Carboxylic acids, iodine + NaHCO3
What exactly does that do?
23 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.i looked up iodolactonization online but didn't exactly get how it works
Answer
i2 forms a triangle intermediate i think
I remember doing this. Cyclic iodinium ion by iodine insertion on the C=C double bond.
Bicarbonate removes acidic proton, so COO(-) acts as nucleophile and opens up the ring. Pyridine (base) and heat eliminates iodine to form double bond.
Wait a sec, are you sure that's the answer?
Elimination should be preferred on the other side imo. Resonance stabilised product
there will be E2 elimination right so there will be a negative charge when we remove the acidic hydrogen and then it removes the iodine hence double bond there. we can do it that way too right?
Yeah, i suppose acidic proton logic works better here
did it this way
Yup
that does indeed show as the answer
wait so how do we know whether to use resonance logic or acidic proton logic :sweaty:
cuz both are in the options
thank u this makes sense
Acidic proton is going to be one step ahead so that
one step ahead as in?
compared to resonance, its just faster?
Any acid-base reaction such as proton abstraction is faster than an electrophile-nucleophile reaction.
ahhh oke i remember now
Yeah
@Opt related doubt
why is (3) the answer (none of the other options are cyclic 🤷)
mechanism is the same (cyclic bromonium)
i'd have expected for it to attack the far end and form a 6 membered ring
rather than a 5 membered one
Ehhhh it might be cus six members rings don't enjoy being both substituted and planar.
Maybe look at a 3D model to see which is closer?
i think its also because NCC is formed there right, at the other end theres no cation stabilization, theres +I effect which stabilises the carbocation and hence we attack there?
hm probably that
that makes sense
thanks yall
+solved @Opt @trin
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