.D. Candidate in Organic Chemistry
Tutor Satisfaction Guarantee

Questions

Subject: Organic Chemistry

TutorMe
Question:

As part of a total synthesis effort to make a natural product, you and your thesis adviser decide that the reduction of an ester to an aldehyde should be part of the synthetic strategy. When you go to your lab you find two reducing agents that can perform the transformation: one is Di-Isobutyl Aluminum Hydride (DIBAL-H), and the other is Lithium Aluminum hydride (LAH). Which reagent is more desirable for the given transformation? Assume that the rest of your molecule is inert to these two reagent, only one equivalent of the reagent is to be used, and the temperature of the reaction is -70 °C.

Inactive
Zachary S.
Answer:

Lets start this discussion off by saying hopefully you have more than these two reducing agents lying around in your lab as these are not the most fun reagents to deal with. Since I cannot draw structures here, I would advise you to Google these structures and get familiar with them. I would also encourage you to write down the chemical reaction of the ester and the reducing agent on the left side of the arrow, and the aldehyde on the right side of the arrow. The first thing I always make my students do when approaching a mechanistic problem is to identify the nucleophile and the electrophile of the reaction. This step is critical. I promise you that organic chemistry will become so much easier if you can identify the "source" of the electrons and the "sink". Electrons flow from areas of high electron density to areas of low electron density. I call this flow of electrons "sink-to-source". Take a moment to identify the nucleophile and the electrophile. Proceed when you have an answer. Hopefully you found clues in the names of the reagents and realized that a hydride is a nucleophilic hydrogen atom with an extra electron in its valence shell, thus giving it an overall negative charge. Dead give away for a nucleophile (95% of the time). That obviously makes the ester the electrophile. Now lets get specific. Where is the electrophilic site on the ester? Well, there are three options: the carbonyl oxygen, the carbonyl carbon, and the ester oxygen. Take a moment to decide the most rational electrophilic spot. Proceed when you have an answer. Both the carbonyl oxygen and the ester oxygen have lone pairs present around them plus increased electronegativity compared to that of carbon which gives them a partial negative charge. Not necessarily the properties of good electrophiles. Furthermore, we know that carbocations can be pretty darn stable under the right conditions which means that carbon "doesn't mind" bearing the partial positive charge; especially because it is adjacent to 2 relatively electronegative oxygens. This makes the carbonyl carbon a rational electrophile. Lets get into the meat of the problem. Now that we have identified our nucleophilic hydride and our electrophilic carbonyl carbon, lets take a look at what happens at when they react with one another. To avoid having a Texas carbon after the hydride attacks, the only logical place for the electrons to move is to the carbonyl oxygen (Everything is bigger in Texas. In this case its a carbon with 5 bonds. We can't be having any of those.).* Now the negatively charged electron rich oxygen can attack the aluminum species and form a complex. Here is where the differences in the reagents come in. When DIBAL-H gets attacked, the hemiacetal that forms is stablized until work-up, at least at -70 °C and when only one equivalent of DIBAL-H is present. Additionally, the bulky and electron donating isopropyl groups also helps stabilize the oxygen-aluminum complex. In the case of LAH, there are multiple potential hydrides present and no electron donating groups to stabilize the oxygen-aluminum complex. What happens in this scenario then is that, even with one equivalent, LAH will continue to reduce the ester all the way down to an alcohol. Back to the question at hand. Which one is better for the reduction of an ester to an aldehyde? The answer is DIBAL-H because it is more mild, and the oxygen-aluminum hemiacetal that forms is stable until work-up. Now if you intend to reduce the ester all the way down to the alcohol, then go nuts with LAH. Although not too nuts. LAH is a dangerous reagent. Also, if you plan to increase the temperature or add excess of the reducing agent, then neither will work. For a visual representation of the mechanism, please visit the following link: https://www.masterorganicchemistry.com/2011/08/26/reagent-friday-di-isobutyl-aluminum-hydride-dibal/ *In the case of DIBAL-H, the carbonyl oxygen actually attacks the organoaluminum compound first before the hydride attacks the carbonyl carbon.

Subject: Chemistry

TutorMe
Question:

You are in the laboratory attempting to run an organic reaction in water. As you monitor the reaction, you notice that your desired product is starting to degrade. Later, it is discovered that one of your reactants generates an acidic byproduct (hydrochloric acid) that you suspect is the cause for product degradation. To test this hypothesis, you want to try to run the reaction in a sodium phosphate buffer (pKa 7.2) to prevent any large changes in pH below a neutral level. Approximately what is the minimum pH of the buffer required to prevent a pH change below 7.0 over the course of the reaction assuming that 20 mmol of acid is generated during the reaction.

Inactive
Zachary S.
Answer:

When I took my introductory chemistry class, I used to look at these problems and go "Oh no! ICE tables!" No need to fear though. Just because you see an acid/base equilibria problem doesn't mean it has to be difficult. Although ICE tables are not that bad once you get used to them. It just takes some practice. In fact, you do not even need an ICE table for this one. Let me explain. Our objective is to find the minimum pH necessary of our buffer to prevent a drop in pH below 7.0. This is because we suspect our product to degrade under acidic conditions (i.e. below 7.0). First, let us write out our balanced chemical equation (charges are listed in parentheses. We will omit spectator ions here as they are not necessary for the problem): H2PO4(-1) + H2O <-----> H3O(+1) + HPO4(-2) Now, we know that dihydrogen phosphate is a weak acid as demonstrated by the chemical equation. Therefore, the hydrogen phosphate on the right side of the equation is our conjugate base. We also know that we are using a buffer. When you hear the word "buffer" the Henderson–Hasselbalch (HH) equation should be popping up in your head! Let us write out the HH equation: pH = pKa + log([conjugate base]/[acid]) Remember, the buffer region of a titration curve is below the equivalence point, or the point where the moles of base in solution are equal to the moles of acid (the point where the slope changes from positive to negative). This means that the buffer region will "resist" changes in pH. At this point, we want to determine the molar ratio of the conjugate base to the acid WHEN THE pH IS AT 7.0. This will serve as the bottom limit molar ratio that the constraints of the problem allow for. Luckily, the pKa of the sodium dihydrogen phosphate was given to us in the problem. 7.0 = 7.2 + log(x) -0.2 = log(x) 10^(-0.2) = 10^(log(x)) 0.63 or 6.3/10 = x Let's make the numbers easy on ourselves and say 6/10, or 3/5. Perfect! Now we know that no matter what, we can only consume conjugate base with the acid generated from the reaction until 3 equivalents of conjugate base remain. Conversely, we cannot generate more than 5 equivalents of dihydrogen phosphate. Since we know that we are generating 20 mmol of hydrochloric acid (or 20 equivalents) during the reaction, the question is what MINIMUM ratio of conjugate base to dihydrogen phosphate in the buffer solution at the BEGINNING of the reaction will be required to achieve a 3:5 ratio of conjugate base:dihydrogen phosphate at the END of the reaction. Reread this sentence again for yourself (Its a mouthful!). Before we answer this, lets use some basic logic regarding the consumption of the conjugate base. We know that HCl is a strong acid (i.e. it ionizes 100%). Therefore, one can assume that the when one equivalent of hydrochloric acid is generated, one equivalent of base is consumed. Additionally, one equivalent of dihydrogen phosphate is generated as a product of the consumption of base. In other words, when we take one equivalent from the conjugate base, we give one equivalent to the dihydrogen phosphate. Here is the equation: HPO4(-2) + HCl <-----> H2PO4(-1) So, what MINIMUM ratio will be required AT THE BEGINNING OF THE REACTION to achieve the end molar ratio of 3:5 conjugate base:dihydrogen phosphate? In this case, it will be 5:3. Lets get into why: Since the amount of HCl generated is 20 mmol (or 20 equivalents) lets multiply our numbers by 10 to make the math easy (our ratio will be 50:30 conjugate base:dihydrogen phosphate). As previously explained, when one equivalent of HCl is generated, one equivalent of base is consumed. I will also reiterate that one equivalent of dihydrogen phosphate is also generated as a result. So, after the first equivalent of HCl is generated the ratio will be 49:31. After the second equivalent of HCl is generated, the ratio will be 48:32. After the third, 47:33. This will repeat until all 20 equivalents of HCl are consumed by conjugate base to give you a final molar ratio of 30:50 conjugate base:dihydrogen phosphate. Ok. Deep breath. Lets summarize what we have done so far. The first step we took was to determine the end molar ratio of conjugate base to dihydrogen phosphate by using the HH equation. This was the lower limit constraint. After we determined this, we calculated the minimum molar ratio of base:acid that would be required at the beginning of the reaction to achieve the desired end molar ratio of 30:50. We determined it was 50:30. One last step (Yay!) to determine the minimum pH of the buffer required to keep the pH of the reaction from falling below 7.0 after the reaction has completed. All we have to do now is plug our numbers back into the HH equation to determine what pH our buffer should be at. pH = 7.2 + log (50/30) pH = 7.42 or approximately 7.4 We did it! Our pH of our buffer needs to be at least 7.4 to avoid falling below 7.0 during the reaction and causing our product to degrade. **********I like to expose students to a different way of thinking than what normally might be taught to help them understand what is happening. When I first started learning chemistry, I would just be shown the math without really understanding how to get there. Here, a standard ICE table may have been used to achieve the same goal. Afterall, this is essentially a titration problem! The only difference is the HCl is being generated as a product from the reaction instead instead of adding it yourself. This actually makes it easier because now the concentration is staying constant throughout. You could have just as well calculated the correct concentration of protons necessary from the ICE table, and converted that to the pH. I essentially just drew out the ideas that goe into the ICE table method. You should approach these kinds of problems in a way that makes sense to YOU. There is no one size fits all when it comes to learning. This is a big part of my philosophy when I teach.

Subject: Basic Chemistry

TutorMe
Question:

Polyrotaxanes (often referred to as molecular necklaces) are supramolecular structures in which multiple cyclic molecules, or rotors, are non-covalently threaded onto a polymer, or axle. You, an expert in making these compounds, have successfully created one as a potential therapy for a disease and are ready for a mouse study (Yay!). You are able to thread 15 copies of a rotor onto the polymer. The rotor is the active component of the molecule for the disease, and the polymer is simply a delivery vehicle for the drug. Being the good scientist you are, you want to compare your synthesized therapy to a proper control (the unthreaded, or "free", cyclic molecule). You want to make sure that the amount of rotor on the polyrotaxane injected into the mouse is the same as the free rotor being injected to accurately compare the control to your therapy. You inject 5 mg of the free rotor into the mouse. How much polyrotaxane (MW 20,000 g/mol) should be injected to achieve an equivalent amount of free rotor?

Inactive
Zachary S.
Answer:

Ok. Take a breath. The first way you should attack this problem is to digest what the problem is asking. Basically, we want the amount of rotor injected from the polyrotaxane to be the same as the amount of free rotor. That means this is a stoichiometry problem! So, what do we know? Well, we know that: 1. There are 15 equivalents of the rotor present on the polymer 2. The molecular weight of the free rotor is 1000 g/mol 3. The molecular weight of the entire polyrotaxane is 20,000 g/mol 4. We injected 5 mg of the rotor into the mouse We want to start the set up of the problem with the amount of free rotor injected because the other numbers are just used as ratios. As we know, chemists like to speak in terms of MOLES, not grams. So lets start by converting the injected amount of free rotor into moles: 0.005 g free rotor × (1 mol free rotor)/(1000 g) = 5 x 10^-6 moles or 5 µmol of free rotor Next, we need to convert the amount of moles of free rotor into moles of polyrotaxane. Since there are 15 units of rotor per 1 unit of polyrotaxane we can do this conversion! Watch: 0.000005 mol rotor × (1 mol polyrotaxane)/(15 mol rotor) = 3.33 x 10^-7 mol polyrotaxane or 0.33 µmol Time to bring it home. Lets put this problem to bed by converting the moles of polyrotaxane in terms of something that we can actually weigh out on a mass balance in grams 0.00000033 mol polyrotaxane × (20,000 g)/(1 mol polyrotaxane)= 66 mg polyrotaxane Therefore, to achieve an equivalent dose of 5 mg free rotor, we would need to inject 66 mg of polyrotaxane :)

Contact tutor

Send a message explaining your
needs and Zachary will reply soon.
Contact Zachary

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.