CHEM 0320 1st Edition Lecture 2 Outline of Last Lecture I. Terminology pertinent to Chapter 14 II. Allylic radicals are reactive intermediates III. Strength of Allylic Carbon Outline of Current Lecture I. Nucleophilic Substitution of Allylic Halides II. If conditions are suitable for carbocation formation, SN1 will occur III. Rates IV. Allylic Anion Material Current Lecture I. Nucleophilic Substitution of Allylic Halides a. NOTE** There are no such things as “naked” protons; they do not exist as H+ but rather as CH3OH2+ (initial proton source), which the epoxide attacks i. Allylic halides readily undergo SN2 reactions with good nucleophiles ii. SN2 with allylic halides are FASTER than SN2 with haloalkanes. iii. This is because of the interaction of the pi system in the SN2 transition state b. Faster = Lower activation energy barrier II. If conditions are suitable for carbocation formation, SN1 will occura. Conditions: i. Weak nucleophile ii. Polar Protic solvent 1. Alcohol, water iii. Good leaving group b. You learned that 1˚ Carbocations (C+) DO NOT form, which is true for alkanes. HOWEVER 1˚ C+ DO FORM for allylic carbocations i. Allylic carbocations are stabilized by resonance ii. Unsymmetrical partial positive dispersion along the allylic C’s:1. Resonance hybrid shows real structure of carbocation and how the positive charge is not shared equally.a. The “bigger,” or more positive, partial positive charge is on the 2˚ (secondary) carbocation, which is more stable then the 1˚ (primary) carbocation (the “smaller” partialpositive charge)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.i. The secondary carbocation handles the positive charge better because of more electron donating groups. III. Rates a. In an example when Allylic Br undergoes SN1 and two carbocations form (primary and secondary), the product that forms from the secondary allylic carbocation has a lower activation energy barrier than the product that forms from the primary allylic carbocation. i. The lower activation barrier is due to the initial secondary carbocation that is formed. ii. The higher activation barrier is due to the initial primary carbocation that is formed. 1. From this information, you can decide between the major and minor products. Major products being those that formed from the more stable allylic carbocation. IV. Allylic Anion Material a. You must compare ACIDITY of allylic H i. To do this, DRAW CONJUGATE BASES separately!!1. The more stable conjugate base will be the stronger acid. To tell which is the more stable conjugate base, know pKa values and “H.I.R.E.S.” a. H: Hyperconjugation? If yes, it’s more stable. b. I: Induction? If yes, it’s more stable. c. R: Resonance? If yes, it’s more stable.d. E: Electronegativity? If yes, it’s more stable. (s vs. p character) e. S: Size? If bigger, it’s more stable.2. Look up pKa table in book for values! Memorize! (MORE ACIDIC = LOWER PKA) ii. If comparing an allylic H to an alkane H, the allylic H will be more stable mainly because it is stabilized by resonance. 1. If the negative charge of the conjugate base is on the C that is part of an alkane chain, then it is unstable. If the negative charge of the conjugate base is on the C that is allylic, it is relatively stablea. Due to resonance distribution of the electrons (prove this using arrow pushingmechanism!) b. NOTE** that in reactions that form allylic anions, CONDITIONS are IMPORTANT! i. Conditions that are capable of supporting high energy carbanion formation:1. nbuLi or TMEDA, acetone, H3O+a. Above conditions are in three steps, respectively ii. Example using simplest allylic substrate and proper conditions.1. Because symmetrical allylic anion formed, only one product is possible.c. Though not shown in picture, you must show resonance in order to tell if allylic anion is SYMMETRICAL or UNSYMMETRICAL! i. Unsymmetrical allylic anions will give you unique products, for which you must write