Unit 11: Alcohols (Alkanols)

Alcohols (or alkanols) are a homologous series of organic compounds containing the hydroxyl (-OH) functional group attached to a saturated carbon atom. Their general formula is $C_nH_{2n+1}OH$.

The presence of the polar -OH group allows alcohol molecules to form hydrogen bonds with each other and with water molecules. This has a profound effect on their physical properties:

• High Boiling Points: Alcohols have significantly higher boiling points than alkanes of similar molar mass. This is because hydrogen bonds are much stronger than the van der Waals forces that are the only intermolecular forces between alkane molecules.
• Solubility in Water: Short-chain alcohols (like methanol, ethanol) are completely soluble in water because the hydroxyl group can form hydrogen bonds with water molecules. However, as the non-polar hydrocarbon chain gets longer, the solubility decreases because the chain disrupts the hydrogen bonding in water.

Ethanol, the most common alcohol, is produced by two main industrial methods:

1. Fermentation

This is a low-technology, anaerobic process where enzymes in yeast break down glucose (from sources like sugarcane or corn) into ethanol and carbon dioxide.
Equation: $C_6H_{12}O_6(aq) \xrightarrow{yeast} 2C_2H_5OH(aq) + 2CO_2(g)$
Conditions: Absence of air, temperature of 35-55°C, and the presence of yeast.
Advantages: Uses renewable resources and low energy.
Disadvantages: It is a slow, batch process that produces an impure aqueous solution of ethanol which then needs to be purified by distillation.

2. Hydration of Ethene

This is a high-technology, continuous flow process where ethene (from crude oil) reacts with steam.
Equation: $C_2H_4(g) + H_2O(g) \rightleftharpoons C_2H_5OH(g)$
Conditions: High temperature (300°C), high pressure (60 atm), and a concentrated phosphoric(V) acid catalyst.
Advantages: It is a fast, continuous process that produces pure ethanol.
Disadvantages: It is energy-intensive and relies on a non-renewable resource (crude oil).

Alcohols are classified as primary ($1^\circ$), secondary ($2^\circ$), or tertiary ($3^\circ$) depending on the number of carbon atoms directly bonded to the carbon atom that carries the -OH group. This classification is important because it determines their reactivity, especially towards oxidation.

• Primary ($1^\circ$) Alcohols: The carbon atom with the -OH group is attached to one or zero other carbon atoms. The -OH group is at the end of a chain.
  Example: Propan-1-ol, $CH_3CH_2CH_2OH$.
• Secondary ($2^\circ$) Alcohols: The carbon atom with the -OH group is attached to two other carbon atoms. The -OH group is within the chain.
  Example: Propan-2-ol, $CH_3CH(OH)CH_3$.
• Tertiary ($3^\circ$) Alcohols: The carbon atom with the -OH group is attached to three other carbon atoms. The -OH group is at a branch point.
  Example: 2-methylpropan-2-ol, $(CH_3)_3COH$.

Reaction with Sodium

Alcohols react with reactive metals like sodium in a redox reaction. The sodium is oxidised, and the hydrogen from the hydroxyl group is reduced, producing hydrogen gas and a salt called a sodium alkoxide. The reaction is less vigorous than the reaction of sodium with water.
Equation: $2CH_3CH_2OH(l) + 2Na(s) \rightarrow 2CH_3CH_2ONa(s) + H_2(g)$
The product, $CH_3CH_2ONa$, is named sodium ethoxide.

Dehydration (Elimination)

Alcohols can be dehydrated to form alkenes by heating them with a catalyst, typically concentrated sulphuric acid ($H_2SO_4$) or hot aluminium oxide ($Al_2O_3$). This is an elimination reaction where a water molecule is removed from the alcohol.
Equation: $CH_3CH_2CH_2OH \xrightarrow{conc. H_2SO_4, heat} CH_3CH=CH_2 + H_2O$

If there are multiple possible positions for the double bond, a mixture of products can be formed.

The iodoform reaction is a specific chemical test used to identify the presence of a particular structural feature in alcohols and other compounds.

The reaction involves warming the substance with iodine ($I_2$) and sodium hydroxide solution ($NaOH(aq)$). A positive result is the formation of a pale yellow precipitate of triiodomethane ($CHI_3$), also known as iodoform.

An alcohol will give a positive iodoform test if it contains the following structural unit:

The oxidation of alcohols is a key reaction that differs based on the classification of the alcohol. A common oxidising agent used is acidified potassium dichromate(VI) ($K_2Cr_2O_7 / H_2SO_4$), which changes colour from orange to green upon reduction. Another is acidified potassium manganate(VII) ($KMnO_4 / H_2SO_4$), which changes from purple to colourless.

• Primary Alcohols: Can be oxidised in two stages.
  1. Gentle heating with the oxidising agent and immediate distillation of the product yields an aldehyde.
     $RCH_2OH + [O] \rightarrow RCHO + H_2O$
  2. Stronger heating under reflux (to prevent the aldehyde from escaping) with excess oxidising agent yields a carboxylic acid.
     $RCH_2OH + 2[O] \rightarrow RCOOH + H_2O$
• Secondary Alcohols: Are oxidised to form a ketone. Ketones are resistant to further oxidation.
  $RCH(OH)R' + [O] \rightarrow RCOR' + H_2O$
• Tertiary Alcohols: Are resistant to oxidation because the carbon atom bearing the -OH group has no hydrogen atoms to be removed. They do not react with standard oxidising agents.

This difference in reactivity provides a chemical test to distinguish between the classes of alcohols.