Introduction to Alcohols
Welcome to the study of alcohols! You have likely encountered alcohols in everyday life, from ethanol in hand sanitizers and fuels to the isopropyl alcohol used in cleaning wipes. In organic chemistry, alcohols are one of the most versatile functional groups because they act as chemical "crossroads," allowing us to convert them into alkenes, halogenoalkanes, aldehydes, ketones, and carboxylic acids.
Don't worry if organic reactions feel a bit overwhelming at first. We will break down their structures, physical properties, and chemical reactions step-by-step.
1. Structure and Classification of Alcohols
Alcohols contain the hydroxyl functional group (\(-\text{OH}\)) attached to a saturated carbon atom. The general formula for a homologous series of aliphatic, saturated alcohols is \(\text{C}_n\text{H}_{2n+1}\text{OH}\) (or \(\text{R}-\text{OH}\)).
Classification of Alcohols
Just like halogenoalkanes, alcohols are classified into three types depending on the environment of the carbon atom attached to the \(-\text{OH}\) group:
1. Primary (\(1^\circ\)) Alcohols:
The carbon carrying the \(-\text{OH}\) group is attached to one other alkyl group (or none, as in methanol, \(\text{CH}_3\text{OH}\)).
Example: Ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\), and Propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\).
2. Secondary (\(2^\circ\)) Alcohols:
The carbon carrying the \(-\text{OH}\) group is directly attached to two other alkyl groups.
Example: Propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\).
3. Tertiary (\(3^\circ\)) Alcohols:
The carbon carrying the \(-\text{OH}\) group is directly attached to three other alkyl groups.
Example: 2-Methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\).
Memory Tip: To quickly classify an alcohol, circle the \(\text{C}-\text{OH}\) carbon and count how many carbons are directly connected to it. 1 carbon = Primary, 2 carbons = Secondary, 3 carbons = Tertiary!
Key Takeaway: The classification (\(1^\circ\), \(2^\circ\), or \(3^\circ\)) determines how an alcohol behaves when it is oxidised.
2. Physical Properties of Alcohols
Boiling Points
Alcohols have significantly higher boiling points than alkanes of similar relative molecular mass (\(M_r\)).
Why? In alkanes, only weak van der Waals forces (London dispersion forces) exist between molecules. In alcohols, the oxygen atom is much more electronegative than hydrogen, creating a polar bond: \(-\text{O}^{\delta-}-\text{H}^{\delta+}\). This allows alcohols to form strong hydrogen bonds between molecules in addition to van der Waals forces. Because hydrogen bonds require much more thermal energy to overcome, the boiling points are higher.
Solubility in Water
Small-chain alcohols (such as methanol, ethanol, and propan-1-ol) are completely miscible (soluble) in water.
Why? The \(-\text{OH}\) group of the alcohol can form hydrogen bonds with water (\(\text{H}_2\text{O}\)) molecules.
However, as the non-polar hydrocarbon chain lengthens (e.g., in hexan-1-ol), the solubility in water decreases rapidly. The long non-polar carbon chain disrupts the hydrogen bonding network of water without forming strong interactions of its own.
Key Takeaway: Hydrogen bonding explains both the unexpectedly high boiling points of alcohols and why small alcohols dissolve easily in water.
3. Reactions of Alcohols
A. Combustion
Alcohols burn cleanly in a plentiful supply of oxygen to produce carbon dioxide and water:
\(\text{CH}_3\text{CH}_2\text{OH} + 3\text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O}\)
Due to this clean combustion and high energy output, ethanol is widely used as a renewable biofuel.
B. Substitution Reactions (Forming Halogenoalkanes)
Alcohols can undergo nucleophilic substitution where the \(-\text{OH}\) group is replaced by a halogen atom (\(-\text{Cl}\), \(-\text{Br}\), or \(-\text{I}\)).
1. Chlorination using Phosphorus Pentachloride (\(\text{PCl}_5\)):
Adding solid \(\text{PCl}_5\) to an alcohol at room temperature produces a chloroalkane, phosphoryl chloride, and hydrogen chloride gas:
\(\text{CH}_3\text{CH}_2\text{OH} + \text{PCl}_5 \to \text{CH}_3\text{CH}_2\text{Cl} + \text{POCl}_3 + \text{HCl(g)}\)
Observation: Steamy/misty white fumes of \(\text{HCl}\) are evolved (which turn damp blue litmus paper red). This is a definitive chemical test for the presence of the \(-\text{OH}\) group.
2. Bromination using \(\text{KBr}\) and concentrated \(\text{H}_2\text{SO}_4\):
Hydrogen bromide (\(\text{HBr}\)) is generated in situ by reacting potassium bromide with concentrated sulfuric acid (\(50\%\)):
\(\text{KBr} + \text{H}_2\text{SO}_4 \to \text{KHSO}_4 + \text{HBr}\)
The \(\text{HBr}\) then reacts with the alcohol under reflux:
\(\text{CH}_3\text{CH}_2\text{OH} + \text{HBr} \to \text{CH}_3\text{CH}_2\text{Br} + \text{H}_2\text{O}\)
3. Iodination using Red Phosphorus and Iodine:
Phosphorus triiodide (\(\text{PI}_3\)) is produced in situ by heating red phosphorus and iodine together (\(2\text{P} + 3\text{I}_2 \to 2\text{PI}_3\)). The alcohol is then warmed with \(\text{PI}_3\):
\(3\text{CH}_3\text{CH}_2\text{OH} + \text{PI}_3 \to 3\text{CH}_3\text{CH}_2\text{I} + \text{H}_3\text{PO}_3\)
C. Elimination (Dehydration) to form Alkenes
An alcohol can lose a molecule of water to form an alkene. This reaction is known as dehydration (or elimination).
Conditions: Heat with concentrated sulfuric acid (\(\text{conc. H}_2\text{SO}_4\)) or pass alcohol vapour over heated aluminium oxide (\(\text{Al}_2\text{O}_3\)) catalyst.
\(\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \text{heat}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}\)
Note: For unsymmetrical alcohols like butan-2-ol, dehydration can yield a mixture of positional isomers (e.g., but-1-ene and but-2-ene) as well as \(E/Z\) stereoisomers.
Key Takeaway: Alcohols undergo substitution to make halogenoalkanes and elimination to make alkenes.
4. Oxidation of Alcohols
Oxidation is one of the most important reactions in AS Chemistry. The standard oxidising mixture used is acidified potassium dichromate(\(\text{VI}\)), written as \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\) (or \(\text{Cr}_2\text{O}_7^{2-} / \text{H}^+\)).
In equations, the oxidising agent is represented simply by the symbol \([\text{O}]\).
Colour Change:
When oxidation occurs, the orange dichromate(\(\text{VI}\)) ions (\(\text{Cr}_2\text{O}_7^{2-}\)) are reduced to green chromium(\(\text{III}\)) ions (\(\text{Cr}^{3+}\)):
Orange \(\to\) Green
1. Oxidation of Primary (\(1^\circ\)) Alcohols
Primary alcohols can be oxidised in two distinct stages:
Stage 1: Partial Oxidation to an Aldehyde
\(\text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \xrightarrow{\text{distillation}} \text{CH}_3\text{CHO} + \text{H}_2\text{O}\)
Technique: Distillation with immediate removal. Aldehydes have lower boiling points than alcohols because they cannot form intermolecular hydrogen bonds. By distilling immediately, the volatile aldehyde boils off before it can be oxidised further.
Stage 2: Complete Oxidation to a Carboxylic Acid
\(\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \xrightarrow{\text{reflux}} \text{CH}_3\text{COOH} + \text{H}_2\text{O}\)
Technique: Heating under Reflux with excess oxidising agent. Reflux prevents volatile vapours from escaping by condensing them back into the reaction flask, ensuring complete oxidation to the carboxylic acid.
2. Oxidation of Secondary (\(2^\circ\)) Alcohols
Secondary alcohols are oxidised to ketones under reflux:
\(\text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \xrightarrow{\text{reflux}} \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O}\)
Propan-2-ol is converted into propanone. Ketones cannot be easily oxidised further because that would require breaking a stable \(\text{C}-\text{C}\) bond.
3. Oxidation of Tertiary (\(3^\circ\)) Alcohols
Tertiary alcohols cannot be oxidised by acidified potassium dichromate(\(\text{VI}\)).
Observation: The solution remains orange (no colour change).
Why? The carbon atom bonded to the \(-\text{OH}\) group does not have a hydrogen atom attached to it (no \(\alpha\)-hydrogen) to be removed during oxidation.
Summary: Oxidation Pathways
• Primary alcohol \(\xrightarrow{[\text{O}], \text{distil}}\) Aldehyde \(\xrightarrow{[\text{O}], \text{reflux}}\) Carboxylic Acid
• Secondary alcohol \(\xrightarrow{[\text{O}], \text{reflux}}\) Ketone
• Tertiary alcohol \(\xrightarrow{[\text{O}]}\) No reaction (remains orange)
Common Pitfalls to Avoid
1. Forgetting water in oxidation equations: Remember that oxidising an alcohol to an aldehyde or ketone always produces \(\text{H}_2\text{O}\) as a co-product alongside the organic molecule.
2. Distillation vs. Reflux: Make sure you specify distillation to obtain an aldehyde from a primary alcohol, and reflux to obtain a carboxylic acid.
3. Incorrect reagent formula: Always specify acidified potassium dichromate(\(\text{VI}\)) or write both \(\text{K}_2\text{Cr}_2\text{O}_7\) and \(\text{H}_2\text{SO}_4\). Just writing potassium dichromate alone is not enough!