Question 3 of 5: Markovnikov Addition of HBr, Amine/Ether Structures & Methanol→Ethanol Synthesis

Question 3: Markovnikov Addition of HBr, Amine/Ether Structures & Methanol→Ethanol Synthesis (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) Markovnikov's rule governs every part here: the proton adds first, to the alkene carbon that already carries more hydrogen, generating the more stable (more substituted) carbocation, which Br⁻ then captures.

(i) CH₃CH₂CH=CH₂ (1-butene): H⁺ adds to the terminal =CH₂, giving a secondary carbocation at C2; Br⁻ attacks there → 2-bromobutane, CH₃CH₂CHBrCH₃ (C₄H₉Br).

(ii) CH₃CH=C(CH₃)₂ (2-methylbut-2-ene): the right-hand alkene carbon already carries two methyl substituents (vs. one on the left), so it forms the more stable tertiary cation; Br⁻ attacks there → 2-bromo-2-methylbutane, CH₃CH₂C(CH₃)₂Br (C₅H₁₁Br).

(iii) 1-Methylenecyclohexane: the ring carbon is the more substituted alkene terminus (two ring C–C bonds vs. the exocyclic carbon's none), so it becomes the tertiary cation and captures Br; the exocyclic carbon picks up the added H, becoming a methyl group:

(iv) 1-Methylcyclopentene: C1 (bearing the methyl) is already the more substituted alkene carbon, so it is where the cation — and then Br — ends up:

(v) CH₃CH=CHCH₃ (2-butene, cis or trans): the alkene is symmetric (each alkene carbon bears one methyl and one H), so Markovnikov's rule gives no regiochemical preference either way — both faces of protonation lead to the same secondary cation → 2-bromobutane (identical product name to part (i), formed as a racemic mixture since a new stereocentre is created at C2).

b) Structures from a molecular formula. (i) C₃H₉N has one degree of unsaturation less than a saturated C₃ skeleton needs for a ring or π-bond (DoU = 0), consistent with a simple amine: propan-1-amine, CH₃CH₂CH₂NH₂ (isopropylamine and trimethylamine are valid alternative isomers of the same formula). (ii) C₄H₁₀O with an ether linkage (no OH, an oxygen bonded to two carbons): diethyl ether, CH₃CH₂–O–CH₂CH₃.

c) Methanol → ethanol, a one-carbon homologation. Methanol has only one carbon, so reaching ethanol requires forming a new C–C bond; the standard three-step route converts methanol to a Grignard reagent and adds it across formaldehyde:

1. Activate the carbon as a leaving group. $$\mathrm{CH_3OH + HBr \longrightarrow CH_3Br + H_2O}$$
2. Form the Grignard reagent (dry ether, Mg turnings). $$\mathrm{CH_3Br + Mg \xrightarrow{\text{dry Et}_2O} CH_3MgBr}$$
3. Add across formaldehyde, then hydrolyse the alkoxide. Grignard addition to HCHO installs one new carbon and, after aqueous acid workup, delivers a primary alcohol: $$\mathrm{CH_3MgBr + HCHO \longrightarrow CH_3CH_2OMgBr \xrightarrow{H_3O^+} CH_3CH_2OH}$$ The carbon that started in methanol (1 C) plus the carbon supplied by formaldehyde (1 C) account for both carbons of the ethanol product.