Aldehyde and ketone reactions


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> [![Reaction scheme](https://i.stack.imgur.com/2XaCb.png)](https://i.stack.imgur.com/2XaCb.png)
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> Reagents to carry out above conversion, *P*, *Q*, *R* respectively are:
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> (**a**) $\ce{H2C=CH-CH2-Br}$, $(\ce{HO-})$, $[\ce{HO-}, Δ]$, Wacker process.
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> (**b**) $\ce{H2C=CH-CH2-Br(HO^-)}$, Wacker process, $\ce{HO-}$, $Δ$.
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> (**c**) Wacker process, $\ce{H2C=CH-CH2-Br(HO^-)}$, $\ce{HO-} (Δ)$.
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> (**d**) Wacker process, $\ce{HO-} (Δ)$, $\ce{H2C=CH-CH2-Br(HO^-)}$.
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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?




Step P is std base-driven alpha alkylation of ketone. Step Q Wacker [details here](http://organicreactions.org/index.php?title=Wacker_oxidation) takes the terminal alkene to methyl ketone then step R is base-catalysed internal aldol [mechanism here](http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/keto_enol/aldolkondensation_aehnliche_reaktionen.vlu/Page/vsc/en/ch/12/oc/keto_enol/intra_aldol/intra_aldol.vscml.html) gives the product shown. So the answer is (b).


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