April 12, 2025, 01:02:02 PM
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Topic: Termination step in free radical substitution  (Read 1185 times)

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Offline Eternal_Fire

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Termination step in free radical substitution
« on: March 02, 2025, 01:09:20 PM »
Hi,
I have this question for my assignment and this is the only question I do not understand. I have attached it below.

Thanks.

Offline Hunter2

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Re: Termination step in free radical substitution
« Reply #1 on: March 02, 2025, 01:17:59 PM »
Which radicals do  you can develop from propane? Which is the more stable radical?
Then do a recombination of them.

Offline Eternal_Fire

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Re: Termination step in free radical substitution
« Reply #2 on: March 02, 2025, 01:22:21 PM »
Do you mean this? How do I combine them?

Offline Hunter2

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Re: Termination step in free radical substitution
« Reply #3 on: March 02, 2025, 01:29:24 PM »
You misunderstood it.
How is bromination as radicals process takes place using propane instead of methane. Here there are no methyl or ethyl radicals.

Offline Eternal_Fire

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Re: Termination step in free radical substitution
« Reply #4 on: March 02, 2025, 01:35:47 PM »
Like this?

Offline Hunter2

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Re: Termination step in free radical substitution
« Reply #5 on: March 02, 2025, 01:39:26 PM »
You skipped  some CH2 groups, but generally on the right Track.
You get only two kind of Propyl, no Ethyl.

Offline Hunter2

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Re: Termination step in free radical substitution
« Reply #6 on: March 02, 2025, 02:20:45 PM »
Which molecules you get?

Offline Eternal_Fire

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Re: Termination step in free radical substitution
« Reply #7 on: March 03, 2025, 07:44:17 AM »
I only get this. Am I missing something?

Offline Hunter2

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Re: Termination step in free radical substitution
« Reply #8 on: March 03, 2025, 08:10:45 AM »
This could be one result, but it's not in the solution.
Why not have a radical on the second C.
Are primary radicals  more stable as secundary radicals?

Offline Eternal_Fire

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Re: Termination step in free radical substitution
« Reply #9 on: March 03, 2025, 12:46:34 PM »
Is this correct now? The first compound in the solution cannot be formed, am I right?

Offline Hunter2

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Re: Termination step in free radical substitution
« Reply #10 on: March 03, 2025, 12:57:04 PM »
Yes you got it.

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