Question

3/4 d) Draw the structure of the MINOR product below. No mechanism needed here. (2 pts) 7) Draw the mechanism of the elimination reaction below (6 pts) EN (1 pt) b) What is the name of the specific mechanism of the reaction? c) What is being eliminated in the reaction? (2 pts) 8) A molecule in the conformation shown below undergoes the same type of elimination reaction as the previous problem. Draw the product of the reaction in full line bond notation (NO ABBREVIATIONS) (4 pts) Et H Me EN Bu Br Ph to each other(1 pt) b) in the Newman projection, the groups being eliminated are to each other (1 pt) c) in the Newman projection, the phenyl and hydrogen groups are 9) Draw the mechanism of the Grignard addition to the following ester. Two equivalents of the Grignard reagent used during the reaction, with an aqueous work up at the end. (10 pts) MgBr

Answer 1

cl Ja Eton a one 6 The mechanism of this particular reaction is E2 elimination reaction. This is step elimination. In this reaction Etzn acts as a strong base & abstracts a an alkene as a product Hp » B-hw.rot BH atom to give the leaving group (el) © In this reaction eliminated. a -el ¢ a gowing B-H H is being 8 Et a H Me H -Me Br 111 Et t-Bu ph ph (sawhorse) (Newmann) t-Bu Br bond rotati from I t-Bu .ph Et pha M E2 .me t-Bu IEL El iie Br (produel) This is also a E2 elimination reaction ☺ The eliminated groups here are -Br & -H

© In the Newmann projection, the phenyl & hydrogen groups are 120° to each other. El ME 4 Por t-Bu 120 ph ti 2 mger regar i Emg (ochs) Bor IC Mgor region OH но (work up) Ester reacts with 2 equivalent of grignard reagent to give product, a tertiang (19) alcohol! a ter tentiary alcohol.