Thanq both of you
You both r correct
1) and 2) all seems correct........
3)a>d>c>b
explanation....
since 2 species are here which bear charge(even when thier octet are complete) they will be relatively unstable.....the positive charge of d) is more dissipated due to more hyperconjugation. than a.....now among b) and c) to compare lets see what happens after it will lose a H+...and c) would be more stable than b).....(see no. of hyperconjugation..in thier H+ lost form....)
please confirm....
3. a>d>c>b ( i think protonated substrates have great affinity to lose H+ and become neutral and +ve charge on O is more unstable than on N as O is more electronegative)
please confirm
In 1,2.. all the circled ones are correct