Lecture Outline Fates of Carbocations

Name ______________________________________ By now we've learned three different reaction mechanisms — SN2, E2, and ionization (SN1/E1). The reactants, reagents, and solvents involved in each of these reactions can seem very similar at first. Toward the end of today's class we're going learn a method for predicting what mechanism(s) is/are likely to occur. The most critical part of that decision-making process is knowing what's required for each of the three mechanisms. So here's your chance to test your understanding — this is your chance to review for your own benefit; we won't pick this one up. Step 1 is to look for a leaving group. What are the leaving groups we've learned? Step 2 is to evaluate the possibility of SN2. (a) What kinds of Cs undergo SN2 reactions? What's the key factor as far as that C is concerned? (b) What nucleophiles will do the job? Specifically, what's the approximate cut-off in basicity for 2nd-row nucleophiles? What about nucleophiles below the 2nd row? Step 3 is to think about the possibility of E2. How strong a base is needed for an E2 reaction? What's the approximate cut-off in basicity? Step 4 is to think about ionization to form a carbocation. (a) What kinds of carbocations can form by ionization? (b) What solvents are required? What are some good ionizing solvents? Lecture outline Fates of carbocations — In general — no matter how they are created — carbocations can: 1. Add a nucleophile. (SN1) 2. Lose H+ (E1) 3. Rearrange to a more stable cation, followed by 1 or 2 (or sometimes 3) Carbocation rearrangements — H H C C C a "1,2-hydride shift" (H moves with e– pair) C R R C C C C a "1,2-alkyl shift" (R moves with e– pair) Rearrangement is fast if it leads to a more stable cation (e.g., 2° —> 3°, strained —> less strained, localized —> delocalized ) — only 1,2-shifts are known; longer range shifts have not been observed R R C C C C — and other high-energy reaction intermediates do not rearrange in this way. Specifically, radicals do not do 1,2 shifts, neither do carbanions, and plain old ordinary stable molecules certainly do not do this!!! Here's an example of an SN1 solvolysis that involves a 1,2-hydride shift. Write the mechanism. Br HOCH3 OCH3 OCH3 + some of this may also be formed — (dep on rxn conditions) major or only product Write the mechanism for this one... OMs OEt OEt HOEt + major or only product some of this may also be formed — (dep on rxn conditions) The ratio of the two SN1 products shown depends on relative rates of rearrangement vs addition of nucleophile ... hard to predict... Here's a general rule that will serve us well: If rearrangement is exothermic, this will be the major rxn pathway, (or the only rxn pathway) Some carbocation rearrangements result in ring expansion or contraction. Although these may look different, they're all just 1,2-shifts! Add the curved arrow to show a standard 1,2-shift of the bold R, then tie the "R" groups together and voilá you've just illustrated a ring expansion. (What two "things" would need to be connected for the shift to result in a ring contraction?) R R R C CH R C CH Here's a cation that does a 1,2-shift of a ring CH2 to make 2-methylcyclopentyl cation. Use curved arrows to show this reaction and draw the cation produced. 2° —> 2°, so why does this rearrangement go? What makes it exothermic? (you may have noticed that the 2° cation produced could rearrange again to make a 3° cation) Here's another example of a 1,2-H shift that might at first appear to be thermoneutral, but it's really exothermic and goes rapidly. Illustrate the rearrangement and draw the cation produced. What provides the thermodynamic "driving force" in this case? SN2 vs E2 vs SN1/ E1 reactions — — how to figure out what happens when (please disregard table 8.5 on p 291 — it's not that complicated. Just use what you already know about the mechanisms, and follow the steps below.) Here's an outline of the critical bits — 1. Is there a good l.g.? yes (go on) 2. Is SN2 possible? no (stop — no rxn) What kind of R–X is required? What kind of nucleophile is needed? yes (fine — go on) no (fine — go on) 3. Is E2 possible? (In addition to or instead of SN2) What kind of base is necessary? Remember, if both SN2 and E2 are possible, faster SN2 => less E2 *** *** yes to 2 or 3 or both (stop — these rxns will usually be faster than ionization) no to 2 and 3 (go on) 4. Is ionization possible? What kind of carbocation? What about solvent? Predict what’s going to happen... What mechanism? Draw the products. CH3CO2– I Et2O I tBuO– DMSO I CH3OH CH3CO2– Br Et2O tBuO– Br DMSO CH3OH Br Extra bonus question. Consider the SN1 reactions of the following bromides in a polar solvent. The first ionizes rapidly to make a 3° cation. If the second were to ionize, the 1° cation produced would rearrange rapidly to the same 3° cation. Does this mean that the second compound should also do a rapid SN1 reaction? Explain. Does similar reasoning also apply to the third compound? Br Br Br Guidelines for deciding whether a reaction will go by SN2, E2, or ionization (SN1 and/or E1) mechanisms Ask yourself the following questions (in order): 1. Is there a good leaving group? If not, no reaction! You need a leaving group for all of these reactions, right? If there's no leaving group, nothing else matters. Halides or sulfonates. (Q: But what if there's a really great nucleophile and the perfect solvent and I heat it up and there's a full moon and I really need to get that nucleophile on there... A: Nothing happens. Nothing. No way, no how.) Bottom line: You cannot use SN2 nor SN1 chemistry to replace an H, an OH, an OR, an NH2, a CH3, or anything else that's not a leaving group, no matter how hard you try!!! 2. Is SN2 possible? OK, we have a leaving group. Now all we need is a reasonably good nucleophile and a cooperative substrate. (a) Nucleophile — this has to be a moderately good base (e.g., (strongest to weakest) HO–, H3N, RCO2– are all ok; F– is ok, but it's pretty weak; H2O and ROH are too weakly basic — these nucleophiles are just too wimpy to react) Remember that nucleophiles below the second row, e.g. RSH, Br–, I–, etc react readily, even though they're extremely weak bases. (b) Substrate (alkyl halide) — this has to be methyl, 1°, or 2° (not 3°!), and watch out for β-branching, which slows the reaction. (c) Solvent — Forget about it. Sure, some solvents are better than others, but in general the choice of solvent is usually not a "make-or-break" issue for this reaction. 3. Is E2 possible? This might happen in addition to or instead of SN2. (a) We need a suitable β-H and a strong base — usually at least as strong as HO– or RO–. (NH3, amines, and cyanide are borderline bases and can do E2 under certain conditions.) (b) If E2 is possible, the ratio of SN2 to E2 depends mainly on how favorable SN2 is. If SN2 is fast, we see little or no E2; if SN2 is slow, we get a significant amount of E2; and if SN2 can't happen (e.g., with 3° R–X), we get only E2. (c) Minor points: Solvent can influence the SN2/E2 ratio — protic solvents tend to enhance E2 since they slow SN2 by hydrogen bonding to the nucleophile, and stronger bases and higher temperatures tend to increase the amount of E2 as well. The temperature effect is due to the fact that entropy favors E2 (2 molecules —> 3) over SN2 (2 molecules —> 2) — If SN2 or E2 (or both) can happen, ionization probably won't. It's normally too slow to compete with the bimolecular SN2 and E2 reactions. (But there are exceptions to this, as we'll see later — e.g., protonated 2° alcohols tend to dissociate to water and carbocations faster than they undergo SN2 attack. And, of course, with very low concentrations of nucleophile/base, the bimolecular reactions would be slowed to the point where ionization may compete. But let's get real here... when we write "NaOEt" over a rxn arrow, we mean a significant amount — enough to do whatever bimolecular reaction it can do — not 0.00001 M.) — If neither SN2 nor E2 can happen, ionization may occur. 4. Ionization So what is required in order for ionization (the first step of the SN1 and E1 reactions)? (a) Formation of a relatively stable carbocation — 3°, 2°, or a delocalized carbocation. (b) A polar solvent. This time solvent matters — protic is better than aprotic, but polarity is the real key. Ionization involves separation of charge, which costs energy. The solvent needs to be polar enough to solvate and thereby stabilize the ions. So which is it, SN1 or E1? This is a tough question. (a) Higher temperatures tend to favor eliminations because these have a more positive ∆S° than the corresponding substitutions (more product molecules are formed). (b) Moderately nucleophilic species will generally give more SN1 than E1. Very nonnucleophilic species will be more likely to do E1. For example, Br– is a good nucleophile, but an extremely weak base, so it does exclusively SN1 (3°-ROH + HBr —> RBr). H2O is less nucleophilic than bromide and slightly more basic; it does mostly SN1 but some E1 also (ca. 80%:20%). In dehydration of alcohols with H2SO4 or H3PO4, the conjugate bases HSO4– and H2PO4– are essentially non-nucleophilic; even though they're also very weak bases, they do exclusively E1.