A Friday SN1 Haiku

Loyal reader Dave sent this in via the comments, and I thought it deserved a wider distribution:

—————

Some of my students challenged me to do the upcoming class in haiku. Since a fractional distillation lab is pretty dull to supervise after everyone is up and running, I put the introduction to SN1 in haiku format. (Each done on a powerpoint slide with a pretty background…)

SN1 Haiku

Leaving group breaks off
Forming carbocation
SN1, first step

very reactive
intermediate species
they need electrons

tertiary good
hyperconjugation helps
resonance does too

add more Nu? No help.
the rate is independent
that’s kinetic proof

climbing two mountains
reaction coordinate
C+ is high pass

how do you decide?
SN1 or SN2?
there are many factors

————-

Thanks Dave. Have a good Friday everyone

Tagged as: fun, substitution

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SN1/SN2/E1/E2 Decision Mind Map

Courtesy of Dave Blackburn, an organic chemistry instructor “who is giving an exam on this today”, here is a mind map of all the different concepts covered in the discussion on SN1, SN2, E1, and E2 reactions.
What’s interesting is that this mind map really boil the “SN1/SN2/E1/E2″ decision ?down to 3 options:
SN2 reactionE2 reactionFormation of a carbocation
If the carbocation is formed, of course, it can go down several different pathways (SN1, E1, possibly preceded by rearrangement).
There are five key categories covered here: substrate, nucleophile, solvent, temperature, and leaving group.?There’s no one place necessary to “start” although beginning with the “substrate” – i.e. ” primary, secondary, or tertiary alkyl halide? ” is always my first choice.
Thanks, Dave!
Click here for larger version

Tagged as: e1, e2, elimination, sn1, SN2, substitution

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Deciding SN1/SN2/E1/E2 (1) – The Substrate

Having gone through the SN1, the SN2, the E1, and the E2 reactions we can now say the following:

Both substitution reactions?and elimination reactions occur with alkyl halides (and related species)?A wide variety of nucleophiles/bases can be used to perform substitution and elimination reactionsA wide variety of solvents can be used in substitution and elimination reactionsWe also have to gauge the importance of factors such as the leaving group and temperature.

This is a lot of different factors to think about. Let’s look at some examples of situations you might encounter:

This is often one of the most difficult parts of organic chemistry for new students: how to weigh multiple (and often contradictory) factors??How do we know which factor is most important? Do we pay attention to the base, substrate, temperature, solvent? How do we go about sorting through a problem like this??As I’ve written before, deciding SN1/SN2/E1/E2 is not completely unlike how a professional bettor might evaluate which sports team is going to win on a given night (does good defence beat good offence? how important is coaching? how important is their recent performance? ) .

In this post and the next few, we’ll walk through one way of thinking about how to evaluate whether a reaction will proceed through SN1/SN2/E1/E2. It’s not 100% foolproof*, but it’s a decent enough framework for our purposes.

It starts by asking questions. In order of importance, I think they are:

The substrateThe nucleophile/baseThe solventThe temperature

It’s also a process of, at least in the beginning,?ruling things out rather than ruling things “in”. In other words, seek to decide what options are?not possible, rather than decide which are possible. It’s a subtle distinction, but a valuable one. Once you’ve crossed certain reactions off the list, you can then start asking yourself which reactions would be most consistent with the reaction conditions.

Before getting specific with each of those 4 questions, let’s start with the most important question you can ask in any situation like the ones above.

The most important step in evaluating any reaction is first to ask yourself “what type of functional group(s) are present in this molecule? This is because the type of functional group will dictate the type of reaction(s) that can occur. Note that in the questions above, all of the starting materials are?alkyl halides or?alcohols. Substitution/elimination reactions are possible for these substrates; many other reaction types (addition, for example) are not.

Question 1: The Substrate

Given that we’re looking at alkyl halides/alcohols, it’s a reasonable expectation that we should evaluate SN1/SN2/E1/E2. The next step is to identify the?type?of alkyl halide we are dealing with.
Look at the carbon that contains the best leaving group. Typically this is Cl, Br, I or some other group that can act as a good leaving group. ?Ask yourself: is this carbon primary, secondary, or tertiary?

Given what we know about SN1, SN2, E1, and E2 reactions, we can say the following:

The “big barrier” to the SN2 reaction is steric hindrance.?The rate of SN2 reactions goes primary > secondary > tertiaryThe “big barrier” to the SN1 and E1 reactions is carbocation stability. The rate of SN1 and E1 reactions proceeds in the order tertiary > secondary > primary.The E2 reaction has no “big barrier”, per se (although later we will have to worry about the stereochemistry)

So how can we apply what we know about each of these reactions to simplify our decision?

If the substrate is?primary, we can rule out SN1 and E1, because primary carbocations are unstable. ?You cannot definitively rule out E2 yet, although I will spill the beans and say that it’s almost certainly going to be SN2 unless you are using a very sterically hindered (“bulky”) base such as tert-butoxide ion (e.g. potassium t-butoxide KOtBu).

If the substrate is?tertiary,?we can rule out SN2, because tertiary carbons are very sterically hindered.

If the substrate is?secondary, we can’t rule out anything (yet).

As you can see, based on the information we’ve evaluated so far, we can’t make a definitive decision on SN1/SN2/E1/E2. We’ll have to look at some other factors before we can make a final decision. Next, we’ll evaluate the role of the nucleophile/base.

—————— Advanced readers only ————–

* One question that comes up a lot is this: are there exceptions? Keeping in mind the two themes of “steric hindrance” and “carbocation stability”, there are edge cases where we can have a particularly sterically hindered primary alkyl halide, or a particularly stable primary carbocation.

For instance, the alkyl halide below (“neopentyl chloride”) is indeed primary, but is so crowded on the carbon adjacent to the primary alkyl halide that it is essentially inert in SN2 reactions. On the SN1/E1 side, the allyl halide below, while primary, can undergo SN1/E1 reactions because the resulting carbocation is stabilized through resonance. As long as you keep in mind the “big barriers” for each reaction, you should be fine.

Tagged as: e1, e2, elimination, sn1, SN2, substitution

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In Summary: Khan Academy Videos For Organic Chemistry

In Summary – Khan Academy for Organic Chemistry

In six previous posts I went through all 73 videos by Khan Academy for organic chemistry. In this final post I wrap up what I’ve learned from watching them, from the perspective of someone who teaches introductory college-level organic chemistry.

Six previous posts:?[1?2?3?4?5?6]

Contents

What’s great about Khan Academy?What does Khan Academy cover?What’s missing?What’s done the best?What are the biggest weaknesses?How useful is Khan Academy for a typical college student learning organic chemistry?Towards version 2.0

1. What’s Great About Khan Academy?

First of all, KA gets one really big thing right: teaching a subject as a series of 10 minute videos that gradually walk through a subject. There is a huge desire for this type of instruction and the value of discovering a popular working model for delivering educational videos cannot be understated.

His style is also very approachable and non-threatening. He doesn’t make you feel stupid. This alone is a huge part of the appeal. The tools he uses are very simple and accessible to everyone. He’s also an excellent illustrator.

Thirdly, I think it’s commendable that Sal Khan just decided to start making videos on a lot of topics without asking anyone’s permission. In some circles, his videos have been unpopular with academics. Here is a particularly lazy criticism of KA. Memo to instructors who complain about KA: you have the tools to make better videos: why aren’t you doing it?

What Does Khan Academy Cover?

I count 73 videos for organic chemistry.

These can be broken down into the following categories:

Nomenclature [27]: Alkanes, alkenes, R/S nomenclature, E/Z nomenclature, alcohols, ethers, epoxides, benzene derivatives, amines, aldehydes, ketones, carboxylic acids, anhydrides, esters, amides, acyl chlorides

Structure and bonding [5] hybrid orbitals, pi bonds, alcohol properties [hydrogen bonding], resonance, Huckel’s rule

Conformations [3] Newman projections [2 videos], chair/boat shapes for cyclohexane

Stereochemistry [4] Intro to chirality, chiral examples (2 videos), enantiomers/diastereomers/constitutional isomers

Reactivity: [8] mechanisms, Markovnikov’s rule, steric hindrance, sn2 stereochemistry, solvent effects, nucleophile strength, nucleophilicity vs. basicity, carboxylic acid derivative reactivity

Reactions: [24] HBr with alkenes, H3O(+) with alkenes, polymerization of alkenes w/ acid, SN2, SN1, E2, E1, free radical alkane chlorination, epoxide ring opening, bromination of aromatics, Friedel Crafts acylation, keto-enol tautomerism, acid-base of carboxylic acids, Fischer esterification, acid chloride formation with SOCl2, amide formation from amine and acyl chloride, aldol reaction.

Addenda [2]

What does a typical introductory organic chemistry course cover that is NOT included in Khan Academy?

If I had to name my top 10 omissions, they would be this:

no discussion of organic acids and base; i.e acidity, basicity, conjugate acid, conjugate base, pKa, how to determine if an acid base reaction is favorableevaluation of resonance forms; what makes one resonance form more stable than another? Not discussed.no discussion of the stereochemistry of the E2 reactiononly 3 alkene addition reactions covered [out of >15] ; no examples with stereochemistry. Missing examples like bromination, hydroboration, epoxidation, etc.no discussion of alkynes or their reactionsno discussion of carbocation rearrangementsno discussion of oxidation/reductiononly one video devoted to the concept of aromaticity, not enough to understand it; only 2 reactions of aromatic derivativesonly 3 videos on reactions of aldehydes, ketones, and carboxylic acid derivativesno discussion of the Diels-Alder or related reactions; no discussion of molecular orbitals

Of the material covered by the KA videos, what’s done the best?

The videos on nomenclature are useful for those who have never encountered these functional groups before. They provide accurate methods for naming these molecules. There was only one major mistake and this was corrected. A few minor mistakes (see notes on individual videos).

The videos on structure and bonding, including hybrid orbitals were done well.

The drawings of Newman projections were done well and have nice descriptions of how to convert a line diagram into a Newman projection.

What are the biggest weaknesses? What could be done better?

The videos on reactions tend to be shallow. There is a lot of imprecise language (e.g. “guy” instead of, say “nucleophile”). Chemical reactivity is? largely explained in an ad hoc fashion rather than in a way that ties back to fundamental concepts of physics. ?There’s an opportunity to explain a lot of organic chemistry by understanding the key factors that stabilize negative and positive charges; that isn’t done here. Arrow pushing is not done correctly. Stereochemistry? is arguably the major theme of first-semester organic chemistry, but its involvement in reactions is almost completely ignored.

What the videos really lack is a consistent description of chemical reactivity rooted in a small number of fundamental physical principles.

How Useful is Khan Academy For A Typical College Student Learning Organic Chemistry?

As they stand right now, KA videos are in no way a standalone alternative to a college-level course in introductory organic chemistry. If you watch all 73 videos, you will in no way have “learned” organic chemistry.

They are a useful supplement for nomenclature, structure and bonding, and some aspects of conformations.

For someone who finds their university lecturer intimidating, the videos on reactions and reactivity can be a light introduction to some of the topics covered in an introductory course.

Towards Version 2.0

Two questions for the future:

1) What is the minimum number of 10-15 minute videos required to teach the core concepts of Org 1 and Org 2 in a manner that would, say, prepare students for a standardized exam like the ACS?

2) What is the optimal order of those videos? Can we test outcomes to see if there is an optimal way in which these topics can be presented?

I look forward to further improvement of the organic chemistry video content at KA.

Tagged as: khan academy, video

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Boring public service announcement

MOC is growing in traffic (yea!) but that means that the hamsters inside my current server are running at the brink of exhaustion. I’m migrating/upgrading to a new server service, so no posts for a few days until this gets settled. Hopefully by next week page load times should be snappier too.

- MGMT

image credit

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Elimination Reactions With Rearrangments

One last (weird) reaction to show you with respect to elimination reactions. Can you see what’s weird about it?

How did that double bond get over?there??Normally when elimination occurs, we remove a hydrogen from the carbon adjacent to the leaving group. But here, something extra has taken place.

Let’s look at all the bonds that form and the bonds that break so we can track down?exactly what happens:

Notice how it differs from a typical elimination reaction? Sure, we’re forming C-C (π), and breaking C-H and C-OH, but we have an extra C-H that forms and an extra C-H that breaks.

This is a sure sign of a rearrangement step!

So what’s going on here?

Well, we start by protonating the alcohol. This allows for water to leave in the next step, which is going to form a carbocation.?Here’s the thing:?the carbocation is?secondary, and we’re adjacent to a?tertiary carbon. So if the hydrogen (and its pair of electrons) were to migrate from C3 in our example to C-2, we’d now have a tertiary carbocation, which is?more stable. Then, a base (water in this example) could remove C-H, forming the more substituted alkene (the Zaitsev product in this case). And that’s how the alkene ends up there.

OK. So that’s one mystery solved.

You might remember that these types of rearrangements can occur in SN1 reactions too. And if you read that post, you might recall that in addition to shifts of hydrogen (“hydride”, because there’s a pair of electrons attached) we can also have alkyl shifts. Here’s a final example. Note – I’ve also made a video of this, you can watch it here.

This pretty much does it for elimination reactions.

In the next series of posts, let’s go though one of the biggest questions students struggle with. Okay, now that we’ve gone through substitution and elimination reactions, HOW DO WE DECIDE WHICH ONE IS GOING TO OCCUR IN EACH SITUATION?

Great question. That’s next.

Next Series, post 1: SN1/SN2/E1/E2 Decision (1) – The Substrate

Tagged as: alkyl shift, e1, eliminations, hydride shift, rearrangement, sn1

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Khan Academy Videos for Organic Chemistry, Part 6

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In this post (the 6th of 6) we finish up going through all the Khan Academy videos for organic chemistry [videos 65-73]. For previous posts in this series, see [1 2 3 4 5]

Video #65 – Keto-Enol Tautomerism
Length: 8:03
Summary: 3-methyl 2-butanone is drawn, and the mechanism for its conversion into its enol form is shown via protonation of the carbonyl oxygen and deprotonation.
Key concepts/Skills:?keto-enol tautomerism, constitutional isomers.
Nitpicky criticisms:?A little bit of vaugeness about where H3O(+) comes from; there shouldn’t be too much in water at pH 7. Might help to say the reaction is assisted by adding acid. Could help to mention what factors favor the keto over the enol, or to give students a typical value for the position of equilibrium [about 104 favoring keto, for instance].
Red flags:?None, this is fine [except for arrows]

Video #66 – Carboxylic Acid Introduction
Length: 8:51
Summary:?A generic carboxylic acid is drawn, and the acid-base reaction between it and water is shown. It’s demonstrated that the stability of the conjugate base is due to resonance. ?Formic acid, acetic acid, and oxalic acid are also drawn.
Key concepts/skills:?acidity, resonance stabilization, carboxylic acids
Nitpicky criticisms:?It’s said that carboxylic acids are more acidic than alcohols. By how much? It would also be good to explicitly say that “the stronger the acid, the weaker the conjugate base”. When the Arhennius definition of an acid as a proton donor is mentioned [1:05] I think he means Bronsted.
Red flags:?Nothing except for the arrow pushing

Video #67 – Carboxylic Acid Naming
Length: 5:26
Summary:?Butanoic acid, hexanoic acid, 3-methyl hexanoic acid, and (E)-3-heptenoic acid are drawn and named.
Key concepts/Skills:?Carboxylic acid nomenclature
Nitpicky criticisms:?It’s fine. One weird thing though – around 5:15 it’s said that “this is only true if you’re assuming that I drew it in the actual 3 dimensional configuration in some way”. Not sure what is meant by this.
Red flags:?None

Video #68 – Fischer Esterification
Length: 17:12
Summary:?Heptanoic acid is drawn with ethanol and H2SO4. Reaction really begins at 5:00. Protonation of carbonl oxygen, addition of ethanol, ethanol losing proton, protonation of OH, and loss of water. Formation of ethyl heptanoate.
Key concepts/Skills:?Equilibrium, carboxylic acids, addition, elimination.
Nitpicky criticisms:?“Fischer”, ahem. The video begins with “let’s think about what might happen if we do this”. Little pet peeve of mine – we can appreciate things looking backward, but it’s difficult to look forward using first principles. The discussion up to 5:00 probably could have been skipped, since it’s just discussing the acidity of H2SO4.?Would be good to mention that acid makes the carbonyl group a better electrophile; also, an opportunity is missed to explain why the nucleophile attacks the carbon instead of the (positively charged) oxygen. It’s said that we’ve seen this “many, many times” in SN2 reactions but this is actually his first time drawing the addition mechanism in carbonyls. It’s good to mention the reaction is in equilibrium, but it would be better to mention *why* equilibrium favors formation of the ester. Finally I know I’m a grouch, but I just don’t like statements like “protons are flying around everywhere” (11:25).
Red flags:?The elimination step at 14:00 is drawn as a bimolecular mechanism, which is incorrect. Deprotonation should be shown as a separate step.

Video #69 – Acid Chloride Formation
Length: 11:48
Summary:?The mechanism for formation of acetyl chloride from acetic acid using SOCl2 is shown.??
Key concepts/skills:?Conversion of carboxylic acids to acid chlorides
Nitpicky criticisms:?Would be good to mention why Cl is a good leaving group. Why do we lose oxygen and not Cl in the elimination step? It would be good to mention that the gas SO2 is formed and this will drive the reaction to completion. The last step shouldn’t have an equilibrium arrow.?Heh, nobody would perform this reaction in a beaker [9:15] ?it stinks too much!
Red flags:?Just the arrow pushing (as always)

Video #70 – Amides, Anhydrides, Esters, and Acyl Chlorides
Length: 8:48
Summary:?The structures of acetamide, N-methyl propanamide, methyl ethanoate, acetic anhydride, ethanoic anhydride, propanoic anhydride, acetyl chloride, and ethanoyl chloride are drawn.
Key concepts/Skills:?Nomenclature of amides, esters, anhydrides, and acid chlorides
Nitpicky criticisms:?None
Red flags:?None

Video #71 – Relative Stability of Amides, Esters, Anhydrides, and Acyl Chlorides
Length: 11:09
Summary:?Acetamide, methyl acetate, acetic anhydride, and acetyl chloride are drawn, along with their resonance forms. The higher reactivity of anhydrides is rationalized as being due to the lower stability of the resonance form.
Key concepts/Skills:?drawing resonance structures, understanding stability of resonance structures.
Nitpicky criticisms:?If nitrogen is electron rich – more electronegative than carbon, then why might it give up its electrons? [2:44]. ?Finally, for the ester, he mentions the real reason why oxygen is less nucleophilic than nitrogen – because it is more electronegative. This is the first time this is mentioned in the whole video series. How does one differentiate between “quite stable” and “just stable”? Would be better just to show an arrow beginning with “most stable” on the left going to “least stable” on the right. Finally, he says “we’ll explore some of the mechanisms in the next few videos”, but in fact it’s just mentioned in one video.

I think ranking leaving group ability is a more powerful way to explain the relative reactivity of carboxylic acid derivatives than the relative stability of resonance forms, but overall this is a useful video.
Red flags:?No resonance structure is drawn for acetyl chloride because it is said that it “has no resonance structure” ; instead of saying it “has no resonance structure”, which is not true, it would be better to say that the resonance form is not “significant”; also, chlorine is less electronegative than oxygen, so the answer is not electronegativity (9:20) but poor orbital overlap.

Video #72 – Amide Formation From Acyl Chloride
Length: 9:01
Summary:?The reaction between butanoyl chloride and dimethylamine is shown to give N,N-dimethylbutanamide.
Key concepts/Skills:?Reactivity of acyl chlorides, nucleophilic acyl substitution.
Nitpicky criticisms:?Another instance of “Let’s think about what might happen….”. The “stability” argument from the previous video is used to explain the reactivity of acyl chlorides relative to amides, but leaving group ability is more valuable. Compare Cl(-) as a leaving group versus Me2N(-)! This explains everything.
Red flags: Just arrow pushing.

Video #73 – Aldol Reaction
Length: 11:41
Summary:?The aldol addition reaction between two aldehydes RCH2CHO in the presence of NaOH is shown.
Key concepts/Skills:?Enolates, addition reactions, aldol reaction, acidity of alpha protons, enolates as nucleophiles.
Nitpicky criticisms:?It’s said that this is a review of previous discussions of enolate ions, but in fact this is the first description of them. It would be better to consistently show the aldehyde hydrogen.
Red flags:?The arrow pushing here makes it look like the carbonyl carbon is the nucleophile, which isn’t the case [6:32].

And that’s it: 73 videos for organic chemistry from Khan Academy. Again, lots of nomenclature videos here, plus a few on reactions. Kind of disappointing that there’s so few on the reactions of aldehydes, ketones, and carboxylic acids; nothing on reduction, oxidation, extensions of enolate chemistry… a lot of work still left to do, in other words. ?In the next post I’ll write an overall summary of these videos that answers the question, “How useful are Khan Academy videos for learning organic chemistry?”.

Tagged as: khan academy, videos

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#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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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.

sp2 Hybrid Orbitals and the Structure of Ethylene

The same kind of orbital hybridization that accounts for the methane structure also accounts for the bonding together of carbon atoms into chains and rings to make possible many millions of organic compounds. Ethane, C2H6, is the simplest molecule containing a carbon–carbon bond.

We can picture the ethane molecule by imagining that the two carbon atoms bond to each other by s overlap of an sp3 hybrid orbital from each. The remaining three sp3 hybrid orbitals on each carbon over- lap with the 1s orbitals of three hydrogens to form the six C]H bonds. The C]H bonds in ethane are similar to those in methane, although a bit weaker—421 kJ/mol (101 kcal/mol) for ethane versus 439 kJ/mol for methane. The C ] C bond is 154 pm long and has a strength of 377 kJ/mol (90 kcal/mol). All the bond angles of ethane are near, although not exactly at, the tetra- hedral value of 109.5°.

The Sections of a Formal Lab Report

The Sections of a Formal Lab Report

Title:

The title of the report should be descriptive but concise. Do not include experimental data or long chemical names. The title should catch the reader’s attention and orient them to the purpose of the study.

Abstract:

An abstract is a brief summary of the most important findings of a scientific experiment. It allows the reader to survey the contents of a paper or lab report quickly, and to decide whether to keep reading. It should clearly identify the purpose and main results of an experiment, and should be clear, concise, and self-contained. An abstract should be between 100 and 200 words in length and, for CHM151Y, should be one single paragraph.

Introduction:

The introduction section of a lab report explains the experiment and provides the reader with relevant background information for the experiment. In some cases, an introduction will include both theoretical and practical background information. For CHM151Y, it should be between 250 and 500 words in length.

Experimental:

The experimental section describes the practical details of the experiment without reporting or interpreting experimental results (save this for the next section). This section should include a description of the experiment that is sufficient in detail such that another scientist could reproduce your results but not so detailed as to provide unnecessary information. Since this is written for people in the field we assume certain knowledge (i.e. we assume the reader knows what a Pasteur pipette is). The experimental section should not be copied from the laboratory manual.

Results and Discussion:

The results of the experiment (i.e. experimental data) should be reported before the discussion in a neat, easy to read format. Tables and figures are an excellent way to report experimental data. The discussion section should follow the results and is where you interpret the results of an experiment. It differs from a results section in the analysis you do. When you present your results, you simply state them as they are, pointing out important ones but not explaining why they are important or what they mean. Sometimes, scientific articles combine their results and discussion into one section. The discussion section of your lab report is where you will demonstrate that you understand the experiment beyond simply being about to follow the instructions to complete it. You will analyze your results and interpret them. This includes both the successful parts of the experiment, and the parts of the experiment that may not have proceeded as expected. If you obtained results that are different from what you expected, you should do your best to explain why this happened. Explain what you know for certain from your interpretation of your results, and state any conclusions that you can draw. For CHM151Y, the results and discussion should be between 500 and 1000 words in length.

Conclusions:

A conclusion provides a clear summary of the main findings of an experiment, their meaning, their relationship to the purpose of the experiment, and their relevance beyond the work presented in the lab report. Together with the introduction, the conclusion frames the lab report. It is often very short in an undergraduate lab report, and for CHM151Y should be between 50 and 100 words in length.

References:

Any idea that is not your own needs to be referenced.

sp3 Hybrid Orbitals and the Structure of Methane

The bonding in the hydrogen molecule is fairly straightforward, but the situ- ation is more complicated in organic molecules with tetravalent carbon atoms. Take methane, CH4, for instance. As we’ve seen, carbon has four valence electrons (2s2 2p2) and forms four bonds. Because carbon uses two kinds of orbitals for bonding, 2s and 2p, we might expect methane to have two kinds of C ] H bonds. In fact, though, all four C ] H bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron. How can we explain this?

An answer was provided in 1931 by Linus Pauling, who showed mathemat- ically how an s orbital and three p orbitals on an atom can combine, or hybrid- ize, to form four equivalent atomic orbitals with tetrahedral orientation. These tetrahedrally oriented orbitals are called sp3 hybrids. Note that the superscript 3 in the name sp3 tells how many of each type of atomic orbital combine to form the hybrid, not how many elec- trons occupy it.

The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer. When an s orbital hybridizes with three p orbitals, the resultant sp3 hybrid orbitals are unsymmetrical about the nucleus. One of the two lobes is larger than the other and can therefore overlap more effec- tively with an orbital from another atom to form a bond. As a result, sp3 hybrid orbitals form stronger bonds than do unhybridized s or p orbitals.

The asymmetry of sp3 orbitals arises because, as noted previously, the two lobes of a p orbital have different algebraic signs, 1 and 2, in the wave func- tion. Thus, when a p orbital hybridizes with an s orbital, the positive p lobe adds to the s orbital but the negative p lobe subtracts from the s orbital. The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction.

When each of the four identical sp3 hybrid orbitals of a carbon atom overlaps with the 1s orbital of a hydrogen atom, four identical C]H bonds are formed and methane results. Each C ] H bond in methane has a strength of 439 kJ/mol (105 kcal/mol) and a length of 109 pm. Because the four bonds have a specific geometry, we also can define a property called the bond angle. The angle formed by each H ] C ] H is 109.5°, the so-called tetrahedral angle. 

Describing Chemical Bonds: Valence Bond Theory

How does electron sharing lead to bonding between atoms? Two models have been developed to describe covalent bonding: valence bond theory and molecular orbital theory. Each model has its strengths and weaknesses, and chemists tend to use them interchangeably depending on the circumstances. Valence bond theory is the more easily visualized of the two, so most of the descriptions we’ll use in this book derive from that approach.

According to valence bond theory, a covalent bond forms when two atoms approach each other closely and a singly occupied orbital on one atom overlaps a singly occupied orbital on the other atom. The electrons are now paired in the overlapping orbitals and are attracted to the nuclei of both atoms, thus bond- ing the atoms together. In the H2 molecule, for instance, the H ] H bond results from the overlap of two singly occupied hydrogen 1s orbitals.

The overlapping orbitals in the H2 molecule have the elongated egg shape we might get by pressing two spheres together. If a plane were to pass through the middle of the bond, the intersection of the plane and the overlapping orbit- als would be a circle. In other words, the H]H bond is cylindrically symmetri- cal. Such bonds, which are formed by the head-on overlap of two atomic orbitals along a line drawn between the nuclei, are called sigma (s) bonds.

During the bond-forming reaction 2 H∙ n H2, 436 kJ/mol (104 kcal/mol) of energy is released. Because the product H2 molecule has 436 kJ/mol less energy than the starting 2 H∙ atoms, the product is more stable than the reactant and we say that the H ] H bond has a bond strength of 436 kJ/mol. In other words, we would have to put 436 kJ/mol of energy into the H]H bond to break the H2 molecule apart into H atoms. [For convenience, we’ll generally give energies in both kilocalories (kcal) and the SI unit kilojoules (kJ): 1 kJ 5 0.2390 kcal; 1 kcal 5 4.184 kJ.]

How close are the two nuclei in the H2 molecule? If they are too close, they will repel each other because both are positively charged, yet if they’re too far apart, they won’t be able to share the bonding electrons. Thus, there is an opti- mum distance between nuclei that leads to maximum stability. Called the bond length, this distance is 74 pm in the H2 molecule. Every cova- lent bond has both a characteristic bond strength and bond length.

Development of Chemical Bonding Theory

By the mid-1800s, the new science of chemistry was developing rapidly and chemists had begun to probe the forces holding compounds together. In 1858, August Kekulé and Archibald Couper independently proposed that, in all organic compounds, carbon is tetravalent—it always forms four bonds when it joins other elements to form stable compounds. Furthermore, said Kekulé, car- bon atoms can bond to one another to form extended chains of linked atoms. In 1865, Kekulé provided another major advance when he suggested that carbon chains can double back on themselves to form rings of atoms.

Although Kekulé and Couper were correct in describing the tetravalent nature of carbon, chemistry was still viewed in a two-dimensional way until 1874. In that year, Jacobus van’t Hoff and Joseph Le Bel added a third dimen- sion to our ideas about organic compounds when they proposed that the four bonds of carbon are not oriented randomly but have specific spatial directions. Van’t Hoff went even further and suggested that the four atoms to which car- bon is bonded sit at the corners of a regular tetrahedron, with carbon in the center.

Note the conventions used to show three-dimensionality: solid lines represent bonds in the plane of the page, the heavy wedged line represents a bond coming out of the page toward the viewer, and the dashed line represents a bond receding back behind the page, away from the viewer. These representations will be used throughout the text.

Why, though, do atoms bond together, and how can bonds be described electronically? The why question is relatively easy to answer: atoms bond together because the compound that results is more stable and lower in energy than the separate atoms. Energy—usually as heat—always flows out of the chemical system when a bond forms. Conversely, energy must be put into the chemical system to break a bond. Making bonds always releases energy, and breaking bonds always absorbs energy. The how question is more difficult. To answer it, we need to know more about the electronic properties of atoms.

We know through observation that eight electrons (an electron octet) in an atom’s outermost shell, or valence shell, impart special stability to the noble- gas elements in group 8A of the periodic table: Ne (2 , 8); Ar (2 , 8 , 8); Kr (2 , 8 , 18, 8). We also know that the chemistry of main-group elements is governed by their tendency to take on the electron configuration of the nearestnoble gas. The alkali metals in group 1A, for example, achieve a noble-gas con- figuration by losing the single s electron from their valence shell to form a cation, while the halogens in group 7A achieve a noble-gas configuration by gaining a p electron to fill their valence shell and form an anion. The resultant ions are held together in compounds like Na1 Cl2 by an electrostatic attraction that we call an ionic bond.

But how do elements closer to the middle of the periodic table form bonds? Look at methane, CH4, the main constituent of natural gas, for example. The bonding in methane is not ionic because it would take too much energy for carbon (1s2 2s2 2p2) either to gain or lose four electrons to achieve a noble-gas configuration. As a result, carbon bonds to other atoms, not by gaining or losing electrons, but by sharing them. Such a shared-electron bond, first pro- posed in 1916 by G. N. Lewis, is called a covalent bond. The neutral collection of atoms held together by covalent bonds is called a molecule.

A simple way of indicating the covalent bonds in molecules is to use what are called Lewis structures, or electron-dot structures, in which the valence- shell electrons of an atom are represented as dots. Thus, hydrogen has one dot representing its 1s electron, carbon has four dots (2s2 2p2), oxygen has six dots (2s2 2p4), and so on. A stable molecule results whenever a noble-gas configura- tion is achieved for all the atoms—eight dots (an octet) for main-group atoms or two dots for hydrogen. Simpler still is the use of Kekulé structures, or line- bond structures, in which a two-electron covalent bond is indicated as a line drawn between atoms.

The number of covalent bonds an atom forms depends on how many addi- tional valence electrons it needs to reach a noble-gas configuration. Hydrogen has one valence electron (1s) and needs one more to reach the helium configu- ration (1s2), so it forms one bond. Carbon has four valence electrons (2s2 2p2) and needs four more to reach the neon configuration (2s2 2p6), so it forms four bonds. Nitrogen has five valence electrons (2s2 2p3), needs three more, and forms three bonds; oxygen has six valence electrons (2s2 2p4), needs two more, and forms two bonds; and the halogens have seven valence electrons, need one more, and form one bond.

Valence electrons that are not used for bonding are called lone-pair electrons, or nonbonding electrons. The nitrogen atom in ammonia, NH3, for instance, shares six valence electrons in three covalent bonds and has its remaining two valence electrons in a nonbonding lone pair. As a time-saving shorthand, nonbonding electrons are often omitted when drawing line-bond structures, but you still have to keep them in mind since they’re often crucial in chemical reactions.