Alcohols, Ethers, Aldehydes and Ketones
Alcohols (classification, properties), carbonyl group, nucleophilic addition, hemiacetals/acetals, alcohol oxidation, reactions of aldehydes/ketones (reduction, Grignard addition, aldol condensation).
Complete Theory
4Alcohols (R—OH) are compounds characterized by the hydroxyl functional group —OH bonded to a saturated (sp³) carbon. They are classified into:
Acidity of alcohols: alcohols are very weak acids (pKₐ ≈ 15–18), but can be deprotonated by strong bases (NaH, NaNH₂, alkali metals) to form alkoxides RO⁻. Alkoxides are strong bases and excellent nucleophiles, used in the Williamson synthesis to produce ethers.
Ethers (R—O—R'): compounds with an oxygen atom bonded to two alkyl or aryl groups. They are polar aprotic solvents (diethyl ether, THF) with low boiling points compared to the corresponding alcohols (no intermolecular H-bonding). They are relatively inert, but the C—O bond can be cleaved by strong acids (HI, HBr).
Real-world applications: ethanol is the active ingredient of alcoholic beverages and a biofuel (E85). Isopropanol is a disinfectant. Methanol is a fuel and industrial precursor (formaldehyde, MTBE). THF and diethyl ether are essential laboratory solvents.
- Primary (1°): R—CH₂OH (e.g., ethanol CH₃CH₂OH)
- Secondary (2°): R₂CH—OH (e.g., isopropanol (CH₃)₂CHOH)
- Tertiary (3°): R₃C—OH (e.g., tert-butanol (CH₃)₃COH)
Acidity of alcohols: alcohols are very weak acids (pKₐ ≈ 15–18), but can be deprotonated by strong bases (NaH, NaNH₂, alkali metals) to form alkoxides RO⁻. Alkoxides are strong bases and excellent nucleophiles, used in the Williamson synthesis to produce ethers.
Ethers (R—O—R'): compounds with an oxygen atom bonded to two alkyl or aryl groups. They are polar aprotic solvents (diethyl ether, THF) with low boiling points compared to the corresponding alcohols (no intermolecular H-bonding). They are relatively inert, but the C—O bond can be cleaved by strong acids (HI, HBr).
Real-world applications: ethanol is the active ingredient of alcoholic beverages and a biofuel (E85). Isopropanol is a disinfectant. Methanol is a fuel and industrial precursor (formaldehyde, MTBE). THF and diethyl ether are essential laboratory solvents.
The carbonyl group (C=O) is one of the most important functional groups in organic chemistry. It is present in aldehydes, ketones, carboxylic acids, and their derivatives. Its structure and reactivity are determined by the polarization of the double bond:
Real-world applications: formaldehyde (HCHO) is used as a disinfectant and in resin production. Acetone ((CH₃)₂C=O) is a universal solvent. The carbonyl group is the central reactive site in countless biological and synthetic reactions.
- Oxygen, more electronegative than carbon, attracts both the and bond electrons, creating a dipole:
- The carbonyl carbon is electrophilic (partial positive charge) and can be attacked by nucleophiles
- The carbonyl oxygen is basic (lone pairs) and can be protonated in acidic medium, increasing the carbon's electrophilicity
- The carbonyl carbon has sp² hybridization: trigonal planar geometry, angles ≈ 120°
Real-world applications: formaldehyde (HCHO) is used as a disinfectant and in resin production. Acetone ((CH₃)₂C=O) is a universal solvent. The carbonyl group is the central reactive site in countless biological and synthetic reactions.
The characteristic reaction of aldehydes and ketones is nucleophilic addition to the carbonyl. General mechanism:
Real-world applications: acetal formation is fundamental for protecting the carbonyl group during reactions involving other parts of the molecule. Grignard reagents are among the most powerful tools in organic synthesis for forming C—C bonds.
- The nucleophile (Nu⁻) attacks the electrophilic carbonyl carbon
- The electrons of C=O shift to oxygen, forming an alkoxide intermediate (RO⁻)
- Protonation of the alkoxide in aqueous medium gives the final alcohol
- Reduction to alcohols: NaBH₄ reduces aldehydes to 1° alcohols and ketones to 2° alcohols. LiAlH₄ (more powerful) does the same. Mechanism: transfer of H⁻ (hydride) as nucleophile.
- Grignard addition (RMgX): Grignard reagents (R—MgX, X=Cl,Br,I) are strongly basic nucleophiles. Addition to formaldehyde gives 1° alcohols, to aldehydes gives 2° alcohols, to ketones gives 3° alcohols.
- Addition of HCN (hydrogen cyanide): forms cyanohydrins (R₂C(OH)CN). Reversible reaction, base-catalyzed. Cyanohydrins are intermediates for -hydroxyacid synthesis.
- An alcohol (R'OH) adds to the carbonyl forming a hemiacetal (R₂C(OH)OR')
- In excess alcohol, the hemiacetal reacts further losing H₂O and forming an acetal (R₂C(OR')₂)
Real-world applications: acetal formation is fundamental for protecting the carbonyl group during reactions involving other parts of the molecule. Grignard reagents are among the most powerful tools in organic synthesis for forming C—C bonds.
The aldol condensation is one of the most important C—C bond-forming reactions in organic chemistry. It involves two molecules of aldehyde or ketone in the presence of base (or acid).
Mechanism in basic medium:
Mechanism in basic medium:
- Enolate formation: the base abstracts an proton (adjacent to the carbonyl), forming an enolate ion, a resonance-stabilized intermediate (negative charge delocalized between C and O).
- Nucleophilic addition: the enolate (nucleophile) attacks the carbonyl of a second aldehyde/ketone molecule, forming a new C—C bond.
- Protonation: the alkoxide intermediate is protonated to give a eta-hydroxyaldehyde or eta-hydroxyketone (the aldol product).
- (Optional) Dehydration: upon heating, the aldol loses H₂O forming an alpha,eta-unsaturated compound (enone or enal).
- Self-condensation: two molecules of the same compound. Example: acetaldehyde → 3-hydroxybutanal (aldol) → 2-butenal (crotonaldehyde).
- Crossed (mixed) condensation: between two different carbonyl compounds. Requires that one of them has no H (e.g., formaldehyde, benzaldehyde) to avoid complex mixtures.
- Cannizzaro reaction: aldehydes without H (e.g., formaldehyde) in strong base undergo disproportionation: one molecule oxidizes to acid, the other reduces to alcohol.
Worked Examples
2Example 1Acetal formation
Given
Benzaldehyde (C₆H₅CHO) + excess ethanol (CH₃CH₂OH) in anhydrous HCl
Find
Reaction product
Explain the mechanism
Step-by-step solution
1Benzaldehyde (C₆H₅—CHO) reacts with two ethanol molecules in acidic medium to form an acetal. This is a carbonyl protecting group reaction.
2Mechanism: (1) protonation of C=O (increases electrophilicity), (2) attack by the first EtOH molecule (nucleophile) → hemiacetal, (3) protonation of the hemiacetal OH and loss of H₂O → O-stabilized carbocation, (4) attack by the second EtOH molecule → acetal.
3Product: benzaldehyde diethyl acetal (C₆H₅CH(OCH₂CH₃)₂). The acetal is stable under basic and neutral conditions, but hydrolyzes in acidic medium (reversible reaction).
✓ Final result: Benzaldehyde diethyl acetal (C₆H₅CH(OC₂H₅)₂) — carbonyl protecting group
Example 2Oxidation of alcohols
Given
Cyclohexanol + K₂Cr₂O₇/H₂SO₄ (Jones reagent)
Find
Oxidation product of cyclohexanol (2° alcohol)
Step-by-step solution
1Cyclohexanol is a secondary alcohol. The oxidizing agent (Cr(VI), chromic acid) oxidizes 2° alcohols to ketones: cyclohexanol → cyclohexanone. The color changes from orange (Crⱽᴵᴵ) to green (Crᴵᴵᴵ).
2The Jones oxidation mechanism involves formation of a chromate ester intermediate, followed by E2 elimination with C=O formation.
3Note: 1° alcohols are oxidized to aldehydes (with PCC, pyridinium chlorochromate, in CH₂Cl₂) or to carboxylic acids (with K₂Cr₂O₇/H₂SO₄). 3° alcohols do not oxidize (no H on the C—OH carbon).
✓ Final result: Cyclohexanol (2° OH) → cyclohexanone (ketone)
Exercises with Solutions
4Exercise 1Alcohol classificationMedium
Problem to solve
Classify the following alcohols as primary, secondary, or tertiary: (a) 1-butanol, (b) 2-butanol, (c) 2-methyl-2-propanol.
Given data
(a) CH₃CH₂CH₂CH₂OH(b) CH₃CH₂CH(OH)CH₃(c) (CH₃)₃COH
Step-by-step solution
1(a) The OH group is bonded to a terminal C (CH₂OH) → the carbon is bonded to one R group and two H → primary.
2(b) The OH is bonded to an internal C (CH(OH)) → the carbon is bonded to two R groups and one H → secondary.
3(c) The OH is bonded to a tertiary C (COH) → the carbon is bonded to three R groups → tertiary.
✓ Final answer: (a) primary, (b) secondary, (c) tertiary
Exercise 2Grignard additionHard
Problem to solve
What product is obtained from the reaction of Grignard reagent CH₃MgBr with 2-butanone (CH₃COCH₂CH₃), followed by acidic hydrolysis?
Given data
CH₃MgBr + CH₃COCH₂CH₃, then H₃O⁺
Step-by-step solution
12-butanone (CH₃—CO—CH₂CH₃) is a ketone. CH₃MgBr is the Grignard reagent (methylmagnesium bromide, strongly nucleophilic).
2The methyl anion (CH₃⁻) attacks the electrophilic carbonyl carbon. The alkoxide intermediate, after acidic hydrolysis, is protonated.
3Product: 2-methyl-2-butanol (CH₃—C(OH)(CH₃)—CH₂CH₃), a tertiary alcohol.
✓ Final answer: 2-methyl-2-butanol (tertiary alcohol, Grignard addition to ketone)
Exercise 3Aldol condensationHard
Problem to solve
Write the self-aldol condensation product of acetaldehyde (CH₃CHO) under basic conditions. What is obtained after heating?
Given data
2 CH₃CHO + OH⁻ (dilute)
Step-by-step solution
1Under basic conditions, acetaldehyde forms an enolate ion: CH₂⁻—CHO (resonance-stabilized). This enolate attacks the carbonyl of a second acetaldehyde molecule.
2The 3-hydroxybutanal (aldol) forms: CH₃CH(OH)CH₂CHO, a eta-hydroxyaldehyde.
3Upon heating (Δ), the aldol loses a water molecule (dehydration, E1cb elimination) forming an alpha,eta-unsaturated compound: 2-butenal (crotonaldehyde) CH₃CH=CHCHO.
✓ Final answer: 3-hydroxybutanal (aldol); dehydration → 2-butenal (crotonaldehyde)
Exercise 4Grignard alcohol synthesisVery Hard
Problem to solve
Propose a synthesis of 2-phenyl-2-propanol (C₆H₅C(OH)(CH₃)₂) using a Grignard reagent.
Given data
Target: 2-phenyl-2-propanol (tertiary alcohol)
Step-by-step solution
12-phenyl-2-propanol has a tertiary OH group bonded to a C with two methyls and one phenyl. We can form it by Grignard addition of methylmagnesium bromide (CH₃MgBr) to acetophenone (C₆H₅COCH₃).
2Reaction: C₆H₅COCH₃ + CH₃MgBr → C₆H₅C(O⁻MgBr⁺)(CH₃)₂, then H₃O⁺ → C₆H₅C(OH)(CH₃)₂.
3Alternative synthesis: addition of phenylmagnesium bromide (C₆H₅MgBr) to acetone ((CH₃)₂CO): (CH₃)₂CO + C₆H₅MgBr → same product. Both approaches are valid.
✓ Final answer: Acetophenone + CH₃MgBr (or acetone + C₆H₅MgBr), then H₃O⁺
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