Kalabamboo
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I am unsure how to approach this question. Can somebody help please?
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username3718068
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I'm assuming you know the what a branched chain and an optical isomer are.
All you have to do is re arrange some atoms/groups of atoms in the molecule to form another molecule that achieves both outcomes.

As per MS: label all the carbons 1-5 starting from left going to right.

1st molecule:
*take a hydrogen off at C4 (leaves CH free radical at Carbon 4 = C4)
*methyl group off C1 (this will leave with CH2 free radical at C2)
now place the H on C1, and CH3 on C4.

2nd molecule:
* take off Br at C3
* take off CH3 at C1
swap the groups for each other
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Kalabamboo
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(Original post by dip0)
I'm assuming you know the what a branched chain and an optical isomer are.
All you have to do is re arrange someatoms/groups of atoms in the molecule to form another molecule that achieves both outcomes.

As per MS: label all the carbons 1-5 starting from left going to right.

1st molecule:
*take a hydrogen off at C4 (leaves CH free radical at Carbon 4 = C4)
*methyl group off C1 (this will leave with CH2 free radical at C2)
now place the H on C1, and CH3 on C4.

2nd molecule:
* take off Br at C3
* take off CH3 at C1
swap the groups for each other
Thanks a lot!

So it's just a bit of trial and error for this question, right?

Also, for the last one in the ms, how can Br form 2 bonds?
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Izzi3
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Br doesn’t form two bonds the carbon on the CH2 is bonded to the Br. Carbon can form 4 bonds one with bromine, 1 with each of the two hydrogen’s and the final bond is with the carbon to form the rest of the chain.
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