Bromination of Methylbutane (AS hard problem)

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#1
Hi all! So I came across this question and I was a bit confused and I'd like to know if someone could help me out

When ethane reacts with bromine, any one of the six hydrogen atoms can be replaced. The
symmetry in the molecule of ethane means that all six of these hydrogen atoms are equivalent
and so there is only one possible product no matter which hydrogen is replaced.
With larger alkanes, there is often the possibility of forming different structural isomers of the
bromoalkane product, depending upon which hydrogen atom is replaced by a bromine atom. In
the bromination of methylbutane, four different bromoalkanes can be formed (labelled A-D).

The question then asks me to state which compound will be formed in least proportion assuming that the bromine will bond with methylbutane in equal proportions, so I said that it was going to be in the carbon that only had 1 hydrogen.

The question then asks to use this assumption to calculate the percentage of each compound formed. I noticed that there are 12 hydrogens in total. So it's 3/12 on the right carbon, 2/12 on the carbon next to it, 1/12 on the carbon that unites 3 carbons and finally, 3/12 for the left carbon, however the mark scheme says that it's 6/12. Could someone please explain why?

Thanks
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#2
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2 years ago
#3
to form A you substitute a H atom on the CH3 groups corresponding to (CH3)2 in (CH3)2CHCH2CH3 by a Br atom
there are 6 such H atoms so 6/12 or 50%
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#4
(Original post by BobbJo)
to form A you substitute a H atom on the CH3 groups corresponding to (CH3)2 in (CH3)2CHCH2CH2 by a Br atom
there are 6 such H atoms so 6/12 or 50%
Why do we count the CH3 from the methyl group at the top though? Shouldn't it be only 3 from the carbon on the left? I'm looking at methylbutane in this way:
CH3CH(CH3)CH2CH3

So A has 3 hydrogens therefore 3/12
B has 1 hydrogen so 1/12
C has 2 hydrogens so 1/12
D has 3 hydrogens so 3/12

Where do the 6 hydrogens come from? Are we supposed to count the 2 CH3 groups from the left and the top as if they were the same? And if so, why?
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2 years ago
#5
(Original post by Crow_M)
Why do we count the CH3 from the methyl group at the top though? Shouldn't it be only 3 from the carbon on the left? I'm looking at methylbutane in this way:
CH3CH(CH3)CH2CH3

So A has 3 hydrogens therefore 3/12
B has 1 hydrogen so 1/12
C has 2 hydrogens so 2/12
D has 3 hydrogens so 3/12

Where do the 6 hydrogens come from? Are we supposed to count the 2 CH3 groups from the left and the top as if they were the same? And if so, why?
Yes, the H atoms on both the 2 CH3 groups have to be counted.

Both will form the compound A (1-bromo-2-methylbutane)

There is free rotation about a C-C bond, making the compound formed if the top methyl group was substituted identical to the one formed if the left methyl group was substituted. If you replace the H on the top methyl group, you get compound A. If you replace the H on the left methyl group, you get compound A again.

B, C, D are correct. Percentages should add up to 100% so this should alert you about a possible mistake
1
#6
(Original post by BobbJo)
Yes, the H atoms on both the 2 CH3 groups have to be counted.

Both will form the compound A (1-bromo-2-methylbutane)

There is free rotation about a C-C bond, making the compound formed if the top methyl group was substituted identical to the one formed if the left methyl group was substituted. If you replace the H on the top methyl group, you get compound A. If you replace the H on the left methyl group, you get compound A again.

B, C, D are correct. Percentages should add up to 100% so this should alert you about a possible mistake
Oh I see! I didn't think about the free rotation from the C-C bond. Thanks for your help!
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