Welcome to Carbonyl Compounds, Fats, and Oils!
Welcome to one of the most practical and visually satisfying areas of organic chemistry! In this chapter, we explore compounds containing the vibrant carbonyl group, \(>\text{C}=\text{O}\), specifically aldehydes and ketones, and then dive into the biochemical and industrial world of fats and oils (triglycerides). Whether you are predicting the outcome of a mirror-forming test in the lab or understanding why butter is solid while olive oil is liquid, these concepts tie directly into real-world phenomena and your OCR B (Salters) exam.
Don't worry if organic mechanisms have felt overwhelming in the past. We will break down every reaction step-by-step, highlight exactly what examiners look for, and use intuitive analogies to make the ideas stick.
1. The Carbonyl Group: Aldehydes and Ketones
Structure and Bonding
Both aldehydes and ketones contain the carbonyl functional group, which consists of a carbon atom double-bonded to an oxygen atom: \(>\text{C}=\text{O}\). The carbon atom is \(sp^2\) hybridised and planar, with a bond angle of approximately \(120^\circ\) around the carbonyl carbon.
How do they differ?
- Aldehydes have the carbonyl group at the end of a carbon chain. The carbonyl carbon is bonded to at least one hydrogen atom: \(-\text{CHO}\). Their systematic names end in -al (for example, ethanal, \(\text{CH}_3\text{CHO}\)).
- Ketones have the carbonyl group in the middle of a carbon chain. The carbonyl carbon is bonded to two other carbon atoms: \(-\text{CO}-\). Their systematic names end in -one (for example, propanone, \(\text{CH}_3\text{COCH}_3\)).
Polarity and Intermolecular Forces
Oxygen is significantly more electronegative than carbon. This pulls the shared electron density in the \(\text{C}=\text{O}\) double bond toward the oxygen atom, making the carbonyl bond strongly polar:
\(\text{C}^{\delta+}=\text{O}^{\delta-}\)
Because of this polarity:
- Carbonyl molecules experience permanent dipole-dipole attractions between molecules, in addition to London (dispersion) forces.
- They do not have an \(\text{O}-\text{H}\) bond, so pure aldehydes and ketones cannot form hydrogen bonds with each other. Hence, their boiling points are higher than non-polar alkanes of similar molar mass, but lower than alcohols or carboxylic acids.
- However, the lone pairs on the \(\text{O}^{\delta-}\) atom can accept hydrogen bonds from water molecules. This explains why small carbonyl compounds (like methanal, ethanal, and propanone) are completely soluble in water.
Key Takeaway: The carbonyl carbon carries a partial positive charge (\(\text{C}^{\delta+}\)), making it vulnerable to attack by electron-rich species called nucleophiles.
2. Distinguishing Aldehydes from Ketones
Because an aldehyde has a hydrogen atom directly attached to the carbonyl carbon (\(-\text{CH}=\text{O}\)), it can be easily oxidised to a carboxylic acid (\(-\text{COOH}\)). Ketones lack this hydrogen atom; oxidising a ketone requires breaking a strong carbon-carbon bond, which does not happen under mild conditions.
We can exploit this difference using three classic laboratory tests:
Test 1: Tollens' Reagent (The Silver Mirror Test)
Reagent: An ammoniacal solution of silver nitrate, containing the complex ion \([\text{Ag}(\text{NH}_3)_2]^+\).
Conditions: Warm gently in a water bath (never heat directly with a naked flame, as organic vapours are flammable and silver fulminate can form upon drying).
With an aldehyde: The aldehyde is oxidised to a carboxylate ion, while the silver ions are reduced to metallic silver: \(\text{Ag}^+ + e^- \rightarrow \text{Ag}\text{(s)}\). A striking silver mirror coats the inside of the test tube (or a dark grey precipitate forms).
With a ketone: No reaction; the solution remains colourless.
Test 2: Fehling's or Benedict's Solution
Reagent: An alkaline solution containing copper(II) ions, \(\text{Cu}^{2+}\), complexed with tartrate (Fehling's) or citrate (Benedict's), giving a clear deep-blue colour.
Conditions: Warm in a hot water bath.
With an aldehyde: The aldehyde is oxidised, and blue \(\text{Cu}^{2+}\) ions are reduced to copper(I) oxide: \(2\text{Cu}^{2+} + 2e^- + 2\text{OH}^- \rightarrow \text{Cu}_2\text{O}\text{(s)} + \text{H}_2\text{O}\). A brick-red precipitate of \(\text{Cu}_2\text{O}\) forms.
With a ketone: No reaction; the solution remains clear blue.
Test 3: Acidified Potassium Dichromate(VI)
Reagent: Potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7\), acidified with dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
Conditions: Heat under reflux or warm gently.
With an aldehyde: Orange dichromate(VI) ions, \(\text{Cr}_2\text{O}_7^{2-}\), are reduced to green chromium(III) ions, \(\text{Cr}^{3+}\). The colour changes from orange to green.
With a ketone: No reaction; the solution remains orange.
Key Takeaway: Aldehydes reduce mild oxidising agents (\(\text{Ag}^+\) to \(\text{Ag}\); \(\text{Cu}^{2+}\) to \(\text{Cu}_2\text{O}\)) because they possess a readily oxidised \(\text{C}-\text{H}\) bond at the carbonyl centre. Ketones do not.
3. Reactions of the Carbonyl Group
A. Reduction to Alcohols
Carbonyl compounds can be reduced back to alcohols using the reducing agent sodium tetrahydridoborate(III), \(\text{NaBH}_4\), in aqueous or ethanolic solution.
- Reduction of an aldehyde produces a primary alcohol:
\(\text{RCHO} + 2[\text{H}] \rightarrow \text{RCH}_2\text{OH}\)
- Reduction of a ketone produces a secondary alcohol:
\(\text{RCOR}' + 2[\text{H}] \rightarrow \text{RCH(OH)R}'\)
In written equations, the reducing agent is conventionally represented as \([\text{H}]\). The active nucleophile supplied by \(\text{NaBH}_4\) is the hydride ion, \(\text{H}^-\), which carries a lone pair of electrons: \(:\text{H}^-\).
B. Addition of Hydrogen Cyanide (Cyanide Ions)
Aldehydes and ketones react with hydrogen cyanide, \(\text{HCN}\), in the presence of a base or alkali metal cyanide (such as \(\text{NaCN}\) or \(\text{KCN}\)), to form hydroxynitriles (compounds containing both an \(-\text{OH}\) and a \(-\text{C}\equiv\text{N}\) group on the same carbon atom).
For example, with ethanal:
\(\text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN}\) (2-hydroxypropanenitrile)
Why is this reaction so important in synthetic chemistry? It increases the length of the carbon skeleton by one carbon atom! The nitrile group can subsequently be hydrolysed into a carboxylic acid (\(-\text{COOH}\)) or reduced into an amine (\(-\text{CH}_2\text{NH}_2\)).
C. The Mechanism: Nucleophilic Addition
Because the \(\text{C}=\text{O}\) bond is polar (\(\text{C}^{\delta+}=\text{O}^{\delta-}\)), the electron-deficient carbon atom attracts nucleophiles (electron pair donors). The reaction proceeds in two clear steps:
Step 1 (Attack): The nucleophile (\(:\text{H}^-\) from \(\text{NaBH}_4\), or \(:\text{C}\equiv\text{N}^-\) from cyanide) uses its lone pair of electrons to form a dative covalent bond with the partially positive carbonyl carbon atom (\(\text{C}^{\delta+}\)). Simultaneously, the \(\pi\)-bond electrons of the \(\text{C}=\text{O}\) group are pushed entirely onto the electronegative oxygen atom, forming a negatively charged intermediate containing an oxide ion (\(:\text{O}^-\)).
Step 2 (Protonation): The negatively charged oxygen atom uses its lone pair of electrons to capture a proton (\(\text{H}^+\)) from a solvent water molecule or \(\text{HCN}\), forming the neutral alcohol or hydroxynitrile product and regenerating the basic catalyst.
Common Exam Pitfall: When drawing this mechanism, make sure your curly arrow starts precisely from the lone pair or negative charge of the nucleophile and points directly to the carbonyl carbon. A second arrow must start from the double bond and point onto the oxygen atom.
D. Testing for the Carbonyl Group: 2,4-Dinitrophenylhydrazine (2,4-DNPH)
To detect whether an unknown liquid contains an aldehyde or a ketone, we use 2,4-dinitrophenylhydrazine (also known as Brady's reagent).
- A positive result is the rapid formation of a bright yellow or orange precipitate (a 2,4-dinitrophenylhydrazone derivative).
- Carboxylic acids and esters do not react with 2,4-DNPH, so this test specifically identifies aldehydes and ketones.
Identifying the Specific Carbonyl Compound:
1. Filter off the orange precipitate under reduced pressure using a Buchner funnel.
2. Purify the solid derivative by recrystallisation from a minimum volume of hot solvent.
3. Dry the crystals and determine their melting point using melting point apparatus.
4. Compare the measured melting point to a database or book of known melting temperatures of 2,4-dinitrophenylhydrazone derivatives to identify the exact original aldehyde or ketone.
Key Takeaway: Carbonyl compounds undergo nucleophilic addition across the \(\text{C}=\text{O}\) bond. 2,4-DNPH confirms the presence of a carbonyl group, while derivative melting points pinpoint the exact compound.
4. Fats and Oils: Chemistry of Triglycerides
Fats and oils are naturally occurring esters found in animal tissue and plant seeds. Chemically, they are triesters of the alcohol propane-1,2,3-triol (commonly called glycerol) and three long-chain carboxylic acid molecules known as fatty acids. These triesters are called triglycerides.
Structure of Glycerol and Fatty Acids
- Glycerol has three \(\text{OH}\) groups: \(\text{CH}_2\text{OH}-\text{CHOH}-\text{CH}_2\text{OH}\).
- Fatty acids are long, unbranched hydrocarbon chains with a terminal carboxyl group, typically containing between 12 and 22 carbon atoms (often with an even number of carbons, such as stearic acid, \(\text{C}_{17}\text{H}_{35}\text{COOH}\), or oleic acid, \(\text{C}_{17}\text{H}_{33}\text{COOH}\)).
When glycerol condenses with three fatty acid molecules, three ester linkages (\(-\text{COO}-\)) are formed, releasing three molecules of water (\(\text{H}_2\text{O}\)):
\(\text{Glycerol} + 3\,\text{Fatty Acids} \rightarrow \text{Triglyceride} + 3\text{H}_2\text{O}\)
Why are Fats Solid and Oils Liquid?
The distinction between a "fat" (solid at room temperature) and an "oil" (liquid at room temperature) comes down to intermolecular forces and molecular shape:
1. Saturated Fatty Acid Chains (Fats):
- Saturated fatty acids contain only single \(\text{C}-\text{C}\) bonds in their hydrocarbon chains.
- These chains are straight and flexible, allowing the triglyceride molecules to pack closely together in an orderly lattice.
- Close packing enables extensive instantaneous dipole-induced dipole (London) forces to form between adjacent chains.
- More thermal energy is required to overcome these intermolecular attractions, leading to higher melting points. Thus, animal fats (like lard and butter) are solid at room temperature.
2. Unsaturated Fatty Acid Chains (Oils):
- Unsaturated fatty acids contain one or more \(\text{C}=\text{C}\) double bonds.
- In naturally occurring vegetable oils, these double bonds almost always have the cis configuration.
- Each cis-double bond introduces a permanent rigid kink (bend) of about \(120^\circ\) in the carbon backbone.
- These kinks prevent the triglyceride molecules from packing neatly together.
- As a result, the distance between chains is greater, making the London dispersion forces significantly weaker.
- Less energy is needed to separate the molecules, resulting in lower melting points. Thus, vegetable oils (such as sunflower or olive oil) are liquid at room temperature.
Key Takeaway: Cis double bonds create kinks that disrupt regular molecular packing, weakening London forces and lowering the melting points of oils relative to saturated fats.
5. Reactions and Analysis of Fats and Oils
A. Hydrolysis of Triglycerides
Like all esters, the ester bonds in triglycerides can be broken by hydrolysis using water in the presence of an acid or a base catalyst:
1. Alkaline Hydrolysis (Saponification):
Heating a fat or oil under reflux with aqueous sodium hydroxide (\(\text{NaOH}\)) breaks the three ester links to yield glycerol and the sodium salts of the fatty acids:
\(\text{Triglyceride} + 3\text{NaOH} \rightarrow \text{Glycerol} + 3\text{RCOO}^-\text{Na}^+\)
These sodium carboxylate salts are what we call soaps! The long hydrophobic hydrocarbon tail dissolves in non-polar grease, while the ionic, hydrophilic head (\(-\text{COO}^-\text{Na}^+\)) interacts with water, allowing grease to be washed away.
2. Acid Hydrolysis:
Heating with hot aqueous dilute acid (such as \(\text{HCl}\) or \(\text{H}_2\text{SO}_4\)) produces glycerol and the free fatty acids:
\(\text{Triglyceride} + 3\text{H}_2\text{O} \xrightleftharpoons{\text{H}^+} \text{Glycerol} + 3\text{RCOOH}\)
B. Hydrogenation (Hardening of Oils)
Unsaturated vegetable oils can be converted into solid or spreadable fats (used in margarines) by catalytic hydrogenation.
- Reagents and Conditions: Hydrogen gas, \(\text{H}_2\), in the presence of a finely divided nickel catalyst at elevated temperature (around \(150^\circ\text{C}\)).
- Chemistry: Hydrogen adds across the \(\text{C}=\text{O}\) non-adjacent \(\text{C}=\text{C}\) double bonds by electrophilic addition, converting unsaturated chains into saturated chains.
- Result: Straightening out the kinked chains allows closer packing, strengthening London forces and raising the melting point (converting liquid oil into semi-solid spreadable fat).
C. Analysis: Measuring Unsaturation (Iodine Value)
The degree of unsaturation in a fat or oil can be determined quantitatively using its iodine value.
- Definition: The iodine value is the mass of iodine, \(\text{I}_2\), in grams that reacts completely with \(100\text{ g}\) of fat or oil.
- Principle: Halogens add across \(\text{C}=\text{C}\) double bonds in a \(1:1\) molar ratio:
\(-\text{CH}=\text{CH}- + \text{I}_2 \rightarrow -\text{CHI}-\text{CHI}-\)
- Interpretation: A higher iodine value means more \(\text{C}=\text{C}\) double bonds are present per molecule, indicating a higher degree of unsaturation.
D. Gas-Liquid Chromatography (GLC) of Fatty Acids
Triglycerides themselves are very large, non-volatile molecules that would decompose if vaporised directly inside a gas chromatograph. To analyse the fatty acid composition of a fat or oil, we carry out transesterification to convert the fatty acid chains into volatile fatty acid methyl esters (FAMEs):
1. The fat or oil is reacted with methanol (\(\text{CH}_3\text{OH}\)) in the presence of an alkaline catalyst (\(\text{NaOH}\) or \(\text{NaOCH}_3\)):
\(\text{Triglyceride} + 3\text{CH}_3\text{OH} \rightarrow \text{Glycerol} + 3\text{RCOOCH}_3\)
2. The volatile methyl esters are injected into a gas-liquid chromatography (GLC) column.
3. In GLC, the mobile phase is an unreactive carrier gas (such as \(\text{He}\) or \(\text{N}_2\)), and the stationary phase is a high-boiling non-volatile liquid coated onto solid support particles inside a coiled column.
4. Separation depends on the relative solubility of each ester in the stationary phase versus its volatility in the gas carrier phase.
5. Each ester emerges at a characteristic retention time, and the area under each peak on the chromatogram is directly proportional to the relative amount of that fatty acid present in the sample.
Key Takeaway: Fats and oils undergo hydrolysis to yield glycerol and fatty acids (or soaps). Converting fatty acids into volatile methyl esters (FAMEs) allows rapid quantitative separation and identification via gas-liquid chromatography.
Quick Review Summary
- Carbonyl Group: Polar \(>\text{C}^{\delta+}=\text{O}^{\delta-}\) double bond, susceptible to nucleophilic addition.
- Distinguishing Aldehydes and Ketones: Aldehydes are oxidised by Tollens' reagent (forming a silver mirror) and Fehling's/Benedict's solution (forming a brick-red precipitate). Ketones give negative results.
- Nucleophilic Addition: \(\text{NaBH}_4\) supplies \(\text{H}^-\) to reduce aldehydes to primary alcohols and ketones to secondary alcohols. Cyanide (\(\text{CN}^-\)) adds to form hydroxynitriles, extending the carbon chain.
- 2,4-DNPH: Forms orange precipitates with aldehydes and ketones; derivative melting points confirm the identity of the specific carbonyl.
- Fats vs Oils: Triglycerides are triesters of glycerol and three fatty acids. Saturated fats pack neatly (higher melting points; solid). Unsaturated oils contain cis kinks preventing close packing (lower melting points; liquid).
- Saponification: Alkaline hydrolysis of triglycerides produces glycerol and soap (carboxylate salts).
- GLC Analysis: Triglycerides are converted into volatile methyl esters (FAMEs) before analysis by gas-liquid chromatography.