#ISOMERISM
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<@&1227988359910260737>
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We aren't considering lone pairs on oxygen, but we do consider the hydrogens on a carbon.
So, even though in both cases the hybridization is the same, in D, the two pairs of hydrogens are in mutually perpendicular planes. In C, there is only one plane in consideration in the first place.
oh
but what about this
Let p_x orbital of middle carbon makes a pi bond with carbon in d, then only p_y orbital of carbon can make pi cond with the other carbon. Also p-z p-z ki overlapping makes sigma bond between carbons.
Now one CH2 group is in the horizontal (x-z) plane one is in the vertical (y-z)
i understand d
but isnt the same with C
yeh
We don't care about if the l.p.s are in the same plane or not
C wale mein there is no hydrogen or stuff attached to O
We just consider atoms in plane
Only lone pairs.
This is fine ig. (Just draw the lone pairs on oxygen tho)
but doesnt this indicate ki yeh non planar hai?
arent the blue and red ones in planes perpendicular to each other?
Yup
And purple is in the plane of the blue orbitals
Oh crap
Wait i made a mistake
toh fir how is the entire thing in the same plane
oo
H, H, C, C, O are in the same plane.
Only the pi bonds and lone pairs are out of plane
And those don't matter.
how is that
matlab this is wrong?
acha
In allenes, the arrangement becomes perpendicular alternatively per d..b.
+solved @flint sable @feral parrot
@pulsar blaze down :S
I'm positive c is non co planar
Oh nvm it is planar somehoe
+solved @flint sable @feral parrot
+solved @flint sable @feral parrot
atp use their names instead of the @ it still works 😠these guys have too many pings
+solved @pulsar blaze
sorry
apna name use krlunga
Lol
+solved @flint sable