(i) Give the possible reason for the relative reactivities when iodide ions in acetone solution reacts with MeBr, EtBr, iso-prBr and t-BuBr under condition where only the S_(N)2 mechanism operates. The relative reactivities were found to be 10,000: 65:0.50:0,0:039. (ii) In an S_(N)2) reaction there relative reactivities are not observed in S_(N)1 condition.

(i) Give the possible reason for the relative reactivities when iodide

1.	(i) Give the possible reason for the relative reactivities when iodide ions in acetone solution reacts with MeBr, EtBr, iso-prBr and t-BuBr under condition where only the S_(N)2 mechanism operates. The relative reactivities were found to be 10,000: 65:0.50:0,0:039. (ii) In an S_(N)2) reaction there relative reactivities are not observed in S_(N)1 condition.
Answer» Solution :In an `S_(N)2` reaction there are five groups attached to the carbon ATOM at which reaction occurs (transition state). Thus, there is crowding in the transition state, and bulkier the groups, the greater will be the compression energy and consequently the reaction will be hindered sterically. Thus from MeBr to t-BuBr the NUMBER of METHYL groups on the central carbon atom increases the steric retardation, therefore relative reactivities are observed. (II) In `S_(N)1` reaction, the INTERMEDIATE state does not contain more than four groups on the central atom, and hence one would expect steric hindrance to be less important in `S_(N)1`. If, however, the molecule contains bulky groups then by ionizing. The molecule can relieve the steric strains. Since the carbonium ion produced is flat (trigonal hybridization) and so there will be steric acceleration.