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@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.Classic question hai ye
yeah i remember doing it, only i forgot how to do it
before hydrolysis its RCOOH and ROH i assume?
the problem im having is the difference between KCN and KCN (alc.) and how and where that matters
Well, I suppose with aqueous KCN, you'd have some H(+) being formed from the water which could protonate your alcohol group.
I assume it's a precaution to prevent elimination
oh
Yeah, that's about it I'd say.
so before adding the KCN would this be RCOOH and RCN?
or RCOCN and RCN?
oh wait there's one more OH group
You have a cyanohydrin on one side, alcohol on the other. Hydrolyse the cyanide, then esterification
The other alcohol won't interfere because, well, your favourite little niche topic, ring strain.
wait so the hydrolysis of the cyanide and the esterification both happen in step 3?
Mhm.
But there is one point I'm iffy about
that's evil 💀
?
also i've forgotten how to reverse engineer aldol
lemme try to find it in notes
ah got it
Well, I'm pretty sure this isn't the case (I'd have to see the energy profile) but you end up with a geminal triol on hydrolysis of your cyanide. Which loses water to give you a -C(+)(OH)-O(-) right? Will that (-) attack the alcohol connected carbon or or does it form a carboxylic acid first and then the alcohol attacks C=O?
Probably the latter ngl. It could also be that there is partial hydrolysis to oxime, at which point OH attacks, giving an amine-oxide, which is further hydrolysed to ester. This is the least likely scenario. I haven't bothered to look up the reaction mechanism.
mind sending that written as a reaction?
Eh, nvm, I'll look it up sometime
alr
thank u
+solved @Opt
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