Alcohols

The alcohols have relatively low volatility due to their ability to form hydrogen bond between alcohol molecules. Uses of alcohols Ethanol is ‘alcohol’ in alcoholic drinks. Ethanol is commonly used as a solvent in the form of methylated spirits. Methanol is used as a petrol additive to improve combustion and is increasing important as a feedstock in the production of organic chemicals; General formula alcohols CnH2n+1OH Classifying Alcohols C O H H H C H H C H H H C H H C H C H H H O H H C C H C H H H H H C H H O H H Propan-1-ol Primary Tertiary alcohols are alcohols where 3 carbon are attached to the carbon adjoining the oxygen Propan-2-ol Secondary methylpropan-2-ol Tertiary Primary alcohols are alcohols where 1 carbon is attached to the carbon adjoining the oxygen Secondary alcohols are alcohols where 2 carbon are attached to the carbon adjoining the oxygen Reactions of alcohols Complete Combustion CH3CH2OH (l) + 3 O2 (g)  2CO2 (g) + 3 H2O(l) The products of complete combustion are CO2 and H2O. In excess oxygen alcohols will burn with complete combustion. The smaller alcohols (up to 3 carbons) are soluble in water because they can form hydrogen bonds with water. The longer the hydrocarbon chain the less soluble the alcohol.

Oxidation reactions of the alcohols Potassium dichromate K2Cr2O7 is an oxidising agent that causes alcohols to oxidise. The exact reaction, however, depends on the type of alcohol, i.e. whether it is primary, secondary, or tertiary, and on the conditions. Partial Oxidation of Primary Alcohols Reaction: primary alcohol  aldehyde Reagent: potassium dichromate (VI) solution and dilute sulphuric acid. Conditions: (use a limited amount of dichromate) warm gently and distil out the aldehyde as it forms: C O C H H H H An aldehyde’s name ends in –al It always has the C=O bond on the first carbon of the chain so it does not need Ethanal an extra number Observation: the orange dichromate ion (Cr2O7 2-) reduces to the green Cr 3+ ion Write the oxidation equations in a simplified form using [O] which represents O from the oxidising agent When writing the formulae of aldehydes in a condensed way write CHO and not COH e.g.CH3CH2CHO N Goalby chemrevise.org 2 propan-1-ol propanal + [O] + H2O CH3CH2CH2OH + [O]  CH3CH2CHO + H2O OH + [O] O + H2O C O C H H C H H H H C O H H H C H H C H H H  Full Oxidation of Primary Alcohols Reaction: primary alcohol  carboxylic acid Reagent: potassium dichromate(VI) solution and dilute sulphuric acid Conditions: use an excess of dichromate, and heat under reflux: (distill off product after the reaction has finished) C C O O H H C H H H H Propanoic acid propan-1-ol Propanoic acid + 2 [O] + H2O Observation: the orange dichromate ion (Cr2O7 2-) reduces to the green Cr 3+ ion CH3CH2CH2OH + 2[O]  CH3CH2COOH + H2O C O H H H C H H C H H H C C O H O H C H H H H OH + 2[O] + H2O O OH Oxidation of Secondary Alcohols Reaction: secondary alcohol  ketone Reagent: potassium dichromate(VI) solution and dilute sulphuric acid. Conditions: heat under reflux C C C O H H H H H H Propanone Ketones end in -one When ketones have 5C’s or more in a chain then it needs a number to show the position of the double bond. E.g. pentan-2-one propan-2-ol Propanone C H H C H C H H H O H H C C C O H H H H H H + [O] + H2O There is no further oxidation of the ketone under these conditions. Observation: the orange dichromate ion (Cr2O7 2-) reduces to the green Cr 3+ ion Tertiary alcohols cannot be oxidised at all by potassium dichromate: This is because there is no hydrogen atom bonded to the carbon with the OH group

Reaction of alcohols with Dehydrating agents Dehydration Reaction: removal of a water molecule from a molecule Reaction: Alcohol  Alkene Reagents: Concentrated Sulphuric or Phosphoric acids Conditions: warm (under reflux) Role of reagent: dehydrating agent/catalyst Type of reaction: acid catalysed elimination C C + H2O H H H C H H H C O H H H C H H C H H H propan-1-ol Propene Some 2o and 3o alcohols can give more than one product, when the double bond forms between different carbon atoms C H C H C H H H C H H H O H H C C C C H H H H H H H H C H H C C C H H H H H H Butan-2-ol can form both alkenes although more but-2-ene would be formed Butan-2-ol But-1-ene But-2-ene But-2-ene could also exist as E and Z isomers

Substitution reactions of Alcohols to form Haloalkanes A mixture of a halide ions with concentrated acid NaCl + H2SO4 can used for substituting a halogen on to an alcohol CH3CH2OH + HCl  CH3CH2Cl + H2O Various other halogenating compounds can be used to substitute the –OH group for a halogen NaCl + H2SO4  NaHSO4 + HCl Reaction: Alcohol  Haloalkane Reagents: Concentrated Sulphuric and sodium halide