[Twickel]

Hi.
I am having trouble with the following questions ( attached). Why are rings being formed and what happens to double bond. I can understand if this was addition of a Br radical but this a reduction of an alkyl halide.

[discodermolide]

Hi.
I am having trouble with the following questions ( attached). Why are rings being formed and what happens to double bond. I can understand if this was addition of a Br radical but this a reduction of an alkyl halide.

Redraw the starting material as I have done. Think about the stability of radicals, primary vs secondary and so on. See if you can come up with an answer.

[Twickel]

I am up to the point where, the  open ring, is the radical ( so the Br has left) and it is going to react with H-SnBu3.

The radical is is not stable, but not sure how it re arranges..
What I did was the double bond, reacts with the radical so, it closes the ring, now the radical is secondary, this new radical then receives a hydrogen?

They didnt teach me this.

Got no idea where the second minor product comes from

[discodermolide]

I am up to the point where, the  open ring, is the radical ( so the Br has left) and it is going to react with H-SnBu3.

The radical is is not stable, but not sure how it re arranges..
What I did was the double bond, reacts with the radical so, it closes the ring, now the radical is secondary, this new radical then receives a hydrogen?

They didnt teach me this.

Got no idea where the second minor product comes from

Basically correct:))
The AIBN generates a Sn radical which abstracts the bromine leaving a primary radical which is not stable. This attacks the C=C to give the 6 membered ring with a primary radical on the C of the methyl group, this attacks Bu3SnH to propagate the chain.
The 7 membered ring is just a by product formed by attack of the radical at the terminal end of the double bond.
I hope that is clear.

[Twickel]

Tahnk you very much, yes it is now.

So whenever it is possible, look for the chance to draw a more stable molecule.

Also when do I draw structures like you have? You did it for  he other Sn2 reaction wit t sulfur. What should I look out for to k shoold draw the starting product differently?

[discodermolide]

Tahnk you very much, yes it is now.

So whenever it is possible, look for the chance to draw a more stable molecule.

Also when do I draw structures like you have? You did it for  he other Sn2 reaction wit t sulfur. What should I look out for to k shoold draw the starting product differently?

Best thing is to make a model. If you can't do that play around with the drawing to see if , for example, your attacking and departing groups can form a ring, 5 or 6 whatever.
Just be careful not to add unwanted atoms like the tosylate example.

[Twickel]

Btw, the radical formed was tertiary not secondary right?
Think I made a mistake i the post

[discodermolide]

Btw, the radical formed was tertiary not secondary right?
Think I made a mistake i the post

No, the first radical formed is primary, this attacks the C=C pi system to give the 6-membered ring and a primary radical which abstracts the H from Bu3SnH to give the major product.

[Twickel]

I thought the primary radical re arranged itself so we get a ring with a tertiary radical?

[discodermolide]

I thought the primary radical re arranged itself so we get a ring with a tertiary radical?

No, I'll draw the step for you give me 5 minutes

Here is what I am talking about. I should have used half arrows but never mind.

[Twickel]

Oh, so the new ring does not give a more stable radical its still primary.

I thought the point of the radical attacking the pi bond was to form a more stable radical.

[discodermolide]

Oh, so the new ring does not give a more stable radical its still primary.

I thought the point of the radical attacking the pi bond was to form a more stable radical.

Not in this case. The ring formation must be very rapid to prevent the formation of a more stable radical, which probably accounts for some formation of the 7 membered ring.
Radical reactions are very fast unless other factors , not  present here, can slow them down. But we won't go into that.

[Twickel]

Thanks, so radicals are n llike carbocations, where a mocule can re arrange itself through shifts to form more stable carbocations.

[discodermolide]

Thanks, so radicals are n llike carbocations, where a mocule can re arrange itself through shifts to form more stable carbocations.

Yes, they are, just not in this case.
A longer lived radical will always seek to re-arrange to a more stable system, akin to carbonic ions.