OMDM doubt hydrocarbons alkenes (continued)
I had posted a similar doubt a few weeks ago, but I couldn't find the relevant problems where I was getting stuck. I have it better framed now, so please help me out with this. It has been bugging me for the longest time.
- 1st image. I opened the cyclic mercurinium ion from the more stable cationic character which I thought to be at the other site due to -I effect of Ph. I thought +R won't be considered since they're not conjugated. But I was wrong.
- 2nd image. Here we don't look at the +M effect of Br. Since it exhibits -I effect, we open the ring from the other site, for more cationic character stability.
- Now what if in a similar condition, we have a system of 2 non conjugated double bonds and 2 conjugated double bonds. What would be the products in each case? This will help clear a few things up for me.

