Optical isomers require an asymmetric carbon atom i.e. four different groups attached to the carbon in question. Are you sure you’ve got the question right?
Original post by SkyRunner61
Optical isomers require an asymmetric carbon atom i.e. four different groups attached to the carbon in question. Are you sure you’ve got the question right?
Yeah I was just going to say - there needs to be a chiral centre somewhere. From your question it looks like this isn't the case!
Original post by SkyRunner61
Optical isomers require an asymmetric carbon atom i.e. four different groups attached to the carbon in question. Are you sure you’ve got the question right?
yes my friend
Original post by papie
How do i draw the optical isomer of this organic molecule with structural formula : (CH3)3CCHO.
If you struggle to visualise them (a lot of people do) then get yourself an organic Molymod set from Amazon. They're very little money and will help you enormously.
Original post by Sidd1
Yeah I was just going to say - there needs to be a chiral centre somewhere. From your question it looks like this isn't the case!
the only other information given is that the molecule that i have to draw the optical isomer for is an isomer of (CH3)3CCHO.. idk if that makes sense
Original post by papie
the only other information given is that the molecule that i have to draw the optical isomer for is an isomer of (CH3)3CCHO.. idk if that makes sense
Oh if it’s draw the optical isomer of an isomer of that that’s actually possible. Like I said, you need a carbon with four different groups attached, so try and see if you can come up with a structure like that, using a broken line to indicate away from you and a wedged like to indicate towards you
The optical isomer just has those two groups switched
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