1q4. trans....because one of the carbons at the middle is sp2 hybridised so planar and there is very less repulsion btw the H-atoms attached to the carbons but in cis that`s not the case
10 Answers
106Q1.
(i) A ... in the resonating str. there is a partial δ- charge on =NH
(ii) A ... more +I effect
Q2. A is least basic ... lp on N is most delocalised due to aromaticity
1
1@Ashish
For Q1) (ii) Isn't the +I effect stabilizing the + charge appearing on N,
making it more stable, hence less basic.
Moreover, don't we consider less availability of lone pair of
e- for donation due to two large CH3 groups
106Oh yeah ... forgot abt the steric effects of the -Me groups.
For 1. (ii) B is more basic as the lone pair on the N is more "available" to react
1qs 4 will b trans because it suits better in the crystal because of its symmetrical structure.
1for the 1st qstn, why is the second one more basic? i mean there are two N groups in the first..both have lone pairs,and by symmetry,the ∂- charge is on the more electro negative n atoms??moreover,there is a methyl group to add +I effect!!
71Try to gather up again from the earlier post by fermat.
13. Ans: True, Because if there is any plane of symmetry... the compound wont be chiral!
11 (i) :N-H= is more dominant EDG than CH3 thus e- density is more at the NH2 group......
1 (ii) A is more basic bcoz e- density at N is increased by the surrounding EWG groups i.e. 2 CH3 groups
4 A is more stable......