Chemistry (9701) | Carbonyl compounds

📚 AS Level Chemistry (9701) Study Notes: Carbonyl Compounds (Topic 17)

Welcome to the exciting world of Carbonyl Compounds! This chapter is crucial for understanding how the $\text{C=O}$ functional group controls reactivity, and how we use simple chemical tests to tell different organic molecules apart.

Don't worry if mechanisms seem complex at first; we will break down the steps using the fundamental concepts of polarity and electron movement.

1. Introduction to Aldehydes and Ketones

Carbonyl compounds all contain the carbonyl functional group, which is a carbon atom double-bonded to an oxygen atom ($\text{C=O}$).

Key Definitions and Nomenclature

The difference between an aldehyde and a ketone depends on what is attached to the carbon in the $\text{C=O}$ group.

• Aldehydes: The carbonyl carbon is bonded to at least one hydrogen atom (RCHO).
  (Example: Ethanal, Propanal)
• Ketones: The carbonyl carbon is bonded to two alkyl (or aryl) groups (RCOR').
  (Example: Propanone, Butanone)

Naming Tip:

• Aldehydes end in the suffix -al. (e.g., Methanal, Butanal).
• Ketones end in the suffix -one. (e.g., Propanone, Pentan-2-one).

Physical Properties: The Role of Polarity

The oxygen atom is much more electronegative than the carbon atom, making the $\text{C=O}$ bond highly polar.

$$ \text{C}^{\delta+} = \text{O}^{\delta-} $$

• This polarity creates permanent dipole-dipole forces between carbonyl molecules.
• Boiling points are higher than alkanes of similar mass (due to these dipole forces).
• Boiling points are lower than corresponding alcohols (alcohols have strong hydrogen bonding, which carbonyls lack, as they don't have an $\text{O-H}$ group).

2. Preparation of Carbonyl Compounds

The standard method for producing aldehydes and ketones is the oxidation of alcohols. The oxidizing agent is typically acidified potassium dichromate(VI) ($\text{K}_2\text{Cr}_2\text{O}_7$) or acidified potassium manganate(VII) ($\text{KMnO}_4$).

A. Producing Aldehydes (Primary Alcohols)

Oxidation of a primary alcohol yields an aldehyde.

Crucial Condition: Aldehydes are easily oxidized further to carboxylic acids. To isolate the aldehyde, you must remove it from the reaction mixture immediately as it forms. This is achieved using distillation.

Reaction Summary:

Primary Alcohol $\xrightarrow{\text{Acidified } \text{K}_2\text{Cr}_2\text{O}_7 \text{ / Distillation}}$ Aldehyde

Example: Ethanol $\to$ Ethanal

B. Producing Ketones (Secondary Alcohols)

Oxidation of a secondary alcohol yields a ketone.

Ketones cannot be easily oxidized further (since there is no $\text{H}$ atom on the carbonyl carbon), so you can use refluxing conditions (heating the mixture without distilling off the product).

Reaction Summary:

Secondary Alcohol $\xrightarrow{\text{Acidified } \text{K}_2\text{Cr}_2\text{O}_7 \text{ / Reflux}}$ Ketone

Example: Propan-2-ol $\to$ Propanone

3. Characteristic Reactions of Carbonyls

The two major types of reactions required at AS level are Reduction and Nucleophilic Addition.

A. Reduction to Alcohols (Addition of $[\text{H}]$)

Reduction is the opposite of oxidation. Carbonyls are reduced back to their corresponding alcohols.

• Aldehydes reduce to primary alcohols.
• Ketones reduce to secondary alcohols.

Reagents and Conditions:

The reducing agents used are powerful hydride donors:

• Sodium tetrahydridoborate ($\text{NaBH}_4$): Used in aqueous or alcoholic solution, usually at room temperature. This is the milder, safer option often used in the lab.
• Lithium tetrahydridoaluminate ($\text{LiAlH}_4$): Much stronger, usually dissolved in dry ether.

We represent the reduction using the symbol $[\text{H}]$:

$$ \text{RCHO} + 2[\text{H}] \to \text{RCH}_2\text{OH} \quad (\text{Primary Alcohol}) $$ $$ \text{RCOR'} + 2[\text{H}] \to \text{RCH}(\text{OH})\text{R}' \quad (\text{Secondary Alcohol}) $$

B. Nucleophilic Addition with Hydrogen Cyanide ($\text{HCN}$)

This reaction is important because it increases the length of the carbon chain (C-C bond formation) and introduces a hydroxyl ($\text{OH}$) group.

Reagent: Hydrogen cyanide ($\text{HCN}$) is added, usually with potassium cyanide ($\text{KCN}$) acting as a catalyst and the source of the strong nucleophile ($\text{CN}^-$ ion).

Conditions: $\text{HCN}$ (or $\text{KCN}$ followed by acid) and heat.

Product: A hydroxynitrile (sometimes called a cyanohydrin).

Example using Propanone:

$$ \text{CH}_3\text{COCH}_3 + \text{HCN} \xrightarrow{\text{KCN catalyst}} \text{CH}_3\text{C}(\text{OH})(\text{CN})\text{CH}_3 $$

Propanone $\to$ 2-hydroxy-2-methylpropanenitrile

Did you know? The hydroxynitrile produced from an aldehyde or unsymmetrical ketone contains a chiral centre, meaning the product is a racemic mixture of two optical isomers.

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4. The Nucleophilic Addition Mechanism

This explains how $\text{HCN}$ (or $\text{CN}^-$) adds to the carbonyl group. Since the carbonyl carbon is electron-deficient ($\delta+$), it is susceptible to attack by nucleophiles.

Mechanism Type: Nucleophilic Addition ($\text{NAD}$) (Syllabus 17.1.3)

Let's look at the reaction with the cyanide ion ($\text{CN}^-$), which is the nucleophile in this reaction.

Step 1: Nucleophilic Attack

The lone pair on the cyanide ion ($\text{CN}^-$) attacks the electron-deficient ($\text{C}^{\delta+}$) carbon atom. Simultaneously, the $\pi$-bond electrons shift onto the oxygen atom, forming an intermediate negatively charged ion called an alkoxide ion.

$$ \text{R}_2\text{C=O} + \text{CN}^- \to [\text{R}_2\text{C}(\text{O}^-)(\text{CN})] $$

(Remember: Use curly arrows to show the movement of electron pairs. The arrow starts at the lone pair/bond and points to the atom it attacks/moves to.)

Step 2: Protonation

The highly reactive alkoxide ion quickly accepts a proton ($\text{H}^+$) from the surrounding solution (usually water or acid) to form the final stable hydroxynitrile product.

$$ [\text{R}_2\text{C}(\text{O}^-)(\text{CN})] + \text{H}^+ \to \text{R}_2\text{C}(\text{OH})(\text{CN}) \quad (\text{Hydroxynitrile}) $$

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5. Identifying Carbonyl Compounds (The Chemist's Toolbox)

You need to know three main chemical tests: one to confirm a carbonyl group, and two to distinguish between aldehydes and ketones, or identify a specific subgroup.

A. The Universal Test: 2,4-DNPH Reagent

(Syllabus 17.1.4)

The 2,4-dinitrophenylhydrazine (2,4-DNPH) test is used to detect the presence of any carbonyl compound (aldehyde or ketone).

• Reagent: 2,4-dinitrophenylhydrazine dissolved in methanol and sulfuric acid.
• Observation: If an aldehyde or ketone is present, a bright yellow or orange precipitate is formed.
• What happens? This is a condensation reaction where water is eliminated to form a dinitrophenylhydrazone derivative.

B. Distinguishing Aldehydes from Ketones (Oxidation Tests)

(Syllabus 17.1.5)

The key chemical difference is that aldehydes are easily oxidized to carboxylic acids, while ketones require powerful oxidising agents that you won't encounter at AS level.

This difference allows us to use mild oxidizing agents to differentiate them:

1. Tollens' Reagent (Silver Mirror Test)

• Reagent: Ammoniacal solution of silver nitrate ($\text{Ag}^+$ in $\text{NH}_3(\text{aq})$).
• Test Result:
  • Aldehyde: Positive. The aldehyde is oxidized, and $\text{Ag}^+$ is reduced to metallic silver ($\text{Ag}$). A distinctive silver mirror forms on the inner surface of the test tube.
  • Ketone: Negative. No reaction, the solution remains clear.

2. Fehling's Reagent (or Benedict's Reagent)

• Reagent: Copper(II) ions ($\text{Cu}^{2+}$) complexed in alkaline solution (deep blue).
• Test Result (Heating required):
  • Aldehyde: Positive. The aldehyde is oxidized, and $\text{Cu}^{2+}$ (blue) is reduced to copper(I) oxide ($\text{Cu}_2\text{O}$). This gives a distinct brick-red precipitate.
  • Ketone: Negative. No reaction, the blue solution remains blue.

C. The Iodoform Test (Alkaline $\text{I}_2(\text{aq})$)

(Syllabus 17.1.6)

This test does not distinguish between aldehydes and ketones, but rather identifies if a molecule contains a specific structural fragment: the methyl carbonyl group ($\text{CH}_3\text{CO}-$).

Reagent: Alkaline aqueous iodine ($\text{I}_2(\text{aq})$) and sodium hydroxide ($\text{NaOH}$).

The Specific Group Tested:

1. Any compound containing the methyl carbonyl group ($\text{CH}_3\text{CO}-$). (e.g., Ethanal, Propanone).
2. Certain secondary alcohols that are immediately oxidized to the $\text{CH}_3\text{CO}-$ group under test conditions ($\text{CH}_3\text{CH}(\text{OH})\text{R}$).

Observation: If the group is present, a pale yellow precipitate of tri-iodomethane ($\text{CHI}_3$) is formed. This precipitate is often referred to as 'iodoform'.

Note: Ethanal is the only aldehyde that gives a positive result for the iodoform test because it is the only one with the $\text{CH}_3\text{CO}-$ structure. All other aldehydes have $\text{R} \text{CH}_2\text{CO}-$ where $\text{R}$ is not $\text{H}$.