Organic Chemistry5

Tapas in Q2 where is the nitrogen group? If it's supposed to be Hoffmann elimination..
In Q3, both substrate and base are bulky, and temperature also suits elimination. So ans is an alkene?
TRUE...base is bulky..HOFFMAN KI JAI LOL :D

@Pritish sorry!!!! x2
edited.x2 (see Q3 now)

As it is a C-C single bond, in Q2, rotation will take place to make H anti to the leaving N⁺Me₃ group. Product will be C(Ph wedge)(CH₃ dashed) = C(CH₃ wedged) (H dashed)

@pritish: Can u state wat the reagents do? pls. (in each q u kno)

See CH₃I is simple. Nitrogen is nucleophilic as long as it has its lone pair, and it will substitute CH₃I in an S_N2 reaction. If the nitrogen atom does not have any protons to lose (to regain its lone pair) it becomes positively charged, that is, a quarternary nitrogen. This fact is exploited because a quarternary nitrogen is a good leaving group. Now usual elimination takes place, E2 style(antiperiplanar) but without Zaitsev Rule.

for Q1: I think:

Q1 Answer:
Ankur u 'seem' to be wrong

Q2: SOLVED (ans is correct)- thx pritish and govind

Q3 Answer: FALSE. Justify???

REG Q3

The less substituted product shown was: