Benadryl Synthesis

The main ingredient in Benadryl, the infamous anti-allergy medicine that also conveniently puts you to sleep, is diphenhydramine. From the name, you can infer that it's got not one, but two phenyl groups in the structure. You can also tell that it's a type of amine. As for the "hydra".. I'm not immediately sure what that means.

It's ironic, I'm sneezing as I write this post. Better get to this synthesis quickly..

As you'd expect (or perhaps not), the structure (click here) is fairly simple. We have our two phenyl groups, and the tertiary amine in the top right.

Step 1: I'd first monobrominate a benzene ring using Br2 and FeBr3, a catalyst. Next, I'd throw in some lithium and form an organolithium reagent.

Step 2: I'd react 2 equivalents of the organolithium with methanoic acid. The first organolithium will abstract a proton from the acid, and the second equiv. will attack the carbonyl carbon. The sp3 intermediate will be frozen in time, until we perform the work-up step that will produce an aldehyde.

Step 3: Work-up - H30+/H20

Step 4: At this point, I would throw in another equivalent of our phenyl organolithium. I'm not going to perform a work-up step to produce the tertiary alcohol though, and you'll see why in a second.

Step 5: Next, I'm going to take an amino acid (shown in the synthesis) and hit it with SOCl2 to turn into an acyl chloride.

Step 6: I'm going to mix the acyl chloride with our product from Step 4. The O- (that was coordinating with Li+) will attack the acyl chloride (hard goes w/ hard) and undergo an addition-substitution type reaction.

Step 7: Next, I'll simply remove the C=O group using hydrazine and base.

Ta-da. We have our diphenhydramine a.k.a. Benadryl.

Note: Perhaps instead of preparing an acyl chloride, I could have just used a haloalkane and proceeded with some SN2 type of substitution that would ultimately save us a step as we wouldn't have any C=O group left to remove. However, my only problem with this idea is that O- is quite hard, while R-X is soft... This is why we typically use softer nucleophiles like organocuprates to perform SN2's. Maybe I'm wrong though, who knows.

Tylenol Synthesis

This synthesis is even easier than Advil, and it takes a mere three steps to complete. Let's get started:

Step 1: Begin with phenol; it's easy to acquire and super cheap. Dump it in some HNO3 and H2SO4, to add an NO2 group on the para position on the ring. Note: you're going to get some ortho substituents too, so you'll obviously have to separate them out.

Step 2: Reduce the NO2 to NH2 using H2, Pd/C.

Step 3: This step is clever... I didn't think of it at first. It's an acetic anhydride - really easy to acquire as well. The NH2 attacks because it's less electronegative (thus a better nucleophile) than the oxygen on the opposite side. My initial thought was to create an N2+ ion and displace it with a nitrogen salt such as NaNH-R. I think using acetic anhydride is classier, though.

Ta-da. Boom - Done.

Advil (Ibuprofen) Synthesis

Super simple.

This is Ibuprofen. You may be familiar with its name. It's the main ingredient in the super-effective Advil tablet which many of you are well-acquainted with.

It's structure can easily be broken down. I've attached a photo of a 4-step synthesis below (click photo to enlarge):

Step 1: Let's start with our basic benzene ring. Adding the first group is easy, all we need is an acyl chloride. The particular acyl chloride that we're using can be made with treating isobutyric acid with SOCl2. We'll mix our benzene ring with this acyl chloride and use AlCl3 as a catalyst, which will make our reactant ultra electrophilic since benzene is stubborn.

Step 2: If you look above, you'll notice that the alkyl group that we want attached doesn't have an oxygen double-bonded to a carbon. We'll get rid of this by performing a Clemmensen Reduction using Zinc and hydrochloric acid. Done.

Step 3: Now, we'll add the next substituent on the opposite side using a friedel-crafts alkylation. The reactant will be a 3-chloropropanol. We have to use AlCl3 again as our catalyst to make this reaction go. Our benzene might still be a little unreactive, even after adding the alkyl group. The only problem with this step is that another group might get added onto the ortho positions, since alkylations are runaway-type reactions... but maybe the steric hindrance will prevent that from happening too fast.

Step 4: Now, all we have to do is oxidize the alcohol into an acid. We can use the king of oxidation, KMnO4 for this step.

That's all it takes to make ibuprofen. Nice and straight forward.

Note: In Step 1, I went with an acylation instead of an alkylation for 2 reasons.
#1: Alkylations are runaway reactions, and we'll get the same substituent all over the ring - we only want it in one spot.
#2: When AlCl3 pulls the Cl group off of our reactant (which happens when you alkylate), the molecule will undergo a rearrangement into a tertiary carbocation... and then our benzene ring will attack the wrong carbon.

Note #2: KNNO4 is used incorrectly in the synthesis, as it would oxidize at both benzylic positions, unwantedly.

Synthesizing THC

THC is the chemical component of marijuana that gives it its unique effect on the body. Below is my proposed synthesis for this molecule. As you look through the synthesis, keep the whole molecular structure, shown above, in mind. You'll get a better understanding of where I'm going with each step. Feel free to click the image below to enlarge it.

Line 1: Start with Benzene. Nitrosylate it twice using H2SO4 and HNO3. Then, perform a friedel crafts acylation. Then, remove the carbonyl group using hydrazine and base to produce product A. 

Line 2: Product A is hydrogenated, converting both NO2 groups to NH2. One group is converted into an N2+ ion using NaNO2 and HCl. The ion is displaced using a lithium enolate -- which would attack from the oxygen, according to the "hard-hard" argument, producing product B.

Line 3: The remaining NH2 group is converted to an N2+ ion using NaNO2 and HCl. The ion is displaced with a hydroxy group from potassium hydroxide. Next, the double bond in the molecule is iodine-ized using hydrogen iodide. Another substituent is added to the least substituted ortho position, between the two oxy-substituents on the ring. See the column on the far right, which shows how to produce the reactant we're going to use. The benzene ring will attack the carbon adjacent to the double bond in our reactant, because conjugation will make SN2 attack at that position easier.

Line 4: The remaining bromine group on the new substituent is treated with magnesium, forming a grignard reagent. The possibility of a 6-membered ring is possible now.

Mechanism: The SP3 carbon bonded with oxygen and iodine is in equilibrium (or at least I would believe so) with the SP2 form, which is produced when oxygen donates its lone pair to carbon, kicking off I- in the process. The reason I say it's in equilibrium is because I- can just attack the electrophilic carbonyl again, taking us back to square one. Let's say I- has been kicked off, and the carbon is SP2 hybridized, and the oxygen has a positive charge. In this state, its really electrophilic. The intramolecular grignard will see this an instantly join the party - forming a favourable 6-membered heterocyclic ring.

And there you have it, that's the synthesis for THC. 

If you look at original molecular structure, you'll notice the stereochemistry of the two hydrogen atoms on the ring. One is up, one is down. I took this account when creating the proper reagent on the far right column. Using Br2 on an alkene produces trans bromines, and if we use one of the two enantiomers, we'll get the right product. Remember, SN2 attack inverts stereochemistry of the carbon it attacks.. you've got to be careful when choosing the enantiomer.

Day 100 - It's Over

Well, that's it. That ends the end of Organic Chemistry 2. The final exam was a beast - a total of nine neuron-wrenching questions that beat every concept out of you for two hours. With the end of every question came a rough transition, like switching gears with a rusty clutch. By the end, you were scavenging around for enough energy to push you through a quick review of your answers.

However, with that being said, it was a clever exam that was, without a doubt, written carefully. I'm confident in my performance and looking forward to getting my results. Regardless of my grade, I'm thrilled with my capabilities in this subject.

Organic Chemistry is unique in its ability to test understanding, problem-solving, and memorization, like no other subject does. It's a bittersweet moment for me. I'm glad to have completed the course.

Congratulations to everyone who wrote today, I wish you all a great break before the next year begins. It's possible I'll continue writing here if a few of us are in CHM361 - Structural Biochemistry.

Update: 68.5/80 on the final exam, the highest in class! 

Day 99 - Into The Grey

How crazy is it that our course is exactly 100 days in duration? For those of you writing the exam tomorrow with 99 problems.. let's hope orgo ain't one.

With an abrupt start to my day (i.e. sleeping at 6:30 am and waking up at 3:30 pm), I'm finally here at the library, about to begin my 8-hour descent into the world of chemistry. 

I reckon I'll be updating this post as the day goes on if I have anything I want to provide you guys with. However, I do have some words of wisdom: 

1. Avoid "mental fixedness"

I struggle with this problem, every. single. time. As soon as I can see a way to solve a problem, my brain literally "locks in" and I run like a horse towards that solution. I lose sight of other ways I could have solved the problem, and sometimes that comes back to bite me in the ass. If you see my exams or midterms, you'll notice that some of my answers are completely different from the topic that the question is about, but they still work. I just see my own solution and snipe it. It's good to be able to see solutions in questions, but I advise you do take a second to "step back" and assess your solution and probe for other possible solutions. Even today this day, I'll check the solutions manual and I'll get slapped in the face with a much more elegant solution and I'm like "Damn! What was I thinking?". This leads me to piece of advice #2: 

2. Read the questions. 

Legit. You can really help yourself out by reading the question topic. If it says "Substitution at Carbonyl with a loss of Oxygen",then you KNOW what kind of mechanisms and products to expect. Use the question titles as tools... they're the railings along the sides of each lane when you go bowling. They're there so you don't run off a cliff. 

---

Last night, after submitting a 17-page report, I wrote down a list of gaps in my understanding. Today I plan on putting on my plumber outfit and sealing those cracks. Then, I'm going to go through the midterms - re-correct my mistakes so they don't happen again. I have to get on the ball with Schotten Bauman. I have to teach myself Birch reductions. I'm going to try out Patrick Gunning's examination from last year using the library's website. Then, I'm going to tackle the last few practice questions posted by Dr. Leigh on blackboard. Possibly also redo some synthesis questions from WASPS. Then, I'm going to quickly go over the slides from each week and make sure I'm on the ball with every slide. Long day ahead. 

Chao!

Stay tuned.

Today's Tune: West Coast Gangsta Rap Instrumental 

6:30 pm: Just finished Gunning's final from last year. It was pretty straightforward, I'd recommend trying it out for practice. You can find it on the library's website. I added two more reagents to my arsenal:
NBS & dil. acid: Used to brominate allylic carbons - we covered this in CHM242.
OsO4 & pyridine: Used to make syn-diols from an alkene.

10:30 pm: Up to this point, I've taught myself how to make Wittig reagents, I've learned how to turn a ketone into an ester using mCPBA, I've re-learned the reagents required to put alcohols on Anti-Markonikov, I've learned how to perform a Birch reduction, and I was slapped in the face with an abrupt reminder that organolithiums and grignards will react with carboxylic acid derivates twice to form tertiary alcohols just because of how reactive they are as nucleophiles. As it turns out, to get a single addition-substitution - you should use organocuprates or perhaps a Weinreib amide, or convert the group to a carboxylic acid and use two equivalents of organolithiums and work it up.. the list is endless it seems. I've also learned the Luche reduction (NaBH4 with CeCl3) to selectively reduce a C=O bond in the presence of a C=C bond, as is in an enone.

Now, I'm going to practice some Birch reduction questions, read up on Benzyne chemistry, finish up some more of Leigh's practice problem sets, and then maybe also do a problem set for grignard's, organolithiums and the like.

5:00 am: Slept. 

Day 95 - Closing In

After a few days of minimal chemistry practice, I decided to tackle last year's final exam paper in a real-life simulation in our library's silent zone.

Much like the way I practiced for this semester's midterm, last year's final, and last year's midterm, I recreate the actual mood of writing the exam.

The only difference this time was that I sipped on a White Mocha Latte as I made my attempt.

I finished the exam after exactly 60 minutes of the allotted 120 minutes. I attribute this speed to having been exposed to one or two questions from this paper beforehand - giving me an edge. In a real life scenario, I estimate that I would be 10-20 minutes slower.

When it's the real deal, you get nervous and your thinking gets spazzy. That's why I never take any caffeine before any kind of assessment. I'm not sure if I'm going to deprive myself of sleep in this final. The past three examinations have been on 4 hours of sleep, and I've seen good results. Maybe I'm leaving marks on the table by depriving myself? Only one way to find out. Let's hope I can get a good night's rest this Tuesday night.

As for the grade, I think I scored close to 100% on last year's final. There were one or two very minor mistakes. I'm hoping this year's final is similar in difficulty. Also, and to be honest, I knew the answers to one or two questions before-hand, since my sneak-peak at the document also included its answers. So maybe reduce that 100% to 90%. Still impressive though.

It's hard to say how I'm going to prepare in the next coming days. I think I'm going to practice some protection chemistry, maybe take a look at the Birch reduction just incase, and I'm not sure what else... I can't say I know everything, but its beginning to feel like it. Not bad.

Today's Tune: Rels Beats - Mecano (Hip Hop Instrumental)