Optical isomers hard exam q Watch

usernamenew
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I dont know how u get p q and r , i drew out c6h12 but i dont know where tp put the dpuble bond so i cant go any further in the question, cant attempt it all
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usernamenew
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(Original post by usernamenew)
I dont know how u get p q and r , i drew out c6h12 but i dont know where tp put the dpuble bond so i cant go any further in the question, cant attempt it all

nit sure if u can see the attached question
It wont let ke send it
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Pigster
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(Original post by usernamenew)
I dont know how u get p q and r , i drew out c6h12 but i dont know where tp put the dpuble bond so i cant go any further in the question, cant attempt it all
What's not to get? You have posted the answer! What don't you get?
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usernamenew
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(Original post by Pigster)
What's not to get? You have posted the answer! What don't you get?
i posted the answer bc i still dont get it!! i have no idea how they got each of those products
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charco
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(Original post by usernamenew)
i posted the answer bc i still dont get it!! i have no idea how they got each of those products
The initial information given tells you that there is one double bond,

P can represent a pair of optical isomers.
This tells you that there is a carbon atom with four different groups.
This is now an exercise in logic.
1. Start with C-H
2. Once you have removed a carbon from C6H12 you have 5 carbon atoms to play with AND you must make three different groups. One of those groups MUST contain C=C, hence you now have 3 carbon atoms to make two different groups. Clearly there is two carbon atoms in one group and one in the other.

CH2=CH-CH(CH3)C2H5
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usernamenew
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(Original post by charco)
The initial information given tells you that there is one double bond,

P can represent a pair of optical isomers.
This tells you that there is a carbon atom with four different groups.
This is now an exercise in logic.
1. Start with C-H
2. Once you have removed a carbon from C6H12 you have 5 carbon atoms to play with AND you must make three different groups. One of those groups MUST contain C=C, hence you now have 3 carbon atoms to make two different groups. Clearly there is two carbon atoms in one group and one in the other.

CH2=CH-CH(CH3)C2H5
thank u so much, in these sorts of questions, should i always begin with C-H?
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charco
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(Original post by usernamenew)
thank u so much, in these sorts of questions, should i always begin with C-H?
Hydrogen is the simplest "group", so I would suggest yes.
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