0
$\begingroup$

Reaction scheme

Reagents to carry out above conversion, P, Q, R respectively are:

(a) $\ce{H2C=CH-CH2-Br}$, $(\ce{HO-})$, $[\ce{HO-}, Δ]$, Wacker process.

(b) $\ce{H2C=CH-CH2-Br(HO^-)}$, Wacker process, $\ce{HO-}$, $Δ$.

(c) Wacker process, $\ce{H2C=CH-CH2-Br(HO^-)}$, $\ce{HO-} (Δ)$.

(d) Wacker process, $\ce{HO-} (Δ)$, $\ce{H2C=CH-CH2-Br(HO^-)}$.

Could someone please explain the R part of this reaction? I can't understand the five-membered cyclization step. Is it aldol condensation or there's something else?

|improve this question
$\endgroup$
2
$\begingroup$

Step P is std base-driven alpha alkylation of ketone. Step Q Wacker details here takes the terminal alkene to methyl ketone then step R is base-catalysed internal aldol mechanism here gives the product shown. So the answer is (b).

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.