Carbonyl Compounds A2 Chemistry Unit 4

What functional group do all carbonyl compounds contain? δ+ δ-

What is the aldehyde functional group?

What is the ketone functional group?

Name the following molecules: Methanal Propanal Methylpropanal Propanone Methylbutanone Pentan-3-one

Describe and explain the boiling point of carbonyl compounds, referring to the type of bonding. Methanal is a gas and the other carbonyl compounds are liquid at room temperature. The molecule are polar, so dipole-dipole forces as well as van der Waals forces exist between them. Intermolecular hydrogen bonding is not possible because neither aldehydes nor ketones have a hydrogen atom that is sufficiently δ+. This causes them to: Have higher boiling points than alkanes and alkenes Have lower boiling points than alcohols

Describe and explain the solubility of carbonyl compounds in water. The lower members of both aldehydes and ketones are soluble in water. This solubility is due to the hydrogen bonding between the lone pair of electron in the δ- oxygen in the carbonyl compound and the δ+ hydrogen in a water molecule. Solubility decreases as the number of carbons increases because the hydrocarbon tail is hydrophobic. δ+ δ- O H

Describe the smell of carbonyl compounds. The lower members of the homologous series of aldehydes have pungent odours. Ketones have much sweeter smells than aldehydes.

Describe the preparation of aldehydes and give the equation for the preparation of ethanal. Aldehydes are prepared by the partial oxidation of a primary alcohol Reagents: Primary alcohol Potassium dichromate (VI) in dilute sulfuric acid Conditions: Temperature of about 60°C Distil of product as it forms Observation: Colour change from orange to green as dichromate (VI) is reduced to chromium (III)

Describe the preparation of ketones and give the equation for the preparation of propanone. Aldehydes are prepared by the oxidation of a secondary alcohol Reagents: Secondary alcohol Potassium dichromate (VI) in dilute sulfuric acid Conditions: Heat under reflux When reaction has finished, distil off product Observation: Colour change from orange to green as dichromate (VI) is reduced to chromium (III)

Describe and explain the differences between aldehydes and ketones with respect to oxidation. Aldehydes are readily oxidised whereas ketones are not. This is due to the presence of the hydrogen atom on the C=O in aldehydes which is very easily oxidised When an aldehyde is oxidised:

Describe the reaction of aldehydes with Fehling’s/Benedict’s solution and give an equation involving ethanal. Both Fehling’s and Benedict’s contain copper (II) complexes. When an aldehyde is warmed with Fehling’s/Benedict’s solution it is oxidised. The solution is alkaline so the salt of the carboxylic acid is produced. The blue solution is reduced to a red precipitate of copper oxide. N.B. With ketones there is no reaction and the solution stays blue.

Describe the reaction of aldehydes with Tollens’ reagent and give an equation involving ethanal. Tollens’ reagent is made by adding a few drops of sodium hydroxide to silver nitrate solution and then dissolving the precipitate in dilute ammonia. When a few drops of Tollens’ are added to the aldehyde and the mixture is warmed gently, the aldehyde is oxidised. Because the solution is alkaline the salt of the carboxylic acid is formed. The colourless Tollen’s reagent is reduced to give a silver mirror. N.B. With ketones there is no reaction and the solution stays colourless.

Describe the reaction of aldehydes with acidified potassium dichromate (VI) and give an equation involving ethanal. When heated with the acidified potassium dichromate the aldehyde is oxidised to a carboxylic acid. The orange potassium dichromate (VI) is reduced to green Cr (III) N.B. With ketones there is no reaction and the solution stays orange.

Describe the reduction of an aldehyde, including the equation for the reduction of ethanal. Reducing agent: Lithium tetrahydridoaluminate (III), LiAlH4 The LiAlH4 acts as a source of H- ions which add on to the δ+ carbon atom – at this point the reactants must be kept dry so it is carried out in ether solution. Next, a solution of aqueous acid is added which protonates the O- formed in the first step. A primary alcohol is formed.

Describe the reduction of an aldehyde, including the equation for the reduction of propanone. Reducing agent: Lithium tetrahydridoaluminate (III), LiAlH4 The LiAlH4 acts as a source of H- ions which add on to the δ+ carbon atom – at this point the reactants must be kept dry so it is carried out in ether solution. Next, a solution of aqueous acid is added which protonates the O- formed in the first step. A secondary alcohol is formed.

Why does LiAlH4 reduce carbonyl compounds but not alkenes? LiAlH4 is a reducing agent that reacts specifically with polar π-bonds.

State and explain the conditions needed for the reaction between hydrogen cyanide and a carbonyl compound. pH 8 – this provides the CN- ion catalyst

Draw the mechanism for the reaction of hydrogen cyanide and ethanal. 2-hydroxy-propanenitrile

Draw the mechanism for the reaction of hydrogen cyanide and propanone. 2-hydroxy-2-methylpropanenitrile

What is the difference between the two reactions? The ketone reaction happens at a slower rate.

Describe the mechanism for the reaction of hydrogen cyanide and a carbonyl compound. The lone pair of electrons on the carbon atom of the CN- ion form a bond with the δ+ carbon atom in the carbonyl compound. At the same time, the π-electrons in the C=O group move to the oxygen atom. The anion formed in the first step removes a proton from an HCN molecule to form the organic product and regenerate the CN- catalyst.

Why is it important that the pH is neither too high or too low? If it is too low, there are not enough CN- ions for the first step. If it is too high there are not enough HCN molecules for the second step.

Describe and explain the optical activity of the product when hydrogen cyanide adds on to a carbonyl compound. The product is a racemic mixture of both enatiomers so it does not rotate the plane of plane-polarised light. This is because the carbonyl compound is planar around the C=O group so the cyanide ion can attack from above or below the plane.

Describe how carbonyl compounds react with compounds containing an H2N- group. The lone pair of electrons on the nitrogen atom acts a nucleophile and forms a bond with the δ+ carbon atom in the C=O group. A water molecule is then lost and a C=N bond is formed.

Describe how 2,4-dinitrophenylhydrazine can be used to test for a carbonyl group. Add a few drops of 2,4-dinitrophenylhydrazine to a solution of the compound. Simple aldehydes and ketones give a yellow precipitate. Aromatic aldehydes and ketones give an orange precipitate.

Give the overall equation for the reaction between a carbonyl compound and 2,4-dinitrophenylhydrazine.

What type of reaction occurs and what type of mechanism is it? Condensation reaction – with the loss of a water molecule Addition-elimination

Describe how you can make and purify a derivative of a carbonyl compound using 2,4-dinitrophenylhydrazine.

Explain how this removes impurities. In step 2 the derivative will recrystallise but the impurities will remain in solution. The third step, washing with cold solvent, will then rinse of any impurities that did crystallise.

How can the purified product be used to identify the original carbonyl compound? Find the melting temperature and compare to known data.

What is the reaction of carbonyl compounds with iodine in alkali called? The iodoform reaction

What type of carbonyl compounds undergo the iodoform reaction? Ethanal and methyl ketones

A pale yellow precipitate forms Describe how to carry out the iodoform reaction, including observations. Warm the organic substance with either: A mixture of iodine and NaOH solution A solution of potassium iodide in sodium chlorate (I) A pale yellow precipitate forms Medical smell

Describe how the iodoform reaction occurs, including equations. Sodium hydroxide solution is added to iodine solution to form iodate (I) ions: These substitute into the –CH3 group next to the C=O group, forming a CI3C=O group. The three halogen atoms and the oxygen atom have an electron-withdrawing effect which weakens and breaks the σ-bond between the two carbon atoms, forming iodoform:

Write the overall equation for the iodoform reaction.