Alcohols

The formula fits the saturated compounds of the type \(\mathrm{C}_n\mathrm{H}_{2n+2}\mathrm{O}\). Four isomers are alcohols and three are ethers (functional isomers):

Butan-1-ol and 2-methylpropan-1-ol are chain isomers; butan-1-ol and butan-2-ol are position isomers. Butan-2-ol also has a chiral carbon (C-2) and exists as two optical isomers.

The -OH group of an alcohol is polar (the oxygen carries a partial negative charge and the hydrogen a partial positive charge), so it forms hydrogen bonds with water molecules:

The energy released by these new bonds compensates for breaking the hydrogen bonds in water and in the alcohol, so lower alcohols (methanol, ethanol, propanols) mix with water in all proportions. As the alkyl chain gets longer, the non-polar (hydrophobic) part resists mixing with water and the solubility decreases.

Hydrocarbon molecules are held only by weak van der Waals (London) forces. Alcohol molecules are held by strong intermolecular hydrogen bonds in addition, and much more energy is needed to separate them.

Ethanol and methoxymethane have the same formula and mass, but ethers cannot form hydrogen bonds with each other, so the ether boils far lower than the alcohol.

For isomeric alcohols the solubility follows the order 3° > 2° > 1°. As branching increases, the non-polar alkyl part becomes more compact and its surface area becomes smaller, so it disturbs the hydrogen-bonded water structure less.

Fermentation is a slow biochemical process in which large organic molecules such as sugar and starch are broken down by the enzymes of micro-organisms (mainly yeast) into smaller compounds such as ethanol and carbon dioxide, usually in the absence of air (anaerobic).

Molasses is the thick brown syrup left after the crystals of sugar are separated from sugarcane juice; it still contains about 50% sucrose. Two enzymes of yeast act in turn: invertase hydrolyses sucrose into glucose and fructose, and zymase changes these into ethanol and carbon dioxide.

Starch is first converted by diastase (from germinated barley) into maltose, then by maltase (from yeast) into glucose, and then by zymase into ethanol:

The fermented liquid (“wash”) has at most about 15% alcohol, because yeast is killed by more alcohol than this. Distillation then gives rectified spirit (95.6%).

Hydroboration-oxidation converts an alkene into an alcohol in two stages.

Step 1: hydroboration. The B-H bonds of borane (diborane, \(\mathrm{B_2H_6}\), or \(\mathrm{BH_3}\)) add across the C=C bond of ethene. Since boron has three B-H bonds, three molecules of ethene add and triethylborane is formed:

Step 2: oxidation. The trialkylborane is oxidised by hydrogen peroxide in alkaline solution (NaOH) to give ethanol and boric acid:

The addition is anti-Markovnikov (boron goes to the carbon with fewer alkyl groups). With ethene the product is the same either way, but propene gives propan-1-ol: \(\mathrm{3CH_3CH=CH_2\xrightarrow{B_2H_6}(CH_3CH_2CH_2)_3B\xrightarrow{H_2O_2/OH^-}3CH_3CH_2CH_2OH}\).

Both are formed by acid-catalysed hydration, which follows Markovnikov’s rule (the -OH goes to the carbon with fewer hydrogens, through the more stable carbocation):

propene gives propan-2-ol (secondary).

2-methylpropene gives 2-methylpropan-2-ol (tertiary).

The Lucas reagent is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride (about 1 : 1 by moles). The alcohol is shaken with the reagent at room temperature and the time taken for turbidity (insoluble alkyl chloride) is noted:

The order of reactivity, 3° > 2° > 1°, follows the stability of the carbocation intermediate.

Oxidation removes the -OH hydrogen and the hydrogen on the same carbon (the alpha hydrogen) to form a C=O bond. A tertiary alcohol, \(\mathrm{R_3C-OH}\), has no hydrogen on the carbon bearing -OH, so a C=O group cannot form without breaking a C-C bond. It therefore resists oxidation under normal conditions and the purple colour of \(\mathrm{KMnO_4}\) does not change.

Ethanol (primary) gives ethanal and then ethanoic acid:

Propan-2-ol (secondary) gives propanone:

In both cases the purple colour of the permanganate disappears.

Acid-catalysed dehydration is an E1 reaction, and its slow step is the formation of a carbocation:

The easier the carbocation forms, the easier the dehydration. A tertiary carbocation is stabilised by three electron-releasing alkyl groups (+I effect and hyperconjugation), so it is more stable than a secondary or a primary one, its formation needs the lowest activation energy and dehydration is fastest.

(The ease of dehydration follows 3° > 2° > 1°. Do not confuse it with oxidation: tertiary alcohols are the hardest to oxidise.)

A primary alcohol is dehydrogenated to an aldehyde (ethanal).

A secondary alcohol is dehydrogenated to a ketone (propanone).

A tertiary alcohol has no alpha hydrogen, so it cannot be dehydrogenated. It is dehydrated instead, and 2-methylpropene is formed.

Reasoning. M is an alcohol (it releases H₂ with sodium). Only a primary alcohol gives an aldehyde on oxidation. The primary alcohol with the formula \(\mathrm{C_3H_8O}\) is propan-1-ol, and its oxidation product is propanal.

Reaction with sodium (hydrogen gas and sodium propoxide):

Oxidation with acidified potassium dichromate (the aldehyde is distilled off as it forms; further oxidation gives propanoic acid):

Reasoning. A ketone is formed by oxidation of a secondary alcohol, so A is a secondary alcohol of formula \(\mathrm{C_4H_{10}O}\): butan-2-ol. Its oxidation product, butan-2-one, has the \(\mathrm{CH_3CO-}\) group and so gives the iodoform test. Dehydration of A follows Saytzeff’s rule and gives the more substituted alkene, but-2-ene.

A small amount of but-1-ene (minor product) is also formed.

Reasoning (working backwards). D is a tertiary alcohol of four carbons, so D is 2-methylpropan-2-ol, \(\mathrm{(CH_3)_3COH}\). It is obtained by acid hydration of an alkene C that follows Markovnikov’s rule, which must be 2-methylpropene (a straight-chain butene would give a secondary alcohol). C is made by dehydrohalogenation of a haloalkane B, which has the branched skeleton, and B is formed from A and \(\mathrm{PCl_5}\). A is therefore the branched primary alcohol, 2-methylpropan-1-ol.

(2-Methylpropan-2-ol itself would also fit the sequence formally, but then the chain of reactions would only return to the starting alcohol, so the primary isomer is the intended A.)

Propan-1-ol (primary): red colour

The product is a nitrolic acid. It has an acidic hydrogen (in the =N-OH group) and dissolves in KOH to form a sodium or potassium salt, which gives a red solution.

Propan-2-ol (secondary): blue colour

The product is a pseudonitrol (a nitro-nitroso compound). It gives a blue solution with KOH.

2-Methylpropan-2-ol (tertiary): colourless

No colour develops on adding KOH.

Nitrous acid attacks the alpha hydrogen (the hydrogen on the carbon that carries the -NO₂ group). In \(\mathrm{R_3C-NO_2}\) that carbon is bonded to three alkyl groups and has no hydrogen, so nitrous acid cannot react and the solution stays colourless.

The two alcohols give different products with nitrous acid. The nitrolic acid from a primary alcohol still has an acidic hydrogen, so it reacts with KOH to form a salt whose anion is red. The pseudonitrol from a secondary alcohol has no hydrogen on that carbon and no acidic hydrogen, so it forms no salt with KOH; the characteristic blue colour is that of its nitroso (C-N=O) group in solution.

Ethanol (a primary alcohol).

Propan-2-ol (a secondary alcohol).

B is ethanoic acid. Carbon dioxide gives a carboxylic acid, not an alcohol.

B is propan-2-ol (secondary alcohol).

B is 2-methylpropan-2-ol (tertiary alcohol).

B is ethanol (primary alcohol).

Lithium aluminium hydride reduces the carboxylic acid to the primary alcohol.

Hydrolysis of the ester gives ethanol and ethanoic acid (with acid) or its sodium salt (with alkali).

A primary amine gives the alcohol and nitrogen gas.

This is Darzen’s reaction; SO₂ and HCl escape as gases, so the chloroethane is pure.

Ethyl hydrogen sulphate is formed (at or below 100 °C).

Diethyl ether is formed by intermolecular dehydration (excess alcohol).

Propene is formed by intramolecular dehydration.

Diethyl ether is formed.

Propene is formed.

Reasoning. Turbidity with the Lucas reagent after about 5 minutes shows that X is a secondary alcohol. Ozonolysis cuts the C=C bond of Y into two carbonyl compounds: W, \(\mathrm{C_2H_4O}\), is ethanal (\(\mathrm{CH_3CHO}\)), so one half of Y is \(\mathrm{CH_3CH=}\). Z, \(\mathrm{C_3H_6O}\), can be propanal or propanone, so two answers are possible:

Answer 1: Z is propanal. Then Y is pent-2-ene, \(\mathrm{CH_3CH=CHCH_2CH_3}\), and X is pentan-2-ol (pentan-3-ol also fits). The major alkene from the dehydration follows Saytzeff’s rule.

Answer 2: Z is propanone. Then Y is 2-methylbut-2-ene, \(\mathrm{(CH_3)_2C=CHCH_3}\), and X is 3-methylbutan-2-ol, a secondary alcohol whose dehydration gives the more substituted alkene (Saytzeff’s rule).

Both sets of compounds satisfy every condition of the question.

A is ethanol, \(\mathrm{CH_3CH_2OH}\) (soluble in water by hydrogen bonding, burns with a pale blue flame, releases H₂ with sodium, and gives the iodoform test because of the \(\mathrm{CH_3CH(OH)-}\) group). Its isomer methoxymethane would not react with sodium. B is ethene.

Conc. \(\mathrm{H_2SO_4}\) acts as a dehydrating agent: it removes the elements of water from ethanol. It is also the acid catalyst: it protonates the -OH group, converting it into the good leaving group \(\mathrm{H_2O}\), which leaves with the neighbouring H⁺ to form the double bond.

Both give a positive iodoform test, so another test is needed. The Lucas test separates them: propan-2-ol (secondary) gives turbidity in about 5 minutes at room temperature, but ethanol (primary) gives no turbidity at room temperature.

(Oxidation also separates them: ethanol gives ethanal, which gives a silver mirror with Tollens’ reagent, whereas propan-2-ol gives propanone, which does not.)

The orange colour of the acidified potassium dichromate changes to green, because orange \(\mathrm{Cr_2O_7^{2-}}\) (chromium +6) is reduced to green \(\mathrm{Cr^{3+}}\). Ethanol is oxidised to ethanoic acid (the smell of vinegar appears):

The teacher’s clues and the matching terms are given below. A few terms answer more than one clue, so the best-fitting term is written for each clue.

Power alcohol (gasohol) is a liquid fuel made by blending absolute ethanol with petrol, in the textbook ratio of about 20% ethanol and 80% petrol (with a little benzene to keep the mixture uniform). Its octane rating is about 90.

Fermentation. Molasses, a by-product of the sugar industry, is diluted, treated with yeast at about 25–35 °C and fermented for 2–3 days. Invertase and zymase convert the sucrose to ethanol (see Short Question 3). The wash is distilled to give rectified spirit (95.6%), and absolute alcohol is obtained from it by azeotropic distillation. In Nepal the by-product molasses of sugar mills is the raw material for industrial ethanol, and traditional drinks such as jaand, tongba, chhyang and raksi are made by fermenting millet, rice, maize or barley.

Oxo process (hydroformylation). An alkene reacts with carbon monoxide and hydrogen at high temperature and pressure in the presence of the catalyst dicobalt octacarbonyl, \(\mathrm{[Co(CO)_4]_2}\), to give an aldehyde, which is hydrogenated to the alcohol with a nickel catalyst:

The product has one carbon more than the alkene. Ethene gives propan-1-ol.