Sn1 carbocation rearrangement challenge problem

Example reaction showing Sn1 mechanism where substrate with six membered ring forms five membered ring product via alkyl shift.

[@syedhassan158]

Why would we have not done the methyl shift and make a tertiary carbocation and the six membered ring is more stable than five membered one?

[@davecampbell892]

Thank-you for the great video :) I have one question, why doesnt the H20 group on the final step act as a leaving group? Only the hydrogen pops off which is different from other videos

[@kristenscopino7485]

I know this is an old question, but I believe it is because if the (-OH2)+ group left it would reform the carbocation. Carbocations are pretty unstable compared to the (-OH2)+ which gets deprotonated really quickly into the stable -OH group.

[@MultiChantal22]

2:28 He said there's a proton transfer, so I assume it's just like any other video (where the lone pair in other H2O attack the proton in the H2O+ part of final product and cause the H to pop out and leave behind the final product). He just writes is as -H+ instead of writing the whole proton transfer mechanism.

[@astic4254]

Think about it...after deprotonation oh the H2O, to further stabilise the molecule of oxygen, the product became an OH attached to an R group (so an alcohol/hydroxyl group). A hydroxyl group is not a good leaving group, as it is a weak base, and to have a strong leaving group it must be a strong base. If it leaves, it will have a formal negative charge, hence becoming an unstable hydroxide anion.

[@tairademisheva3230]

doesnt heat make the mechanism E1?

[@anupamsingh9251]

Why would we have not done the methyl shift and make a tertiary carbocation and the six membered ring is more stable than five membered one?

[@l052_jeeveshpathak8]

i also have the same question that if we were not given the final product then we would make a six membered ring.