Question 1 of 5: Functional-Group Feasibility, Dehydration & Hydration

Question 1: Functional-Group Feasibility, Dehydration & Hydration (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) Feasibility analysis. Two constraints pin this down before any structure is drawn: the degree of unsaturation $$\mathrm{DoU=\dfrac{2(5)+2-10}{2}=1}$$ (exactly one ring or one π-bond somewhere in the molecule), and the fact that the formula contains zero nitrogen atoms. Each part below is answered by checking whether the named functional group can be built without violating either constraint.

1. (i) Ester — PRESENT, possible. An ester's C(=O)O linkage supplies exactly DoU=1 on its own, so a saturated ester is an exact fit at C₅H₁₀O₂. Example: methyl butanoate, CH₃CH₂CH₂C(=O)OCH₃.
   an ester consistent with C5H10O2 — the C=O supplies the one degree of unsaturation
2. (ii) Phenol — ABSENT, impossible. A phenol is by definition an –OH bonded directly to an aromatic ring, and the smallest possible aromatic ring is benzene, which alone needs six carbons and DoU=4 (three ring π-bonds). The given formula has only five carbons and DoU=1 total, so there are neither enough carbons nor enough unsaturation to build any benzene ring — a phenol is structurally impossible here.
3. (iii) Ether — PRESENT, possible. A C–O–C ether linkage carries no unsaturation of its own, so the single required DoU can come from elsewhere (a C=C) while both oxygens sit in ether linkages. Example: 2-methoxyethyl vinyl ether, CH₂=CH–O–CH₂CH₂–O–CH₃ (the vinyl C=C supplies DoU=1; both oxygens are ether oxygens, not carbonyls).
4. (iv) Amide — ABSENT, impossible. Every amide, R–C(=O)–NR′R″, requires a nitrogen atom bonded to the carbonyl carbon. C₅H₁₀O₂ contains no N at all, so no amide of any kind can be drawn — ruled out purely by elemental composition, independent of the DoU argument.
5. (v) Aldehyde — PRESENT, possible. An aldehyde's C=O supplies DoU=1, leaving the second oxygen free to be a plain hydroxyl elsewhere in the chain (a hydroxy-aldehyde). Example: 5-hydroxypentanal, OHC–CH₂CH₂CH₂–CH₂OH.
   a hydroxy-aldehyde consistent with C5H10O2 — the CHO supplies DoU=1, the second O is a plain –OH

b) A dehydration and a hydration. Part (i) starts from an alcohol and removes water (acid-catalysed E1 elimination); part (ii) starts from an alkene and adds water (acid-catalysed Markovnikov addition) — the two are mechanistic mirror images of each other.

(i) 1-Methylcyclohexan-1-ol, H₂SO₄, 50°C. This is a tertiary alcohol, so protonation of the –OH followed by loss of water gives a tertiary carbocation directly (no hydride/methyl shift needed) at C1. Two β-hydrogens are available for the E1 elimination step: the endocyclic hydrogens on C2/C6, or the exocyclic hydrogens on the CH₃ group itself.

1. Protonate and ionise. H₂SO₄ protonates the hydroxyl, which then leaves as water, generating the tertiary cyclohexyl cation at C1: $$\mathrm{C_7H_{14}O \xrightarrow{H^+} [\text{C}_7\text{H}_{13}]^+ + H_2O}$$
2. Zaitsev elimination (major product). Loss of an endocyclic β-H (from C2) gives the more substituted, trisubstituted endocyclic alkene — the thermodynamically favoured product:
   Zaitsev product — the more substituted (trisubstituted) endocyclic alkene
   $$\mathrm{\boxed{\text{1-methylcyclohexan-1-ol} \xrightarrow{H_2SO_4,\ 50^\circ C} \text{1-methylcyclohexene (major)} + H_2O}}$$
3. Hofmann-type elimination (minor product). Loss of an exocyclic β-H (from the CH₃ group instead) gives the less-substituted exocyclic alkene:
   Hofmann-type product — the less-substituted exocyclic alkene
   $$\mathrm{\text{1-methylcyclohexan-1-ol} \xrightarrow{H_2SO_4,\ 50^\circ C} \text{methylenecyclohexane (minor)} + H_2O}$$

Both products are genuine E1 outcomes from the same tertiary cation; Zaitsev's rule predicts 1-methylcyclohexene as the major product because a trisubstituted alkene is more stable than a disubstituted (exocyclic) one.

(ii) Styrene + H₂O. Protonation of the vinyl group can in principle place the new C–H at either alkene carbon, but only one choice puts the resulting positive charge on the carbon directly attached to the ring, where it is stabilised by resonance into the aromatic π-system (a benzylic cation). Markovnikov's rule therefore places –OH on the benzylic carbon, not the terminal CH₂:

$$\mathrm{C_6H_5CH{=}CH_2 + H_2O \xrightarrow{H^+,\ \Delta} \boxed{C_6H_5CH(OH)CH_3}}$$

Product: 1-phenylethanol.

c) Combustion of propan-1-ol. Every carbon ends up as CO₂ and every hydrogen as H₂O; balance oxygen last. Propan-1-ol (C₃H₈O) gives an odd total oxygen requirement per molecule, so the whole equation is doubled to clear the fraction: