Nucleophilic Addition to Carbonyls, Tollens'/Fehling's Tests, Esters & Acylation
Aldehydes, ketones, carboxylic acids, esters, acyl chlorides, and acid anhydrides all share the strongly polar carbonyl functional group (C^δ+=O^δ-). Because aldehydes and ketones lack a good leaving group on the carbonyl carbon, they undergo nucleophilic addition reactions with hydride ions (:H⁻ from NaBH₄) to form alcohols and with cyanide ions (:CN⁻ from KCN/H⁺) to form chiral hydroxynitriles as optically inactive racemic mixtures. In contrast, acyl chlorides and acid anhydrides possess electronegative leaving groups, enabling rapid nucleophilic addition-elimination (acylation) reactions with water, alcohols, ammonia, and primary amines at room temperature, essential for the industrial manufacture of aspirin. Examiners test every curly arrow in both mechanisms alongside Tollens' and Fehling's chemical tests and alkaline soap/biodiesel hydrolysis.
- Oxidation of Alcohols & Distinguishing Aldehydes from Ketones
- Primary alcohols are oxidised by acidified potassium dichromate(VI) (K₂Cr₂O₇ / H₂SO₄, colour change orange Cr₂O₇²⁻ to green Cr³⁺) to aldehydes (distil off immediately) and further to carboxylic acids (heat under reflux with excess oxidising agent). Secondary alcohols are oxidised to ketones; tertiary alcohols resist oxidation. Aldehydes are oxidised to carboxylic acids whereas ketones cannot be oxidised: 1. Tollens' reagent (ammoniacal silver nitrate, [Ag(NH₃)₂]⁺): warmed with an aldehyde, Ag⁺ is reduced to a metallic silver mirror (Ag(s)); ketones give no change. 2. Fehling's solution: blue Cu²⁺(aq) is reduced by an aldehyde to a brick-red Cu₂O(s) precipitate; ketones give no change.
- Nucleophilic Addition to Aldehydes & Ketones (:H⁻ and :CN⁻)
- 1. Reduction with NaBH₄ (sodium tetrahydridoborate(III) in aqueous/methanolic solution, source of hydride nucleophile :H⁻): reduces aldehydes to primary alcohols and ketones to secondary alcohols. 2. Reaction with KCN followed by dilute acid (HCl/H₂SO₄, safer than toxic gaseous HCN) to form a 2-hydroxy-nitrile (cyanohydrin): Step 1: Curly arrow from lone pair on :C⁻≡N to δ⁺ C of C^δ+=O^δ-, and curly arrow from the C=O double bond to the δ⁻ O atom, forming an alkoxide intermediate (R-CH(CN)-O⁻). Step 2: Curly arrow from lone pair on :O⁻ of the intermediate to H⁺ (or H-CN), forming the hydroxynitrile (R-CH(OH)CN).
- Optical Isomerism & Racemic Mixtures in Carbonyl Addition
- Optical isomers (enantiomers) are non-superimposable mirror images around a chiral carbon centre (a carbon bonded to four different groups); one enantiomer rotates plane-polarised light clockwise (+), the other by an equal angle anticlockwise (-). When :CN⁻ attacks an aldehyde (except methanal) or an unsymmetrical ketone, the C=O group is trigonal planar (120°). Therefore, the :CN⁻ nucleophile has an equal (50:50) probability of attacking the planar carbonyl carbon from above or below the plane, producing an equimolar (50:50) racemic mixture (racemate) of both enantiomers, which has zero net optical activity because the equal and opposite rotations cancel out.
Curly-Arrow Organic Mechanisms (Nucleophilic Substitution & Addition)
3 Golden Rules for Curly Arrows:
1. Start at electrons: Every curly arrow MUST start at a lone pair of electrons or the middle of a covalent bond (never start an arrow on a δ+ or positive charge).
2. Show dipoles: Always label δ+ and δ− on polar bonds like C–Br or H–Br.
3. Balance charges: If a neutral molecule loses a leaving ion (:Br⁻), the intermediate carbocation must carry a formal + charge.
Key Concepts & Exam Definitions (6)
- Oxidation of Alcohols & Distinguishing Aldehydes from Ketones
- Primary alcohols are oxidised by acidified potassium dichromate(VI) (K₂Cr₂O₇ / H₂SO₄, colour change orange Cr₂O₇²⁻ to green Cr³⁺) to aldehydes (distil off immediately) and further to carboxylic acids (heat under reflux with excess oxidising agent). Secondary alcohols are oxidised to ketones; tertiary alcohols resist oxidation. Aldehydes are oxidised to carboxylic acids whereas ketones cannot be oxidised: 1. Tollens' reagent (ammoniacal silver nitrate, [Ag(NH₃)₂]⁺): warmed with an aldehyde, Ag⁺ is reduced to a metallic silver mirror (Ag(s)); ketones give no change. 2. Fehling's solution: blue Cu²⁺(aq) is reduced by an aldehyde to a brick-red Cu₂O(s) precipitate; ketones give no change.
- Nucleophilic Addition to Aldehydes & Ketones (:H⁻ and :CN⁻)
- 1. Reduction with NaBH₄ (sodium tetrahydridoborate(III) in aqueous/methanolic solution, source of hydride nucleophile :H⁻): reduces aldehydes to primary alcohols and ketones to secondary alcohols. 2. Reaction with KCN followed by dilute acid (HCl/H₂SO₄, safer than toxic gaseous HCN) to form a 2-hydroxy-nitrile (cyanohydrin): Step 1: Curly arrow from lone pair on :C⁻≡N to δ⁺ C of C^δ+=O^δ-, and curly arrow from the C=O double bond to the δ⁻ O atom, forming an alkoxide intermediate (R-CH(CN)-O⁻). Step 2: Curly arrow from lone pair on :O⁻ of the intermediate to H⁺ (or H-CN), forming the hydroxynitrile (R-CH(OH)CN).
- Optical Isomerism & Racemic Mixtures in Carbonyl Addition
- Optical isomers (enantiomers) are non-superimposable mirror images around a chiral carbon centre (a carbon bonded to four different groups); one enantiomer rotates plane-polarised light clockwise (+), the other by an equal angle anticlockwise (-). When :CN⁻ attacks an aldehyde (except methanal) or an unsymmetrical ketone, the C=O group is trigonal planar (120°). Therefore, the :CN⁻ nucleophile has an equal (50:50) probability of attacking the planar carbonyl carbon from above or below the plane, producing an equimolar (50:50) racemic mixture (racemate) of both enantiomers, which has zero net optical activity because the equal and opposite rotations cancel out.
- Carboxylic Acids, Esterification & Acid vs Alkaline Ester Hydrolysis
- Carboxylic acids (-COOH) are weak acids that effervesce with carbonates (2RCOOH + Na₂CO₃ → 2RCOONa + H₂O + CO₂(g)). Reversible Fischer esterification (RCOOH + R'OH ⇌ RCOOR' + H₂O) uses concentrated H₂SO₄ catalyst heated under reflux. Ester hydrolysis: 1. Acid hydrolysis (dilute HCl/H₂SO₄ under reflux): reversible equilibrium forming carboxylic acid + alcohol. 2. Alkaline hydrolysis (hot aqueous NaOH under reflux): irreversible (goes to 100% completion) forming the carboxylate salt (RCOO⁻Na⁺) + alcohol (acidify with HCl afterwards to liberate free RCOOH).
- Triglyceride Saponification, Glycerol & Biodiesel (Transesterification)
- Vegetable oils and animal fats are triesters of propane-1,2,3-triol (glycerol) and long-chain fatty acids. 1. Saponification: heating a triglyceride under reflux with aqueous NaOH hydrolyses the three ester bonds to produce propane-1,2,3-triol (glycerol, used in cosmetics/pharmaceuticals as a humectant/solvent due to 3 -OH groups forming H-bonds) and 3 moles of sodium carboxylate salts (RCOO⁻Na⁺, which act as soaps). 2. Biodiesel: reacting vegetable oil triglycerides with methanol (CH₃OH) in the presence of a KOH catalyst converts them into glycerol and a mixture of methyl esters of long-chain fatty acids (biodiesel).
- Acylation: Nucleophilic Addition-Elimination & Aspirin Synthesis
- Acyl chlorides (RCOCl) and acid anhydrides ((RCO)₂O) react vigorously with nucleophiles (H₂O → carboxylic acid; ROH → ester; 2NH₃ → primary amide + NH₄Cl; 2R'NH₂ → N-substituted secondary amide + R'NH₃⁺Cl⁻). Mechanism for RCOCl + NuH (:OH₂, :OHR', :NH₃, :NH₂R'): Step 1 (Addition): Curly arrow from lone pair on nucleophile to δ⁺ C of C=O, and curly arrow from C=O double bond to O atom, forming a tetrahedral intermediate with :O⁻ and -Nu⁺H. Step 2 (Elimination): Curly arrow from lone pair on :O⁻ back to the C-O bond (reforming C=O), a curly arrow from the C-Cl bond to Cl (releasing :Cl⁻), and a curly arrow from the Nu-H bond onto Nu⁺ (releasing H⁺). In aspirin synthesis, ethanoic anhydride is preferred over ethanoyl chloride because it is cheaper, less corrosive, reacts less violently with water, and produces non-toxic ethanoic acid rather than corrosive, toxic HCl(g) fumes.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State the reagent, conditions, and observations used to distinguish between propanal and propanone.
Why can NaBH₄ reduce aldehydes and ketones to alcohols, but cannot reduce alkenes (C=C)?
Explain why the reaction of ethanal (CH₃CHO) with KCN followed by dilute HCl produces an optically inactive product even though 2-hydroxypropanenitrile has a chiral centre.
Why do chemists use potassium cyanide (KCN) followed by dilute sulfuric acid instead of hydrogen cyanide gas (HCN) when synthesising hydroxynitriles?
State the four organic products formed when ethanoyl chloride (CH₃COCl) reacts separately with: (1) Water, (2) Ethanol, (3) Ammonia, and (4) Methylamine.
State four reasons why ethanoic anhydride is used instead of ethanoyl chloride in the industrial manufacture of aspirin.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Outline the mechanism for the reaction of propanal (CH₃CH₂CHO) with potassium cyanide (KCN) followed by dilute hydrochloric acid, name the product according to IUPAC rules, and explain why the product does not rotate the plane of plane-polarised light. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Name of mechanism: Nucleophilic addition; IUPAC name of product: 2-hydroxybutanenitrile [1 mark]. 2. Step 1 (Nucleophilic attack): Show the C^δ+=O^δ- dipole on propanal; draw a curly arrow from the lone pair on the carbon atom of :C⁻≡N to the δ⁺ carbonyl carbon, and a curly arrow from the C=O double bond to the δ⁻ oxygen atom [2 marks]. 3. Step 2 (Protonation): Draw the alkoxide intermediate CH₃CH₂CH(CN)O⁻ with a lone pair on the :O⁻ atom, and draw a curly arrow from that lone pair to an H⁺ ion, forming 2-hydroxybutanenitrile [1 mark]. 4. The carbonyl group (C=O) in propanal is trigonal planar (bond angle ~120°) [1 mark]. 5. The :CN⁻ nucleophile attacks from above and below the plane of the C=O group with equal probability, producing an equimolar (50:50) racemic mixture of the two enantiomers, whose equal and opposite rotations of plane-polarised light cancel out [1 mark].
- Q2
Outline the mechanism for the reaction between ethanoyl chloride (CH₃COCl) and methanol (CH₃OH), name the mechanism, and write the overall equation for the reaction. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Name of mechanism: Nucleophilic addition-elimination; overall equation: CH₃COCl + CH₃OH → CH₃COOCH₃ + HCl [1 mark]. 2. Step 1 (Addition): Draw a curly arrow from a lone pair on the oxygen atom of methanol (CH₃ÖH) to the δ⁺ carbonyl carbon of CH₃COCl, and a curly arrow from the C=O double bond to the carbonyl oxygen atom [1 mark]. 3. Correctly draw the tetrahedral intermediate: CH₃-C(O⁻)(Cl)(O⁺(H)CH₃), showing the negative charge with a lone pair on :O⁻ and the positive charge on the oxygen atom of the attached methanol group [1 mark]. 4. Step 2 (Elimination): Draw a curly arrow from the lone pair on :O⁻ back into the C-O bond to reform the C=O double bond, and a curly arrow from the C-Cl bond to the Cl atom to eliminate :Cl⁻ [1 mark]. 5. Draw a curly arrow from the O-H bond onto the positively charged O⁺ atom to eliminate H⁺, yielding methyl ethanoate (CH₃COOCH₃) + HCl [1 mark].
- Q3
A student was given three unlabelled liquid samples known to be propanoic acid, propanal, and propan-2-ol. Describe a sequence of simple test-tube reactions (stating reagents, conditions, and observations) that would positively identify all three liquids. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Add aqueous sodium carbonate (Na₂CO₃) or sodium hydrogencarbonate (NaHCO₃) to a sample of each liquid at room temperature: only propanoic acid will effervesce (fizz), releasing CO₂ gas that turns limewater cloudy [1 mark]. 2. Warm the remaining two liquids in a water bath with Tollens' reagent (or Fehling's solution) [1 mark]. 3. Propanal will be oxidised and form a silver mirror on the inside of the test tube (or a brick-red precipitate with Fehling's), whereas propan-2-ol will show no change [1 mark]. 4. To positively confirm propan-2-ol (a secondary alcohol), warm it with acidified potassium dichromate(VI) (K₂Cr₂O₇ / H₂SO₄): the solution will turn from orange to green as propan-2-ol is oxidised to propanone [1 mark].
- Q4
A vegetable oil triglyceride is formed from propane-1,2,3-triol and three molecules of hexadecanoic acid (C₁₅H₃₁COOH). Write the structural equation for the conversion of this triglyceride into biodiesel using methanol, state the catalyst required, and explain why alkaline hydrolysis of the same triglyceride with NaOH produces soap. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Catalyst for biodiesel transesterification: Potassium hydroxide (KOH) or sodium hydroxide (NaOH) [1 mark]. 2. Equation for biodiesel formation: CH₂(OOC-C₁₅H₃₁)-CH(OOC-C₁₅H₃₁)-CH₂(OOC-C₁₅H₃₁) + 3CH₃OH → CH₂(OH)-CH(OH)-CH₂(OH) + 3C₁₅H₃₁COOCH₃ (propane-1,2,3-triol + 3 moles of methyl hexadecanoate) [2 marks]. 3. Heating the triglyceride with excess aqueous NaOH irreversibly hydrolyses the ester linkages to form propane-1,2,3-triol and sodium hexadecanoate (C₁₅H₃₁COO⁻Na⁺), which acts as a soap because it has a non-polar hydrophobic hydrocarbon tail (C₁₅H₃₁-) that dissolves grease and an ionic hydrophilic carboxylate head (-COO⁻Na⁺) that dissolves in water [1 mark].
Common Exam Questions (4)
What is the most common error students make when drawing the nucleophilic addition-elimination mechanism?
Trying to do the whole mechanism in a single step (like SN2) without drawing the tetrahedral intermediate. You must first break the C=O π-bond to form the tetrahedral intermediate containing :O⁻ and Cl, and then in the second step reform the C=O double bond from the :O⁻ lone pair while kicking out the :Cl⁻ leaving group.
Does the reaction of propanone (CH₃COCH₃) with KCN/H⁺ produce optical isomers?
No. Propanone is a symmetrical ketone (both alkyl groups are methyl, -CH₃). The product, 2-hydroxy-2-methylpropanenitrile ((CH₃)₂C(OH)CN), has two identical -CH₃ groups attached to C2, so C2 is not a chiral centre (not bonded to 4 different groups) and no optical isomers exist.
Why are carboxylic acids soluble in water for short chain lengths (up to C4), and why do they have higher boiling points than alcohols of the same Mr?
Both the C=O and O-H groups in -COOH can form strong hydrogen bonds with water molecules (though solubility decreases as the non-polar hydrocarbon chain gets longer). In pure liquid or gas phase, two carboxylic acid molecules form two hydrogen bonds with each other to create stable hydrogen-bonded dimers, doubling the effective Mr.
How do you purify an organic solid like aspirin by recrystallisation, and how do you test its purity?
Recrystallisation: 1. Dissolve impure solid in the minimum volume of hot solvent. 2. Filter hot through fluted filter paper to remove insoluble impurities. 3. Cool in an ice bath so pure crystals form (soluble impurities stay in solution). 4. Filter under reduced pressure (Buchner funnel), wash with cold solvent, and dry. Test purity by measuring its melting point: a pure solid has a sharp melting point that matches the data book value; impurities lower the melting point and broaden the melting range.
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