Access to Science - Chemistry Alcohols

Learning aims Classify alcohols as primary, secondary or tertiary and predict the products of oxidation. Show how alkenes can be formed from alcohols by acid-catalysed elimination reactions (mechanism not required). Use a simple chemical test, such as Fehling’s solution of Tollen’s reagent, to distinguish between aldehydes and ketones.

Types of alcohols (1) C H OH CH3OH methanol primary C H OH CH3CH2OH Primary alcohols up to 1 other carbon attached to the carbon that bonds to the OH group

Types of alcohols (2) C CH3 H OH propan-2-ol secondary C CH3 OH Secondary alcohols have 2 other carbons attached to the carbon which bonds to the OH C CH3 OH methylpropan-2-ol tertiary Tertiary alcohols have 3 other carbons attached to the carbon which bonds to the OH

Isomers of C4H9OH 1 C H OH CH3CH2CH2 butan-1-ol primary C H CH3 CH3CH2 secondary

Isomers of C4H9OH 2 C H OH CH3CH CH3 2-methylpropan-1-ol primary C CH3 tertiary

Preparation of alcohols CH3CH2Br + NaOH CH3CH2OH + NaBr bromoethane ethanol CH3CHCH3 + NaOH Br CH3CHCH3 + NaBr OH 2-bromopropane propan-2-ol What sort of reaction is this?

Put the diagram in at the bottom of p2 just to remind you! OXIDATION A simple definition of oxidation is the addition of oxygen or the removal of hydrogen. Alcohols are oxidised to differing extents depending on their structural types. The alcohol is warmed with potassium or sodium dichromate(VI), K2Cr2O7 or Na2Cr2O7, in sulphuric acid. The oxidising agent is the dichromate(VI) ion, Cr2O72-. Oxidation requires at least 1 -hydrogen atom. (hydrogen atom attached to same carbon as the OH group)

Oxidation of 1o, 2o and 3o alcohols Primary alcohols are oxidised first to aldehydes and then to carboxylic acids. Secondary alcohols are oxidised to ketones only. Tertiary alcohols are not oxidised at all. During oxidation orange dichromate(VI) is reduced to green chromium(III).

Oxidation of primary alcohols CH3 OH C H CH3 O C H + [O] + H2O ethanol ethanal (an aldehyde) Oxidation first removes hydrogen to give an aldehyde, further oxidation of aldehydes by addition of oxygen can occur to give the acid CH3 O C H CH3 O C OH + [O] ethanoic acid

Oxidation of secondary alcohols CH3 OH C H CH3 O C + [O] + H2O propan-2-ol The dichromate ion oxidises the alcohol by removal of hydrogen propanone (a ketone)

Where does the hydrogen go and the oxygen come from 3CH3CH2OH + Cr2O72- + 8H+ → 3CH3CHO + 2Cr 3+. + 7H2O The hydrogen forms water with some of the oxygen ions on potassium dichromate 3CH3CH2OH + 2Cr2O72- + 16H+ → 3CH3COOH + 4Cr3+ + 11H2O The reaction is complicated and best understood in this format.

Oxidation of alcohols (1) CH3 CH C OH H CH2 CH3 CH C O H CH2 CH3 CH C O OH CH2 3-methylbutan-1-ol (primary) 3-methylbutanal (aldehyde) 3-methylbutanoic acid CH3 CH C OH H CH3 CH C O further oxidation does not occur 3-methylbutan-2-ol (secondary) 3-methybutan-2-one (ketone)

Oxidation of alcohols (2) CH3 C OH CH2CH3 Oxidation does not occur 2-methylbutan-2-ol (tertiary) C CH OH CH3 CH2CH3 H C CH O CH3 CH2CH3 H C CH O CH3 CH2CH3 HO 2-methylbutan-1-ol (primary) 2-methylbutanal (aldehyde) 2-methylbutanoic acid

Elimination C OH H C H + H2O ethanol ethene H2SO4 + H2O ethanol ethene When heated with concentrated sulphuric acid alcohols lose the elements of water to form an alkene (dehydration).

a) Protonation of the OH group ? C O-H H xx C O-H H + H+ A hydrogen ion from the sulphuric acid joins on to one of the lone pairs of electrons on the oxygen atom. An intermediate is formed with the positive charge on the oxygen atom. This is positive charge results from the removal of the electron pair from oxygen into a bond with hydrogen, the oxygen nucleus is now overall +ve

b) Loss of water C O-H H C H + H2O The carbon-oxygen bond breaks releasing a molecule of water because electron density towards oxygen leads to a weak bond, and breaking it allows oxygen to return to a neutral state. Another intermediate is formed with a positive charge on the carbon atom, because carbon is now missing 2 electrons from the bond lost to water.

Loss of a proton C H C H + H+ A carbon-hydrogen bond breaks releasing a hydrogen ion. The electrons are attracted to the positive charge on the other carbon and a double bond is formed. The hydrogen ion can be returned to sulphuric acid, which donated it in the first place, so the sulphuric acid acts a catalyst – remains unchanged at the end of the reaction.

Tests for aldehydes and ketones

Brady’s tests for carbonyl C=O bond First, you can just use it to test for the presence of the carbon-oxygen double bond. You only get an orange or yellow precipitate from a carbon-oxygen double bond in an aldehyde or ketone. With Brady’s reagent an aldehyde or ketone forms a 2,4 dinitrophenylhydrazone, which can be seen as a yellow/orange precipitate. You can use it to help to identify the specific aldehyde or ketone.

What’s happening? Fehling's solution contains copper(II) ions (blue) Aldehydes reduce the complexed copper(II) ion to copper(I) oxide (yellow – red) RCHO + 2Cu2+ +5OH− → RCOO− + Cu2O + 3H2O Tollens' reagent - precipitate of silver(I) oxide, addtion of ammonia solution to redissolve the precipitate forms the ammine complex [Ag(NH3)2]+. Aldehydes reduce the diamminesilver(I) ion to metallic silver. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. 2Ag(NH3)2+ + 2RCHO + 3OH−→ 2Ag + RCOO− + 4NH3+2H2O