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16.1 Alcohols

Syllabus
9701–2028–2029
Topic
16.1
Level
AS

Learning objectives

16.1.1Preparation of alcohols• Recall preparation of alcohols- (a) electrophilic addition of steam to an alkene, using H2O(g) and H3PO4 catalyst- (b) reaction of alkenes with cold dil. acidified potassium manganate( VII) to form a diol- (c) substitution of a halogenoalkane using NaOH(aq) and heat- (d) reduction of an aldehyde or ketone using NaBH4 or LiAl H4- (e) reduction of a carboxylic acid using LiA l H4- (f) hydrolysis of an ester using dil. acid or dil. alkali and heat16.1.2Describe• Describe:- (a) the reaction with oxygen (combustion)- (b) substitution to form halogenoalkanes, e.g. by reaction with HX(g); or with KCl and concentrated- H2SO4 or conc. H3PO4; or with PCl 3 and heat; or with PCl5; or with SOCl 2- (c) the reaction with Na(s)- (d) oxidation with acidified K2Cr2O7 or acidified KMnO4 to:- (i) carbonyl compounds by distillation- (ii) carboxylic acids by refluxing (primary alcohols give aldehydes which can be further oxidised to carboxylic acids, secondary alcohols give ketones, tertiary alcohols cannot be oxidised)- (e) dehydration to an alkene, by using a heated catalyst, e.g. Al2O3 or a conc. acid- (f) formation of esters by reaction with carboxylic acids and conc. H2SO4 as catalyst as exemplified by ethanol16.1.3Classify alcohols as primary, secondary• One alcohol group: (a) classify alcohols as primary, secondary and tertiary alcohols, to include examples with more than; (b) state characteristic distinguishing reactions, e.g. mild oxidation with acidified K 2Cr2O7, colour change; from orange to green16.1.4CH3CH(OH)- alcohol identification• Deduce the presence of a CH3CH(OH)– group in an alcohol, CH3CH(OH)–R, from its reaction with alkaline I2(aq) to form a yellow precipitate of tri-iodomethane and an ion, RCO2 –16.1.5Acidity of alcohols compared with water• Explain the acidity of alcohols compared with water

Select an alcohol preparation from the starting functional group

Starting material Reagent and conditions Alcohol product
alkene H₂O(g), H₃PO₄ catalyst one –OH added by hydration
alkene cold, dilute, acidified KMnO₄ diol, with –OH on both former C=C carbons
halogenoalkane NaOH(aq), heat –X replaced by –OH
aldehyde NaBH₄ or LiAlH₄ primary alcohol
ketone NaBH₄ or LiAlH₄ secondary alcohol
carboxylic acid LiAlH₄ primary alcohol
ester dilute acid or dilute alkali, heat alcohol plus carboxylic acid, or carboxylate in alkali

Work from the precursor's functional group, then preserve its carbon accounting. CN-free reduction and substitution routes retain the carbon skeleton; ester hydrolysis splits at the C–O bond and yields the alcohol from the group originally bonded to the single-bonded ester oxygen.

NaBH₄ reduces aldehydes and ketones but is not the listed reagent for carboxylic-acid reduction; use LiAlH₄ there. Cold dilute manganate(VII) forms a diol, whereas hot concentrated conditions cleave C=C.

Alcohol reactions change the O–H bond, C–O bond or oxidation level

Reaction Reagent and conditions Main organic outcome
complete combustion O₂, ignition CO₂ + H₂O
substitution HX(g); KCl + concentrated H₂SO₄/H₃PO₄; PCl₃ + heat; PCl₅; or SOCl₂ halogenoalkane
sodium Na(s) sodium alkoxide + H₂
dehydration heated Al₂O₃ or concentrated acid alkene + H₂O
esterification carboxylic acid, concentrated H₂SO₄ catalyst ester + H₂O

2 ROH+2 Na→2 RONa+HX2\ce{2ROH + 2Na -> 2RONa + H2}

Alcohol class Acidified K₂Cr₂O₇ or KMnO₄ with distillation With reflux / further oxidation
primary aldehyde is distilled off carboxylic acid
secondary ketone ketone; no further oxidation under these conditions
tertiary no reaction no reaction

Distillation removes the volatile aldehyde before it is further oxidised; reflux returns vapour to the flask so a primary alcohol or aldehyde remains with oxidant long enough to form the carboxylic acid. Acidified dichromate changes orange to green when it is reduced during oxidation.

Do not use oxidation outcome alone without alcohol class and conditions. A tertiary alcohol lacks the H on the carbon bearing –OH needed for these oxidation patterns.

Classify each –OH-bearing carbon and confirm oxidisable classes

Carbon neighbours of the C–OH carbon Class Mild oxidation outcome Acidified K₂Cr₂O₇ observation
1 primary, 1° aldehyde, then carboxylic acid orange → green
2 secondary, 2° ketone orange → green
3 tertiary, 3° no oxidation under these conditions remains orange

For every alcohol group, locate its C–OH carbon and count only the carbon atoms directly bonded to that carbon. A molecule with several –OH groups is classified site by site, because different –OH-bearing carbons can have different classes.

In propane-1,2-diol, CH₃CH(OH)CH₂OH, the –OH on carbon 1 is primary while the –OH on carbon 2 is secondary. The molecule therefore contains both a primary and a secondary alcohol group; “diol” does not replace those local classifications.

The dichromate colour change distinguishes an oxidisable primary/secondary group from a tertiary group, but it does not by itself distinguish primary from secondary; the oxidation product or structure is also needed.

A yellow iodoform precipitate reveals CH₃CH(OH)–

Warm the alcohol with alkaline aqueous iodine. A yellow precipitate of triiodomethane, CHI₃, is a positive result for an alcohol containing CH₃CH(OH)–R (including R = H in ethanol).

The alcohol is first oxidised to a compound containing CH₃CO–. Successive iodination and alkaline cleavage then produce solid CHI₃ and the carboxylate ion RCO₂⁻, so the observation traces back to the required alcohol motif.

CHX3CH(OH)R+4 IX2+6 OHX−→CHIX3(s)+RCOX2X−+5 IX−+5 HX2O\ce{CH3CH(OH)R + 4I2 + 6OH- -> CHI3(s) + RCO2- + 5I- + 5H2O}

Propan-2-ol, CH₃CH(OH)CH₃, gives CHI₃ and CH₃CO₂⁻. Propan-1-ol lacks CH₃CH(OH)– and gives no yellow precipitate; ethanol is the special primary-alcohol positive case.

This is not a general test for every alcohol or every molecule containing CH₃. The methyl group must be connected through the oxidisable CH(OH) centre shown.

Alkyl electron donation makes alcohols weaker acids than water

ROH+HX2O⇌ROX−+HX3OX+\ce{ROH + H2O <=> RO- + H3O+}

Acid strength depends on the stability of the conjugate base left after O–H bond cleavage. An alkyl group has a positive inductive effect and pushes electron density towards oxygen, intensifying the negative charge on RO⁻ relative to OH⁻.

Because an alkoxide ion is less stabilised than hydroxide, formation of RO⁻ is less favourable than formation of OH⁻: ordinary alcohols are weaker acids than water and their ionisation equilibrium lies very far to the left.

2 ROH+2 Na→2 RONa+HX2\ce{2ROH + 2Na -> 2RONa + H2}

Reaction with sodium shows that the O–H proton can be removed; it does not make an alcohol a strong acid. Alcohols do not significantly neutralise aqueous hydroxide because the acid–base equilibrium favours ROH and OH⁻.