Some facts and principles to know cold while learning organic chemistry

 

Somehow this blog got mentioned in a list of the 25 best sites for college chemistry majors (http://www.onlinedegrees.org/25-best-chemistry-blogs-for-college-students/).  Classes are just starting up, so I want to get this out quickly.  It isn’t graven in stone, and additions will be put in from time to time, along with date they were put in, so you don’t have to read through the whole thing to find the new stuff.
It is impossible to learn organic chemistry by memorizing it.  First,  there is just too much of it.  Second, it isn’t hard to make (and be tested on) an organic molecule that has never been made before.  So it is crucial to be able to apply principles to new situations (which is one reason why I think anyone thinking of becoming an MD should be able to do it). 
Nonetheless, a certain number of facts and principles simply must be learned (memorized if you will) and become second nature.   Get them down cold and you’ll be surprised how much easier organic chemistry becomes.  
 I’m presently rereading the first 1000 pages of Clayden, Greeves, Warren and Wothers “Organic Chemistry” Oxford University Press 2001, so this post will be a work in progress. Additions will be surrounded by the date they’re added.   Clayden et. al. is  a great book and full of principles and explanations for them (as far as they can be given).  One of  the few shortcomings of  Clayden’s book  is that it never supplies these  princples about the transition metals, so you are just given an arbitrary collection of facts about them that you must know.  If you don’t like this, be warned — you’ll find a lot of this sort of thing in medical school — because such principles aren’t available even to the most knowledgeable doc.
Another site to look at is  http://masterorganicchemistry.wordpress.com/  — it didn’t make the list of 25, but I think it’s quite good. 
Here goes:
The more electronegative an atom is the lower in energy its atomic orbitals are.  Why? Across a row electronegativity correlates with charge of the nucleus, so more highly charged nuclei (to the right) are more electronegative.  The more electronegative an atom is, the closer the electron orbital is to the nucleus, this is of lower energy because there is less separation of positive and negative charges.  

(4 Sep ’10) Memorizing numbers is hard, but the following is easy.  The first row of the periodic table will soon become your friend  Li Be B C N O F Ne.  The electronegativities start at Li 1.0 and pretty go up by .5 as you go across the row.  C is 2.5 which explains why the bonds it forms with the other elements are covalent — they’re just not that far apart in electronegativity.  Recall the numbers, and you’ll be able to assign the relative polarity {which atom is relatively positive and which atom is relatively negative in each bond} and how covalent the bond is to any other element down farther in the table, once you look up its electronegativty.  Try it with Silicon, Magnesium  etc. etc.  (4 Sep ’10)

S orbitals get closer to the nucleus than p orbitals (which actually have a node at the nucleus), so they are of lower energy.  So sp hybridized orbitals are lower in energy than sp2 than sp3.

Carbanions are more stable the more s character the orbital the two electrons are in (makes sense electrostatically, and negative electrons are on average closer to the nucleus in an s orbital than a p orbital).  Clayden p. 214.

Since alkyl groups are electron donating (I’m not sure why) primary carbanions are more stable than secondary carbanions, and secondary carbanions are more stable than tertiary carbanions.   Just the reverse holds for carbocations.

So the pKa’s are   
methane is 48
benzene is 43
acetylene is 25 (because the anion on an sp hybridized carbon is more stable)

(26 Sep ’10 )Substituted double bonds are more stable Clayden p. 489.  You don’t have to know the reason, but it is interesting nonetheless.  It’s something we used to call hyperconjugation. Clayden has a nice discussion here (although in different terms) — interaction of an antibonding pi* orbital with filled orbitals of parallel C-H and C-C bonds 

(28 Sep ’10) More substituted double bonds are also more nucleophilic than less substituted ones  Clayden 507. This is because the energy level of the  highest HOMO is increased by the electron donating effects of the attached alkyl groups.  I’m not sure why this makes them more stable, in fact I’d think raising an energy level would make a molecule less stable.

Nonbonding electrons in molecules (say NH3) are ALWAYS higher in energy than bonding electrons Clayden p.118
Electrophiles have either (1) an empty atomic orbital (2) a low-energy antibonding orbital.  This is why Br2 is an electrophile despite those 8 electrons around each atom — there is a low lying antibonding orbital.  Why is the antibonding orbital in Br2 of such low energy?  Because the Br-Br sigma bond is weak.   The C-C bond is strong so its antibonding orbital is of higher energy.   Why? Because molecular orbital theory tells us that the antibonding molecular orbital must go up in energy by at least as much as the bonding molecular orbital goes down in energy (all this is relative to the energy of the two unbonded atomic orbitals).    This is only true if we’re speaking of identical atoms, because here the atomic orbitals are of the same energy.  

Clayden p. 119 The most important antibonding orbital is the pi* orbital — is doesn’t have anything in it — rather Zenlike don’t you think?  As Clayden notes, a step on a staircase is still there whether or not anyone is stepping on it. Cute !

Hard nucleophile — these have high charge density because they are low in the periodic table where orbitals are small and close to the nucleus and not screened by other electrons.  Hard nucleophiles tend to go where the positive charge is greatest.  Examples are anions like F-, OH-, RO-, SO4–, Cl- , NH3, RMgBr, RLi  Clayden 237, 411.

Soft nucleophile — these have low charge density.  They react by putting their electrons into low level antibonding orbitals.  Examples are I-, RS-, C=C bonds.  Given the choice between a dipolar bond (like the carbonyl group) and a double bond — they choose the double bond.  

Electrophiles can be categorized the same way — H+ is as hard as they come, Br2 is soft .  The carbon of a C=0 group is hard Clayden 238

Hard nucleophiles tend to react with hard electrophiles, soft with soft (I’m not clear why this is true, but it is).  This is why H20 reacts with C=O in alpha beta unsaturated compounds, while Br2 reacts with the double bond.  Clayden 238

25 Sep ’10 Mnemonics: Med students love them, and the more scatological the better — Never Lower Tillie’s Pants, Mother Might Come Home — for the 8 bones of the wrist.  I’ll put in a few that I’ve found helpful.    The neurologist in me compels me to tell you that scatological stuff is easier to remember because it has an emotional kick.  Where were you and what were you doing when you first heard about 9/11?
I never could keep enantiomer and diastereomer straight.  Enantiomers are mirror images of each other, diastereomers are not.  English speakers will find it far easier to pronounce MIrrorEnantiomer than MirrorDiastereomer.  The rest of you are on your own.
R and S in the Cahn Ingold Prelog system:   The numbers 1, 2 and 3 are in a circle, with 1 on top.  The configuration is R if 2 appears to the Right of the 1 in a circle.
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Comments

  • James  On September 4, 2010 at 12:31 pm

    So if I understand (and summarize) correctly your advice is the following:
    1) understand the effects and consequences of electronegativity
    2) understand the principles of molecular orbitals (HOMO and LUMO) and how they affect reactivity
    3) understand the differences between hard and soft acids and bases.

    If you want to understand the basics of chemical reactivity it’s hard to think of better advice than that.

  • luysii  On September 4, 2010 at 1:38 pm

    Exactly: To the uninitiated, organic chemistry appears to be a blizzard of unrelated facts. Students who have gotten this far usually have pretty good memories and their natural tendency will be to try to memorize everything (which they’ll find after a month or two is impossible).

    The point of this post (which will be expanded from time to time) is tell them what they really have to know — some of it involves memorization. If they have to fill up their memory with something, this is the stuff to put in first.

  • Wavefunction  On September 5, 2010 at 2:04 pm

    As my uncle (a pharmacist) once put it, “Organic chemistry is just like math…only simpler”.

  • James  On September 7, 2010 at 3:41 am

    Finding the line between “what to memorize” and ‘when to focus on underlying patterns” is something I think about a lot and still don’t have a great answer to. For those who lack the superhuman memory abilities of Luysii’s friend who learned Hungarian in 2 weeks, there’s a point where pattern recognition of underlying concepts becomes more efficient. Like with a language, the vocabulary of ochem is definitely rote memorization (names of functional groups, terms/definitions, formulae, etc) whereas grammar (which would be reactions, in the language=ochem metaphor) requires pattern recognition. Memorizing the dozens of reactions in carbonyl chemistry without a conceptual roadmap is similar to memorizing lists of conjugated French verbs without learning the pattern – making a very inefficient use of their time.

  • luysii  On September 7, 2010 at 10:49 am

    My friend didn’t have a superhuman memory. His father was a doc who forced him into medicine. As I recall, he didn’t do particularly well in organic chemistry. His ability to remember word sounds and what they meant didn’t extend to memorizing or learning other things. It was a very specific talent.

  • James  On September 14, 2010 at 11:15 pm

    Being able to learn Hungarian in two weeks seems superhuman to me – it’s in such a weird linguistic family. I’ve had a really hard time learning Hebrew, despite taking four classes in it.

  • CFan  On October 6, 2010 at 7:05 am

    Hi Retread,

    I have read your posts on chemblog and have been inspired by the fact that you have put in so much of efforts to learn (re-learn?) Maths, Organic Chem (if I remember you were reading Wade and posting your comments). I am in a position similar to yours where I am reading Organic Chem from scratch (Clayden’s book alongwith Morrison’s are the bibles for intro to OC, IMO) and have found (realized?) the necessity of Maths to understand Physical Chemistry particularly (more so when we have a new discipline named Phy Org Chem these days). I have also noted some doubts/ stuff which I didn’t understand in CLayden and was not sure how to get them addressed (should I send it to Clayden, will he have the time to go through yada yada). Glad I found that you are doing the same. I am trying to develop some proof writing skills to read/ understand Apostol/ Spivak. I can email you more about the stuff which I need to ask about Math (not relevant to this post).

    —–
    Since alkyl groups are electron donating (I’m not sure why)…

    (26 Sep ’10 )Substituted double bonds are more stable Clayden p. 489. You don’t have to know the reason, but it is interesting nonetheless. It’s something we used to call hyperconjugation. Clayden has a nice discussion here (although in different terms) — interaction of an antibonding pi* orbital with filled orbitals of parallel C-H and C-C bonds

    Hyperconjugation (or No bond resonance) is what it used to be called (in Morrison and Boyd for example). Clayden calls it sigma (bond) conjugation as the mechanism by which alkyl groups release electrons – the section on stability of tertiary carbocations (chapter 17), a pink/(or some such colored) box where he ties in the general notion that electrons release by alkyl groups through their +I effect with the mechanism of release them by sigma conjugation.

    —-
    (28 Sep ’10) More substituted double bonds are also more nucleophilic than less substituted ones Clayden 507. This is because the energy level of the highest HOMO is increased by the electron donating effects of the attached alkyl groups. I’m not sure why this makes them more stable, in fact I’d think raising an energy level would make a molecule less stable.
    —-

    Yeah, I had (have?) this doubt as well. But my explanation (and correct me if I am wrong) is that when we try to check out nucleophilicity of multi-substituted double bonds, we use the filled sigma Bonding orbital with the filled pi-BMO (higher in energy)which is the original HOMO. When the overlap occurs and 2 new filled orbitals result, the higher one becomes the HOMO => nucleophilicity increases in the multi-substituted double bond case.

    Regarding the stability, I’ve seen the filled sigma Bonding orbital being overlapped with the unfilled pi-star ABMO (higher in energy). This gives a lower filled BMO => increase in stability of the resulting molecule. BTW, the LUMO would have increased in energy => lower electrophilicity of the multi-substituted double bond case => higher nucleophilicity (which we know from the treatment in the above paragraph.

    This argument is also used to explain the increased stability of amide bond as well – overlap of NBMO (lone pair on N) with the pi-star ABMO of the C=O bond.

    I was trying to group all these cases where he has used the MO theory based arguments for proving the increased stability/ nucleophilicity behavior etc.

    Till now, there are 4 cases –

    1. Amide linkage’s extra stability
    2. more subst. Carbocation’s extra stability
    3. more subst. double bond’s extra stability
    4. more subst. double bond’s extra nucleophilicity (in a similar fashion the alpha effect in case of hydroperoxides anion as well).

    I guess that in the case of stability they take the BMO ( C=C pi bond, C-C, C-H sigma bond)/ NBMO (lone pair on N) with the filled electrons and try to see that by overlapping with an ABMO (C=C or C=O pi star)/ NBMO (the p orbital), if it’d be possible to lower the energy of the filled orbital’s electron pair. IF yes, then there is increased stability.

    And in the final case, it is the overlap of the two BMOs.

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