Question

Questions 1-5 are to be answered from the following possibilities: CH.CH HOCH, CH.CH 1. What is (are) the product(s) formed in the following reaction: CHEBE a1 .5.8 c. 2 d. 10 c. 1.58 .2.5,8 & 1,2,5,8 h. 1,5,8,10 2. What is (are) the product(s) formed in the following reaction: i. 1.2.5,8,10 j. 1, 2,10 O HO CH, CH 1 0% acetone 2 i. 1.2.5,8,10 j. 1,2,10 al b. 5,8 . 2 d. 10 .1.5.8 2,5,8 g. 1,2,5,8 h. 1,5,8,10 3. What is (are) the product(s) formed in the following reaction: HCI 1.3,7 ; 3,6 a. 3.9 6.3 6.6 0.9 0.6,9 17 24 h 3,4 4. What is (are) the product(s) formed in the following reaction: + oH a: 3.9 6.3 0.6 0.9 0.69 17 34 34 i 3,7 ; 3.6 5. What is (are) the major product(s) formed in the following reaction: HC HC CHESNA ? 100% acetone 23,9 b.3 0.6 0.9 0.69 17 34 h 3,4 13,7 j. 3,6

Answer 1

1) KOH, DMSO - strong base with polar aprotic solvent on the tertiarybenzyl halide.
elimination occurs by E2 mechanism giving rise to alkenes.
So the answer is formation of 1.5,8 .OPTION e

2)with 90% H2O and 10% acetone(polar protic solvent ,poor Nucelophile and weak base), the reaction is SN1 along with some elimination by E1.
thus the products are 1,2,5,8,10 OPTION i

3) A secondary benzyl halide with strong base /Nucleophile Ch3O- . So SN2/E2 is possible.
The products are 3,9 OPTION a

4) Asecondary benzyl halide with strong bulky base /weak Nu . thus only E2 possible.
the products are only 3
OPTION b

5) A secondary benzyl alcohol with strong nucleophile/weak base SH- in weakly polar aprotic solvent., only SN2 possible .

Product is 4

OPTION g