Freeradical Substitution & UV Light Help

Watch
x3502
Badges: 4
Rep:
?
#1
Report Thread starter 1 year ago
#1
Hi Guys,

Wonder if anyone could possibly help me with a question regarding Alkanes and free radical substitution with halogens.

I am having trouble describing the reaction mechanism between propane and bromine in the presence of UV light, and explaining how the presence of the UV light allows this reaction to take place.

Any help would be very gratefully received!
0
reply
Pigster
Badges: 20
Rep:
?
#2
Report 1 year ago
#2
(Original post by x3502)
Hi Guys,

Wonder if anyone could possibly help me with a question regarding Alkanes and free radical substitution with halogens.

I am having trouble describing the reaction mechanism between propane and bromine in the presence of UV light, and explaining how the presence of the UV light allows this reaction to take place.

Any help would be very gratefully received!
This should help: http://tiny.cc/mx6hlz
0
reply
username5211340
Badges: 14
Rep:
?
#3
Report 1 year ago
#3
So essentially what you have is UV light breaking up a Bromine molecule to give you two free radicals that are by their nature, very reactive. You should know that free radicals have a single unpaired electron. Think of it as the reverse of forming a covalent bond, you're separating the two atoms and thus, the electron they provide to each other goes back to them.

1) Br - Br ----> Br. + Br.

And then, one of these free radicals can proceed to attack a hydrocarbon (propane in this case) giving the following:

2) CH3CH2CH3 + Br. ------> CH3CH2CH2. + HBr

Then what you have is the hydrocarbon free radical attacking a Br2 molecule to give you this product:

3) CH3CH2CH2. + Br2 -------> CH3CH2CH2Br + Br.

As you can see, the free radicals been regenerated, so it can go on to attack more hydrocarbon molecules, keep in mind that you can have forms of reactions too, I've chosen this one because its the simplest to demonstrate.

Do note that there are three steps to radical reactions you should know, the first being initiation, where the halogen molecule splits, then you have propagation, which is a list of the middle steps you can have, and lastly you have termination where two free radicals will join to give you a neutral product.

Hope it helps!
Last edited by username5211340; 1 year ago
2
reply
username5211340
Badges: 14
Rep:
?
#4
Report 1 year ago
#4
(Original post by Pigster)
This should help: http://tiny.cc/mx6hlz
wasn't really funny tbh...
0
reply
x3502
Badges: 4
Rep:
?
#5
Report Thread starter 1 year ago
#5
(Original post by Chemastronomical)
wasn't really funny tbh...
Thank you so much for your help, and 100% agree, didn't quite get the humour!
0
reply
Pigster
Badges: 20
Rep:
?
#6
Report 1 year ago
#6
(Original post by x3502)
Thank you so much for your help, and 100% agree, didn't quite get the humour!
My point was, there are countless webpages explaining this topic. Even a quick search on TSR got me this: https://www.thestudentroom.co.uk/sea...mine+mechanism

Rather than going, "I don't get it", a more specific request for help would probably have got more specific assistance.

It all just struck me as rather lazy.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How do you prefer to get careers advice?

I like to speak to my friends and family (18)
8.74%
I like to do my own research online using careers specific websites (125)
60.68%
I like speaking to the careers advisors at school, college or uni (30)
14.56%
I prefer to listen watch videos or listen to podcasts of people in my chosen career (29)
14.08%
Something else (let us know in the thread) (4)
1.94%

Watched Threads

View All