Lecture Outline Substitution and Elimination Reactions

Name ______________________________________ Text section 8.1 provides an introduction to the SN2 reaction. 1 (a) First, a little practice with curved arrows... lone pair in, bond pair out – add the curved arrows and draw the resulting products (or resonance structure). H H N O H CH3 H O CH3 Br O H (b) Notice that the electron flow is the same in each example. Make sure that your arrows are all going in the correct direction. (c) Make sure that you've kept track of every electron pair (you may want to add the lone pair that I've omitted in the last two examples), and that the formal charges are correct. Note that in each case (but the first) the charge ends up in a better place than where it started. You're already familiar with resonance and with acid-base reactions. The third rxn above is an SN2, and the 4th is a nucleophilic attack on a carbonyl. (In this example it leads to reduction of the carbonyl group.) (d) Keeping in mind this same electron flow, and resisting the urge to do anything more complicated, draw the products of the following two SN2 reactions. In each case the electrons flow from the nucleophile (electron pair donor) toward the leaving group (halogen). Br S + CH3 I + NH (e) Please check that you didn't do anything more than shift two pairs of electrons! Lecture outline Substitution and Elimination Reactions Text Ch 8 and 9 SN2 reaction — + Y: R—X nucleophile Y—R + X: alkyl halide (usually) leaving group (lg) E2 reaction — H + B: C BH C + C + C X base SN1 and E1 reactions — add'n of nucleophile (HY: or Y: – ) R X – X: R Y R carbocation loss of H+ alkene Lecture outline (note that this differs from the text's order) SN2 — nucleophilic subst. (nucleophile replaces halogen) E2 — elimination (H and halogen are lost to make alkene) SN1 & E1 — subst. and elim via a carbocation Summary — which happens when? X: SN2 examples (add the lone pairs and curved arrows, and draw the products) CH3 O + CH2 CH3 Br CH2 Br + CH3 S H CH3 CH3 NH2 I Cl O H 1. Kinetics and mechanism For Y:– + R–X —> Y–R + X:– rate = k[Y:–][R–X] transition state: [Yd– - - - - - R - - - - - Xd–]‡ rxn coordinate diagram: this is consistent with a onestep (i.e. concerted), bimolecular rxn mechanism 2. Stereochemistry — in principle, the reaction could take two stereochemically distinct courses ... a. "frontside" attack by the nucleophile CH3CH2 CH3CH2 C H CH3 Br + C H CH3 OH !– OH ‡ Br !– b. "backside" attack by the nucleophile CH3CH2 HO C H CH3 + Br Draw the products of these rxns... HO Br O CH3O Cl O EtO Cl CH3—I O !– HO CH2CH3 !– C Br H CH 3 ‡ Another experiment that establishes this stereochemical course is reaction of an optically active iodoalkane with radioactive iodide ion. Two things can be measured independently — the change in optical activity and the incorporation of radioactivity. R + I R I* I* + I What would you expect for rate of loss of optical activity if the reaction were to occur with (a) retention of configuration? (b) with inversion of configuration? (c) with 50:50 retention/inversion? 3. Effect of alkyl halide structure a. Substitution at the C to which the halogen is attached (the α-C) R–Br + R—Br acetone I (solvent) R–I relative rate constant CH3—Br CH3CH2—Br (1°) (CH3)2CH—Br (2°) (CH3)3C—Br (3°) To generalize and simplify this, we can say: methyl and 1° haloalkanes ... 2° haloalkanes ... 3° haloalkanes ... + Br reason — steric hindrance to backside attack ... R R C C X R H H X R 3° 1° b. Branching at next C (β-C) — C# C C! X " R–Br + I acetone R–I (solvent) R—Br (all 1°) + Br relative rate constant CH3CH2—Br Br Br Br H H CH3 C CH3 H H 5 C X CH3 Note that vinylic and aryl halides do not do SN2 rxns. reacts 10 times more slowly than C CH3 X