For b) of the following question, the solution says: But is it not possible for

Question

For b) of the following question,

the solution says:

But is it not possible for the hydroxide to release a proton from the other allylic position and thereby form a different carboanion?

Also for c) of the same question, the solution says:

why wouldn't the ethanol then attack the carbocation since this seems like an SN1 reaction?

Illustrate by means of appropriate structures (including all relevant resonance forms) the initial species formed by breaking the weakest C-H bond in 1 hyphen butene; treating 4 hyphen methylcyclohexene with a powerful base (e.g., butyllithium-TMEDA); heating a solution of 3 hyphen chloro- l - methylcyclopentene in aqueous ethanol. Step 3 of 4 Treating 4 hyphen methylcyclohexene with a powerful base. 4 - methylcyclohexene, when treated with a powerful base, releases proton from an alpha carbon to form a carbanion. The resonance structure of the carbanion is shown below. Heating a solution of 3-chloro-lmethylcyclopentene in aqueous ethanol. This will form a carbocation and that carbocation is stabilized by resonance structures as shown below. 4 - chloro - 1 hyphen methylcyclopentene

Explanation / Answer

in the second mechanism as we see that the reaction is SN1 the ETOH will atttack the carbocation hen it reaches its most stable position

as in a SN1 reaction the rds step involves on the CARBOCATION and not the nucleophile which ius goin to attack on the CARBOCATION

so in rds it rearranges it to the most stable state

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