Intermolecular Aldol
This is an aldol reaction, verified as per solution. My question is, which hydrogen will be taken away (where will negative charge form)? And why?
13 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.More specifically, how does one know if the second-last or third-last hydrogen will be abstracted
Heck, why not the ones in the already existing 5 membered ring?
This is what the solution says
It's always the most acidic hydrogen which is abstracted by the base. If you consider the one that is chosen in the solution, you can see that the conjugate base(carbanion) is stabilised by both the Carbonyl group to it's immediate left, as well as by the -I effect of the ester.
As to why it's not the other carbon α to the carbonyl group is because the = O on the pentane ring is in the plane of the pentane ring while our carbanion is out of the plane of the ring, so there isn't electron cloud displacement.
Ah ok, how do you judge whether something is in plane w.r.t another thing
Well, carbonyl groups are always planar. Substitutents on planar rings (less than six carbons usually, sometimes six) are always out of plane.
Also is steric hindrance a factor in base abstraction (in general)
Oh ight
I'll look up a visual for that
Rarely. Only in cases of bulky bases such as tertiary butoxides, and even so, only if very sterically hindered.
Alright
Thank ya
Aldehydes has got me in for a loop today
Steric hindrance during proton abstraction from an sp3 carbon is, as far as I can remember, only for the E1cb mechanism
I don't recall specifically learning an E1cb mechanism
I'll look it up
+solved @Opt
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