#ChemCoach : Online Organic Chemistry Tutor

Early morning in Jerusalem, 2010

For any readers who are interested in chemistry as a career, blogger SeeArrOhh of Just Like Cooking is running a blogging carnival called #ChemCoach where chemists talk about their careers and how they got there. If you’re at all interested in the huge variety of jobs that people with a chemistry background do, I highly recommend checking out the main post here?, which links to over 20 entries by a variety of very interesting people.?This is my entry.

My current job:?I work independently as an online organic chemistry tutor. Basically I run a small business.

What I do in a standard “work day”. It depends on the time of year, but I spend a considerable portion of most days in Skype chats with students who are taking organic chemistry classes. Summers are pretty quiet on the tutoring front, but the months from October through May are often pretty full. ?I’ve worked as much as 11 hours in one day, but I find that the maximum I can tutor in one day and stay sane is about 8 hours.

When I’m not tutoring I’m writing articles for my blog, or developing study materials that I think will be useful for organic chemistry students. I spend a lot of time in ChemDraw!

What kind of schooling/training/experience helped me get there? My background is in total synthesis. I have a Ph.D. in organic chemistry and worked as a postdoctoral fellow for four years (two separate postdocs). My first postdoc was a very traditional total synthesis postdoc at a high powered institution. The second postdoc I took in Jerusalem in order to finally be able to live with my wife (we did 6.5 years of long distance before that – long story).

I finished my first postdoc just in time for the recession. ?We moved to Jerusalem in late ’08 and the recession really changed my focus. Not only did it make “getting a job” harder, I saw how precarious even the careers of my friends with “safe” jobs in pharma could be with plant closings and mergers. I didn’t want to be in the position where an individual could walk into my office and fire me or tell me I had to move cities. My solution was to start my own business on the side. ?In ’09 I started a spectacularly unsuccessful scientific editing service called WriteChem that never got a client. Later that year I was doing some organic chemistry tutoring in person, but the language barrier ?meant that I didn’t have many students. Sometime around January 2010 I saw Chatroulette and saw how easy it was to connect to people all over the world with online video. So it struck me – why not tutor people online through Skype?

In order to attract people to tutor I thought I’d need a website. ?I remember seeing Tenderbutton?and Not Voodoo?in ’06 and thinking that someone needed to build a site that combined the best features of them. Four years later, nobody had really done it yet, so that “someone” turned out to be me. I figure that if I created something truly valuable to people, everything will eventually work out. And so far it has.

The best thing about my work is that ?it allows me to work independently, have flexible hours, and live anywhere. And I get to meet all these great people.

How does chemistry inform my work? I talk and write about chemistry every hour of the day. It probably helps my teaching that many of the reactions I talk to students about I’ve run personally. Beyond undergraduate tutoring, it’s fun to get inquiries from people I never would have expected, like graduate students, chemists in private industry and in law enforcement (I had one person ask me about doing a series of sessions for a city crime lab). ?On another note, ?I find that the methodical habits and attention to detail that were valuable for being a successful chemist transfer well to running a small business. I still run “experiments” but they’re of a different kind.

A unique, interesting or funny anecdote about my career: You never know what you’re going to get with a Skype chat. I’ve worked with women breast-feeding their crying babies (camera off) , people who keep chickens, people in airports.
Living in Jerusalem was fun. In the last few months we lived just outside the gates of the Old City. In order to hit the North American evening crowd, I’d wake up at 1:30 am, tutor from 2am to 8am [I'd quit my postdoc by this point], go for a run through the Old City, sleep until noon, write / tutor for the next several hours, eat dinner, and go to bed at 9pm.
I’d be tutoring at 4am (that’s early evening EST)? and the Call To Prayer would suddenly blare out from each of the three mosques in the adjoining valley. That always got a reaction. Sessions would also get interrupted by the noises from fighting street cats. I actually miss that now.

For more posts in the #chemcoach series, check out SeeArrOh’s blog

Tagged as: stories, teaching, tutoring

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Comparing the E1 and SN1 Reactions

Imagine you’re starting with the alcohol on the left and you’d like to get to the alkene on the right.

What bonds are formed and broken here? We’re forming C-C (π), we’re breaking C-H, and we’re breaking C-OH. It’s an elimination reaction.

Notice a problem here? We need to have HO(-) as a leaving group. If you’ll recall, strong bases [like HO(-) ] are terrible leaving groups – which makes the E1 pathway unlikely. So what if we tried to use a strong base, maybe trying to promote an E2 reaction? Well, that would be even worse – we’d likely deprotonate OH before the C-H, and you can imagine that we’d have to have O(2-) as a leaving group here. Not good!

That means that the reaction, as written, is very unlikely to happen.

Yet, there is something very simple that we can do to make this reaction work. We’d need to have a better leaving group (a weaker base). How can we do this?

Add acid!

If we add a strong acid, we turn OH into H2O+, the conjugate acid is a better leaving group. Now, water can leave, forming a carbocation; and then a base can break the C-H bond, forming the alkene.
Notice that this is now a classic E1 reaction. The rate is going to be dependent on the stability of the carbocation. This one is tertiary, so it should proceed at a reasonably high rate.

A question arises here. What’s going to act as the base? As it stands, a C-H bond adjacent to a carbocation has an extremely high acidity (at least below -2, if you follow pKa). That means that just about any weak base (water, or the conjugate base of the acid) is sufficient to deprotonate the carbon. It’s possible that more than one species can act as a base here. I’ve shown water removing the proton, but it’s not unreasonable to show the conjugate base of the acid removing the proton in most circumstances.

Now comes one of the things about organic chemistry that often causes trouble for students. For one of the first times in our discussions here, we’re dealing with a situation where we can have?competing reactions.

Let’s back up. The E1 reaction goes through a carbocation, correct? Well, if you’ll recall, so does the SN1 reaction. We’ve already seen examples where a carbocation was formed from an alcohol by adding a strong acid like HCl, HBr, or HI, and we ended up with the alkyl halide. Why???The halide ions are decent nucleophiles under the reaction conditions.

So how can we stack the deck in favor of the E1 process? Use a strong acid with a conjugate base that is a poor nucleophile.?The usual choice is H2SO4. The HSO4(-) ion is a relatively poor nucleophile due to the negative charge of the oxygen being distributed throughout the molecule (resonance). Two other acids you might see for this purpose are p-toluenesulfonic acid (p-TsOH), which is essentially a cousin of H2SO4, and phosphoric acid (H3PO4).

Also, don’t forget that elimination reactions are favored by heat.

In summary, if you’d like E1 to predominate over SN1: choose an acid with a weakly nucleophilic counterion [H2SO4, TsOH, or H3PO4], and heat.

If you’d like SN1 to predominate over E1, choose an acid like HCl, HBr, or HI.

We’re almost done talking about elimination reactions. Next post – we’ll talk about rearrangements.

Next Post: The E1 Reaction With Rearrangements

Tagged as: base, conjugate acid, e1, e2, elimination, leaving groups, nucleophile

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Success Stories: How Stu Aced Organic Chemistry

I recently got an email from a reader, Stu (not his real name) who told me he got an A in Organic Chemistry 1, and was on pace to do the same in Org 2. I asked him if he’d be willing to share with the readers of this blog how he did it. I think you’ll agree that this is a story of tremendous dedication, hard work, and attention to detail. The rest I’ll leave to Stu [note: emphasis and subtitles have been added by me]
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How I Got An A In Organic 1
I guess I should start off with quantifying what I did in organic 1 and so far in organic 2, then I can describe how I did it.
So, this summer in organic 1 we had a quiz (mostly review stuff from gen chem, i.e. acid base chemistry, lewis structures, etc:10%), HW (25%), 2 tests (20% each) and a final (25%).
Each of the tests had 5 (out of 100) extra credit points, no extra credit for the quiz, just a hair of extra credit for the HW, which was through Sapling. The extra credit was a nice boost, making available 105 points but only grading 100 of those. So, any scores that I mention are uncurved but do include the 5 extra credit point
I got a 100% on everything except for the first exam. My initial grade was an 83%, but I submitted for a regrade and got an 87%. More on this later..
So far in Organic 2, we just took our first exam last week and I scored a 99.5%.
Now how I did it:
1. Course Preparation
I don’t really have one overarching strategy, and I wouldn’t say that my strategy is the most elegant or efficient, but I would say that I am getting there. Especially in the beginning, in was more of “siege” tactics than anything.
My one bad habit (?) with classes is getting ahead of the class pace, sometimes learning all of the subject during a semester break. So, knowing about the reputation that organic chemistry had, I decided to “run into the burning building”, for lack of better words, and tackle it head on rather than fear it. I guess you could say that my first step was to get into the right mindset.
Next, I started early, before the semester started. This way, I was able to stay about 4-5 chapters ahead of the lecture. What I would do is pre-read the chapter to know what was going on, then read it again slowly and doing all of the problems. If there was a word or concept I didn’t know, I would not move forward until I understood and mastered it. I sketched out just about every mechanism and molecule I could find. I have notebooks full of molecules and mechanisms. This was a serious time investment.
So anyway, I probably spent about 10-20 hours on each chapter, depending on the difficulty. A lot of this was time well spent, although some of it was wasted since later on in lecture the teacher would emphasis certain parts while skipping others, want us to know about certain reactions/mechanisms but not even mention others. Although my approach was not efficient, I did reach a level pretty close to mastery on those concepts once I stepped into lecture. This allowed me to fully understand everything the teacher was talking about, and to ask relevant questions.
2. Exam Preparation
About a week or two before an exam, I would take the list of “suggested book problems” that our teacher gave us and would rework those problems for the relevant chapters. Keep in mind that I had already done these problems, but by this point it was sometimes 3-4 weeks previously. Any problems I had trouble with would go onto a list and I would keep reworking them until they got eliminated from the list.
The book “Organic Chemistry as a Second Language” was a HUGE help, especially for IR and NMR. I only skimmed the textbook chapters on these subjects, as this book was more pointed and relevant.
3. Organizing Notes and Reactions
A day or two after lectures, I would compile all of my notes. Concepts would do into an outline form Word document. Reactions go into my reaction notebook. Here is a description of my reaction notebook:
4 sections:
?A chronological list of reactions with starting material, conditions, and products. Any reactions whose mechanisms were discussed in class have a *A list of mechanismsA list of reactions organized by common starting product, i.e. Reactions starting with Alkenes, starting with Primary R-OH, Secondary R-OH, Epoxides, Oxidizing Primary R-OH, etcA list of reactions organized by common ending product (retro) i.e. Reactions ending with Alkenes, Primary R-OH, Alkanes, etc
The reactions in lists 1,3,4 are all cross indexed. For example, hydroboration might be listed as #28 in each list. This makes it easy to identify errors, omissions, demo molecules that could be improved. Most of the molecules were ones given in lecture. Some of them used prototype molecules. I did this if I wanted to demonstrate a particular stereochemistry concept, or showing how different reagents lead to different products with the same molecule, etc.
My main study of reactions was with lists 3+4. I have found that identifying reactants and products is usually pretty easy if you know one or the other and the conditions, but remembering all of the conditions can be a challenge, so that is what I focused on.
4. Learning Reactions
What I would do this these lists is get a piece of paper, cover up all of the conditions, and see how many I could write out on the paper. Any that I missed would get an X next to them. I would keep going through the lists, repeating just the incorrect ones, until all of the Xs were gone. Then I would go through each group again. If I could do them all perfectly, that group got a tick mark. Then, wait a while, come back to it. If I could still do the group perfectly again, another tick mark. Once I got to 5 tick marks for each group, remembering all of the reactions was no sweat.
Me and a friend of mine traded synthesis problems for a while. I found that I learned a lot just by writing them and seeing all of the different ways I could move functional groups in 2 steps, 3 steps, etc.
This reaction notebook has been an awesome help, but unfortunately it wasn’t until I was quite a way into Organic 1 that I figured it out. Better late than never!
5. Keeping Track of Mistakes
Another thing I did was I kept a list of every mistake I did on every exam. This included conceptual, technical, and cognitive mistakes.? For example, I did horrible on my first exam, considering the amount of time I invested in it. I wasn’t lacking in the first two categories, but I made a lot of cognitive mistakes. I was a little too overconfident, so much so that I didn’t notice a lot of mistakes I made even though I checked my work a few times. Also, since I wasn’t used to how organic tests were formatted, that threw me off a little bit. I made some really dumb mistakes.
There was one mistake I remember where the question asked to outline the free radical addition of HBr to an alkene. I figured that I would be a hot shot and detail the mechanism, even though that wasn’t asked for. Also, I wasn’t entirely sure how much detail was required when it said “outline”, so I decided to play it safe and prove that I knew what I was talking about. Well, I detailed everything, except that I forgot to draw out the starting alkene!!!!!? This definitely went on the list. The moral of the story was to learn from every mistake. Also, there were a few stereochemistry concepts (particularly meso) that I still wasn’t quite strong enough on, so I had to figure those out, quickly.
6. Final Words of Advice
Another piece of advice is to not get behind and to try to understand everything as it is presented. If something looks foreign, it must be made familiar immediately. Look stuff up, do extra practice problems. I have done probably thousands of problems…way too many than I would even like to think of.
I feel like I am a little more efficient now. When I go into class, I have still read all of the chapter and have done most of the practice problems, or at least the recommended ones. Sometimes, if I don’t do all of the problems in a chapter, I will do ones that are similar to the recommended ones, since they have appeared on tests.? I don’t try to memorize every reactions before lecture, since I don’t know which ones the teacher will want us to know. Memorizing all of them would just be wasted time.
Also, I always consider homework to be “free points” in every class that I take. There is so much time and so many resources available that getting anything else than close to 100% is almost illogical to me.
Anyway, hope this helps. I think that should be the bulk of my approach.
-”Stu”
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Want to share your success story (or even a failure story) ? Send me an email via the “Feedback” button above

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Ribosome Regulates Viral Protein Synthesis, Revealing Potential Therapeutic Target

Viruses can be elusive quarry. RNA viruses are particularly adept at defeating antiviral drugs because they are so inaccurate in making copies of themselves. With at least one error in every genome they copy, viral genomes are moving targets for antiviral drugs, creating resistant mutants as they multiply. In the best-known example of success against retroviruses, it takes multiple-drug cocktails to corner HIV and narrow its escape route.

THE HYDROGENATION OF ALKENES

THE HYDROGENATION OF ALKENES

This page looks at the reaction of the carbon-carbon double bond in alkenes with hydrogen in the presence of a metal catalyst. This is called hydrogenation. It includes the manufacture of margarine from animal or vegetable fats and oils.

The Shape of Molecules

The Shape of Molecules

    The three dimensional shape or configuration of a molecule is an important characteristic. This shape is dependent on the preferred spatial orientation of covalent bonds to atoms having two or more bonding partners. Three dimensional configurations are best viewed with the aid of models. In order to represent such configurations on a two-dimensional surface (paper, blackboard or screen), we often use perspective drawings in which the direction of a bond is specified by the line connecting the bonded atoms.In most cases the focus of configuration is a carbon atom so the lines specifying bond directions will originate there. As defined in the diagram on the right, a simple straight line represents a bond lying approximately in the surface plane. The two bonds to substituents A in the structure on the left are of this kind. A wedge shaped bond is directed in front of this plane (thick end toward the viewer), as shown by the bond to substituent B; and a hatched bond is directed in back of the plane (away from the viewer), as shown by the bond to substituent D. Some texts and other sources may use a dashed bond in the same manner as we have defined the hatched bond, but this can be confusing because the dashed bond is often used to represent a partial bond (i.e. a covalent bond that is partially formed or partially broken). The following examples make use of this notation, and also illustrate the importance of including non-bonding valence shell electron pairs (colored blue) when viewing such configurations .

Methane		Ammonia		Water

Bonding configurations are readily predicted by valence-shell electron-pair repulsion theory, commonly referred to as VSEPR in most introductory chemistry texts. This simple model is based on the fact that electrons repel each other, and that it is reasonable to expect that the bonds and non-bonding valence electron pairs associated with a given atom will prefer to be as far apart as possible. The bonding configurations of carbon are easy to remember, since there are only three categories.

Configuration	Bonding Partners	Bond Angles	Example
Tetrahedral	4	109.5º
Trigonal	3	120º
Linear	2	180º

In the three examples shown above, the central atom (carbon) does not have any non-bonding valence electrons; consequently the configuration may be estimated from the number of bonding partners alone. For molecules of water and ammonia, however, the non-bonding electrons must be included in the calculation. In each case there are four regions of electron density associated with the valence shell so that a tetrahedral bond angle is expected. The measured bond angles of these compounds (H₂O 104.5º & NH₃ 107.3º) show that they are closer to being tetrahedral than trigonal or linear. Of course, it is the configuration of atoms (not electrons) that defines the the shape of a molecule, and in this sense ammonia is said to be pyramidal (not tetrahedral). The compound boron trifluoride, BF₃, does not have non-bonding valence electrons and the configuration of its atoms is trigonal. 

The best way to study the three-dimensional shapes of molecules is by using molecular models. Many kinds of model kits are available to students and professional chemists. Some of the useful features of physical models can be approximated by the model viewing applet Jmol. This powerful visualization tool allows the user to move a molecular stucture in any way desired. Atom distances and angles are easily determined. To measure a distance, double-click on two atoms. To measure a bond angle, do a double-click, single-click, double-click on three atoms. To measure a torsion angle, do a double-click, single-click, single-click, double-click on four atoms. A pop-up menu of commands may be accessed by the right button on a PC or a control-click on a Mac while the cursor is inside the display frame.

One way in which the shapes of molecules manifest themselves experimentally is through molecular dipole moments. A molecule which has one or more polar covalent bonds may have a dipole moment as a result of the accumulated bond dipoles. In the case of water, we know that the O-H covalent bond is polar, due to the different electronegativities of hydrogen and oxygen. Since there are two O-H bonds in water, their bond dipoles will interact and may result in a molecular dipole which can be measured. The following diagram shows four possible orientations of the O-H bonds.

The bond dipoles are colored magenta and the resulting molecular dipole is colored blue. In the linear configuration (bond angle 180º) the bond dipoles cancel, and the molecular dipole is zero. For other bond angles (120 to 90º) the molecular dipole would vary in size, being largest for the 90º configuration. In a similar manner the configurations of methane (CH₄) and carbon dioxide (CO₂) may be deduced from their zero molecular dipole moments. Since the bond dipoles have canceled, the configurations of these molecules must be tetrahedral (or square-planar) and linear respectively.
The case of methane provides insight to other arguments that have been used to confirm its tetrahedral configuration. For purposes of discussion we shall consider three other configurations for CH₄, square-planar, square-pyramidal and triangular-pyramidal. 

Substitution of one hydrogen by a chlorine atom gives a CH₃Cl compound. Since the tetrahedral, square-planar and square-pyramidal configurations have structurally equivalent hydrogen atoms, they would each give a single substitution product. However, in the trigonal-pyramidal configuration one hydrogen (the apex) is structurally different from the other three (the pyramid base). Substitution in this case should give two different CH₃Cl compounds if all the hydrogens react. In the case of disubstitution, the tetrahedral configuration of methane would lead to a single CH₂Cl₂ product, but the other configurations would give two different CH₂Cl₂ compounds. These substitution possibilities are shown in the above insert.

Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur

In addition to forming single and double bonds by sharing two and four elec- trons, respectively, carbon also can form a triple bond by sharing six electrons. To account for the triple bond in a molecule such as acetylene, H O C q C O H, we need a third kind of hybrid orbital, an sp hybrid. Imagine that, instead of combining with two or three p orbitals, a carbon 2s orbital hybridizes with only a single p orbital. Two sp hybrid orbitals result, and two p orbitals remain unchanged. The two sp orbitals are oriented 180° apart on the x-axis, while the remaining two p orbitals are perpendicular on the y-axis and the z-axis.

When two sp carbon atoms approach each other, sp hybrid orbitals on each carbon overlap head-on to form a strong sp–sp s bond. At the same time, the pz orbitals from each carbon form a pz–pz p bond by sideways overlap, and the py orbitals overlap similarly to form a py–py p bond. The net effect is the sharing of six electrons and formation of a carbon–carbon triple bond. The two remain- ing sp hybrid orbitals each form a s bond with hydrogen to complete the acety- lene molecule.

As suggested by sp hybridization, acetylene is a linear molecule with H]C]C bond angles of 180°. The C]H bonds have a length of 106 pm and a strength of 558 kJ/mol (133 kcal/mol). The C-C bond length in acetylene is 120 pm, and its strength is about 965 kJ/mol (231 kcal/mol), making it the shortest and strongest of any carbon–carbon bond.