When rubbing alcohol and water are at room temperature, which one boils first
2 answers:
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Answer:
See answer for details
Explanation:
In this case, we have the following substract:
Ph-CH2 - O - CH2CH3
That's the benzyl ethyl ether.
When this compound reacts with a concentrated Solution of HI, as we have an acid medium, we will have an SN1 reaction, where the Hydrogen will attach to the oxygen and the iodine will go to the most stable carbon, in this case, the carbon next to the ring is the most stable (because of the double bond of the ring, charges can be stabilized better this way), so product A will be the benzyl with the iodine and product B, will be the alcohol:
A: Ph - CH2 - I
B: CH3CH2OH
When B reacts again with HI, it's promoting another SN1 reaction, where the OH substract the H from the HI, the OH2+ will go out the molecule, leaving a secondary carbocation (CH2+) and then, the Iodine (I-) can go there via SN1 and the final product would be:
C: CH3CH2I
See picture for mechanism:
