[INeedSerotonin]

Hello

I have this exercise that asks me how many different products one can obtain with the reaction 2-methyl-butane + Cl2.

The answer is 4, but I can see 6 different places where Cl can exchange with H. For example:



Please, can you guys guide me here? I'm lost. I tried searching about halogenation of alkanes, but I still cannot understand where on Earth "4" comes from.

Thanks

[Borek]

Actually any of 12 hydrogen atoms present can be attacked, but there will be only 4 _different_ products.

You have already realized it doesn't matter which of the three hydrogen atoms in the methyl group is replaced (at least I assume that's what you meant when you counted atoms in methyl groups only once). Now, there is a -CH₂- group present. Does it matter which hydrogen reacts?

Do you see where it is going?

[INeedSerotonin]

Thanks, Borek! I'm still kind of lost, though, so I think I haven't grasped the reasoning behind it yet 

I tried to see my book's resolution, and I noticed it ignored the methyl groups (as you mentioned), but I still cannot understand why.

I would have guessed that, in the -CH₂- group, it doesn't really matter where the Cl goes, since the sigma bond allows rotation. I remember a teacher telling me that the sigma + pi bond doesn't allow rotation, but the sigma-only bond does. So it doesn't matter where it attacks, we can just rotate it 180°.

... or maybe this doesn't have anything to do with this exercise and I'm just totally lost.

But as for all the rest, I really don't know. I would still have answered "5", maybe, because now I count the two H in -CH₂- as one.

[Babcock_Hall]

One general strategy in these problems is to name two products that you think may or may not be identical.  If you name them correctly, you can decide whether or not they are the same thing.

[MNIO]

see this image


don't forget about (1) free rotation around the C-C bonds, (2) if you can copy an image and rotated it around so that it's identical to the original, the H's are the same

[INeedSerotonin]

Thank you, guys! I'm starting to understand.

Couldn't the Cl also replace one of the hydrogens from the methyl group? Like CH2Cl? Can I say there is no halogenation on substituents?

[chenbeier]

That is already mentioned. In the pictures all  3 H are exchanged to show they are equal, but of course first 1 will exchanged.

[Enthalpy]

"The reaction of 2-methyl-butane and Cl₂" should be written more accurately. We suppose it's a single substitution, but it could be a full combustion, several substitutions, with partial sooting, and so on, and then the products are innumerable.