username2975726
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"Outline the following steps in the mechanism for the reaction of methane with flourine to form fluoromethane, CH3F"

"Initiation
1st Propagation
2nd Propagation
Termination"

I've tried:

Initiation : F2 --> [email protected] + [email protected]

1st Prop : CH4 + [email protected] --> [email protected] + HF
2nd Prop : [email protected] + F2 --> CH3F + [email protected]

Termination : ?? :/

I've messed up on the propagation since I've already formed CH3F so I can't do the termination. What did I do wrong?

Thanks
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ChemistryWebsite
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Your propogation steps are fine and will give you the desired Product CH3F.

Termination is when any 2 free radicals come togethor.
In your example this could be methyl free radical and fluorine free radical, also producing CH3F. Don't worry that you have aleady got this in your propogation step.
You can also terminae with 2 x fluorine free radicals or 2 x methyl free radicals.

The vast majority of the CH3F produced comes from the propogation steps that repeat many times before a termination step occurs.
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username2975726
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(Original post by TutorsChemistry)
Your propogation steps are fine and will give you the desired Product CH3F.

Termination is when any 2 free radicals come togethor.
In your example this could be methyl free radical and fluorine free radical, also producing CH3F. Don't worry that you have aleady got this in your propogation step.
You can also terminae with 2 x fluorine free radicals or 2 x methyl free radicals.

The vast majority of the CH3F produced comes from the propogation steps that repeat many times before a termination step occurs.
oooh ok

so for termination i can just put [email protected] + [email protected] = CH3F without it relating to the previous step?
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ChemistryWebsite
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Yes. It is 2 different ways to get the same product.
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deathnote123
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Propagation -
A good tip for the propagation steps is to know that both equations cancel out (free radicals cancel) to give the reaction given in the question. These equations are known as chain reactions due to the recycling of the X• (X= F, Cl, Br, I). So in addition to that, know that the product of the 2nd equation is the free radical you introduced in the 1st equation.
This step is all good well done!

Termination -
Usually, three reactions occur, two of which are the same free radicals reacting with each other, and one of which is where two different free radicals react. Out of these three, one has to produce a by-product.
So let's see...
F• + •F → F2
CH3• + •CH3 → C2H6 (Ethane ~ By-product)
CH3• + •F → CH3F

As you can see, two of the products formed, are found in the main reaction whilst, one is a new product. In the exam, they will usually ask you to display the by-product by that I mean you should show all the bonds.
I hope this helped you, please give me feedback
Thanks
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Pigster
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You have done nothing wrong.

Prop. steps involve 1 rad + 1 molecule -> 1 rad + 1 molecule.

Termination steps involve 2 rad -> 1 molecule.
e.g. [email protected] + [email protected] -> CH3F
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username2975726
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(Original post by TutorsChemistry)
Yes. It is 2 different ways to get the same product.
(Original post by deathnote123)
Propagation -
A good tip for the propagation steps is to know that both equations cancel out (free radicals cancel) to give the reaction given in the question. These equations are known as chain reactions due to the recycling of the X• (X= F, Cl, Br, I). So in addition to that, know that the product of the 2nd equation is the free radical you introduced in the 1st equation.
This step is all good well done!

Termination -
Usually, three reactions occur, two of which are the same free radicals reacting with each other, and one of which is where two different free radicals react. Out of these three, one has to produce a by-product.
So let's see...
F• + •F → F2
CH3• + •CH3 → C2H6 (Ethane ~ By-product)
CH3• + •Cl → CH3Cl

As you can see, two of the products formed, are found in the main reaction whilst, one is a new product. In the exam, they will usually ask you to display the by-product by that I mean you should show all the bonds.
I hope this helped you, please give me feedback
Thanks
thanks so much, really helpful ☺
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deathnote123
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Thank you so much. I actually made a mistake in the last step I used Cl instead of F so my apologies..
If you need help with anything else please let me know
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HateOCR
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You did the question correctly. Termination is where any two radicals react to form a more stable molecule
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ChemistryWebsite
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Also worthy of note is that you get far more than these products. The reaction will continue, and there is no reason why a second hydrogen will be substituted. This happens in propogation step when a fluorine free radical collides with CH3F to give CH2F free radical ; then this free radical collides with F2 to give CH2F2 and regenerate the fluorine free radical again.
In fact all the hydrogens on the alkane can be substituted. This means there are a variety of alkyl free radicals (with 0, 1, 2 or 3 fluorines) so there are a large amount of possible termination products.

Also worthy of note:
Only a tiny proportion of the halogen molecules actually cleave to give free radicals. At any moment through the reaction there are a small number of free radicals present within the population of molecules. This makes it very unlikely that free radicals will collide, there are few terminations. The free radicals collide repeatedly with molecules so there are numerous propogation reactions.
As free radicals are very reactive the propogation steps happen quickly and easily, hence the "chain reaction".
Last edited by RK; 1 year ago
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charco
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The F-F bond is extremely weak and dissociates into fluorine free-radicals in the presence of visible light at all wavelengths below (760nm) infrared.

This, coupled with the very high C-F bond enthalpy makes the fluorination very fast and highly exothermic.

It's not typical.
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