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Carbonyls: Nucleophilic Addition and Tests

Explain nucleophilic addition to carbonyls and interpret their tests.

  • GCE A-Level H2 Chemistry 9476-2027

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Carbonyl questions combine mechanism language (why the carbonyl carbon is electrophilic) with deduction from tests. This lesson covers nucleophilic addition at C = O and the exam-safe test logic: 2,4-DNPH first, then Tollens’/Fehling’s to identify aldehydes.

This topic gets cleaner if you cross-check Organic Mechanisms: Curly Arrows, Electrophiles, Nucleophiles while navigating from the Organic Chemistry hub.

Definitions (Must Know)

A. Carbonyl group

A carbonyl group is C = O.

B. Aldehyde vs ketone

  • Aldehyde: carbonyl at the end of a chain, RCHO
  • Ketone: carbonyl within a chain, RCOR

C. Nucleophilic addition

In nucleophilic addition, a nucleophile attacks the δ + carbonyl carbon and the π bond breaks, forming new σ bonds.

D. 2,4-DNPH (Brady’s reagent)

2,4-DNPH is a reagent that gives an orange/yellow precipitate with aldehydes and ketones (carbonyl compounds).

E. Tollens’ reagent

Tollens’ reagent oxidises aldehydes, producing a silver mirror/grey silver precipitate.

F. Fehling’s solution

Fehling’s solution is reduced by aldehydes, giving a blue → brick-red precipitate change.

Key Ideas (What Earns Marks)

  • The carbonyl bond is polar: oxygen is δ-, carbon is δ +, so carbonyl carbon is electrophilic.
  • Aldehydes are generally more reactive than ketones (less steric hindrance and less electron donation).
  • Core test: 2,4-DNPH gives an orange/yellow precipitate with aldehydes and ketones.
  • Aldehydes are reducing agents: they give positive Tollens’/Fehling’s tests; ketones do not (under these conditions).
Quick Recall (Test Table)

| Reagent | Positive observation | Conclusion | | :--- | :--- | :--- | | 2,4-DNPH | orange/yellow precipitate | carbonyl present (aldehyde or ketone) | | Tollens’ | silver mirror / grey Ag(s) | aldehyde present | | Fehling’s | blue → brick-red precipitate | aldehyde present |

Detailed Explanations

A. Why the carbonyl carbon is attacked (the key causal chain)

Because the C = O bond is polar (O is δ- and C is δ +), therefore nucleophiles attack the carbonyl carbon (electrophilic centre).

B. Nucleophilic addition of HCN (forming hydroxynitriles)

Overall (example: ethanal): CH₃CHO + HCN → CH₃CH(OH)CN

What to say:

  • CN⁻ is the nucleophile that attacks the carbonyl carbon.
  • The reaction is often done using KCN with dilute acid to generate HCN in situ (conditions phrasing may vary in mark schemes).

Workflow (mechanism language, simplified):

  1. CN⁻ attacks the carbonyl carbon and the π bond breaks to give an alkoxide.
  2. The alkoxide is protonated (from HCN / acid) to give the hydroxynitrile.

Mini example: Ethanal gives CH₃CH(OH)CN (a hydroxynitrile).

C. Carbonyl tests

1) 2,4-DNPH (Brady’s reagent)

  • Positive result: orange/yellow precipitate
  • Conclusion: carbonyl present (aldehyde or ketone)

2) Tollens’ reagent

  • Positive for aldehydes: silver mirror / grey silver precipitate
  • Negative for ketones (typically)

3) Fehling’s solution

  • Positive for aldehydes: blue → brick-red precipitate
  • Negative for ketones (typically)

D. Workflow: identifying an unknown carbonyl from tests (exam method)

  1. Do 2,4-DNPH: if positive → carbonyl present.
  2. Do Tollens’ or Fehling’s:
    • positive → aldehyde
    • negative → ketone
  3. State observation first, then conclusion (mark-scheme style).

Example: CH₃CHO + [O] → CH₃COOH

This is why aldehydes give positive Tollens’/Fehling’s tests: they are easily oxidised.

F. Formation and reduction

Primary alcohols form aldehydes when oxidised and distilled; continued oxidation under reflux forms carboxylic acids. Secondary alcohols form ketones. Aldehydes and ketones are reduced to alcohols by LiAlH₄ in dry ether followed by water, or by H₂ with a nickel catalyst:

RCHO + 2[H] → RCH₂OH RCOR' + 2[H] → RCH(OH)R'

G. The tri-iodomethane test

Warm the compound with iodine in alkaline solution. A pale-yellow precipitate of CHI₃ is positive. The test is given by ethanal and by methyl ketones with the form CH₃COR. Alcohols with the form CH₃CH(OH)R also respond because they are oxidised to methyl ketones under the test conditions; ethanol is the special primary-alcohol case because it forms ethanal.

Worked Examples

Modelled example 1

HCN Addition to Propanone

Core

Problem

Write an equation for the addition of HCN to propanone.
Study the worked solution

  1. Locate the reaction centre

    Method

    Use the polar C=O group.

    Reason

    The carbonyl carbon is electron deficient and undergoes nucleophilic addition.

    Working

    CH₃COCH₃

  2. Add H and CN

    Method

    Replace C=O by C–OH and add CN to the same carbon.

    Reason

    Addition incorporates both parts of HCN without changing the carbon skeleton.

    Working

    CH₃COCH₃ + HCN → CH₃C(OH)(CN)CH₃

Guided practice 2

Fehling’s Test for Ethanal

About 4 min

Problem

State the observation when ethanal is warmed with Fehling’s solution.

Select the complete observation

Initial colour
Positive result

Hints

Hint 1: test family

Ethanal is an aldehyde.

Hint 2: full observation

State both the initial solution colour and the precipitate formed.

View solution step by step

  1. Identify the positive aldehyde test

    Method

    Warm ethanal with the blue reagent.

    Reason

    Ethanal reduces the copper(II) species in Fehling’s solution.

    Working

    Blue solution initially.

  2. State the visible result

    Method

    Observe a brick-red precipitate.

    Reason

    The precipitate is the characteristic positive Fehling result.

    Working

    Blue solution → brick-red precipitate.

Common misconception 3

Use Both Carbonyl Test Results

Find and correct the mistake

Learner claim

An unknown gives an orange precipitate with 2,4-DNPH but no reaction with Tollens’ reagent. A learner calls it an aldehyde because 2,4-DNPH was positive. Correct the deduction.

Separate class and subclass evidence

2,4-DNPH establishes
Conclusion

View solution step by step

  1. Use the broad test

    Method

    Infer that a carbonyl group is present.

    Reason

    2,4-DNPH reacts with both aldehydes and ketones.

    Working

    Orange precipitate → aldehyde or ketone.

  2. Use the discriminating test

    Method

    Exclude an aldehyde and deduce a ketone.

    Reason

    With fresh reagent and suitable warming, an aldehyde should reduce Tollens’ reagent. Among the supplied aldehyde-or-ketone alternatives, a negative result therefore supports the ketone.

    Working

    Carbonyl + Tollens negative → ketone.

Challenge 4

Deduce an Aldehyde from a Test Sequence

Minimal support

Evidence integration

An unknown compound gives an orange precipitate with 2,4-DNPH and a silver mirror with Tollens’ reagent. Deduce the functional group and justify each inference.

Build the two-stage inference

First test establishes
Silver mirror identifies

Hints

Hint 1: broad evidence

Use 2,4-DNPH to establish the broad functional-group family.

Hint 2: discriminator

Then use the silver mirror to distinguish aldehyde from ketone.

View solution step by step

  1. Establish a carbonyl

    Method

    Use the orange 2,4-DNPH precipitate.

    Reason

    This is positive evidence for a carbonyl compound.

    Working

    Unknown is an aldehyde or ketone.

  2. Deduce the subclass

    Method

    Use the silver mirror to identify an aldehyde.

    Reason

    Tollens’ reagent is reduced by an aldehyde but not a ketone in this test sequence.

    Working

    Functional group: aldehyde.

Common Mistakes

  • Claiming 2,4-DNPH distinguishes aldehydes from ketones (it doesn’t; it detects carbonyls).
  • Forgetting the observation for Tollens’ (silver mirror) or Fehling’s (brick-red precipitate).
  • Writing “addition polymerisation” language here: carbonyls typically undergo nucleophilic addition, not electrophilic addition.
  • Writing the conclusion without the observation (tests are “see → deduce” questions).

Use the Organic Chemistry topic check to practise and check your understanding.

Exam Tips

  • If asked to identify an unknown carbonyl:
    • use 2,4-DNPH first (is there a carbonyl at all?)
    • then use Tollens’/Fehling’s to decide aldehyde vs ketone
  • If asked “why is the carbonyl carbon attacked?”, mention δ + carbon due to polar C = O bond.
  • Write the carbonyl functional group clearly: aldehyde -CHO vs ketone > C = O (internal).

Mind Stretchers

Mind stretcher 1Extension

Suggest one reason why aldehydes are generally more reactive than ketones in nucleophilic addition.

Show Hint

Use the test observations to classify first; only then use structural evidence.

Show Answer

Mark scheme (any one):

  • Aldehydes have less steric hindrance around the carbonyl carbon.
  • Ketones have two alkyl groups which donate electron density and reduce the δ + on the carbonyl carbon.

Mind stretcher 2: Separating detection from identificationExtension

Question. An unknown gives an orange precipitate with 2,4-DNPH and a silver mirror with Tollens’ reagent. State what each result establishes and what remains unknown.

Show Hint

One test detects C=O; the other distinguishes an aldehyde from a ketone.

Show Answer

The 2,4-DNPH result supports a carbonyl compound. The Tollens’ result identifies it as an aldehyde rather than a ketone. Its exact carbon skeleton is still unknown without further evidence.

Syllabus and review details

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