Alcohols

Alcohols contain the hydroxy group, $\ce{-OH}$, bonded to a saturated carbon atom. Ethanol is the best known: a solvent, a fuel and the alcohol in drinks. Alcohols sit at the centre of the AS organic map, because they can be made from alkenes, halogenoalkanes, carbonyl compounds, carboxylic acids and esters, and converted into halogenoalkanes, alkenes, carbonyl compounds, carboxylic acids and esters. This note covers how alcohols are made, their physical properties, and their reactions with sodium, with halogenating agents, by dehydration and by esterification, together with the acidity of alcohols compared with water. Their oxidation, which distinguishes primary, secondary and tertiary alcohols, is covered in Oxidation of alcohols.

Structure and classification

The general formula of the alcohols is $\ce{C_{n}H_{2n+1}OH}$. The O–H bond is polar ($\ce{O^{\delta-}-H^{\delta+}}$) and so is the C–O bond ($\ce{C^{\delta+}-O^{\delta-}}$), and the oxygen carries two lone pairs. The C–O–H bond angle is about $104.5^\circ$.

class	OH carbon bonded to	example
primary	one carbon (methanol: none)	$\ce{CH3CH2CH2OH}$ propan-1-ol
secondary	two carbons	$\ce{CH3CH(OH)CH3}$ propan-2-ol
tertiary	three carbons	$\ce{(CH3)3COH}$ 2-methylpropan-2-ol

Physical properties: hydrogen bonding

Alcohols form hydrogen bonds between molecules: the $\delta+$ hydrogen of one O–H is attracted to a lone pair on the oxygen of another molecule.

compound	$M_r$	boiling point / $^\circ\text{C}$	strongest intermolecular force
propane, $\ce{CH3CH2CH3}$	44	$-42$	instantaneous dipole–induced dipole
ethanol, $\ce{CH3CH2OH}$	46	78	hydrogen bonding

• High boiling points. Hydrogen bonds are much stronger than instantaneous dipole–induced dipole forces, so more energy is needed to separate alcohol molecules. Ethanol boils $120\ ^\circ\text{C}$ higher than propane despite a similar $M_r$.
• Boiling point rises along the series (methanol $65\ ^\circ\text{C}$, ethanol $78\ ^\circ\text{C}$, propan-1-ol $97\ ^\circ\text{C}$, butan-1-ol $117\ ^\circ\text{C}$), because the instantaneous dipole–induced dipole forces increase with the number of electrons.
• Solubility in water. Short-chain alcohols (methanol, ethanol, propanol) mix with water in all proportions, because they form hydrogen bonds with water molecules. As the hydrocarbon chain gets longer, more of the molecule cannot hydrogen bond, so solubility falls: hexan-1-ol is only slightly soluble.

Making alcohols

Some notes on the reducing agents:

• Sodium tetrahydridoborate(III), $\ce{NaBH4}$, is mild and safe. It is used in aqueous or alcoholic solution and reduces aldehydes and ketones, but not carboxylic acids or esters.
• Lithium tetrahydridoaluminate(III), $\ce{LiAlH4}$, is much more powerful. It reduces aldehydes, ketones and carboxylic acids. It reacts violently with water, so it is used in dry ether, with water or dilute acid added afterwards.
• Both act as a source of hydride ions, $\ce{H-}$, which are nucleophiles; in equations they are written as $\ce{[H]}$.

Industrially, ethanol is made either by hydration of ethene (from crude oil) or by fermentation of sugars (a renewable route, but slower and giving a dilute solution that must be distilled).

Reactions of alcohols

Combustion

Alcohols burn with a clean, almost colourless (pale blue) flame. Ethanol is used as a fuel, often blended with petrol. Balancing: for $\ce{C_{n}H_{2n+1}OH}$, remember the oxygen already in the alcohol.

$$\ce{2CH3CH2CH2OH + 9O2 -> 6CO2 + 8H2O}$$

Reaction with sodium

Sodium reacts with the O–H bond. The hydrogen atom is replaced by sodium, and hydrogen gas is released:

$$\ce{2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2O^-Na+ + H2(g)}$$

The product, sodium ethoxide, is an ionic compound containing the ethoxide ion, $\ce{CH3CH2O-}$. Evaporating the excess ethanol leaves a white solid.

Observations: effervescence (bubbles of hydrogen, which burns with a squeaky pop), the sodium sinks (ethanol is less dense than sodium; in water, sodium floats) and slowly disappears, and the mixture gets warm. The reaction is much gentler than sodium with water.

Any compound with an O–H group reacts with sodium this way, including water and carboxylic acids. This is used as a test for an OH group: no reaction means no O–H (for example in an ether or ester).

The acidity of alcohols compared with water

Both water and alcohols can lose $\ce{H+}$ from an O–H group, so both are (very weak) acids:

$$\ce{H2O <=> OH- + H+}$$ $$\ce{CH3CH2OH <=> CH3CH2O- + H+}$$

A consequence: the ethoxide ion is a stronger base than the hydroxide ion, because it attracts $\ce{H+}$ more strongly.

Substitution to halogenoalkanes

The OH group can be replaced by a halogen atom using any of the reagents in Halogenoalkanes. The reaction with $\ce{PCl5}$ at room temperature gives steamy white fumes of HCl, a test for an OH group:

$$\ce{CH3CH2CH2OH + PCl5 -> CH3CH2CH2Cl + POCl3 + HCl}$$

Dehydration to alkenes

Removing a water molecule from an alcohol gives an alkene. The OH comes off one carbon and an H comes off a neighbouring carbon:

$$\ce{CH3CH2CH2OH ->[Al2O3][heat] CH3CH=CH2 + H2O}$$

Conditions: pass the alcohol vapour over heated aluminium oxide, or heat the alcohol with excess concentrated sulfuric acid (about $170\ ^\circ\text{C}$ for ethanol) or concentrated phosphoric acid. Phosphoric acid gives a cleaner product, because concentrated sulfuric acid also oxidises some of the alcohol and chars it. When the OH is in the middle of an unsymmetrical chain, a mixture of alkenes forms (see Alkenes).

Esterification

An alcohol reacts with a carboxylic acid, when heated with a few drops of concentrated sulfuric acid as catalyst, to form an ester and water. This is a condensation reaction and is reversible:

$$\ce{CH3COOH + CH3CH2OH <=> CH3COOCH2CH3 + H2O}$$

Ethanoic acid and ethanol give ethyl ethanoate. Details, naming and hydrolysis are in Esters.

Worked examples

Practice

Answers

1. (a) Steam, phosphoric acid catalyst, about $300\ ^\circ\text{C}$ and $6\ \text{MPa}$. (b) NaOH(aq), heat under reflux. (c) $\ce{NaBH4}$ (or $\ce{LiAlH4}$ in dry ether). (d) Cold, dilute, acidified potassium manganate(VII).
2. Ethanol molecules form hydrogen bonds with each other (between the $\delta+$ H of O–H and a lone pair on another O). Propane has only instantaneous dipole–induced dipole forces. Hydrogen bonds are much stronger, so far more energy is needed to separate ethanol molecules.
3. $\ce{2CH3CH2CH2OH + 2Na -> 2CH3CH2CH2O^-Na+ + H2}$. Sodium propoxide.
4. The ethyl group is electron-donating, so it increases the negative charge density on the oxygen of the ethoxide ion, making $\ce{CH3CH2O-}$ less stable than $\ce{OH-}$. Ethanol is therefore less likely to lose $\ce{H+}$ than water.
5. Pass the vapour over heated $\ce{Al2O3}$ (or heat with excess concentrated $\ce{H2SO4}$ or $\ce{H3PO4}$). $\ce{CH3CH(OH)CH3 -> CH3CH=CH2 + H2O}$.
6. Pent-1-ene (H removed from C1), and pent-2-ene (H removed from C3) as both cis-pent-2-ene and trans-pent-2-ene.
7. $\ce{CH3CH2COOH + CH3OH <=> CH3CH2COOCH3 + H2O}$; heat with a few drops of concentrated sulfuric acid; methyl propanoate.
8. $\ce{C4H9OH + 6O2 -> 4CO2 + 5H2O}$ (oxygen: right side $8 + 5 = 13$; left $1 + 12 = 13$).
9. (a) Either works; $\ce{NaBH4}$ is safer. $\ce{CH3COCH2CH3 + 2[H] -> CH3CH(OH)CH2CH3}$. (b) $\ce{LiAlH4}$ (in dry ether) is needed. $\ce{CH3CH2CH2COOH + 4[H] -> CH3CH2CH2CH2OH + H2O}$.
10. W has no O–H (no reaction with Na) and is reduced by $\ce{NaBH4}$, so it is an aldehyde or ketone, $\ce{C3H6O}$. X, $\ce{C3H8O}$, is an alcohol (fizzes with Na). X dehydrates to propene, so it is propan-1-ol or propan-2-ol. Steam adds to propene mainly by Markovnikov's rule to give propan-2-ol, which is Y; Y is an isomer of X, so X is propan-1-ol, $\ce{CH3CH2CH2OH}$. A primary alcohol comes from reducing an aldehyde, so W is propanal, $\ce{CH3CH2CHO}$. Y is propan-2-ol, $\ce{CH3CH(OH)CH3}$.