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The Course of the E2 Reaction Can Be Influenced by the Stability of the Alkenes Produced

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The Course of the E2 Reaction Can Be Influenced by the Stability of the Alkenes Produced Name ______________________________________ The course of the E2 reaction can be influenced by the stability of the alkenes produced. This is discussed in section 9.4 of the text. One factor that affects alkene stability is alkyl substitution. The more alkyl groups attached to double bond Cs the more stable the alkene. The stabilization amounts to about 1.5 - 2 kcal/mol per alkyl group. 1 Rank the following (constitutionally) isomeric alkenes from most stable to least stable. Another factor we're already familiar with is steric strain (see p 181). Strain is also present when a double bond is trans in too small a ring. The smallest trans cycloalkene that is stable at room temperature is trans-cyclooctene. Make a model of cyclohexene and see if you can twist the double bond into the trans configuration. Similarly, double bonds to a bridgehead C of a small or moderate-sized bicyclic molecule like the one below are extremely strained. 2 Make a model of the compound at right. You won't be able to build this with two sp2 Cs for the double bond Cs — you'll need an sp3 C for the one at the bridgehead. Aside from the fact that one C is forced to be pyramidal, what's the problem with this π-bond? 3 We learned something about the stability of dienes earlier in the course. That will be important as well. Review that if necessary, and rank the dienes below in order of stability. 4 How many HW points do you think you have? _______ Lecture outline E2 reaction — Elimination, bimolecular. This reaction has second-order kinetics, rate = k[RX][B–], consistent with a concerted (i.e. one-step) reaction. Shown below is the simplest structural unit required for the E2 reaction. This is sometimes referred to as a "β-elimination". RO H C C X 1. Base — strong base required... RO–, HO– or stronger – – (certainly not weaklings like H2O, RCO2 , Cl , etc) amines are borderline (conj acid pKa ≈ 10 is the approximate cut-off) 2. l.g. — very weak base (halide ion or sulfonate ester) 3. Regiochem — if several β-Hs are present, several alkenes may be formed CH3 Br RO C CH CH3 H CH3 A B – – – a. Small bases (HO , CH3O , EtO , etc) can easily attack all β-Hs, product stability determines product ratio — in this case, the major product is the more substituted alkene ("Zaitsev's rule") b. Bulky bases (e.g., tBu–O–) preferentially attack the most sterically accessible β-Hs (this usually means 1° > 2° > 3°) — in this case, the major product is the less substituted alkene ("Hofmann's rule") 4. Stereochem — concerted elimination can happen only if C–H and C–X (l.g.) bonds are coplanar. This is because C(sp3) orbitals must become C(p) orbitals with the proper alignment to make the new π-bond. H X C C C C syn (eclipsed) H C C C C X anti H no rxn — orbs perpendicular C C X ! "-bond can't form The preference for anti elimination is demonstrated by reactions such as the following.... Br not EtO tBu tBu tBu To make sense of this, let’s start by (a) drawing the three staggered conformers of the alkyl bromide. (We'll complete the structures in class.) (b) What's the relationship between the H and Br that need to be eliminated to form the new π-bond? (c) When H and Br are lost and what’s left flattens out, what does it look like? Br H Br H Br H tBu tBu tBu E2 rxns go via a syn arrangement of H and X only if anti is not possible, e.g., which product is formed in this case? (Deuterium, D (= 2H) is an isotope of H with one proton and one neutron in its nucleus.) H EtO D or H Br ??? D H H H H and Br have ! = 0° D and Br have ! = 120° Why isn't the bridgehead H and the Br eliminated? — in acyclic cmpds, free rotation around σ-bonds usually allows every β-H to get anti to the l.g. in some conformation Here's another case that nicely illustrates the stereoelectronic requirements for the E2 reaction. How can we explain the following observations? OTs EtO and CH CH3 CH3 3 OTs EtO only CH CH3 3 Competition between E2 and SN2 — In many cases, both occur. The key is the SN2 rate, since this is more sensitive to structural "issues" (esp steric effects) than the E2 rate. — 3° haloalkanes Br HO — 2° haloalkanes Br OEt EtO + — 1° haloalkanes EtO Br OEt + — but β-branching slows SN2; so E2 competes EtO Br OEt + — How would you expect the ratio to change in the last case if EtO– were replaced with tBuO–? .
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