Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 15.8 Reactions of carbonyls-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) – Unit 4 – 15.8 Reactions of carbonyls-Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) – Unit 4 – 15.8 Reactions of carbonyls-Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.
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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics
15.8 Reactions of Carbonyl Compounds
(i) Reactions with Fehling’s/Benedict’s Solution, Tollens’ Reagent and Acidified Dichromate(VI)
Aldehydes are readily oxidised because the carbonyl carbon is bonded to a hydrogen atom. Ketones are generally resistant to oxidation under mild conditions.
Aldehydes
- Oxidised to carboxylic acids.
- Positive results with oxidation tests.
Ketones
- Not oxidised by mild oxidising agents.
- Negative results with these tests.
1. Fehling’s Solution / Benedict’s Solution
Reagent
- Alkaline copper(II) solution.
Observation with Aldehydes
- Blue solution forms a:
Brick-red precipitate
due to formation of \( \mathrm{Cu_2O} \).
Observation with Ketones
- No change.
General Oxidation Equation
\( \mathrm{RCHO + [O] \rightarrow RCOOH} \)
2. Tollens’ Reagent
Reagent
- Ammoniacal silver nitrate solution.
Observation with Aldehydes
- Formation of a:
Silver mirror
due to deposition of silver metal.
Observation with Ketones
- No reaction.
Importance
- Distinguishes aldehydes from ketones.
3. Acidified Dichromate(VI) Ions
Reagent
- \( \mathrm{K_2Cr_2O_7/H^+} \)
Observation with Aldehydes
- Orange solution changes to:
Green
Observation with Ketones
- No change.
Oxidation Equation
\( \mathrm{RCHO + [O] \rightarrow RCOOH} \)
Why Ketones Resist Oxidation
- No hydrogen attached directly to carbonyl carbon.
- Oxidation would require breaking C–C bonds.
(ii) Reduction with Lithium Tetrahydridoaluminate(III)
Reagent
\( \mathrm{LiAlH_4} \) in dry ether (ethoxyethane)
\( \mathrm{LiAlH_4} \) is a strong reducing agent supplying hydride ions (\( \mathrm{H^-} \)).
In equations, reduction may be represented by:
\( \mathrm{[H]} \)
Reduction of Aldehydes
- Aldehydes are reduced to:
Primary alcohols
General Equation
\( \mathrm{RCHO + 2[H] \rightarrow RCH_2OH} \)
Example
\( \mathrm{CH_3CHO + 2[H] \rightarrow CH_3CH_2OH} \)
Reduction of Ketones
- Ketones are reduced to:
Secondary alcohols
General Equation
\( \mathrm{RCOR’ + 2[H] \rightarrow RCH(OH)R’} \)
Example
\( \mathrm{CH_3COCH_3 + 2[H] \rightarrow CH_3CH(OH)CH_3} \)
Conditions
- Dry ether solvent required because \( \mathrm{LiAlH_4} \) reacts violently with water.
Summary Table
| Reaction | Aldehyde | Ketone |
|---|---|---|
| Fehling’s/Benedict’s | Brick-red precipitate | No reaction |
| Tollens’ reagent | Silver mirror | No reaction |
| Acidified dichromate(VI) | Orange → green | No reaction |
| \( \mathrm{LiAlH_4} \) | Primary alcohol | Secondary alcohol |
Key Features
- Aldehydes are easily oxidised.
- Ketones resist mild oxidation.
- \( \mathrm{LiAlH_4} \) reduces carbonyl compounds to alcohols.
- Oxidation tests distinguish aldehydes from ketones.
Example 1:
Compound X gives a silver mirror with Tollens’ reagent and forms a green solution with acidified dichromate(VI). Identify the functional group present.
▶️ Answer/Explanation
The compound is oxidised by both reagents.
This behaviour is characteristic of an aldehyde.
Therefore, compound X contains an aldehyde group.
Example 2:
State the product formed when propanone is reduced using \( \mathrm{LiAlH_4} \).
▶️ Answer/Explanation
Propanone is a ketone.
Reduction converts ketones into secondary alcohols.
Product:
\( \mathrm{propan\text{-}2\text{-}ol} \)
(iii) Reaction with HCN in the Presence of KCN
Aldehydes and ketones undergo nucleophilic addition reactions with hydrogen cyanide, \( \mathrm{HCN} \), to form hydroxynitriles.
Role of Reagents
- \( \mathrm{KCN} \) provides the nucleophile:
\( \mathrm{CN^-} \)
- \( \mathrm{HCN} \) provides \( \mathrm{H^+} \).
Why Carbonyl Compounds React
- Carbonyl bond is polar.
- Oxygen is more electronegative than carbon.
\( \mathrm{C^{\delta +}=O^{\delta -}} \)
Therefore, the carbonyl carbon is electrophilic.
Mechanism (Nucleophilic Addition)
Step 1: Nucleophilic Attack
- Lone pair on \( \mathrm{CN^-} \) attacks carbonyl carbon.
- \( \mathrm{\pi} \)-bond electrons move to oxygen.
Step 2: Protonation
- Alkoxide ion gains \( \mathrm{H^+} \) from HCN.
- Hydroxynitrile forms.
General Equation
\( \mathrm{RCHO + HCN \rightarrow RCH(OH)CN} \)
Example
\( \mathrm{CH_3CHO + HCN \rightarrow CH_3CH(OH)CN} \)
Optical Activity Evidence
- Carbonyl carbon is planar.
- \( \mathrm{CN^-} \) can attack from either side equally.
- If a chiral centre forms, both enantiomers form.
- Product becomes a racemic mixture.
(iv) Reaction with 2,4-Dinitrophenylhydrazine (2,4-DNPH)
2,4-DNPH is used as a qualitative test for aldehydes and ketones.
Purpose of the Test
- Detect presence of carbonyl group.
- Identify unknown carbonyl compounds using melting point data.
Observation
- Formation of an:
Orange precipitate
due to formation of a 2,4-dinitrophenylhydrazone derivative.
Importance of Melting Point
- Different derivatives have different melting points.
- Measured melting point compared with data table to identify compound.
Limitations
- Cannot distinguish aldehydes from ketones.
- Only confirms carbonyl group presence.
(v) Iodoform Test
The iodoform test identifies compounds containing the:
\( \mathrm{CH_3CO-} \)
group.
Reagents
- Iodine
- Aqueous alkali (e.g. NaOH)
Positive Result
- Formation of a:
Yellow precipitate
of iodoform:
\( \mathrm{CHI_3} \)
Compounds Giving Positive Test
- Ethanal
- Methyl ketones
- Secondary alcohols containing:
\( \mathrm{CH_3CH(OH)-} \)
Reason
- These compounds can form the \( \mathrm{CH_3CO-} \) group under reaction conditions.
Example
\( \mathrm{CH_3COCH_3} \) gives positive iodoform test.
Summary Table
| Reagent | Positive Observation | Purpose |
|---|---|---|
| \( \mathrm{HCN/KCN} \) | Hydroxynitrile formed | Nucleophilic addition |
| 2,4-DNPH | Orange precipitate | Detect carbonyl group |
| Iodine/alkali | Yellow \( \mathrm{CHI_3} \) precipitate | Detect \( \mathrm{CH_3CO-} \) group |
Key Features
- \( \mathrm{HCN} \) reactions proceed by nucleophilic addition.
- Planar carbonyl groups can produce racemic mixtures.
- 2,4-DNPH detects aldehydes and ketones.
- Iodoform test identifies methyl ketones and ethanal.
Example 1:
Explain why addition of HCN to ethanal produces a racemic mixture.
▶️ Answer/Explanation
The carbonyl carbon in ethanal is planar.
\( \mathrm{CN^-} \) can attack from either side equally.
A chiral centre forms, producing both enantiomers in equal amounts.
Therefore, a racemic mixture is formed.
Example 2:
Compound X forms an orange precipitate with 2,4-DNPH and a yellow precipitate with iodine in alkali. Suggest a possible functional group present.
▶️ Answer/Explanation
The 2,4-DNPH test shows a carbonyl compound is present.
Positive iodoform test indicates a \( \mathrm{CH_3CO-} \) group.
Therefore, compound X is likely a methyl ketone.
