Question 5 of 5: Ester/Alcohol Reactions, Dehydration & Hydrogenation of Butene Isomers

Question 5: Ester/Alcohol Reactions, Dehydration & Hydrogenation of Butene Isomers (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.

a) i) Acid-catalysed ester hydrolysis (the reverse of Fischer esterification, an equilibrium): propyl acetate is cleaved back to its acid and alcohol.

$$\mathrm{CH_3C(=O)OCH_2CH_2CH_3 + H_2O \underset{\text{heat}}{\overset{H^+}{\rightleftharpoons}} CH_3COOH + CH_3CH_2CH_2OH}$$

ii) Acid-catalysed dehydration of a primary alcohol (E1/E2, conc. H₂SO₄, heat) removes H₂O to form the alkene; with only two possible β-hydrogens (all on C2) there is no regiochemical choice:

$$\mathrm{CH_3CH_2CH_2OH \xrightarrow[\text{catalyst}]{H_2SO_4} CH_3CH{=}CH_2 + H_2O}$$

iii) Base-catalysed ester hydrolysis (saponification) is irreversible — NaOH consumes the carboxylic acid product as its carboxylate salt as fast as it forms, driving the reaction to completion (unlike (i)'s acid-catalysed equilibrium):

$$\mathrm{CH_3C(=O)OCH_2CH_2CH_3 + NaOH \xrightarrow{\text{heat}} CH_3COO^-Na^+ + CH_3CH_2CH_2OH}$$

iv) Anti addition of Br₂ across the alkene (bromonium-ion mechanism), no Markovnikov selectivity issue since the two new C–Br bonds form on the same symmetric pair of carbons regardless:

$$\mathrm{(CH_3)_2C{=}CH_2 + Br_2 \longrightarrow (CH_3)_2CBr{-}CH_2Br}$$

v) Nitration of benzene (electrophilic aromatic substitution; the NO₂⁺ electrophile is generated in situ from HNO₃/H₂SO₄):

$$\mathrm{C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O}$$

b) i) Dehydration of 3-methyl-2-butanol, (CH₃)₂CH–CH(OH)–CH₃. Protonation and loss of water at C2 first gives a secondary carbocation; a hydride shift from the adjacent C3–H converts it to a more stable tertiary cation at C3 before elimination occurs — the same "secondary → tertiary hydride shift" trap seen whenever a carbocation forms next to a more substituted carbon. Eliminating (Zaitsev, most-substituted alkene) from the shifted tertiary cation gives the major product:

$$\mathrm{(CH_3)_2CH{-}CH(OH){-}CH_3 \xrightarrow{H_2SO_4,\ -H_2O} \underbrace{(CH_3)_2C{=}CH{-}CH_3}_{\text{2-methyl-2-butene (major)}}}$$

A minor product, 3-methyl-1-butene (CH₂=CH–CH(CH₃)–CH₃), also forms from direct elimination out of the original (unshifted) secondary cation; both are confirmed as valid C₅H₁₀ dehydration products (alcohol minus exactly one H₂O).

ii) Dehydration of ethanol — with only two carbons, there is only one possible alkene, so no regiochemistry question arises at all:

$$\mathrm{CH_3CH_2OH \xrightarrow{H_2SO_4,\ \text{heat}} CH_2{=}CH_2 + H_2O}$$

c) Hydrogenation of three C₄ unsaturated hydrocarbons. All three starting materials are different (a terminal alkene, a cis-internal alkene, and a symmetric internal alkyne), yet catalytic hydrogenation destroys every piece of positional and geometric information that distinguished them — all three converge on the same alkane, butane (a genuine result, not a simplification):

$$\mathrm{CH_3CH_2CH{=}CH_2 \ (\text{1-butene}) + H_2 \xrightarrow{Pd/Ni} CH_3CH_2CH_2CH_3}$$

$$\mathrm{CH_3CH{=}CHCH_3 \ (\textit{cis}\text{-2-butene}) + H_2 \xrightarrow{Pd/Ni} CH_3CH_2CH_2CH_3}$$

$$\mathrm{CH_3C{\equiv}CCH_3 \ (\text{dimethylacetylene, 2-butyne}) + 2\,H_2 \xrightarrow{Ni} CH_3CH_2CH_2CH_3}$$