Back-to-School Trick Questions

September 14th, 2010

The kids are back in school, which means it’s time to mess with their heads.  Since the first week of undergrad orgo is usually devoted to a review of electronic structure and bonding, why not try this on for size:

Fake Problem Set #1.  (Insert fake instructions)  NOTE:  Draw all resonance structures where appropriate.

1)  The protons shown in green for 2,4-pentanedione (I) are roughly 10^40 times more acidic that those on n-pentane (II).  Provide an explanation. 

pentanedione and pentane

2)  The protons on benzene (III, C6H6) are roughly 10^7 times more acidic than those on n-pentane.  Provide an explanation.

benzene

3)  Boric acid (V) is a weak acid in water.  It is roughly a million times more acidic than methanol (IV).  Provide an explanation for the acidity of boric acid (by drawing chemical structures).

 boric acid and methanol

* All acidities based on pKa values in water (either measured or extrapolated) in  Evans’ pKa table.

Answers after the jump…

Answers

1.  The first part of this question is a fastball to set up the curves.  (The major effect here is stabilization of the conjugate base of I by the electron delocalization described in the resonance structures shown below.  The negative charge on the conjugate base of II is not similarly delocalized.) 

answer 1

2.  Now that you’ve got the students thinking about resonance effects, you are bound to sucker more than a few into writing nonsensical resonance structures to explain the increased acidity of benzene relative to aliphatic hydrocarbons:

answer 2

The moral of the story?  Just because you can push electrons around Lewis structures with a pencil doesn’t mean that doing so makes any sense.   The lone pair on the conjugate base of benzene is in an orbital orthogonal to the pi-system of the ring, so there’s no delocalization of the charge by a resonance effect.   Also, early on in the class, more than a few kids forget about the implied hydrogens on organic structures.  Drawing in the five implied hydrogens would make them realize the “alternate” resonance structures have Texas carbon issues.

The correct explanation for the increased acidity of benzene vs. alkanes is that the electron density is placed in an orbital with more s character (sp2 vs. sp3), and the electron-withdrawing (inductive) effect of the nearby sp2 (vs. sp3) carbons also helps stabilize the charge.

3.  If the kids are still fixated on resonance—or if they’re thinking back to nitric acid and carbonic acid from gen chem—you are probably bound to encounter answers where boric acid is treated as a Bronsted acid (H+ donor, top reaction, in blue).  The “real way” boric acid behaves as an acid in water is first by acting as a Lewis acid (bottom reaction, in black).  In fairness to the undergrads, I’d bet >25% of graduate students around the country would not be able to correctly explain the acidity of boric acid.

answer 3

 

Now…go terrorize your students and coworkers.


27 Responses to “Back-to-School Trick Questions”

  1. Matt Says:

    Boric acid is an acid because its name is boric acid. QED

  2. excimer Says:

    1 and 2 are reasonable, but 3 is pure evil for an undergrad class. I… kind of love it.

  3. Dr Polarity Says:

    Switch a carbon to some weird looking heteroatom, and all logic goes out the window in most undergrad classes…

  4. Hap Says:

    Embarrassingly enough, I didn’t know 3). I’m…somewhere out of grad school. Oops.

  5. Stephen Bahl Says:

    I got the first two one and probably not the whole explanation for the second one, but the third one stumped me. Are “Texas carbon issues” structures where carbon has too many bonds? What’s the story behind that term?

  6. Chemjobber Says:

    @Stephen: Everything’s bigger in Texas, including valences. :-)

    If I would have answered a version of #3 during an interview talk, I think I would have gotten the job. Dammit.

  7. Paul Says:

    @Stephen: Yup. A “Texas” carbon is a carbon in a Lewis structure that is drawn with more than 4 structural pairs (i.e., bonds + lone pairs). In the incorrect blue resonance form in question 2, the carbon with the negative charge has 5 structural pairs: 2 sigma bonds to carbon, 1 (implied) sigma bond to hydrogen, 1 pi bond, and 1 lone pair.

    Interestingly, there is no Wikipedia article for Texas carbon, but there is one for the town Carbon, Texas.

  8. excimer Says:

    …and then, if that’s not big enough for you, there’s the Canada carbon…

  9. James Says:

    Good point about “pitch selection” Paul, that’s a good trick. I wonder how the order in which questions are asked affects the results.

    Excimer, why “Canada carbon”? I never understood that.

  10. provocateur Says:

    the kid’s telling ,me to prove it for ‘2’ & ‘3’.
    Until then he is sticking to resonance!

  11. excimer Says:

    Canada carbon: 6 bonds to carbon. Everything’s bigger in Texas, but Canada is so big it’s beyond fathoming.

  12. Ron Says:

    This reminds me of another post that was on the old ChemBark site, found here. I love posts like this 😀

  13. Paul Says:

    Oh man…I love that allene question. EVERYONE falls for it.

  14. Ron Says:

    What about this fill-in-the-reagents question? In my experience, most undergraduates will fill in an eight-step synthesis because they don’t recognize the symmetry of the starting material and product.

  15. Paul Says:

    @Ron: That’s a good one. I shall add that to my repository of pain.

  16. Wavefunction Says:

    A related question is the Fries rearrangement which also throws dem undergrads off balance.

  17. Mitch Says:

    For #3 you should include H3O+ in the final step for the sake of charge and mass balance sanity and clarity.

  18. Chris at UW Says:

    Here’s one undergrad entering the first quarter of o-chem who will be a little better prepared for this question should it come up.

  19. Five Key Factors That Influence Acidity « Master Organic Chemistry Says:

    […] Watch out though – it isn’t enough for a π system to simply be adjacent to a proton – the electrons of the conjugate base have to be in an orbital which allows for effective overlap (for a dastardly trick question in this vein that routinely stymies Harvard premeds, look here.) […]

  20. Matt Says:

    I think this is sad, or at least the satisfaction that you can “get” students with them is sad. Look, if you paid some martial artist to teach you said art, and all that person did was throw you to the ground and beat the #$@$ out of you, then flex over you as if they’d proved something…you’d be upset. And they would be idiots for thinking there was any glory in beating the crap out of the unskilled and ignorant.

    This is the academic equivalent of that. Again, not the hard questions, the smugness that so few can answer them.

    If you, with years of experience, can’t fool even bright students seeing the material for the first time, you really are in sad shape. I’ve had professors and known professors who think like that, who somehow get their jollies by keeping score against their students. Psychiatric counseling is needed there.

    The challenge is to so capture students with why the material is interesting, and so fascinate them with its intricacies, that you manage to move large numbers of them to the point where they CAN answer questions like these. Any utter imbecile can teach nothing and fool the whole class; it takes enormous talent and work to stimulate and teach people to where hard questions become easy. C.f. Feynman’s lectures in physics.

  21. Paul Says:

    Matt, you’ve got the wrong end of the stick.

    But fear not, I’ll straighten you out this weekend with the next post.

  22. ChemBark » Blog Archive » The Purpose of Trick Problems Says:

    […] Matt—one of several, it’s not this Matt—got his knickers in a twist about the last set of trick orgo problems that I posted.  His rant is copied below. I think this is sad, or at least […]

  23. Paul Says:

    Good…I’m only running three weeks behind.

  24. 5 Key Factors That Affect Acidity in Organic Chemistry — Master Organic Chemistry Says:

    […] Watch out though – it isn’t enough for a π system to simply be adjacent to a proton – the electrons of the conjugate base have to be in an orbital which allows for effective overlap (for a dastardly trick question in this vein that routinely stymies Harvard premeds, look here.) […]

  25. WWWTP? – Nomenclature Edition | ChemBark Says:

    […] readers of ChemBark know that I love trick questions, and structures like the ones above make for great nomenclature practice. Many […]

  26. Haynes Says:

    So whats the deal with boric acid? Does it have to do with its hybridization compared to methanol, or is it something wacky with its p-orbital. I am curious cause I just want to understand why it reacts the way it does.

    Thanks again for the great question

  27. Curious student Says:

    The answer to the third question is beyond my brain’s thinking capability(which is not so low) ( it is probably because i’m in 12th grade )…But can someone PLEASE explain because i wanna badly know why boron reacts that way?…i’ll be grateful..thanks


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