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C6H14 + Cl2 in UV light Below are 5 products, say which one cannot be a product, prove this by showing how each of the other pr...Asked by Ava
C6H14 + Cl2 in UV light
Below are 5 products, say which one cannot be a product, prove this by showing how each of the other products were formed from the equation above.
1-chlorohexane
2-chlorohexane
2,2-dichlorohexane
dodecane
hex-1-ene
Below are 5 products, say which one cannot be a product, prove this by showing how each of the other products were formed from the equation above.
1-chlorohexane
2-chlorohexane
2,2-dichlorohexane
dodecane
hex-1-ene
Answers
Answered by
DrBob222
Ava. I'm not an organic chemist and I don't know the answer. I googled
"free radical halogenation of alkanes" and came up with this. It may or may not help.
http://en.wikipedia.org/wiki/Free-radical_halogenation
"free radical halogenation of alkanes" and came up with this. It may or may not help.
http://en.wikipedia.org/wiki/Free-radical_halogenation
Answered by
Devron
I do not know who provided you the names, but 1-chlorohexane is not valid: it should just be Chlorohexane. Chlorohexane and dodecane are the two primary products from the termination step in free radical halogenation, as well as HCl, therefore you can rule those two answers out. It is plausible to produce a mono-substituted product, so I would have to say that you can get a di-substituted product. But I have never heard of an alkene being produce from a radical halogenation reaction. Therefore, I will go with hex-1-ene as the correct answer choice. I can not post a link but, I believe that you can google the mechanism to show how all of the products will be produced.
Answered by
Ava
It did help. Thank you very much :-)
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