Question

4. Consider the following reaction: OC(CH3); (CH3)3COH Draw the byproducts of the reactions and use curved arrows to show the movement of electrons. a) b) What happens to the reaction ra i. The solvent is changed to DMF The concentration of-OC(CH3)3 is decreased ii. The base is changed to OH iv. The halide is changed to CHaCH2CH2CH2CH(Br)CH3 v. The leaving group is changed to

Answer 1

(A) It is E_2 (Elimination bimolecular) reaction No by product formed here b) (i) If the solvent is changed to DMF, then SN_2 reaction will occur instead of 'E_2' Because DMF is a polar aprotic solvent It means DMF has no H - bonding \r\n (DMF) So Nu^- is remain unaffected as a result SN_2 reaction will occur (ii) Basically 'E_2' is a bimolecular reaction It's note depends upon the concentration of nucleophile Rate alpha [Nu^-] [Rx] So as the concentration of - OC (CH_3) is decreased the rate of reaction is \r\n also decreased (iii) Attack at C- atom leads Substitution and attack at H - atom leads elimination Since strong base attack proton, so the more the basic the nucleophile, more likely that elimination is going to replace substitution Here when base is changed to^- OH, sN_2 favoures over E_2 reaction as^- OH is less basic than - OC (CH_3)_3