Answer:
Here's what I get
Explanation:
(a) Intermediates
The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).
(b) Relative Stabilities
The relative stabilities decrease in the order shown.
N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.
(c) Relative reactivities
The relative reactivities would be
C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃
Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as
Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo
SN¹ and
E¹ elimination reaction due to the formation of
stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as
nucleophile it will produce
ether and when it acts as a
base it will produce
unsaturated compound. The reaction along with both products is shown below,
Jasmine have 20 fish because4 times 5 equals 20
<span>No. Zn can only have oxidation states of 1+ or 2+. For this compound to be able to exist, the Zn would have to have an oxidation state of 2- to counteract the 2+ from the 2 sodium ions. Sodium ions each have a +1 charge, so the 2 sodium ions would carry a +2 charge. In order for the compound to exist, the net charge between the 2 sodium and 1 zinc atoms would need to be 0.</span>
