Destroyer OC question

[Member 139582]

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Question: Consider the monochlorination of 2-methylbutane. How many products are expected if stereoisomers were counted?

a - 1
b - 2
c - 6
d - 4
e - 5

The answer is c, which I got, but in the answer key, it shows "I is chiral, and thus has an enantiomer"
How is this carbon chiral?? It has 2 hydrogens to it lol. Is this a mistake? I thought carbon III was the only chiral carbon, and that there were 5 places for cholorination. Please let me know what you guys think.

Thanks in advance.

 

[emminent]

No it's right; someone else posted a question about that. Maybe I did too but I forgot. You need to look at the tertiary carbon being chiral. Do not solely focus on the C w/ the Cl attached to it only.

 

[pjyi]

#1 Carbon = 1 (no chiral center)
#2 Carbon = 2 (it can attack from either side)
#3 Carbon = 2 (it can attack from either side)
#4 Carbon = 1 (no chiral center)
-----------------------------------------------------
Total = 6

 

[Member 139582]

ohh gotchu. Yea I was only thinking of the C with the Cl. Yea now I understand, the tertiary carbon is chiral. Thanks man.

 

[emminent]

Sorry pjyi but that's wrong man. Trust me on this one. Figure I has 2 because you look at the tertiary carbon not the carbon that has Cl attached to it.

 

[pjyi]

Chlorination has less selectivity so it can occur at any of the four carbons although there will be more reactions at #2 Carbon
Bromination on the other hand, will only occur at #2 Carbon because it has much higher selectivity towards tertiary carbons. Therefore would have 2 stereoisomers
Here is a picture

I was wrong!

 

[emminent]

You need to have a Hydrogen on carbon two and secondly there's two methyl groups on C2 so having two of the same group makes that C2 not chiral.

 

[pjyi]

LOL I was totally wrong, you're right
I don't know what I was thinking
Yeah I think there should be only 5 products
Thanks emminent

 

[emminent]

It's cool dude just gotta help each other out against the real enemy... the DAT