//M0//QN1//CS//DL0//EQ

Alcohols, phenols, and ethers are organic compounds containing oxygen. Alcohols have a hydroxyl (-OH) group attached to a saturated carbon atom, phenols have the -OH group attached to an aromatic ring, and ethers consist of an oxygen atom connected to two alkyl or aryl groups. These compounds exhibit distinct physical and chemical properties and are widely used in industries, laboratories, and everyday life. Ethanol (C2H5OH) is a widely used alcohol. It is synthesized by fermenting glucose (C6H12O6) under anaerobic conditions with the help of the enzyme zymase.(a) Write the reaction for the fermentation of glucose to produce ethanol.(b) Why does ethanol have a higher boiling point than butane?(c) Explain why ethanol is soluble in water.(d) Describe the reaction when ethanol reacts with sodium.

//X

(a) C6H12O6 2C2H5OH + 2CO2
(b) Ethanol forms strong intermolecular hydrogen bonds, whereas butane exhibits only weak Van der Waals forces.
(c) The -OH group in ethanol can form hydrogen bonds with water molecules, making it highly soluble.
(d) 2C2H5OH + 2Na → 2C2H5ONa + H2

//M0//QN2//CS//DL0//EQ

Dehydration of alcohols involves the removal of a water molecule to form either alkenes or ethers. The reaction is facilitated by protonic acids like concentrated H2SO4, H3PO4, or catalysts such as anhydrous ZnCl2 or Al2O3. Formation of Alkenes: When primary alcohols are heated with concentrated H2SO4 at 433–443 K, they undergo intramolecular dehydration to form alkenes. Secondary and tertiary alcohols dehydrate under milder conditions, and the order of dehydration ease is: 3° > 2° > 1°. Saytzeff’s Rule: The major product is the alkene with the more substituted double bond. Formation of Ethers: Primary alcohols, when heated with a protic acid like H2SO4 at

413K, undergo intermolecular dehydration to form dialkyl ethers.

Chemical Reactions:
CH3CH2OH CH2 = CH2 + H2O
2CH3CH2OH CH3CH2–O–CH2CH3 + H2O
(a) What products are formed during the dehydration of ethanol at 433K and 413K?(b) Why is dehydration of tertiary alcohols easier than that of primary alcohols?(c) State the rule followed during alkene formation by dehydration.(d) Write the reaction for the dehydration of propan-2-ol at 443K.

//X

(a) At 433K: Ethene (C2H4) is formed.
At 413K: Diethyl ether (C2H5OC2H5) is formed.
(b) Tertiary alcohols form more stable carbocations, which facilitates the dehydration process.
(c) Dehydration of alcohols follows Saytzeff’s rule, where the major product is the more substituted alkene.
(d) CH3CH(OH)CH3 CH3CH = CH2 + H2O

//M0//QN3//CS//DL0

Williamson’s synthesis is a reliable method to prepare symmetrical and unsymmetrical ethers via an SN2 mechanism. It involves the reaction of an alkoxide ion with a primary alkyl halide. Secondary and tertiary alkyl halides are not suitable because they undergo elimination, producing alkenes. Ethers can be cleaved by hydrogen halides to form alcohols and alkyl halides. The alkyl halide formed corresponds to the alkyl group with fewer carbon atoms due to lower steric hindrance. For unsymmetrical ethers, cleavage in polar medium proceeds via a carbocation intermediate.(1) Williamson’s Synthesis preparation of ethyl methyl ether: CH3ONa + C2H5Cl → CH3OC2H5 + NaCl(2) Cleavage of ethers with hydrogen halides:(a) Cleavage of symmetrical ether: CH3OCH3 + HI → CH3OH + CH3I(b) Cleavage of unsymmetrical ether (tertiary butyl ethyl ether):(CH3)3COC2H5 + HBr → (CH3)3–C–Br+C2H5OH(a) Why are primary alkyl halides preferred in Williamson’s synthesis?(b) Write the products of cleavage of dimethyl ether with HI.(c) What are the products of cleavage of tertiary butyl ethyl ether with HBr in polar medium?(d) Why does the cleavage of unsymmetrical ether proceed via the carbocation mechanism?

//X

(a) Primary alkyl halides undergo SN2 reactions efficiently, while secondary and tertiary alkyl halides tend to undergo elimination, leading to alkene formation instead of ethers.
(b) The cleavage of dimethyl ether produces methanol and methyl iodide.
CH3 – O – CH3 + HI → CH3OH + CH3I
(c) The products are tertiary butyl bromide and ethyl alcohol.
(d) In a polar medium, the reaction favours carbocation formation. The more stable tertiary carbocation from the bulky group directs the reaction pathway, leading to the formation of tertiary butyl bromide.

//M0//QN4//CS//DL0//EQ

Phenol reacts with NaOH to form sodium phenoxide, which reacts with CO2 under pressure and is acidified to produce salicylic acid. (a) Why does phenol require NaOH for Kolbe’s reaction?(b) What is the role of CO2 in this reaction?(c) Why is salicylic acid more acidic than phenol?(d) Why does the reaction produce predominantly ortho products?

//X

(a) NaOH converts phenol into sodium phenoxide, increasing nucleophilicity for reaction with CO2.
(b) CO2 acts as an electrophile, reacting with sodium phenoxide to form ortho-sodium salicylate.
(c) The electron-withdrawing carboxyl group stabilizes the phenoxide ion, enhancing acidity.
(d) The phenoxide ion directs the electrophile to the ortho position due to resonance effects.

//M0//QN5//CS//DL0

A pharmaceutical company manufactures antiseptic products containing isopropyl alcohol, phenol and cresols. To improve efficiency, researchers study the acidity, reactivity, solubility, dehydration and oxidation properties of several alcohols and phenols. They observe the following experimental results:

Phenol react with NaOH but ethanol does not.
Alcohols undergo dehydration in presence of concentrated H2SO4 at high temp.
Tertiary alcohols dehydrate faster than secondary and primary.
Phenol forms brominated product easily with Br2 water, giving a white precipitate of 2, 4, 6-tribromophenol.
Oxidation of alcohols depends on their.
1° Alcohol → Aldehyde → Acid 2° Alcohol → Ketone 3° Alcohol → No oxidation Researchers also compare acidity values Compound pka Ethanol 16 Water 15 - 7 Phenol 10(a) Why does phenol react with NaOH while ethanol does not? Explain using acidity and resonance.(b) Arrange the following in increasing order of dehydration rate:
1° Alcohol, 2° alcohol, 3° alcohol
(c) Why does phenol undergo electrophilic bromination easily, even without a catalyst?(d) Predict the products formed when isopropyl alcohol (2°) undergoes oxidation using acidified K2Cr2O7.

//X

(a) Phenol (pka 10) is much more acidic than ethanol (pka 16). So, phenoxide ion is stabilized by resonance. And ethoxide ion has no resonance stabilization. Thus, phenol forms phenoxide with NaOH; ethanol cannot.
(b) 3° > 2° > 1°
(c) Phenoxide ion strongly activates the ring:
(i) O– donates electrons (+ M)
(ii) Ring becomes highly reactive
(iii) Electrophilic substitution occurs fast, giving 2, 4, 6 - tribromophenol
(d) Isopropyl alcohol is a secondary (Keton) Product: Acetone (propanone).

//M0//QN6//CS//DL0

A perfume and flavor manufacturing company uses ethers and phenolic derivatives to prepare fragrances. They analyze the behavior of ethers in:

Acid cleavage, autoxidation, ether cleavage by HI / HBr, Williamson synthesis, boiling point variation. Experimental results: (i) Diethyl ether react with HI +O give ethanol + ethyl iodide. (ii) Anisole (methoxybenzene) gives phenol + methyl iodide with HI (iii) Ethers have lower boiling points than alcohols of comparable molecular mass. (iv) Phenol show stronger hydrogen bonding than alcohols. (v) Williamson synthesis works best with primary alkyl halides not tertiary.(a) Explain why ethers have lower boiling point than alcohol.(b) Write the cleavage products when diethyl ether react with excess HI.(c) Why does anisole produce phenyl and methyl iodide when treated with HI?(d) Why is Williamson ether synthesis unsucessful with tertiary alkyl halides?

//X

(a) Ether cannot form intermolecular H-bond whereas alcohol can. Thus ether molecules are less strongly held → BP is low.
(b) C2H5 – O – C2H5 + HI → C2H5I + C2H5OH
With excess HI
C2H5OH + HI → C2H5I + H2O
(c) In anisole, phenyl - oxygen bond is strong due to resonance. and O – CH3 bond is weaker.
C6H5 – O – CH3 + HI → C6H5OH + CH3I
Phenol form because the aromatic – O-bond is not cleaved.
(d) Tertiary alkyl halides undergo elimination (E2) instead of substitution.
Strong bases used in Williamson method course.
R3C – X → R2C = CR2 + HX
Hence ether cannot form.
Alcohol: Organic compound with an –OH group attached to an sp³ carbon (R–OH).
Phenol: Aromatic compound with –OH bonded directly to a benzene ring (Ar–OH); more acidic due to resonance.
Ether: Compound containing R–O–R′ (or Ar–O–R); neutral, less polar than alcohols, prepared by Williamson synthesis or acid dehydration.
1°, 2°, 3° Alcohols: Classified by the carbon carrying –OH (primary, secondary, tertiary), which determines oxidation and dehydration behaviour.
Benzylic Alcohol: Alcohol in which –OH is attached to a carbon next to an aromatic ring; highly reactive due to resonance stabilization.
Allylic/Vinylic: Allylic = carbon next to C=C; vinylic = carbon directly attached to C=C (vinylic halides are least reactive in substitution).
Williamson Ether Synthesis: SN2 formation of ethers by reacting an alkoxide (R–O⁻) with a primary alkyl halide (R′–X).
Grignard Reagent: Organomagnesium halide (R–MgX) used to form alcohols by addition to carbonyl compounds under anhydrous conditions.
Oxidation of Alcohols: 1° alcohols → aldehydes → acids; 2° → ketones; 3° resist oxidation under mild conditions.
PCC/KMnO4: PCC gives controlled oxidation (1° → aldehyde), whereas KMnO₄/dichromate fully oxidize (1° → acid).
Dehydration of Alcohols: Acid-catalysed elimination forming alkenes; ease: 3° > 2° > 1° (E1/E2 depending on substrate).
Kolbe Reaction: Phenoxide ion reacts with CO2 (electrochemical) to form ortho/para salicylic acid derivatives.
Reimer–Tiemann Reaction: Phenol forms o-hydroxybenzaldehyde using CHCl3 and NaOH (formylation reaction).
Acidity Trend: Phenols are far more acidic than alcohols due to resonance-stabilized phenoxide; EWGs increase acidity.
SN1 vs SN2 (R–OH → R–X): SN1 favoured by 3° carbocations (racemization), SN2 favoured by 1° substrates (inversion).