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.
1. 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.
2. 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.
If hyperconjugation were the reason, still what we did in the 2nd case can't be explained.
3. 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.
11 Replies
@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.First things first
The second case can be explained easily
Now why is that?
The +M effect solely relies on the fact that a lone pair is donated to the adjacent atom
And the adjacent atom in our case isn’t really bearing that positive charge
Secondly, could you please show me the exact molecule you are concerned about when you non conjugated and conjugated double bonds?
Ah right that makes sense. But then why are we considering the +M effect of Ph in the first case? There's no positive charge on the adjacent atom there as well, right?
just some simple case like this
i just wanna know the products. It'll help clear a few doubts
So why wouldn't we consider the -I effect of phenyl
That sorta positive charge would be stabilised by that phenyl ring electron cloud
'cause it's in conjugation
In the first one, you'll attack on the allylic site (more stable positive)
In the second one, you'll the site away from the other double bond (double bonds are electron withdrawing)
You would but the +M (not exactly +M though but it's the conjugation in the transition state which increases its stability but some handwaving can be done here) would dominate the -I
Okay.. I get it now
Thanks, this clears it up
Thank you so much for your time
My doubt's finally been resolved
phew
+solved @Enamine
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