[catarina1998]

What are the reaction mechanisms (like sn1, sn2, e1, e2) for the following reactions?
. Bromine (in CCl₄) with hexene, without light (i think this has to do with heterolytic fission?)
. Bromine (in Ccl₄) with hexene, with light (maybe this is about homolytic fission idk)
. Bromine water in hexene (and it doesnt specify whether is with or without light)

[catarina1998]

I didnt understand your answer
I was hoping someone could tell me what carbocations are formed etc in a very simple way. 

[Borek]

I didnt understand your answer

Don't worry, it was a spammer quoting abstract of some random paper on the subject ( J. Chem. Soc., Perkin Trans. 2, 1983, 1079-1086).

[Babcock_Hall]

Not all polar reactions proceed through carbocations.  What is another possibility?

[catarina1998]

I didnt understand your answer

Don't worry, it was a spammer quoting abstract of some random paper on the subject ( J. Chem. Soc., Perkin Trans. 2, 1983, 1079-1086).

ok

Not all polar reactions proceed through carbocations.  What is another possibility?

I dont know 
anyway i asked this question because i had a lab report to turn in 2 days ago and i wasn't sure about those reaction mechanisms. Are these correct?

[Babcock_Hall]

I see some problems in the first chemical process, the one at the top of the page.  One, the reactant is hexane, not hexane (which is what you drew).  Two, bromine is neutral, but what you drew looks like bromide ion, which has a negative charge (I could be misinterpreting what you wrote).  Three, if bromide ion removed a proton from hexane (which isn't likely), then the carbon that remained would be negatively charged, not positively charged.

Two of the three reactions (the one in CCl₄ without light and the one in water) proceed through the same intermediate, but the intermediate leads to different products in the two different solvents.  The alkene and Br₂ react to form this intermediate by displacing one bromine atom in the form of bromide ion.  The positively charged intermediate that forms in the first step has three atoms, the two carbon atoms from the double bond of hexane and one bromine.  Could you try to draw it?  You can draw it in such a way that all atoms have a Lewis octet of electrons.  Once you get this intermediate, the two products become very easy to see.

[catarina1998]

Does the first reaction exist (bromine in ccl4 without light + hexane)? I'm asking because i was googling it and i found a site that said that hexanes dont react with bromine in the dark, and since this was for a lab report, im not sure if i should have seen a reaction or not (i saw it but maybe i did something wrong)... (And I can't find the reaction mechanism for that one  )

I tried to draw again the other reactions (bromine in ccl4 without light + hexene, and bromine water + hexene). Are they correct?

[Babcock_Hall]

You're close.  However, a free carbocation is not formed; instead, a bromonium ion is formed.  I often only draw one resonance form of the bromonium ion, but the other two are helpful in understanding the reaction with water.  http://courses.chem.psu.edu/chem210/mol-gallery/bromonium/bromonium.html

[catarina1998]

I tried to draw them again, now with the bromonium ion. 

[Babcock_Hall]

The bottom reaction looks fine.  The top bromonium ion just needs an alkyl chain in place of one of the hydrogens.

[catarina1998]

Ok, I corrected it in my paper now. 
Thanks for helping me with this stuff

[organic92]

. Bromine (in CCl4) with hexene, without light (i think this has to do with heterolytic fission?)
--> NO radical involved so ---> Typical Electrophilic addition to alkenes.
. Bromine (in Ccl4) with hexene, with light (maybe this is about homolytic fission idk)
--> Radical can be involved. Br· adds to terminal position of hexene and 2º radical is formed, another Br· terminates the reaction.
. Bromine water in hexene (and it doesnt specify whether is with or without light)
--> Revise halohydrin formation reactions.