TBR O. Chem Question 4.9

[mophead4]

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So the question asks, "How many mono-chlorinated structural isomer products are possible when 2,5-dimethylhexane undergoes free radical cholorination?" The answer is 3, but I'm not sure why. Their reasoning is that because of the mirror plane through carbon 3-to-carbon 4, there are only 3 unique carbons. Are they only considering carbons 1,2, and 3 and carbons 4, 5, and 6 unique carbons and negating the methyl groups? Sorry I know this is probably a very straight-forward question. More than anything I guess I'm just wondering how you know which carbons are halogenated during free-radical halogenation. Thanks in advance 🙂

 

[TieuBachHo]

If you flip 1,2,3 into 4,5,6 you would have the same structural isomers right? That's what they mean UNIQUE carbons. When this alkane, 2,5-dimethylhexane, undergoes free radical chlorination, there are only three UNIQUE possible combinations. The radical is either added to 1,2,3 or 4,5,6 (1,2,3 or 4,5,6 are of the same configuration if you just flip them over). The methyl groups are placed in 2 and 5 which means that they are not of important factor here (Draw it out, then you'll see why). The methyl groups are assentially the same thing if you add the free radical to carbon #1 or #6. SO there are only 3 possbile UNIQUE radical addition.

SIDE NOTE: You should also know if the question ask for the most favored/majority products in radical addition rxn in alkane. The most stable carbon (most substituted carbon) takes precedent.

 

[geeyouknit]

. More than anything I guess I'm just wondering how you know which carbons are halogenated during free-radical halogenation. Thanks in advance 🙂

In free radical halogenation, I think the halogen can attach to primary, secondary, and tertiary carbons (how many other carbons are attached to that carbon). All 3 are possible, but the most substituted carbons are favored.
Tertiary > secondary > primary in terms of which product is more favored.