CHEM 0320 1st Edition Lecture 1Outline of Last Lecture I. N/AOutline of Current Lecture I. Terminology pertinent to Chapter 14 II. Allylic radicals are reactive intermediates III. Strength of Allylic Carbon Current LectureI. Terminology pertinent to Chapter 14 a. Alkenyl carbon = vinyl carbon i. Vinyl H attached to vinyl carbon ii. Alkenyl C is the alkene C 1. Sp2 C bonded to another sp2 Cb. Allylic carbon are bonded to an alkenyl carbon i. There can be more than 1. ii. For reactions, it is important to look at allylic hydrogens1. Hydrogens that are bonded to allylic C, because they undergo reactions.c. Allylic Halide (i.e. Cl) can be attached to allylic C II. Allylic radicals are reactive intermediates a. Allylic radicals contain an sp2 C with a half-filled P orbitali. Due to resonance, 2 C’s bear the burden of the half-filled P orbitalsb. Allylic C-H bonds are WEAKER than alkane C-H bonds.i. In the hemolytic cleavage of an alkane C-H bond, hyperconjugation is the stability argument. ii. In hemolytic cleavage of an allylic C-H bond, resonance is the stability argument, NOT hyperconjugation. 1. Bond dissociation energy (BDE) is SMALLER for breaking allylic C-Hdue to resonance stabilization of allylic radicals that is not possiblein alkane radicals. a. Resonance distribution is more important than hyperconjugation 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.c. Forming monohalogenated productsvs. vicinal di-halo product i. Alkene + X2/hv  vicinal di-halo product 1. WHY? Because you are given a stoichiometric amount of halide, where light is IGNORED. ii. Alkene + LOWCONC. X2/radical initiator halogenated alkene 1. Radical initiator = NBS2. Keeping low concentration of X2 is difficult, so CCl4 is needed. iii. Mechanism for Allylic Bromonation: how to trap allylic radicals 1. Initiation step: a. Br – Br + hv  2 Br radical i. Single headed arrow showing 1 e- goes to 1 Br and the other e- goes to the second Br (homolytic cleavage) 2. Propagationstep 1:a. 1 e- from Br radical meets 1 e- from allylic H, forming H – Br (H abstraction)b. Other e- goes back to the allylic C, forming allylicradical.i. MUST WRITE RESONANT FORM OF ALLYLIC RADICAL: to tell difference, name them! 1. SYMMETRICAL vs. UNSYMMETRICAL products: a. Symmetrical resonant allylic radicals will form same product (one) b. Unsymmetrical resonant allylic radicals will form different products (two or more) – must write all! c. When you name “them,” you are naming each product after Br has trappedintermediate. If unique products arise, the intermediates are unsymmetrical. 3. Propagationstep2: a. Single e- from allylic radical meets single e- from Br – Br radical (single headed arrows), and other e- from Br – Br goes back to the other Br, forming allylic bromide and regenerated Br radical.III. Strength of Allylic Carbon a. C-H bonds at two positions do not have strength. 1˚ allylic C vs. 2˚ C differ:i. 2˚ allylic C is weaker and more stable, 3˚ is weakest, most stable 1. For more complex substrates, many regioisomers can form.b. Can we make allylic radical even when we have one or more allylic C’s?i. NO IF YOU HAVE NO ALLYLIC HYDROGENS1. Radical allylic bromonation to generate allylic radical cannot takeplace without initial allylic Ha. **Allylic Carbons may not necessarily be bonded to a H. Only Allylic H undergo NBS b. When you see NBS, think “radical allylic bromonation” i. When you form allylic radical, immediately form resonant structures because in reality, allylic radicals are resonance stabilized. ii. After forming resonant structures, trap each structure (or regioisomer) with Br – Br (molecular Br), and see if product for each regioisomer is the same. 1. If the same, you write down one product. If the products are unique, write down ALL of