Branched-chain isomer?
Watch this threadPage 1 of 1
Skip to page:
Kalabamboo
Badges:
14
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
0
reply
username3718068
Badges:
11
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report
#2
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
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
1
reply
Kalabamboo
Badges:
14
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
(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
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

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?
0
reply
Izzi3
Badges:
3
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report
#4
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.
1
reply
X
Page 1 of 1
Skip to page:
Quick Reply
Back
to top
to top