Alcohols

Alcohols are organic compounds containing one or more hydroxyl ($-\mathrm{OH}$) groups bonded to an $sp^3$ hybridised carbon atom. The general formula for a mono-alcohol is $\mathrm{R-OH}$. The $-\mathrm{OH}$ functional group is polar and capable of hydrogen bonding, which dominates the physical and much of the chemical behaviour of alcohols.

Classification

Alcohols are classified by the number of alkyl groups attached to the carbon bearing the hydroxyl group:

• Primary ($1^\circ$): The $-\mathrm{OH}$ carbon is bonded to one alkyl group (and two hydrogens). General structure: $\mathrm{RCH}_2\mathrm{OH}$. Examples: methanol, ethanol, propan-1-ol.
• Secondary ($2^\circ$): The $-\mathrm{OH}$ carbon is bonded to two alkyl groups (and one hydrogen). General structure: $\mathrm{R}_2\mathrm{CHOH}$. Examples: propan-2-ol, butan-2-ol.
• Tertiary ($3^\circ$): The $-\mathrm{OH}$ carbon is bonded to three alkyl groups (and no hydrogens). General structure: $\mathrm{R}_3\mathrm{COH}$. Examples: 2-methylpropan-2-ol (tert-butanol).

Physical Properties

Boiling Points

Alcohols have much higher boiling points than alkanes of similar molecular mass due to hydrogen bonding between the $-\mathrm{OH}$ groups. Each molecule can both donate and accept hydrogen bonds, creating an extensive intermolecular network.

Compound	$M_r$	Boiling Point
$\mathrm{CH}_4$	16	$-162^\circ\mathrm{C}$
$\mathrm{CH}_3\mathrm{OH}$	32	$65^\circ\mathrm{C}$
$\mathrm{C}_2\mathrm{H}_6$	30	$-89^\circ\mathrm{C}$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$	46	$78^\circ\mathrm{C}$

Solubility

Short-chain alcohols (methanol, ethanol, propan-1-ol) are miscible with water in all proportions. The $-\mathrm{OH}$ group forms hydrogen bonds with water, overcoming the hydrophobic effect of the alkyl chain. As chain length increases, the hydrophobic character dominates and solubility decreases sharply.

Oxidation of Alcohols

Oxidation of alcohols is the most synthetically important reaction of this class. The products depend on the class of alcohol and the reaction conditions.

Oxidising Agent

Acidified potassium dichromate(VI) ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 / \mathrm{H}_2\mathrm{SO}_4$) is the standard oxidising agent. It is orange and is reduced to $\mathrm{Cr}^{3+}$ (green). The colour change from orange to green provides a visual indication that oxidation has occurred.

Primary Alcohols

Primary alcohols are oxidised in two stages:

$$\mathrm{RCH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}]} \mathrm{RCHO} \xrightarrow{[\mathrm{O}]} \mathrm{RCOOH}$$

Stage 1: Alcohol to aldehyde. The aldehyde is the initial product. To isolate the aldehyde, use distillation (gentle heating, collecting the product as it forms at its lower boiling point). The aldehyde is more volatile than the alcohol and distils off before it can be further oxidised.

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\mathrm{distillation}} \mathrm{CH}_3\mathrm{CHO} + \mathrm{H}_2\mathrm{O}$$

Stage 2: Aldehyde to carboxylic acid. To oxidise all the way to the carboxylic acid, use reflux (heating under a condenser so the product returns to the reaction mixture). The aldehyde remains in contact with the oxidising agent and is further oxidised.

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\mathrm{reflux}} \mathrm{CH}_3\mathrm{COOH} + \mathrm{H}_2\mathrm{O}$$

An alternative for producing aldehydes without over-oxidation is to use PCC (pyridinium chlorochromate) in dichloromethane, which selectively oxidises primary alcohols to aldehydes.

Secondary Alcohols

Secondary alcohols oxidise directly to ketones:

$$\mathrm{R}_2\mathrm{CHOH} \xrightarrow{[\mathrm{O}]} \mathrm{R}_2\mathrm{C}=\mathrm{O} + \mathrm{H}_2\mathrm{O}$$

Ketones resist further oxidation under these conditions because there is no hydrogen atom on the carbonyl carbon that can be removed (C--C bond cleavage requires much harsher conditions).

$$\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3 \xrightarrow{[\mathrm{O}]} \mathrm{CH}_3\mathrm{COCH}_3$$

Tertiary Alcohols

Tertiary alcohols are not oxidised by acidified dichromate(VI). The carbon bearing the $-\mathrm{OH}$ group has no hydrogen atoms (all three substituents are alkyl groups), so the initial oxidation step cannot occur. The orange solution remains orange, which provides a useful negative test.

$$(\mathrm{CH}_3)_3\mathrm{COH} + [\mathrm{O}] \to \mathrm{no\ reaction}$$

Mechanism of Alcohol Oxidation

The oxidation proceeds via a chromate ester intermediate:

1. The alcohol attacks the chromium(VI) centre, displacing a water molecule and forming a chromate ester.
2. A base (water or $\mathrm{HSO}_4^-$) abstracts the proton from the $-\mathrm{OH}$ group.
3. The C--H bond on the same carbon breaks heterolytically, transferring a hydride ($\mathrm{H}^-$) equivalent to the chromium. The C=O double bond forms as the Cr(VI) is reduced to Cr(IV).
4. Chromium(IV) disproportionates or is further reduced to Cr(III).

Dehydration (Elimination to Alkenes)

Alcohols undergo elimination of water when heated with a strong acid catalyst (concentrated $\mathrm{H}_2\mathrm{SO}_4$ or $\mathrm{H}_3\mathrm{PO}_4$) at $170$--$180^\circ\mathrm{C}$:

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \xrightarrow{\mathrm{H}_2\mathrm{SO}_4,\,170^\circ\mathrm{C}} \mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O}$$

Mechanism

1. Protonation of the hydroxyl group by the acid catalyst, converting $-\mathrm{OH}$ into $-\mathrm{OH}_2^+$ (a good leaving group, since water is a weak base).

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{H}^+ \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}_2^+$$

2. Loss of water to form a carbocation (E1 mechanism for tertiary alcohols; E2 for primary).

3. Loss of a proton ($\mathrm{H}^+$) from an adjacent carbon, forming the C=C double bond.

When multiple alkenes are possible, Zaitsev's rule applies: the more substituted alkene predominates.

Worked Example. Dehydration of butan-2-ol:

$$\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3 \xrightarrow{\mathrm{H}^+,\,\Delta} \mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_3\,(\mathrm{major}) + \mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3\,(\mathrm{minor})$$

But-2-ene (more substituted, disubstituted alkene) is the major product; but-1-ene (monosubstituted) is the minor product.

Esterification

Alcohols react with carboxylic acids in the presence of a strong acid catalyst (concentrated $\mathrm{H}_2\mathrm{SO}_4$) to form esters:

$$\mathrm{RCOOH} + \mathrm{R}'\mathrm{OH} \rightleftharpoons \mathrm{RCOOR}' + \mathrm{H}_2\mathrm{O}$$

This is an equilibrium reaction. The position of equilibrium can be shifted to the right by:

• Using an excess of one reactant (usually the cheaper alcohol).
• Removing the water as it forms (e.g. by using a Dean-Stark apparatus).

Example:

$$\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \rightleftharpoons \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$$

Ethanoic acid + ethanol $\rightleftharpoons$ ethyl ethanoate + water.

Mechanism

1. Protonation of the carboxylic acid on the carbonyl oxygen.
2. Nucleophilic attack of the alcohol on the protonated carbonyl carbon, forming a tetrahedral intermediate.
3. Proton transfer from the alcohol oxygen to one of the hydroxyl oxygens.
4. Elimination of water from the tetrahedral intermediate.
5. Deprotonation to yield the ester.

Nucleophilic Substitution of Alcohols

Alcohols are poor substrates for direct nucleophilic substitution because $-\mathrm{OH}$ is a poor leaving group. However, the hydroxyl group can be converted into a better leaving group:

Protonation: In acidic conditions, $-\mathrm{OH}$ is protonated to $-\mathrm{OH}_2^+$, making water the leaving group. This allows substitution to proceed.

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{H}_2\mathrm{O}$$

Conversion to a halogenoalkane: Using phosphorus tribromide ($\mathrm{PBr}_3$), phosphorus pentachloride ($\mathrm{PCl}_5$), or thionyl chloride ($\mathrm{SOCl}_2$):

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{PBr}_3 \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{H}_3\mathrm{PO}_3$$ $$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{SOCl}_2 \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl} + \mathrm{SO}_2 + \mathrm{HCl}$$

Thionyl chloride is preferred because the byproducts ($\mathrm{SO}_2$ and $\mathrm{HCl}$) are gases that escape, driving the reaction to completion and simplifying purification.

Alcohols as Intermediates in Organic Synthesis

Alcohols are central to organic synthesis because they can be converted into virtually every other functional class. The following summary shows the key transformations:

Target functional group	Reagent/conditions from alcohol
Alkene	Dehydration with concentrated $\mathrm{H}_2\mathrm{SO}_4$ at $170^\circ\mathrm{C}$
Aldehyde	Distillation with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ (primary alcohol only)
Carboxylic acid	Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ (primary alcohol only)
Ketone	Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ (secondary alcohol)
Halogenoalkane	$\mathrm{PBr}_3$, $\mathrm{SOCl}_2$, or conc. $\mathrm{HBr}$
Ester	Carboxylic acid + concentrated $\mathrm{H}_2\mathrm{SO}_4$
Ether	Williamson ether synthesis (deprotonation with Na, then $\mathrm{R}'\mathrm{Br}$)

Worked Example: Multi-Step Synthesis

Propose a synthesis of butanoic acid from but-1-ene.

Step 1: Addition of $\mathrm{HBr}$ (electrophilic addition):

$$\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{CH}_3$$

This gives 2-bromobutane (Markovnikov product). For the linear product, use anti-Markovnikov addition (radical mechanism with organic peroxide initiator):

$$\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3 + \mathrm{HBr} \xrightarrow{\text{peroxides}} \mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$$

Step 2: Substitution to alcohol (aqueous $\mathrm{NaOH}$, SN2):

$$\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{NaOH} \to \mathrm{HOCH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$$

Step 3: Oxidation to carboxylic acid (reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$):

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\mathrm{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH}$$

Williamson Ether Synthesis

Ethers can be synthesised by the Williamson ether synthesis, which is an SN2 reaction between an alkoxide ion and a halogenoalkane:

$$\mathrm{R}^-\mathrm{ONa} + \mathrm{R}'\mathrm{Br} \to \mathrm{R}^-\mathrm{O}^-\mathrm{R}' + \mathrm{NaBr}$$

The alkoxide is generated by reacting the alcohol with sodium metal:

$$2\mathrm{ROH} + 2\mathrm{Na} \to 2\mathrm{RO}^-\mathrm{Na}^+ + \mathrm{H}_2$$

Choice of reactants: The alkoxide should be derived from a primary alcohol, and the halogenoalkane should also be primary, to avoid competing elimination reactions. Using a tertiary halogenoalkane would give predominantly elimination (alkene) rather than substitution (ether).

Worked Example. Synthesise ethoxyethane (diethyl ether) from ethanol:

$$2\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + 2\mathrm{Na} \to 2\mathrm{CH}_3\mathrm{CH}_2\mathrm{O}^-\mathrm{Na}^+ + \mathrm{H}_2$$ $$\mathrm{CH}_3\mathrm{CH}_2\mathrm{O}^-\mathrm{Na}^+ + \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{OCH}_2\mathrm{CH}_3 + \mathrm{NaBr}$$

Identification of Alcohols: Classification Test

The Lucas test distinguishes between primary, secondary, and tertiary alcohols:

Reagent: Lucas reagent (anhydrous $\mathrm{ZnCl}_2$ in concentrated $\mathrm{HCl}$).

Alcohol type	Observation	Reaction
Tertiary ($3^\circ$)	Immediate cloudiness	SN1 (fast carbocation formation)
Secondary ($2^\circ$)	Cloudiness within 1--5 minutes	SN1 (slower)
Primary ($1^\circ$)	No cloudiness at room temperature	Very slow; may need heating

The cloudiness is caused by the formation of an insoluble halogenoalkane. Tertiary alcohols react fastest because the carbocation intermediate is most stable.

Common Pitfalls

1. Distillation vs reflux. Distillation isolates the aldehyde; reflux produces the carboxylic acid. Confusing these conditions leads to the wrong product.

2. Using the wrong oxidising agent. $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ (orange to green) and $\mathrm{KMnO}_4$ (purple to colourless) are both strong oxidising agents. Tollen's reagent (silver mirror) and Fehling's solution (brick red precipitate) are mild oxidising agents that only oxidise aldehydes (not ketones).

3. Assuming all alcohols can be oxidised. Tertiary alcohols are resistant to oxidation because there is no $\alpha$-hydrogen to remove. No colour change is observed with $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$.

4. Forgetting that $\mathrm{PBr}_3$ and $\mathrm{SOCl}_2$ produce different halogenoalkanes. $\mathrm{PBr}_3$ gives bromoalkanes; $\mathrm{SOCl}_2$ gives chloroalkanes. Using $\mathrm{SOCl}_2$ when you need a bromoalkane is a common error.

5. Incorrect use of the Lucas test. The Lucas test only works for water-soluble alcohols. If the alcohol is insoluble, cloudiness will not be observed even if the reaction occurs.

Dehydration of Alcohols to Alkenes

Alcohols undergo acid-catalysed dehydration to form alkenes when heated with concentrated sulphuric or phosphoric acid:

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3 \xrightarrow{\mathrm{H}_2\mathrm{SO}_4,\,170^\circ\mathrm{C}} \mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_3 + \mathrm{H}_2\mathrm{O}$$

Mechanism (E1):

1. Protonation of the alcohol to form an alkyloxonium ion ($\mathrm{ROH}_2^+$), making $-\mathrm{OH}_2^+$ a good leaving group.
2. Loss of water to form a carbocation (rate-determining step).
3. Loss of a proton ($\mathrm{H}^+$) from an adjacent carbon, forming the C=C double bond.

Zaitsev's rule: The major product is the more substituted alkene (the one with more alkyl groups on the C=C). For butan-2-ol, the major product is but-2-ene (disubstituted), not but-1-ene (monosubstituted).

Carbocation rearrangements: If a less stable carbocation can rearrange to a more stable one (via hydride shift or alkyl shift), it will. For example, dehydration of butan-1-ol may initially give a primary carbocation, which can undergo a hydride shift to form a secondary carbocation, producing a mixture of but-1-ene and but-2-ene.

Ester Formation

Alcohols react with carboxylic acids in the presence of a strong acid catalyst (typically concentrated $\mathrm{H}_2\mathrm{SO}_4$) to form esters:

$$\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \rightleftharpoons \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$$

Conditions: Heat under reflux with a few drops of concentrated $\mathrm{H}_2\mathrm{SO}_4$.

Mechanism:

1. Protonation of the carbonyl oxygen of the carboxylic acid.
2. Nucleophilic attack by the alcohol oxygen on the carbonyl carbon.
3. Proton transfer and loss of water.
4. Deprotonation to give the ester.

The reaction is an equilibrium. To drive it to the right, use an excess of one reactant or remove the water as it forms (e.g. using a Dean-Stark apparatus).

Uses of Esters

• Solvents -- ethyl ethanoate is a common organic solvent (nail polish remover).
• Flavourings and fragrances -- many esters have pleasant fruity odours (e.g. isoamyl acetate smells of banana, methyl salicylate of wintergreen).
• Plasticisers -- added to PVC to make it flexible.
• Biodiesel -- fatty acid methyl esters (FAME) produced by transesterification of vegetable oils.

Hydrolysis of Esters

Esters can be hydrolysed back to the carboxylic acid and alcohol:

• Acid hydrolysis: Reflux with dilute acid (reversible; gives carboxylic acid + alcohol).
• Base hydrolysis (saponification): Reflux with aqueous $\mathrm{NaOH}$ (irreversible; gives carboxylate salt + alcohol). The carboxylate can be acidified to recover the carboxylic acid.

$$\mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{NaOH} \to \mathrm{CH}_3\mathrm{COO}^-\mathrm{Na}^+ + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$$

Alcohols in Green Chemistry

Bioethanol

Ethanol produced by fermentation of sugars is a renewable fuel source:

$$\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \xrightarrow{\text{yeast}} 2\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 2\mathrm{CO}_2$$

The ethanol produced by fermentation is approximately 95% pure (azeotrope with water). Absolute ethanol (100%) requires further drying with molecular sieves or calcium oxide.

Advantages and Disadvantages of Bioethanol as a Fuel

Advantage	Disadvantage
Carbon neutral (the $\mathrm{CO}_2$ released was recently absorbed by the crop)	Land use competition with food production
Renewable	Lower energy density than petrol
Biodegradable	Engine modifications may be needed for high blends
Reduces dependence on fossil fuels	Fermentation is slow; limited production capacity

2. Oxidation of tertiary alcohols. Tertiary alcohols are not oxidised by $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 / \mathrm{H}^+$. The absence of a colour change is the key diagnostic.

3. Using $-\mathrm{OH}$ as a leaving group in nucleophilic substitution. $-\mathrm{OH}$ is a poor leaving group. Always protonate first (acidic conditions) or convert to a halide.

4. Forgetting that esterification is an equilibrium. Unless you remove water or use excess reagent, the yield is limited by the equilibrium position.

5. Ignoring Zaitsev's rule in dehydration. For unsymmetrical secondary and tertiary alcohols, the major elimination product is the more substituted alkene.

Practice Problems

Problem 1

Describe a chemical test to distinguish between propan-1-ol, propan-2-ol, and 2-methylpropan-2-ol using acidified potassium dichromate(VI).

Solution:

• Propan-1-ol: Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. The orange solution turns green. The product is propanoic acid. (If distilled instead, propanal is produced -- it can be detected by its smell and by Tollens' or Fehling's test.)

• Propan-2-ol: Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. The orange solution turns green. The product is propanone. Propanone gives no reaction with Tollens' reagent or Fehling's solution, distinguishing it from the aldehyde produced from propan-1-ol.

• 2-Methylpropan-2-ol: Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. No colour change (solution remains orange). Tertiary alcohols are not oxidised.

Problem 2

Ethanol can be converted to bromoethane by three different methods. Write equations for each and identify the mechanism.

Solution:

Method 1: $\mathrm{HBr}$ (concentrated), heat.

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{H}_2\mathrm{O}$$

Mechanism: $\mathrm{S_N}2$ (primary substrate, protonated $-\mathrm{OH}_2^+$ as leaving group).

Method 2: $\mathrm{PBr}_3$.

$$3\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{PBr}_3 \to 3\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{H}_3\mathrm{PO}_3$$

Mechanism: $\mathrm{S_N}2$ (primary substrate, bromide ion as nucleophile).

Method 3: $\mathrm{SOCl}_2$ (produces chloroethane, not bromoethane -- so this method is not valid for bromoethane). For chloroethane:

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{SOCl}_2 \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl} + \mathrm{SO}_2 + \mathrm{HCl}$$

Therefore, only Methods 1 and 2 produce bromoethane from ethanol.

Problem 3

A student attempts to convert 2-methylpropan-2-ol to 2-methylpropene by heating with concentrated sulphuric acid. The reaction produces a mixture of products. Explain the formation of the minor product.

Solution:

The expected product is 2-methylpropene (via E1 elimination):

$$(\mathrm{CH}_3)_3\mathrm{COH} \xrightarrow{\mathrm{H}^+} (\mathrm{CH}_3)_3\mathrm{COH}_2^+ \to (\mathrm{CH}_3)_3\mathrm{C}^+ + \mathrm{H}_2\mathrm{O} \to (\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CH}_2 + \mathrm{H}^+$$

The minor product is di-isopropyl ether, formed by SN1 substitution:

$$(\mathrm{CH}_3)_3\mathrm{C}^+ + (\mathrm{CH}_3)_3\mathrm{COH} \to (\mathrm{CH}_3)_3\mathrm{COCH}_2\mathrm{C}(\mathrm{CH}_3)_3$$

The tertiary carbocation can either lose a proton (elimination, major) or be attacked by another alcohol molecule (substitution, minor). At higher temperatures, elimination is favoured.

Problem 4

Design a synthesis of ethyl ethanoate from ethanol, using no other carbon-containing reagents. State all reagents and conditions.

Solution:

Ethyl ethanoate ($\mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3$) requires an ethanoic acid component and an ethanol component.

Step 1: Oxidise ethanol to ethanoic acid (reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$):

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\mathrm{reflux}} \mathrm{CH}_3\mathrm{COOH}$$

Step 2: Esterification (heat ethanoic acid with ethanol and a few drops of concentrated $\mathrm{H}_2\mathrm{SO}_4$ under reflux):

$$\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \rightleftharpoons \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$$

The reaction is an equilibrium. To drive it forward, use an excess of ethanol or remove water as it forms.

Worked Examples: Alcohols in Depth

Example 1: Oxidation of a Primary Alcohol -- Step-by-Step

Oxidation of butan-1-ol to butanoic acid.

Stage 1: Alcohol to aldehyde.

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\text{distillation}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO} + \mathrm{H}_2\mathrm{O}$

Conditions: Distillation with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. The aldehyde (butanal, b.p. $75^\circ\mathrm{C}$) distils off as it forms, before it can be further oxidised. The orange dichromate turns green.

Stage 2: Aldehyde to carboxylic acid.

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO} \xrightarrow{[\mathrm{O}],\,\text{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} + \mathrm{H}_2\mathrm{O}$

Conditions: Reflux with excess acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. The aldehyde remains in the reaction mixture and is further oxidised to butanoic acid.

Mechanism of Stage 1:

1. The alcohol oxygen attacks the chromium(VI) centre, displacing water and forming a chromate ester.
2. A base removes the $\alpha$-proton, and the C--H electrons transfer as a hydride to the chromium, forming the C=O bond.
3. Chromium(VI) is reduced to chromium(IV), which disproportionates to chromium(III).

Example 2: Dehydration Mechanism with Carbocation Rearrangement

Dehydration of 3,3-dimethylbutan-1-ol with concentrated $\mathrm{H}_2\mathrm{SO}_4$.

Step 1: Protonation: $\mathrm{(CH}_3)_3\mathrm{CCH}_2\mathrm{CH}_2\mathrm{OH} + \mathrm{H}^+ \to \mathrm{(CH}_3)_3\mathrm{CCH}_2\mathrm{CH}_2\mathrm{OH}_2^+$

Step 2: Loss of water forms a primary carbocation (unstable):

$\mathrm{(CH}_3)_3\mathrm{CCH}_2\mathrm{CH}_2\mathrm{OH}_2^+ \to \mathrm{(CH}_3)_3\mathrm{CCH}_2\mathrm{CH}_2^+ + \mathrm{H}_2\mathrm{O}$

Step 3: Hydride shift from the adjacent carbon to form a more stable tertiary carbocation:

$\mathrm{(CH}_3)_3\mathrm{CCH}_2\mathrm{CH}_2^+ \to \mathrm{(CH}_3)_2\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{CH}_2\mathrm{CH}(\mathrm{CH}_3)_2$

Step 4: Loss of a proton gives the alkene:

$\mathrm{(CH}_3)_2\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{CH}_2\mathrm{CH}(\mathrm{CH}_3)_2 \to \mathrm{(CH}_3)_2\mathrm{C}=\mathrm{CHCH}(\mathrm{CH}_3)_2 + \mathrm{H}^+$

The product is 2,3,3-trimethylbut-1-ene. The carbocation rearrangement (hydride shift) occurs because the tertiary carbocation is much more stable than the primary carbocation.

Example 3: Williamson Ether Synthesis Planning

Target: Ethoxybenzene (phenetole) from phenol.

The Williamson ether synthesis requires an alkoxide and a halogenoalkane. Phenol is acidic enough to be deprotonated by $\mathrm{NaOH}$:

$\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{NaOH} \to \mathrm{C}_6\mathrm{H}_5\mathrm{O}^-\mathrm{Na}^+ + \mathrm{H}_2\mathrm{O}$

Then reaction with bromoethane (SN2):

$\mathrm{C}_6\mathrm{H}_5\mathrm{O}^- + \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} \to \mathrm{C}_6\mathrm{H}_5\mathrm{OCH}_2\mathrm{CH}_3 + \mathrm{Br}^-$

Important: The halogenoalkane must be primary to avoid elimination. If we used 2-bromopropane, elimination would compete strongly with substitution.

Example 4: Esterification Yield Calculation

In the preparation of ethyl ethanoate, $6.00\,\mathrm{g}$ of ethanoic acid ($M = 60.05\,\mathrm{g/mol}$) is reacted with excess ethanol in the presence of concentrated $\mathrm{H}_2\mathrm{SO}_4$. After purification, $4.20\,\mathrm{g}$ of ethyl ethanoate ($M = 88.11\,\mathrm{g/mol}$) is obtained. Calculate the percentage yield.

$n(\text{ethanoic acid}) = \frac{6.00}{60.05} = 0.0999\,\mathrm{mol}$

Theoretical moles of ethyl ethanoate (1:1 ratio): $0.0999\,\mathrm{mol}$

Theoretical mass: $0.0999 \times 88.11 = 8.80\,\mathrm{g}$

$\text{Percentage yield} = \frac{4.20}{8.80} \times 100 = 47.7\%$

The yield is less than 50% because esterification is a reversible equilibrium. Using excess ethanol or removing water would improve the yield.

Example 5: Multi-Step Synthesis with Alcohol Intermediates

Target: Propanoic acid from propene.

Step 1: Markovnikov addition of HBr:

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CHBrCH}_3$

Step 2: Substitution with aqueous NaOH (SN2 on a secondary halide -- gives a mixture with some elimination, but substitution predominates in aqueous conditions):

$\mathrm{CH}_3\mathrm{CHBrCH}_3 + \mathrm{NaOH}(aq) \to \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$

Step 3: Oxidation of propan-2-ol with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ under reflux:

$\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3 \xrightarrow{[\mathrm{O}],\,\text{reflux}} \mathrm{CH}_3\mathrm{COCH}_3$

This gives propanone (a ketone), not propanoic acid. Ketones cannot be further oxidised by $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$.

Corrected route: Use anti-Markovnikov addition to get the primary alcohol:

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \xrightarrow{\text{peroxides}} \mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3$

$\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{NaOH}(aq) \to \mathrm{HOCH}_2\mathrm{CH}_2\mathrm{CH}_3$

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}],\,\text{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$

This gives propanoic acid as required. The key was using the anti-Markovnikov addition to obtain the primary alcohol.

Example 6: Lucas Test Interpretation

Three unlabelled bottles contain butan-1-ol, butan-2-ol, and 2-methylpropan-2-ol. Describe the Lucas test results.

Procedure: Add $2\,\mathrm{cm}^3$ of Lucas reagent (anhydrous $\mathrm{ZnCl}_2$ in concentrated $\mathrm{HCl}$) to $1\,\mathrm{cm}^3$ of each alcohol at room temperature.

Results:

Alcohol	Observation	Time	Mechanism
2-methylpropan-2-ol	Immediate cloudiness	$< 10\,\mathrm{s}$	SN1 (tertiary carbocation forms rapidly)
Butan-2-ol	Cloudiness develops	$1$--$5\,\mathrm{min}$	SN1 (secondary carbocation forms more slowly)
Butan-1-ol	No cloudiness	$> 30\,\mathrm{min}$	SN2 is very slow in acidic, non-aqueous medium

The cloudiness is caused by the formation of the insoluble chloroalkane.

Example 7: Conversion of Alcohol to Halogenoalkane -- Reagent Comparison

Convert ethanol to a halogenoalkane using three different reagents.

Method 1: Concentrated HBr (heat under reflux).

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{H}_2\mathrm{O}$

Product: Bromoethane. Mechanism: SN2 (protonated $-\mathrm{OH}_2^+$ is the leaving group).

Method 2: $\mathrm{PBr}_3$ (room temperature).

$3\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{PBr}_3 \to 3\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{H}_3\mathrm{PO}_3$

Product: Bromoethane. Mechanism: SN2 (bromide from $\mathrm{PBr}_3$ is the nucleophile).

Method 3: $\mathrm{SOCl}_2$ (pyridine, room temperature).

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{SOCl}_2 \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl} + \mathrm{SO}_2 + \mathrm{HCl}$

Product: Chloroethane (not bromoethane). Mechanism: SN2. Advantage: gaseous byproducts escape, driving the reaction to completion.

Practical Techniques for Alcohol Chemistry

Required Practical: Oxidation of a Primary Alcohol (AQA RP 7)

Objective: To oxidise ethanol to ethanoic acid and determine the enthalpy change.

Safety: Acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ is toxic, corrosive, and an oxidising agent. Chromium(VI) compounds are carcinogenic. Wear gloves, eye protection, and work in a fume cupboard.

Procedure:

1. Add $20\,\mathrm{cm}^3$ of ethanol to a round-bottom flask.
2. Slowly add $40\,\mathrm{cm}^3$ of acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ solution.
3. Reflux the mixture for 30 minutes. The orange solution turns green.
4. Distil the product, collecting the fraction that boils at $118^\circ\mathrm{C}$ (ethanoic acid).
5. Test the distillate with $\mathrm{Na}_2\mathrm{CO}_3$: effervescence ($\mathrm{CO}_2$) confirms a carboxylic acid.

Purification: The crude ethanoic acid can be purified by fractional distillation.

Distillation vs Reflux -- When to Use Each

Technique	Purpose	Apparatus
Distillation	Separate a volatile product from the reaction mixture	Round-bottom flask, still head, condenser, receiver
Reflux	Heat a reaction mixture without losing volatile components	Round-bottom flask, condenser pointing downward

Rule of thumb: If the product is volatile and you want to isolate it, use distillation. If the reaction needs prolonged heating without loss of reactants or solvent, use reflux.

For the oxidation of primary alcohols:

• To isolate the aldehyde: distil the product as it forms (lower b.p. than the alcohol).
• To produce the carboxylic acid: reflux (keep the aldehyde in the mixture for further oxidation).

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Describe how you would distinguish between butan-1-ol, butan-2-ol, and 2-methylpropan-2-ol using acidified potassium dichromate(VI). State the observations for each.

Mark Scheme:

5 marks:

Butan-1-ol: Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. Orange solution turns green. Product is butanoic acid (primary alcohol oxidises to carboxylic acid under reflux) (2 marks for observation + product).

Butan-2-ol: Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. Orange solution turns green. Product is butanone (secondary alcohol oxidises to ketone) (2 marks for observation + product).

2-methylpropan-2-ol: Reflux with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$. No colour change (solution remains orange). Tertiary alcohols are not oxidised (1 mark for observation).

Q2 (6 marks)

Explain the mechanism for the acid-catalysed dehydration of butan-2-ol to form but-2-ene. Include the role of the acid catalyst.

Mark Scheme:

6 marks:

• Protonation: $\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{H}^+ \to \mathrm{CH}_3\mathrm{CH}(\mathrm{OH}_2^+)\mathrm{CH}_2\mathrm{CH}_3$ (1 mark). The acid catalyst protonates the $-\mathrm{OH}$ group, converting it to $-\mathrm{OH}_2^+$, a good leaving group.
• Loss of water (rate-determining step): Formation of a secondary carbocation (1 mark).
• Deprotonation: Loss of a proton from an adjacent carbon, forming the C=C double bond (1 mark).
• Zaitsev's rule: The major product is but-2-ene (more substituted alkene) rather than but-1-ene (1 mark).
• The acid catalyst is regenerated (1 mark for showing $\mathrm{H}^+$ on the product side).
• Curly arrow mechanism (1 mark).

Q3 (5 marks)

A student prepares ethyl ethanoate from ethanoic acid and ethanol using concentrated $\mathrm{H}_2\mathrm{SO}_4$ as a catalyst. The student obtains a yield of 55%.

(a) Write the equation for the reaction. (1 mark)

(b) Suggest two ways the student could increase the yield. (2 marks)

(c) Explain why the yield can never reach 100% under these conditions. (2 marks)

Mark Scheme:

(a) $\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \rightleftharpoons \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$ (1 mark).

(b) Use excess of one reactant (e.g. ethanol) (1 mark). Remove water as it forms using a Dean-Stark apparatus or molecular sieves (1 mark).

(c) The reaction is an equilibrium, so some reactants always remain (1 mark). At equilibrium, the rates of the forward and reverse reactions are equal (1 mark).

Q4 (4 marks)

Explain why $\mathrm{SOCl}_2$ is preferred over concentrated $\mathrm{HCl}$ for converting an alcohol to a chloroalkane in organic synthesis.

Mark Scheme:

4 marks:

• $\mathrm{SOCl}_2$ produces gaseous byproducts ($\mathrm{SO}_2$ and $\mathrm{HCl}$) that escape from the reaction mixture (1 mark).
• This drives the reaction to completion (Le Chatelier's principle) (1 mark).
• The product is obtained in higher purity without the need for separation from aqueous reagents (1 mark).
• Concentrated $\mathrm{HCl}$ gives an equilibrium (reversible reaction) that requires removal of water to drive to completion (1 mark).

Q5 (5 marks)

Propose a mechanism for the reaction of ethanol with ethanoyl chloride to form ethyl ethanoate. Explain why this reaction is preferred over the acid-catalysed esterification of ethanoic acid with ethanol.

Mark Scheme:

5 marks:

• Nucleophilic attack: The oxygen lone pair of ethanol attacks the electrophilic carbonyl carbon of ethanoyl chloride (1 mark).
• Tetrahedral intermediate forms as the $\pi$ electrons move onto the oxygen (1 mark).
• The $\mathrm{Cl}^-$ is expelled as the C=O reforms (1 mark).
• $\mathrm{Cl}^-$ removes a proton from the $-\mathrm{OH}_2^+$ group, yielding ethyl ethanoate and $\mathrm{HCl}$ (1 mark).
• This reaction is preferred because it is irreversible (no equilibrium) and proceeds rapidly at room temperature without a catalyst, giving a higher yield (1 mark).

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tip

Diagnostic Test Ready to test your understanding of Alcohols? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Alcohols with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.