Investigate the structural formulae, properties, classification (primary, secondary, tertiary), oxidation reactions and production by hydration of alkenes for alcohols up to C8

A 4 mark answer needs both alcohols classified, both oxidation products named, both equations, and the colour change.

Propan-1-ol is a primary alcohol ($-OH$ on a terminal carbon). It oxidises in two steps: first to an aldehyde, then on further oxidation to a carboxylic acid.

Under reflux with excess acidified dichromate, the final product is propanoic acid. To isolate propanal, distil it off as it forms.

Propan-2-ol is a secondary alcohol ($-OH$ on a middle carbon). It oxidises in one step to a ketone and no further (the carbon has no $H$ left to lose):

Observation. Acidified dichromate is orange ($Cr_2O_7^{2-}$) and turns green ($Cr^{3+}$) as it is reduced. Both alcohols give the same colour change; the products differ.

Markers reward (1) correct classification, (2) the two-step pathway for the primary, (3) the single-step ketone for the secondary, (4) the visual colour change.

Both molecules have similar molar masses (ethanol 46 g/mol, ethane 30 g/mol), so dispersion forces alone cannot account for the 167 degrees C difference.

Ethanol contains a polar $O-H$ bond. The hydrogen, bonded to highly electronegative oxygen, can hydrogen-bond to the oxygen of a neighbouring ethanol molecule. Hydrogen bonds are roughly ten times stronger than dispersion forces.

Ethane has only C-H and C-C bonds, both nearly non-polar, so the only intermolecular force is weak dispersion. The molecules separate easily, giving a very low boiling point.

To boil ethanol, the energy supplied must overcome hydrogen bonds and dispersion. Hence the dramatically higher boiling point.

Markers reward (1) identifying the polar $O-H$ and hydrogen bonding, (2) stating that ethane only has dispersion forces, (3) linking IMF strength to boiling point.

A 3-mark classify needs the correct label AND the justification (carbon count) for each.

(a) $CH_3CH_2CH_2CH_2OH$ (butan-1-ol)

The C-OH carbon is bonded to only one other carbon. Primary.

(b) $(CH_3)_2CHOH$ (propan-2-ol)

The C-OH carbon is bonded to two other carbons (the two methyl groups). Secondary.

(c) $(CH_3)_3COH$ (2-methylpropan-2-ol)

The C-OH carbon is bonded to three other carbons (three methyl groups). Tertiary.

Marking criteria: 1 mark per correct classification with its carbon-count justification; a label with no justification earns half credit at most.

(a) Primary alcohol to carboxylic acid. Reagent: excess acidified potassium dichromate, $K_2Cr_2O_7 / H_2SO_4$. Condition: reflux. Colour change: orange to green.

(b) Secondary alcohol to ketone. Reagent: acidified potassium dichromate, $K_2Cr_2O_7 / H_2SO_4$ (or acidified permanganate). Condition: reflux (or gentle heating; no distillation trick is needed because the ketone cannot oxidise further). Colour change: orange to green (or purple to colourless for permanganate).

Marking criteria: 1 mark for correct reagent and condition in (a), 1 mark for correct reagent and condition in (b), 1 mark for both correct colour changes.

Step 1: write the balanced equation.

Step 2 (a): moles of glucose.

Step 3 (b): moles and mass of ethanol (mole ratio glucose : ethanol = 1 : 2).

Step 4 (c): moles and volume of $CO_2$ (mole ratio glucose : $CO_2$ = 1 : 2, so moles of $CO_2$ equal moles of ethanol).

Final answers (3 s.f.): $n(\text{glucose}) = 0.200$ mol; ethanol $= 0.400$ mol, $18.4$ g; $CO_2 = 9.92$ L.

Marking criteria: 1 mark for the correctly balanced equation, 1 mark for moles of glucose, 1 mark for correctly applying the 1:2 mole ratio to ethanol, 1 mark for the mass of ethanol with correct sig figs, 1 mark for applying the 1:2 ratio to $CO_2$, 1 mark for the final gas volume with correct sig figs and units. Note this is a THEORETICAL yield; real fermentation gives less due to side reactions and the broth stopping around 15% ethanol by volume.

(a) Description

Solubility is high and roughly constant (fully miscible, off the top of the practical scale) for methanol through propan-1-ol, then falls steeply and continuously from butan-1-ol through to octan-1-ol, which is only sparingly soluble (a few tenths of a gram per 100 g water).

(b) Explanation

Every alcohol in the series has the same hydrogen-bonding hydroxyl group, so the polar, water-favouring part of the molecule is constant. As the carbon chain lengthens, the non-polar alkyl tail grows, and its dispersion-force interactions with the surrounding non-polar tails of other alcohol molecules (and its disruption of water's hydrogen-bond network) become more significant relative to the single OH group. Beyond about four to five carbons, the hydrophobic tail dominates the molecule's overall behaviour, so solubility in water drops sharply.

(c) Prediction

Nonan-1-ol (C9) would plot below (further right and lower than) octan-1-ol. Adding one more $CH_2$ unit further increases the non-polar character of the molecule relative to its single OH group, so solubility continues to decrease with each additional carbon in the chain.

Marking criteria: 1 mark for describing the plateau at short chain lengths, 1 mark for describing the steep decline at longer chain lengths, 1 mark for linking the trend to the constant OH group versus the growing non-polar tail, 1 mark for correctly identifying dispersion/hydrophobic effects as the cause, 1 mark for a correctly justified prediction for nonan-1-ol.

Method. Add a few drops of X to acidified potassium dichromate solution and warm gently in a water bath. Observe the colour. If a colour change is seen, take a small sample of the resulting mixture (containing the oxidation product) and add it to freshly prepared Tollens' reagent in a clean test tube, then warm gently in a water bath without shaking.

Expected observations and conclusions.

• If the dichromate stays orange: X does not oxidise, so X is tertiary. Tollens' test is not required.
• If the dichromate turns green AND the Tollens' test gives a silver mirror: the oxidation product is an aldehyde, so X is primary.
• If the dichromate turns green AND the Tollens' test gives no silver mirror (solution stays clear/grey, no coating): the oxidation product is a ketone, so X is secondary.

Marking criteria: 1 mark for the correct first reagent, condition and what colour change is being tested for, 1 mark for correctly using the oxidation product (not X itself) in the Tollens' test, 1 mark for the tertiary conclusion, 1 mark for the primary conclusion (green plus silver mirror), 1 mark for the secondary conclusion (green plus no silver mirror).

This is a 7-mark EVALUATE: markers reward a judgement supported by a comparison across multiple named criteria, not a plain description of each method.

Band 6 PLAN.

• Thesis: Markovnikov hydration of propene is the more suitable industrial route specifically for a SECONDARY alcohol solvent (propan-2-ol), because fermentation of glucose by yeast produces almost exclusively ethanol (a primary alcohol) via a fixed biochemical pathway and cannot be redirected to make propan-2-ol at all; the two processes are therefore not interchangeable for this target product.
• Criterion 1, product identity and yield: hydration of propene gives propan-2-ol directly via Markovnikov addition ($CH_3CH=CH_2 + H_2O \xrightarrow{H_2SO_4} CH_3CH(OH)CH_3$) in high conversion per pass under the right conditions (dilute $H_2SO_4$ catalyst, heat, pressure); fermentation is biologically constrained to ethanol from glucose and self-limits at around 15% ethanol as yeast are poisoned by the product, capping the achievable concentration regardless of feedstock quantity.
• Criterion 2, purity: the hydration route from petrochemical propene gives a relatively pure product needing only distillation; fermentation broth is dilute and contaminated with yeast cells, unreacted sugars and other metabolites, requiring more extensive purification (fractional distillation, sometimes dehydration) to reach solvent-grade purity.
• Criterion 3, feedstock and environmental impact: propene is derived from fossil-fuel cracking, so hydration depends on a non-renewable feedstock and its overall carbon footprint tracks oil price and refining emissions; fermentation uses a renewable, biologically-fixed carbon feedstock (glucose from crops) and can be considered closer to carbon-neutral, but only produces ethanol, not the secondary alcohol required here.
• Judgement: for the SPECIFIC target of a secondary alcohol solvent, hydration of propene is the only viable industrial option of the two; fermentation is environmentally preferable in general but is simply the wrong reaction for this product, so it cannot be assessed as an alternative on equal terms.

Model paragraph (excerpt). Although fermentation is often praised as the more sustainable route to industrial alcohols, it cannot supply a secondary alcohol such as propan-2-ol at all, since yeast metabolism converts glucose specifically to ethanol via a fixed enzymatic pathway. Markovnikov hydration of propene, by contrast, is regiochemically suited to producing propan-2-ol directly, placing the hydroxyl group on the more substituted carbon of the alkene in a single catalysed step. On yield and purity grounds, hydration also outperforms fermentation, which self-limits near 15% ethanol and requires costly purification of a dilute, contaminated broth. The environmental trade-off, a non-renewable feedstock for hydration against fossil-fuel-free glucose for fermentation, is real but secondary to the fact that only one of the two processes can deliver the required product.

Marker's note: top-band answers (1) explicitly compare across at least three named criteria (yield, purity, feedstock/environment), (2) correctly recognise that fermentation cannot produce a secondary alcohol from glucose and use this as the decisive point, (3) include a correctly regiochemical equation for the hydration route, and (4) close with an explicit, justified judgement rather than a balanced non-answer.