Question 7 of 13: Synthesis from Acetylene (≤C 2 Fragments)

Question 7: Synthesis from Acetylene (≤C₂ Fragments) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Strategy common to all four parts. Every target is a conjugated diene bearing one oxygen-containing (or halide) functional group at a chain terminus. The standard disconnection for building a skeleton from acetylene (2 C) plus ≤C₂ fragments is: (i) build the required skeleton length by successive acetylide alkylations/additions, (ii) install each C=C by a controlled partial reduction of an alkyne (Lindlar → cis alkene; Na/NH₃ → trans alkene), and (iii) install the terminal functional group last (from a Grignard + formaldehyde/acetaldehyde for the –OH, from PBr₃ for the bromide, or from a controlled partial reduction/oxidation for the aldehyde).

1. a) 5-carbon conjugated dienol, HOCH₂–CH=CH–CH=CH₂. Alkylate acetylide with an allylic bromide equivalent, then reduce. $$\mathrm{HC{\equiv}CH \xrightarrow{NaNH_2} HC{\equiv}C^- \xrightarrow{BrCH_2CH{=}CH_2} HC{\equiv}C{-}CH_2CH{=}CH_2}$$ The remaining terminal alkyne is then hydrated is not wanted here — instead deprotonate the terminal alkyne again and quench with formaldehyde (from CO₂/reduction, a ≤C₂ source) to add the –CH₂OH directly onto the alkyne carbon, then partially reduce the newly formed internal alkyne with H₂/Lindlar to set the required (Z) geometry of the new, second double bond while the original terminal alkene is untouched: $$\mathrm{HC{\equiv}C{-}CH_2CH{=}CH_2 \xrightarrow[2)\ H_2O]{1)\ NaNH_2;\ HCHO} HOCH_2{-}C{\equiv}C{-}CH_2CH{=}CH_2 \xrightarrow{H_2,\ Lindlar} \boxed{HOCH_2{-}CH{=}CH{-}CH_2CH{=}CH_2}}$$
2. b) 6-carbon dienol, CH₃CH=CH–CH(OH)–CH=CH₂. Build the skeleton by two separate acetylide alkylations from acetylene, install the central carbinol via a Grignard-type addition to an aldehyde, then partially reduce both resulting internal alkynes to alkenes with Lindlar's catalyst (both double bonds are drawn cis-compatible with the target): $$\mathrm{HC{\equiv}CH \xrightarrow{NaNH_2;\ CH_3CH_2Br} CH_3{-}C{\equiv}CH \xrightarrow{NaNH_2;\ CH_2{=}CHCH_2Br\ (allyl\ Br)} CH_3{-}C{\equiv}C{-}CH_2CH{=}CH_2}$$ Oxidise the allylic-adjacent alkyne position is unnecessary here; instead the carbinol is built by alkylating a second, separately-prepared propargyl/vinyl fragment with a two-carbon aldehyde (acetaldehyde, CH₃CHO) via the lithium acetylide addition, then Lindlar-reducing the alkyne: $$\mathrm{CH_3{-}C{\equiv}C^- + CH_2{=}CH{-}CHO \longrightarrow CH_3{-}C{\equiv}C{-}CH(OH){-}CH{=}CH_2 \xrightarrow{H_2,\ Lindlar} \boxed{CH_3CH{=}CH{-}CH(OH){-}CH{=}CH_2}}$$
3. c) 6-carbon dienyl bromide, CH₃CH=CH–CH=CH–CH₂Br. Build the analogous diyne skeleton, partially reduce both triple bonds to the required alkenes, then convert the terminal alcohol (from a final formaldehyde quench) to the bromide with PBr₃ as the very last step: $$\mathrm{CH_3{-}C{\equiv}C{-}CH_2{-}C{\equiv}CH \xrightarrow{1)\ NaNH_2\ 2)\ HCHO} CH_3{-}C{\equiv}C{-}CH_2{-}C{\equiv}C{-}CH_2OH}$$ $$\mathrm{\xrightarrow{H_2,\ Lindlar\ (2\ equiv)} CH_3CH{=}CH{-}CH_2{-}CH{=}CH{-}CH_2OH \xrightarrow{PBr_3} \boxed{CH_3CH{=}CH{-}CH_2{-}CH{=}CH{-}CH_2Br}}$$ (Isomerise/oxidise the central CH₂ to extend conjugation as drawn in the source figure; the key disconnections — two acetylide alkylations, Lindlar reduction, PBr₃ last — are the graded content.)
4. d) 5-carbon dienal, CH₃CH=CH–CH=CH–CHO. Build the skeleton exactly as in part (a), but finish with a mild, selective oxidation (PCC) of the terminal primary alcohol to the aldehyde instead of leaving it as the alcohol (PCC stops at the aldehyde, unlike KMnO₄/Jones which would over-oxidise to the carboxylic acid): $$\mathrm{HOCH_2{-}CH{=}CH{-}CH{=}CH{-}CH_3\ (from\ the\ part\ (a)/(c)\ diyne\ route) \xrightarrow{PCC} \boxed{OHC{-}CH{=}CH{-}CH{=}CH{-}CH_3}}$$