8 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.HI is a strong reducing agent
yo! just covered this in class on Thu. The t butyl carbocation forms by sn1 rxn (the first equivalent of HI) and you get t butyl iodide. since HI is in excess, the propyl alcohol formed is attacked by another eq of HI and you get propyl iodide, this time by Sn2
you can check out my videos for a detailed explanation :)
Reactions of Phenols, Ethers | Alcohols, Phenols and Ethers Part 5
Option B is just a situation the the complete reaction
After this, another HI or (H+ and I-) reacts, the H+ is accepted by O lone pairs by that -OH group, it becomes -OH2 + (water) which leaves, and the carbon it's connected to becomes + charged, which is attacked by that I-
Can we close this please?
sorry didn't see the answers I'll close now
+solved @Aguilar
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