Question 1 of 5: Functional-Group Feasibility & Alkene/Arene Hydration

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

1. (i) Alcohol — PRESENT, possible. A hydroxyl group by itself needs no ring or π-bond, so nothing stops one (or two) –OH groups from appearing as long as the required DoU=1 is supplied by something else in the skeleton (a ring, here). Example: cyclopentane-1,2-diol — the five-membered ring accounts for the one degree of unsaturation, and both oxygens are hydroxyls.
   an alcohol (diol) consistent with C5H10O2 — the ring 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 already needs six carbons (and DoU=4, from the three ring π-bonds). The given formula has only five carbons total, so there are not enough carbons left to build any benzene ring at all — a phenol is structurally impossible at C₅.
3. (iii) Ether — PRESENT, possible. Like the alcohol, a C–O–C ether linkage carries no unsaturation of its own, so the single required DoU can again come from elsewhere (a C=C here) while both oxygens sit in ether linkages. Example: 2-methoxyethyl vinyl ether, CH₂=CH–O–CH₂CH₂–O–CH₃ (a vinyl ether and a methyl ether in the same five-carbon chain; the vinyl C=C supplies DoU=1).
4. (iv) Amide — ABSENT, impossible. Every amide, R–C(=O)–NR′R″, requires a nitrogen atom bonded to the carbonyl carbon. The molecular formula C₅H₁₀O₂ contains no N at all, so no amide of any kind can be drawn — this one is ruled out purely by elemental composition, independent of the DoU argument.
5. (v) Carboxylic acid — PRESENT, and in fact the exact match. A saturated, straight-chain monocarboxylic acid has the general formula C_nH_2nO₂ (the C=O and C–OH of the –COOH group together account for exactly DoU=1, using both oxygens). At n=5 that formula is exactly C₅H₁₀O₂ — no other adjustment is even needed. Example: pentanoic acid (valeric acid), CH₃CH₂CH₂CH₂COOH.
   a carboxylic acid consistent with C5H10O2 — C_nH_2nO₂ is the exact saturated mono-acid formula

b) Two hydrations. Both reactions are acid-catalysed (Markovnikov) additions of water across a C=C; the only question in each case is which alkene carbon gets the new –OH.

(i) Ethylene + H₂O. Ethylene's two alkene carbons are equivalent (symmetric), so there is no regiochemical choice to make:

$$\mathrm{H_2C{=}CH_2 + H_2O \xrightarrow{H_2SO_4,\ 100^\circ C} CH_3CH_2OH}$$

Product: ethanol.

(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 — water's OH ends up on the more substituted/more stable-cation carbon — therefore places –OH on the benzylic carbon, not the terminal CH₂:

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

Product: 1-phenylethanol.