Alcohols (Full) PDF

# The alcohol homologous series Alcohols contain the -OH functional group, known as the hydroxyl group. The hydroxyl group is responsible for both the physical and chemical properties of the alcohols. The simplest alcohol, methanol, $CH_3OH$, is used as a high-performance fuel because of its effic...

# The alcohol homologous series Alcohols contain the -OH functional group, known as the hydroxyl group. The hydroxyl group is responsible for both the physical and chemical properties of the alcohols. The simplest alcohol, methanol, $CH_3OH$, is used as a high-performance fuel because of its efficient combustion. Methanol is also an important chemical feedstock - the starting material in many industrial syntheses. It can be converted into polymers, paints, solvents, insulation, adhesives, and many other useful products. The second member of the alcohol homologous series, ethanol, is used primarily in alcoholic drinks and as a fuel, and also finds use as a solvent and a feedstock. ## Naming alcohols You will recall that the suffix -ol is added to the stem name of the longest carbon chain. The position of the alcohol functional group in the chain is indicated using a number. # Physical properties When you compare the physical properties of alcohols with alkanes of the same number of carbon atoms (e.g., methanol and methane), there are some interesting differences. - Alcohols are less volatile, - have higher melting points, and - greater water solubility than the corresponding alkanes. The differences become much smaller as the length of the carbon chain increases. These differences can be explained by considering the polarity of the bonds in both the alkanes and alcohols, and the effect that these bonds have on the strength of the intermolecular forces. - The alkanes have non-polar bonds because the electronegativity of hydrogen and carbon are very similar. - The alkane molecules are therefore non-polar. - The intermolecular forces between non-polar molecules are very weak London forces. - Alcohols have a polar O-H bond because of the difference in electronegativity of the oxygen and hydrogen atoms. - Alcohol molecules are therefore polar. - The intermolecular forces will be very weak London forces but there will also be much stronger hydrogen bonds between the polar O-H groups. ## Volatility and boiling points The difference in the boiling points of the alkanes and alcohols can be seen in figure 3. In the liquid state, intermolecular hydrogen bonds hold the alcohol molecules together. These bonds must be broken in order to change the liquid alcohol into a gas. This requires more energy than overcoming the weaker London forces in alkanes, so alcohols have a lower volatility than the alkanes with the same number of carbon atoms. ## Solubility in water A compound that can form hydrogen bonds with water is far more water-soluble than a compound that cannot. Alkanes are non-polar molecules and cannot form hydrogen bonds with water. Alcohols such as methanol and ethanol are completely soluble in water, as hydrogen bonds form between the polar -OH group of the alcohol and the water molecules (figure 4). As the hydrocarbon chain increases in size, the influence of the -OH group becomes relatively smaller, and the solubility of longer-chain alcohols becomes more like that of hydrocarbons - solubility decreases. # Classifying alcohols Alcohols can be classified as primary, secondary, or tertiary. This classification depends on the number of hydrogen atoms and alkyl groups attached to the carbon atom that contains the alcohol functional group. ## Primary alcohols Methanol and ethanol, the two simplest alcohols in the alcohols homologous series, are both primary alcohols. In a primary alcohol the -OH group is attached to a carbon atom that is attached to two hydrogen atoms and one alkyl group. (Methanol, with three hydrogen atoms and no carbon atoms attached, is an exception that is still classified as a primary alcohol.) ## Secondary alcohols In a secondary alcohol the -OH group is attached to a carbon atom that is attached to one hydrogen atom and two alkyl groups. Propan-2-ol and pentan-3-ol are both examples of secondary alcohols. ## Tertiary alcohols In a tertiary alcohol the -OH group is attached to a carbon atom that is attached to no hydrogen atoms and three alkyl groups. 2-Methylpropan-2-ol and 2-methylbutan-2-ol are both examples of tertiary alcohols. # Combustion of alcohols Alcohols burn completely in a plentiful supply of oxygen to produce carbon dioxide and water. The equation shows the complete combustion of ethanol. $C_2H_5OH(l) + 3O_2(g) \longrightarrow 2CO_2(g) + 3H_2O(l)$ The reaction is exothermic, releasing a large quantity of energy in the form of heat. As the number of carbon atoms in the alcohol chain increases the quantity of heat released per mole also increases. # Oxidation of alcohols Primary and secondary alcohols can be oxidised by an oxidising agent. The usual oxidising mixture is a solution of potassium dichromate (VI), $ K_2Cr_2O_7 $, acidified with dilute sulfuric acid, $H_2SO_4$. If the alcohol is oxidised, the orange solution containing dichromate (VI) ions is reduced to a green solution containing chromium(III) ions (figure 1). $Cr_2O_7^{2-} \longrightarrow Cr^{3+}$ dichromate (VI) ions chromium(III) ions ## Oxidation of primary alcohols Primary alcohols can be oxidised to either aldehydes or carboxylic acids. The product of the oxidation depends on the reaction conditions used because aldehydes are themselves also oxidised to carboxylic acids. ### Preparation of aldehydes On gentle heating of primary alcohols with acidified potassium dichromate, an aldehyde is formed. To ensure that the aldehyde is prepared rather than the carboxylic acid, the aldehyde is distilled out of the reaction mixture as it forms. This prevents any further reaction with the oxidising agent. The dichromate (VI) ions change colour from orange to green, as shown in figure 1. The equation in figure 2 shows the oxidation of butan-1-ol, a primary alcohol, to form the aldehyde butanal, using acidified potassium dichromate (VI). Notice that [O] is used to indicate the oxidising agent. ### Preparation of carboxylic acids If a primary alcohol is heated strongly under reflux, with an excess of acidified potassium dichromate (VI), a carboxylic acid is formed. Use of an excess of the acidified potassium dichromate (VI) ensures that all of the alcohol is oxidised. Heating under reflux ensures that any aldehyde formed initially in the reaction also undergoes oxidation to the carboxylic acid. You can see that the conditions of the oxidation of a primary alcohol, such as whether a reagent in excess the conditions and the technique used, influence the product formed. - When preparing the aldehyde, use distillation to remove the aldehyde from the reaction mixture. - When preparing the carboxylic acid, heat the alcohol under reflux. ## Oxidation of secondary alcohols Secondary alcohols are oxidised to ketones. It is not possible to further oxidise ketones using acidified dichromate (VI) ions. To ensure the reaction goes to completion, the secondary alcohol is heated under reflux with the oxidising mixture. The dichromate (VI) ions once again change colour from orange to green. The oxidation of propan-2-ol to propanone is shown in figure 4. ## Oxidation of tertiary alcohols Tertiary alcohols do not undergo oxidation reactions. The acidified dichromate (VI) remains orange when added to a tertiary alcohol. # Dehydration of alcohols Dehydration is any reaction in which a water molecule is removed from the starting material. An alcohol is heated under reflux in the presence of an acid catalyst such as concentrated sulfuric acid, $H_2SO_4$, or concentrated phosphoric acid, $H_3PO_4$. The product of the reaction is an alkene. Dehydration of an alcohol is an example of an elimination reaction. Alcohols react with hydrogen halides to form haloalkanes. When preparing a haloalkane, the alcohol is heated under reflux with sulfuric acid and a sodium halide the hydrogen bromide is formed in situ (in place). $NaBr(s) + H_2SO_4(aq) \longrightarrow NaHSO_4(aq) + HBr(aq)$ The HBr formed reacts with the alcohol to produce the haloalkane.