Question 6 of 13: Retrosynthesis — Acetylide + Alkyl Halide
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-12, Organic Chemistry — December 2017. 3 hours, closed-book
examination (no calculator required); NOTES on page 1 state that TEN (10) questions constitute a
complete exam paper and only the first 10 as they appear in the answer book are marked, but this
sitting prints 13 numbered questions — every question and sub-part below is answered in full.
Reference texts: McMurry, Organic Chemistry, 9th ed. (acid–base
strength of drugs, pharmacokinetics/lipophilicity, β-lactam reactivity, SN2
stereochemistry, Williamson-ether-type syntheses, alkyne alkylation, IR/NMR structure
elucidation, radical vs. ionic HBr addition, electrophilic aromatic substitution & synthesis
design, acid strength/resonance & induction, polymer/monomer identification). Every molecular
formula, mass balance, and stereochemical (R/S) assignment below.
Approach. Every internal (or terminal) alkyne here is disconnected at the
C–C bond immediately next to the triple bond, on the side that gives the more sensible
(least hindered, primary) alkyl halide electrophile — acetylide alkylation is an
SN2 reaction, so a primary (never secondary/tertiary, which would eliminate
instead) alkyl halide must be used as the electrophile.
a) HC≡C–CH2CH2CH(CH3)2
(5-methylhex-1-yne). The triple bond is terminal, so the only sensible disconnection is
at the C1–C2 vs. C2–C3 bond keeping the acetylide as the small, terminal fragment:
cut the C2–C3 bond. This requires the acetylide of acetylene itself
(HC≡C−, from HC≡CH + NaNH2) reacting with the primary
halide 1-halo-3-methylbutane, (CH3)2CHCH2CH2X
(isoamyl halide):
$$\mathrm{HC{\equiv}C^- + (CH_3)_2CHCH_2CH_2X \xrightarrow{S_N2} HC{\equiv}C{-}CH_2CH_2CH(CH_3)_2 + X^-}$$
b) CH3–C≡C–C(CH3)2CH2CH3
(2,3-dimethylpent-3-yne, drawn with the triple bond internal). One side of the triple
bond (C(CH3)2CH2CH3) is a tertiary carbon
— its halide could never be alkylated by SN2 (it would only eliminate under
these strongly basic conditions). The disconnection must therefore be made on the
methyl side: acetylide
CH3CH2C(CH3)2C≡C− (from
2,3-dimethylpent-1-yne, itself made by alkylating acetylide with the tertiary-adjacent primary
halide first) reacting with methyl iodide (CH3I) as the necessarily
primary electrophile:
$$\mathrm{CH_3CH_2C(CH_3)_2{-}C{\equiv}C^- + CH_3I \xrightarrow{S_N2} CH_3{-}C{\equiv}C{-}C(CH_3)_2CH_2CH_3 + I^-}$$
c) Cyclohexyl–C≡C–CH2CH2CH3
(1-pentynylcyclohexane). Cyclohexyl is a secondary centre and, like a tertiary halide,
cannot serve as the SN2 electrophile (competing E2 dominates on a secondary ring
carbon under strongly basic acetylide conditions). The disconnection is therefore made on the
propyl side: cyclohexylacetylide (from ethynylcyclohexane +
NaNH2) reacting with 1-halopropane (a primary halide):
$$\mathrm{C_6H_{11}{-}C{\equiv}C^- + CH_3CH_2CH_2X \xrightarrow{S_N2} C_6H_{11}{-}C{\equiv}C{-}CH_2CH_2CH_3 + X^-}$$
