= Solution
The first equivalent of <titanocene methylidene> methylenates the <ester>: the <carbonyl> <oxygen> and <titanium> form an <oxatitanacyclobutane>, whose fragmentation gives \b[$\mathrm{PhC(OMe)=CH_2}$] and a <titanium>–<oxygen> product. The second equivalent adds across this electron-rich double bond. The initial adduct is a <titanacyclobutane> with cyclic connectivity $\mathrm{Cp_2Ti-CH_2-C(Ph)(OMe)-CH_2-}$, closing from the last methylene to <titanium>. The <ester>'s phenyl-substituted <carbon> lies opposite Ti; the two new methylenes originate from the two methylidene equivalents.
Beta-methoxy elimination opens that metallacycle to \b[B, $\mathrm{Cp_2Ti(OMe)[CH_2C(Ph)=CH_2]}$], the <allyltitanium reagent>. The C–OMe bond breaks, OMe transfers to Ti, one Ti–C bond becomes the new C=C bond, and the other Ti–C bond remains as the allylic <nucleophile>. The cyclic adduct is therefore a precursor to the opened reagent B; treating B as merely the one-equivalent <enol ether> loses the second reagent <carbon>.
The <aldehyde> <oxygen> coordinates to <titanium>. Allylic <carbon> transfer to its <carbonyl> <carbon>, accompanied by movement of the C=O pi <electrons> to <oxygen>, gives the <titanium> <alkoxide> $\mathrm{PhCH_2CH_2CH(O{-}Ti)CH_2C(Ph)=CH_2}$. Acidic aqueous workup cleaves O–Ti and protonates <oxygen>, giving \b[$\mathrm{PhCH_2CH_2CH(OH)CH_2C(Ph)=CH_2}$]. A six-membered allyl-transfer arrangement is a convenient mechanistic model. No enantioselectivity is imposed by these achiral reagents.
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