optical isomerism
is this optically active? if yes then how
can we even check optical activity in bondline structure ?
18 Replies
@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Notice that there are 2 chiral carbon (2nd and 4th) but the whole structure itself is symmetric about the middle
I think this is a meso compound
Optically inactive
You actually can't say
Really?
Until you have the wedge structure
Oh is it the R and S configuration thing
Because the two chlorines on the left and write could be on different sides of the plane, so you won't have symmetry
What if it was just planar
Which I think is the assumption when just given a bond line structure
Actually it can't be
Nvm
Alr
VSEPR just vanished from my head for a sec there
yeah gotchu
First one has plane of symmetry.
Second one doesn't.
so all in all just from bondline structure we cant determine if compound is optically active or inactive
ok so
to make optical isomers of a compound is it necessary for it to be optically active ??
i dont understand the relation between optical activity and formulation of optical isomers
you generally dont, but since it is obvious here that there is sp³ carbons. So their are many non planar atoms, hence, there doesn't exist a Plane of symmetry.
Can u pls clear this?
I don't understand your question
+solved @Opt
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