**Alcohols** contain one or more hydroxyl (–OH) groups directly attached to carbon atoms of an **aliphatic (saturated or unsaturated) system**. Example: CH₃OH (methanol), CH₃CH₂OH (ethanol).
**Phenols** contain one or more hydroxyl (–OH) groups directly attached to carbon atoms of an **aromatic system**. Example: C₆H₅OH (phenol), methylphenol (cresol).
**Ethers** are compounds formed by replacing the hydrogen atom of the –OH group of an alcohol or phenol with an **alkyl (R–) or aryl (Ar–) group**. Example: CH₃OCH₃ (dimethyl ether), C₂H₅OC₆H₅ (ethyl phenyl ether).
**Based on number of hydroxyl groups:**
**Based on carbon hybridization (sp³ C–OH bonds):**
**Allylic alcohols** — –OH group attached to an sp³ carbon **adjacent to a carbon-carbon double bond** (C=C–CH(OH)–). Example: CH₂=CH–CH₂OH (prop-2-en-1-ol). Show enhanced reactivity due to conjugation with C=C.
**Benzylic alcohols** — –OH group attached to an sp³ carbon **next to an aromatic ring** (Ar–CH₂OH). Example: C₆H₅CH₂OH (benzyl alcohol or phenylmethan-1-ol). Show enhanced reactivity due to resonance stabilization of benzylic carbocation.
**Note:** Vinylic alcohols (–OH attached to sp² carbon of C=C) and vinyl alcohols are **unstable and tautomerize to carbonyl compounds** (CH₂=CH–OH → CH₃CHO).
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**Rule:** Replace the final 'e' of the alkane name with the suffix **'ol'**.
**Numbering:** Number the longest carbon chain starting from the end **nearest to the –OH group** to give the hydroxyl group the lowest possible number.
**Position indicators:** Use numbers to indicate positions of –OH group and other substituents.
**Polyhydric alcohols:** Retain the final 'e' of alkane and add the ending 'ol'. Use multiplicative prefixes (di-, tri-, tetra-) to indicate number of –OH groups, with position numbers separated by commas.
**Examples:**
**Rule:** Select the **larger alkyl/aryl group as the parent hydrocarbon**. The smaller group becomes a prefix using the term **'alkoxy'** or **'aryloxy'** (e.g., –OCH₃ is methoxy, –OC₂H₅ is ethoxy, –OC₆H₅ is phenoxy).
**Addition:** Insert the 'oxy' prefix at the position where the oxygen is attached to the parent hydrocarbon.
**Symmetrical ethers:** Use prefix **'di'** before the alkyl group. Example: C₂H₅–O–C₂H₅ → Diethyl ether or **ethoxyethane**.
**Unsymmetrical ethers:** Name alphabetically in common nomenclature; in IUPAC, use the larger group as parent.
**Examples:**
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#### (i) Acid-Catalyzed Hydration
**Reaction:** Alkenes react with water in the presence of **H₂SO₄ or H₃PO₄** (acid catalyst) to form alcohols.
**General:** R–CH=CH₂ + H₂O → (catalyst) → R–CHOH–CH₃
**Example:** CH₃–CH=CH₂ + H₂O → (H⁺/catalyst) → CH₃–CHOH–CH₃ (propan-2-ol)
**Regioselectivity:** For **unsymmetrical alkenes, follows Markovnikov's rule** — the H adds to the carbon with more H atoms, and OH adds to the carbon with fewer H atoms (more substituted carbon).
**Mechanism (Three-Step):**
**Step 1 (Protonation):** H₂O + H⁺ → H₃O⁺ (formation of hydronium ion)
**Step 2 (Nucleophilic Attack):** Water molecule (nucleophile) attacks the **carbocation** from the rear, forming a **C–O bond** and displacing the positive charge.
**Step 3 (Deprotonation):** The protonated alcohol (H₃O⁺ bonded to C) loses a proton (H⁺) to water, yielding the final alcohol and regenerating the H⁺ catalyst.
**Key Point:** Carbocation stability determines regioselectivity. More substituted carbocation is more stable, so OH adds to that carbon.
**Exam Tip:** Primary alcohols from primary alkenes are obtained in very low yield due to **primary carbocation's extreme instability**. Secondary and tertiary alcohols form readily.
#### (ii) Hydroboration–Oxidation
**Reaction:** Alkenes react with **diborane (BH₃)₂** (or borane in THF) followed by oxidation with **H₂O₂ in aqueous NaOH** to form alcohols.
**General:** R–CH=CH₂ + (BH₃)₂ → R–CHBH₂–CH₃ (trialkyl borane intermediate) → **(ii) H₂O₂/NaOH** → R–CHOH–CH₃ + B(OH)₃
**Example:** CH₃–CH=CH₂ + (BH₃)₂ → (step 1) → (CH₃–CH₂–CH₂–)₃B (tri-n-propylborane) → **(ii) H₂O₂/NaOH** → CH₃–CH₂–CH₂–OH (propan-1-ol)
**Regioselectivity (Anti-Markovnikov):** The **boron atom adds to the carbon with MORE hydrogen atoms**. The H adds to the carbon with FEWER hydrogen atoms. This produces the **opposite regiochemistry to Markovnikov's rule**.
**Stereochemistry:** Addition is **syn** (both B and H add to the same face of the C=C bond), and the subsequent oxidation **retains the configuration**.
**Advantage:** Excellent yield, anti-Markovnikov regioselectivity for primary alcohols, no carbocation rearrangement.
**Exam Context:** Used to prepare primary alcohols from terminal alkenes where acid-catalyzed hydration would give secondary alcohols.
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#### (i) Reduction of Aldehydes and Ketones
**Reaction with H₂ (Catalytic Hydrogenation):**
R–CHO + H₂ → (Pt/Pd/Ni catalyst) → R–CH₂OH (primary alcohol from aldehyde)
R₂CO + H₂ → (Pt/Pd/Ni catalyst) → R₂CHOH (secondary alcohol from ketone)
**Mechanism:** Hydrogen atoms add across the C=O double bond via a surface-catalyzed reaction. Metal catalyst dissociates H₂ into atomic H, which transfers to C and O.
**Reaction with NaBH₄ (Sodium Borohydride):**
R–CHO + NaBH₄ → (in EtOH/H₂O) → R–CH₂OH (primary alcohol)
R₂CO + NaBH₄ → (in EtOH/H₂O) → R₂CHOH (secondary alcohol)
**Mechanism:** NaBH₄ acts as a source of hydride ion (H⁻). The hydride nucleophilically attacks the electrophilic carbonyl carbon (C=O has C^δ⁺ and O^δ⁻). The resulting alkoxide ion is protonated by water during workup.
**Advantages of NaBH₄:** Safer and milder than LiAlH₄, compatible with esters and carboxylic acids (does NOT reduce them), works in aqueous/alcoholic solvents.
**Reaction with LiAlH₄ (Lithium Aluminum Hydride):**
R–CHO + LiAlH₄ → (in dry ether, then H₂O workup) → R–CH₂OH (primary alcohol)
R₂CO + LiAlH₄ → (in dry ether, then H₂O workup) → R₂CHOH (secondary alcohol)
**Mechanism:** LiAlH₄ is a strong reducing agent; H⁻ from Al–H bonds attacks the carbonyl carbon. Requires strictly anhydrous conditions and inert atmosphere.
**Advantages:** Stronger reducing agent than NaBH₄; reduces aldehydes, ketones, carboxylic acids, esters, amides, acid chlorides.
**Product Rule:**
#### (ii) Reduction of Carboxylic Acids and Esters
**From Carboxylic Acids (using LiAlH₄):**
R–COOH + (i) LiAlH₄ (ii) H₂O → R–CH₂OH (primary alcohol)
**Example:** CH₃CH₂COOH + LiAlH₄ → CH₃CH₂CH₂OH (propan-1-ol)
**Mechanism:** LiAlH₄ first deprotonates the carboxylic acid (very acidic; pKa ≈ 5), then attacks the C=O of the carboxylate ion with H⁻, followed by reduction of the intermediate to primary alcohol.
**Important:** NaBH₄ does NOT reduce carboxylic acids; LiAlH₄ is required.
**From Esters (Commercial Industrial Route):**
**Route 1 (Laboratory):** R'–O–CO–R + LiAlH₄ → (anhydrous ether) → R–CH₂OH + R'–OH
Example: CH₃–CO–O–CH₂CH₃ (ethyl acetate) + LiAlH₄ → CH₃–CH₂OH + CH₃–CH₂–OH (both form ethanol as a product, complex mixture)
**Route 2 (Industrial — via esterification then reduction):**
**Example:** CH₃CH₂COOH + R'OH → (H⁺) → CH₃CH₂COO–R' + H₂O → (H₂/Ni) → CH₃CH₂CH₂OH + R'OH
**Advantage of ester route:** Avoids expensive LiAlH₄; uses cheaper H₂ gas with recyclable catalysts.
**Product:** Ester reduction yields a **primary alcohol** from the acyl group (R–CO–O–R' → R–CH₂OH).
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**Grignard reagent:** R–Mg–X or Ar–Mg–X (e.g., CH₃–MgBr = methylmagnesium bromide), prepared by reacting an alkyl/aryl halide with Mg metal in dry ether.
**Reactions with Aldehydes and Ketones:**
General mechanism: Grignard is a **strong nucleophile and strong base**. The R⁻ (carbanion) attacks the electrophilic carbonyl carbon (C^δ⁺).
**Step 1 (Nucleophilic Addition):** R–MgX + R'–CHO → R–CH(O⁻MgX)–R' (adduct formation)
**Step 2 (Hydrolysis):** R–CH(O⁻MgX)–R' + H₂O → R–CH(OH)–R' + Mg(X)(OH) or MgXOH
**Products depend on the aldehyde/ketone used:**
**(i) With Methanal (HCHO):**
H–CHO + R–MgX → (i) H–CH(O⁻MgX) (ii) H₂O → **H–CH₂–OH (primary alcohol)**
Example: HCHO + CH₃–MgBr → CH₃–CH₂OH (ethanol, primary)
**(ii) With Other Aldehydes (R'–CHO, where R' ≠ H):**
R'–CHO + R–MgX → (i) R–CH(O⁻MgX)–R' (ii) H₂O → **R–CH(OH)–R' (secondary alcohol)**
Example: CH₃–CHO + C₂H₅–MgBr → CH₃–CH(OH)–C₂H₅ (butan-2-ol, secondary)
**(iii) With Ketones (R'₂CO):**
R'₂CO + R–MgX → (i) R–C(O⁻MgX)(R')₂ (ii) H₂O → **R–C(OH)(R')₂ (tertiary alcohol)**
Example: (CH₃)₂CO + C₆H₅–MgBr → (CH₃)₂C(OH)–C₆H₅ (2-methylpropan-2-ol, tertiary, also called phenylpropanone or 2-methyl-2-phenylpropan-1-ol)
**Key Point (Exam Important):**
**Conditions:** Strictly anhydrous; performed in dry ether (Et₂O or THF); Grignard is extremely moisture-sensitive and reactive.
**Limitation:** Grignard reagents react with any compound containing an active hydrogen (water, alcohols, carboxylic acids, amines, alkynes). Cannot be used in the presence of these groups.
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**Reaction:** Chlorobenzene (or other haloarenes) is treated with **molten NaOH at high temperature and pressure** (typically **623 K, ~300 atm**).
C₆H₅–Cl + NaOH → (623 K, 300 atm) → C₆H₅–O⁻Na⁺ + NaCl + H₂O (sodium phenoxide formed)
C₆H₅–O⁻Na⁺ + HCl (dilute) → C₆H₅–OH + NaCl (phenol released by acidification)
**Mechanism:** **Nucleophilic aromatic substitution (SNAr mechanism):**
**Step 1 (Addition):** The aromatic ring is activated toward nucleophilic attack by the electron-withdrawing Cl (which is attached to the sp² aromatic carbon). The nucleophilic OH⁻ attacks the aromatic carbon bearing Cl, forming a **Meisenheimer complex (negatively charged σ-adduct)** with a σ bond between OH⁻ and C, and the delocalized negative charge distributed over the ring.
**Step 2 (Elimination):** Cl⁻ leaves (as the leaving group), and the aromatic π system is restored, yielding phenoxide ion (C₆H₅–O⁻).
**Conditions required:** The reaction is slow and requires high temperature and pressure because aromatic C–X bonds are very strong (no carbocation intermediate like in SN1/SN2). Only halo**benzenes** or activated haloarenes (with electron-withdrawing groups like NO₂) undergo this reaction readily.
**Limitation:** Alkyl groups on the benzene ring (electron-donating) make the reaction slower; electron-withdrawing groups (like NO₂) activate the ring and accelerate the reaction.
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**Reaction:**
**Step 1 (Sulfonation of Benzene):**
C₆H₆ + H₂SO₄ (conc.) or oleum → C₆H₅–SO₃H (benzenesulfonic acid)
**Step 2 (Fusion with molten NaOH):**
C₆H₅–SO₃H + NaOH (molten, heated) → C₆H₅–O⁻Na⁺ (sodium phenoxide) + Na₂SO₃
**Step 3 (Acidification):**
C₆H₅–O⁻Na⁺ + HCl (dilute) → C₆H₅–OH + NaCl
**Mechanism of Step 2:** The –SO₃H group is a strong **electron-withdrawing group** (deactivating) via inductive effect. In molten NaOH at high temperature, nucleophilic OH⁻ attacks the aromatic ring (now more activated due to electron withdrawal by SO₃H), displacing the –SO₃H group as SO₃²⁻/SO₃H⁻.
**Advantage:** The reaction is cleaner than haloarene method because the leaving group (SO₃H) is better than Cl⁻. Also, the –SO₃H group **strongly activates the aromatic ring toward nucleophilic substitution** (ortho and para).
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**Reaction:** Aromatic primary amines are converted to phenols via diazonium salts.
**Step 1 (Diazotization):**
R–NH₂ (aromatic amine) + NaNO₂ + HCl (cold, 273–278 K) → R–N₂⁺Cl⁻ (diazonium chloride salt) + NaCl + H₂O
Example: C₆H₅–NH₂ (aniline) + NaNO₂ + HCl → C₆H₅–N₂⁺Cl⁻ (benzene diazonium chloride) + NaCl + H₂O
**Step 2 (Hydrolysis to Phenol):**
**Method A (Heating in water):**
R–N₂⁺Cl⁻ + H₂O → (warm, heat) → R–OH (phenol) + N₂↑ + HCl
C₆H₅–N₂⁺Cl⁻ + H₂O → (heat) → C₆H₅–OH (phenol) + N₂↑ + HCl
**Method B (Dilute acid):**
R–N₂⁺Cl⁻ + dilute H₂SO₄ or HCl → (heat) → R–OH + N₂↑ + H⁺
**Mechanism:** The diazonium salt is highly electrophilic and unstable. Water (or H₂O formed from dilute acid) attacks the N₂⁺ group (the positive charge makes N an excellent leaving group), displacing N₂ gas (very stable, excellent leaving group) and forming a carbocation, which is rapidly trapped by water to give the phenol.
Alternatively: R–N₂⁺ + H₂O → R⁺ + N₂↑ → R⁺ + H₂O → R–OH + H⁺
**Exam Important:** This reaction is **highly stereospecific and does NOT involve rearrangement**. The –N₂⁺ group is an excellent leaving group because N₂ is a very stable, inert gas.
**Advantage:** Very versatile; allows introduction of OH on an aromatic ring at specific positions by starting with the appropriately substituted aniline.
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**Reaction (Cumene Process or Isopropylbenzene Hydroperoxide Process):**
**Step 1 (Oxidation of Cumene to Cumene Hydroperoxide):**
C₆H₅–CH(CH₃)₂ (cumene) + O₂ (air) → C₆H₅–C(CH₃)₂–O–OH (cumene hydroperoxide) + byproducts
Performed at moderate temperature (~323 K) with oxygen/air in the presence of catalysts (or no catalyst).
**Step 2 (Acid-catalyzed rearrangement — Baeyer-Villiger rearrangement variant):**
C₆H₅–C(CH₃)₂–O–OH + dilute acid (H₂SO₄) → C₆H₅–OH (phenol) + CH₃–CO–CH₃ (acetone) + H₂O
**Mechanism (Step 2):** The hydroperoxide is protonated by the dilute acid, forming C₆H₅–C(CH₃)₂–O⁺–OH. A **phenyl group (C₆H₅)** migrates from the central carbon to the adjacent oxygen atom, with simultaneous cleavage of the O–O bond. This produces **phenyl acetate** (C₆H₅–O–CO–CH₃) as an intermediate, which is immediately **hydrolyzed by water to phenol and acetic acid**. The acetic acid further reacts with acetone equivalents or rearranges to acetone.
**Overall:** C₆H₅–C(CH₃)₂–O–OH → C₆H₅–OH + (CH₃)₂CO
**Industrial Significance:** This is the **major industrial method** for phenol production worldwide because:
**Exam Context:** Be aware this is the preferred industrial route; understand the mechanism of rearrangement.
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**General Trend:** Boiling points increase with increasing number of carbon atoms (increase in molar mass and van der Waals forces).
**Effect of Branching:** Branching decreases boiling points because it reduces the surface area available for intermolecular van der Waals forces. For example:
**Hydrogen Bonding Effect:**
Alcohols and phenols have **unusually high boiling points** compared to other organic compounds of similar molar mass because of **extensive intermolecular hydrogen bonding**.
Structure of hydrogen bond in alcohols: **R–O–H···O–H–R** (or **R–O–H···O–R'** for ethers). The H bonded to the highly electronegative O in one molecule is attracted to the lone pairs of O in a neighboring molecule.
**Comparison (molar masses similar, but BPs very different):**
The intermediate BP of dimethyl ether (between propane and ethanol) shows the importance of H-bonding from the –O–H group as a donor.
**Note on Phenols:** Phenols also have high boiling points due to H-bonding, though often lower than alcohols of similar molar mass because the aromatic ring reduces molecular flexibility and affects H-bonding geometry.
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**Principle:** "**Like dissolves like**" — alcohols and phenols are polar compounds and dissolve in the polar solvent water.
**Mechanism:** Alcohols and phenols form **hydrogen bonds with water molecules.**
Structure: **R–O–H···O–H₂** (alcohol H-bonds to water's O lone pairs) and **H₂O–H···O–R** (water H-bonds to alcohol's O lone pairs).
**Solubility Trend:**
**Examples of solubility in water at room temperature:**
**Phenols:** Generally more soluble in water than expected for simple hydrocarbons, but less soluble than alcohols of similar molar mass because the aromatic ring is bulky and hydrophobic. Phenol itself dissolves to ~8.3
Q1. Which of the following compounds is correctly named according to IUPAC nomenclature?
Answer: A — The longest chain is 4 carbons; numbering from the end nearest to −OH gives −OH at position 2, hence butan-2-ol (IUPAC). Option B is wrong because position 3 is farther from the −OH; option C uses 'hydroxy' as prefix which is incorrect for primary functional groups; option D is a common name, not IUPAC.
Q2. An alcohol has the structure: CH₂=CH−CH₂−CH₂−OH. Which statement correctly describes this compound?
Answer: C — The −OH is attached to sp³ C adjacent to the C=C (allylic position), making this an allylic alcohol. Option A ignores the unsaturation location; option B confuses vinylic alcohols (−OH on C=C) with allylic alcohols; option D is irrelevant as there is no aromatic ring.
Q3. The IUPAC name of HO−CH₂−CH₂−OH is:
Answer: B — For polyhydric alcohols, the 'e' of the alkane is retained and 'diol' is added with locants; the correct IUPAC name is ethane-1,2-diol. Option A omits the hyphen; option C is a common name; option D is incorrect nomenclature.
Q4. Classify the following alcohol: (CH₃)₂CH−OH
Answer: B — In (CH₃)₂CH−OH, the carbon bearing −OH is bonded to two other carbons (two methyl groups), making it a secondary carbon and secondary alcohol. Option A (primary) would be R−CH₂−OH; option C (tertiary) would be R₃C−OH.
Q5. Which of the following is an unsymmetrical ether?
Answer: C — An unsymmetrical ether has different alkyl or aryl groups attached to oxygen. Options A and B are symmetrical (dimethyl and diethyl ethers); option D is also symmetrical (diphenyl ether); only CH₃−O−C₂H₅ (ethyl methyl ether) has different groups.
Q6. The common name for C₂H₅−O−C₂H₅ is:
Answer: B — The compound has two identical ethyl groups attached to oxygen, so the common name is diethyl ether. Option A refers to a different compound; option C is incomplete; option D uses non-standard terminology.
Q7. Which pair of names refers to the same compound?
Answer: B — Propan-2-ol (IUPAC) is CH₃CH(OH)CH₃, which is the same as isopropyl alcohol (common name). Option A: propan-1-ol is n-propyl alcohol, not isopropyl; option C: butan-1-ol is n-butyl; option D: tert-butyl is 2-methylpropan-2-ol.
Q8. A compound has two −OH groups at positions 1 and 3 on a benzene ring. Which name is correct?
Answer: D — The IUPAC name is 1,3-benzenediol (positions 1 and 3 on benzene); meta-benzenediol refers to 1,3-disubstitution; the common name is resorcinol. All three naming methods are correct and accepted.
Q9. Assertion: Vinylic alcohols are stable compounds that can be isolated and studied. Reason: The C−OH bond in vinylic alcohols is strong due to sp² hybridization of the carbon.
Answer: C — Vinylic alcohols (enols) are highly unstable and immediately tautomerize to more stable aldehydes or ketones through keto-enol tautomerism. The assertion is false because these compounds cannot be easily isolated under normal conditions.
Q10. HOTS: Consider three compounds: (1) CH₃CH₂OH, (2) C₆H₅OH, and (3) HC≡C−OH. Rank them in order of increasing acidity and explain the basis for your ranking.
Answer: B — Acidity increases as the carbon bearing −OH becomes more sp-hybridized (greater s-character stabilizes the negative charge). Alkynes (sp C) are most acidic (~10⁻²⁵ pKa region for terminal alkynes), phenols (sp²) are weakly acidic (~10 pKa), and alcohols (sp³) are essentially neutral. Option D is incomplete; vinylic alcohols are unstable and do not persist to demonstrate acidity.
What is the key difference between an alcohol and a phenol in terms of carbon hybridization?
Alcohols have −OH attached to sp³ hybridized carbon (aliphatic), while phenols have −OH attached to sp² hybridized carbon (aromatic ring).
IUPAC nomenclature rule for alcohols: how do you name CH₃CH₂CH(OH)CH₃?
Identify the longest chain containing −OH, number from the end nearest to −OH (giving −OH position 2), name as butan-2-ol.
Define a primary alcohol with an example.
A primary alcohol has the −OH group attached to a carbon bonded to only one other carbon (R−CH₂−OH); example: CH₃CH₂OH (ethanol).
What is an allylic alcohol? Give an example.
An allylic alcohol has −OH attached to sp³ carbon adjacent to a C=C double bond; example: CH₂=CH−CH₂−OH (allyl alcohol).
Distinguish between symmetrical and unsymmetrical ethers.
Symmetrical ethers have identical alkyl or aryl groups on both sides of oxygen (R−O−R), while unsymmetrical ethers have different groups (R−O−R').
Name the compound C₂H₅−O−C₂H₅ by both common and IUPAC methods.
Common name: diethyl ether; IUPAC name: ethoxyethane (treating one ethyl as main chain and other as ethoxy substituent).
Why are vinylic alcohols (C=C−OH) unstable?
Vinylic alcohols tautomerize immediately to aldehydes or ketones because the enol form is thermodynamically unfavourable due to poor orbital overlap.
Classify CH₃CH(OH)CH₂CH₃ and state whether it is primary, secondary or tertiary.
This is butan-2-ol, a secondary alcohol, because −OH is attached to a secondary carbon bonded to two other carbons.
What functional group difference makes phenols acidic while alcohols are essentially neutral?
In phenols, the aromatic ring delocalizes the negative charge of the phenoxide anion through resonance, stabilizing it; alcohols lack this resonance stabilization.
Name the common name for HO−CH₂−CH₂−OH and its IUPAC name.
Common name: ethylene glycol; IUPAC name: ethane-1,2-diol (two −OH groups at positions 1 and 2 of a two-carbon chain).
Define a primary alcohol and write the IUPAC name of CH₃CH₂CH₂OH. [2 marks]
Primary alcohol = −OH attached to carbon bonded to only one other carbon (R−CH₂−OH). For IUPAC: identify longest chain (3 carbons), replace final 'e' with 'ol', number from −OH end (position 1).
Explain why phenols are more acidic than alcohols. (Show the role of resonance in your explanation.) [5 marks]
Phenols ionize to phenoxide anion: C₆H₅−OH ⇌ C₆H₅−O⁻ + H⁺. The negative charge on oxygen is delocalized into the aromatic π-system through resonance, stabilizing the conjugate base. Alcohols lack this aromatic ring, so the alkoxide anion (R−O⁻) cannot be resonance-stabilized, making loss of H⁺ much less favorable. Write resonance structures to show charge distribution in phenoxide.
A compound has the molecular formula C₅H₁₀O₂ and contains two −OH groups. One −OH is on a primary carbon and the other is on a secondary carbon. Draw the structure and give both the common and IUPAC names. Also classify this compound as monohydric, dihydric, or polyhydric. [6 marks]
Dihydric alcohol with 5-carbon main chain: primary −OH at end (position 1), secondary −OH at position 2 or higher. Structure: HO−CH₂−CH(OH)−CH₂−CH₂−CH₃ or similar. IUPAC: pentane-1,2-diol (number from primary −OH end). Common: 1,2-pentanediol or 2-hydroxypentanol (older style). Classification: dihydric because it contains exactly 2 −OH groups.
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