JEE Chemistry Alcohols Phenols and Ethers Notes Set 04

Alcohols: Definition and Classification

An alcohol has an OH group on a saturated (sp3) carbon. If the OH sits on a carbon of a C=C bond the compound is an enol, and if it is on a benzene ring the compound is a phenol. Benzyl alcohol (OH on a side-chain carbon) behaves like an ordinary alcohol.
By the number of OH groups, alcohols are monohydric, dihydric or trihydric. A monohydric alcohol is primary (1°) as \( RCH_2OH \), secondary (2°) as \( R_2CHOH \) and tertiary (3°) as \( R_3COH \).

Preparation of Alcohols

1. Hydrolysis of alkyl halides: warming with aqueous alkali or moist silver oxide replaces the halogen by OH through nucleophilic substitution. With a tertiary halide and strong, concentrated alkali, elimination to the alkene is the main reaction, so aqueous K2CO3 or moist Ag2O is preferred.

\[ \ce{R-X + NaOH ->[H2O] R-OH + NaX} \]
\[ \ce{(CH3)3C-Br ->[conc. NaOH] (CH3)2C=CH2 (major) + (CH3)3C-OH (minor)} \]

2. From alkenes.
(a) Acid-catalysed hydration: cold concentrated H2SO4 adds across the double bond following Markovnikov's rule to give an alkyl hydrogen sulphate, which on boiling with water gives the alcohol (propene gives propan-2-ol).
(b) Hydroboration-oxidation: \( B_2H_6 \) in THF, then \( H_2O_2 \) with OH-, adds water in an anti-Markovnikov way, so propene gives propan-1-ol.
(c) Oxymercuration-demercuration: \( Hg(OAc)_2 \) in THF-water, then \( NaBH_4 \) in alkali, adds water with Markovnikov orientation and without any rearrangement of the carbon skeleton. The OH comes from water and the new H from NaBH4.

3. Reduction of aldehydes and ketones to 1° and 2° alcohols.
Catalytic hydrogenation (H2 with Ni, Pt or Pd) reduces the C=O group and also any C=C bond. LiAlH4 (in ether) and NaBH4 reduce C=O but leave an isolated C=C untouched. NaBH4 is the milder reagent, reducing aldehydes, ketones and acid chlorides; LiAlH4 also reduces carboxylic acids and esters to alcohols and amides to amines. LiAlH4 can also reduce a C=C that is conjugated with a phenyl ring, as in cinnamaldehyde.
Bouveault-Blanc reduction uses sodium in ethanol. Meerwein-Ponndorf-Verley reduction uses aluminium isopropoxide in isopropyl alcohol and converts ketones into secondary alcohols.

\[ \ce{CH3CH=CHCHO + 2[H] ->[LiAlH4, ether] CH3CH=CHCH2OH} \]
\[ \ce{CH2=CHCHO + 4[H] ->[H2, Ni] CH3CH2CH2OH} \]

4. Hydrolysis of esters by acid or alkali gives the alcohol and a carboxylic acid. The reaction breaks the acyl-oxygen bond. The alcohol, being more volatile, is separated by distillation.

5. From ethers: heating with dilute H2SO4 under pressure hydrolyses an ether to alcohols. With concentrated HI a symmetric ether gives ROH and RI; a mixed ether gives the smaller alkyl group as iodide (unless one group is tertiary). Refluxing the products with aqueous KI converts the alkyl iodide to the alcohol as well.

6. From primary amines (only) with nitrous acid: \( RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O \). Methylamine is an exception: it gives methyl nitrite or dimethyl ether instead of methanol.

7. Reduction of carboxylic acids and derivatives: ease of catalytic hydrogenation is RCOCl > RCOOR > RCOOH, but it needs very high pressure and temperature. LiAlH4 reduces acids to primary alcohols and reduces an ester to two alcohols, one from the acyl part and one from the alkoxy part.

8. From Grignard reagents. The alcohol obtained depends on the carbonyl compound used:

9. Industrial methods.
Methanol (wood spirit) is made from water gas: \( C + H_2O \rightarrow CO + H_2 \), then \( CO + 2H_2 \rightarrow CH_3OH \) at about 300 °C and 200 atm over a ZnO/Cr2O3/CuO catalyst. It can also come from the destructive distillation of wood or from oxidation of methane over copper at 100 °C and 200 atm.
Ethanol is made by fermentation. Invertase turns molasses (sucrose) into glucose and fructose, and zymase converts both to ethanol and CO2. Starch is first hydrolysed by diastase and maltase to glucose before the same step.
The oxo process (alkene + CO + H2 with a cobalt catalyst, then H2/Ni) gives primary alcohols such as propan-1-ol from ethene.

Rectified spirit is 95.87 percent ethanol and 4.13 percent water, a constant-boiling (azeotropic) mixture boiling at 78.13 °C, so absolute alcohol cannot be obtained by simple distillation. Denatured (methylated) spirit is rectified spirit made undrinkable with methanol, pyridine or acetone. Power alcohol is petrol mixed with alcohol in about 4 : 1 ratio with benzene or ether.

Physical Properties of Alcohols

1. Lower alcohols dissolve readily in water by hydrogen bonding, and solubility falls as the carbon chain grows. For isomers solubility follows 1° > 2° > 3°.
2. Boiling points are far higher than those of alkanes or ethers of similar mass because of intermolecular hydrogen bonding. Among isomeric alcohols, 1° > 2° > 3°.
3. Lower alcohols form crystalline addition compounds such as \( CaCl_2\cdot 4C_2H_5OH \) and \( MgCl_2\cdot 6C_2H_5OH \), so alcohols cannot be dried over anhydrous CaCl2.
4. Methanol is highly toxic (it can cause blindness); ethanol is much less toxic and is a hypnotic.

Chemical Reactions of Alcohols

Because oxygen is electronegative, both the O-H and the C-O bonds are polarised. Cleavage of O-H explains the acidic behaviour (reactivity order CH3OH > 1° > 2° > 3°). Cleavage of C-O needs the OH to be converted into a better leaving group (H2O after protonation), which is why alcohols react with halogen acids but not with halide ions alone.

1. Reaction with active metals gives alkoxides and hydrogen: \( 2ROH + 2Na \rightarrow 2RONa + H_2 \). Reactivity order is CH3OH > 1° > 2° > 3°. Alcohols do not react with NaOH, and water decomposes an alkoxide back to the alcohol.
Acidity: water is slightly more acidic than most alcohols, giving H2O > CH3CH2OH > (CH3)2CHOH > (CH3)3COH. The bulky tert-butoxide ion is poorly solvated, so tert-butyl alcohol is the weakest acid. Relative acidity: H2O > ROH > RC≡CH > NH3 > RH > H2, and relative basicity of the conjugate bases runs the other way: H- > R- > NH2- > RC≡C- > RO- > OH-. Alcohols also react with a Grignard reagent to release an alkane (\( ROH + CH_3MgBr \rightarrow CH_4 + ROMgBr \)) and with diazomethane in the presence of BF3 to give methyl ethers.

2. Esterification.
With carboxylic acids (acid catalysed, reversible) the reaction goes by acyl-oxygen cleavage, as shown by labelling the alcohol oxygen with O-18, which ends up in the ester. Reactivity: for alcohols CH3OH > 1° > 2° > 3°; for acids HCOOH > CH3COOH > RCH2COOH > R2CHCOOH > R3CCOOH. Tertiary alcohols tend to eliminate instead.
With acid chlorides or anhydrides the reaction is irreversible, so yields are high. It can be used to count OH groups of a polyhydric alcohol: number of OH groups = (molar mass of the acetylated product - molar mass of the alcohol) / 42.
Sulphonyl chlorides (for example TsCl or MsCl) in the presence of a base give sulphonate esters. Inorganic acids give inorganic esters, such as alkyl nitrates with HNO3, dimethyl carbonate with carbonic acid, and methyl hydrogen sulphate or dimethyl sulphate with H2SO4.

3. Conversion into alkyl halides.
(a) With HX and anhydrous ZnCl2: reactivity of the acid is HI > HBr > HCl, and of the alcohol 3° > benzyl, allyl > 2° > 1° > CH3OH. Tertiary, allyl, benzyl and most secondary alcohols react by the SN1 path through a carbocation, so rearrangement is possible. Methyl and primary alcohols react by SN2.
(b) With PCl3, PCl5 or red P with Br2 or I2: no rearrangement. \( 3ROH + PCl_3 \rightarrow 3RCl + H_3PO_3 \) and \( ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl \).
(c) With thionyl chloride: \( ROH + SOCl_2 \rightarrow RCl + SO_2 + HCl \). Without a base the mechanism is SNi, so the configuration at a chiral carbon is retained. In the presence of pyridine the reaction is SN2 and the configuration is inverted.

4. Dehydration. Heating with excess concentrated H2SO4 or H3PO4 at 170 °C, or passing vapour over alumina at 350 °C, gives alkenes following Saytzeff's rule. Ease of dehydration is 3° > 2° > 1°. With excess alcohol at about 140 °C (or alumina at 260 to 280 °C) an ether is formed instead.

\[ \ce{CH3CH2OH ->[conc. H2SO4 (excess), 170 °C] CH2=CH2 + H2O} \]
\[ \ce{2CH3CH2OH ->[conc. H2SO4, 140 °C] CH3CH2OCH2CH3 + H2O} \]

5. With ammonia: alcohol vapour and NH3 passed over alumina or thoria at 360 °C give a mixture of 1°, 2° and 3° amines.
6. Reduction: \( ROH + 2HI \xrightarrow{\text{red P}} RH + I_2 + H_2O \).

7. Oxidation. This distinguishes the three classes of alcohol.
Primary alcohols give aldehydes and then carboxylic acids (acidified KMnO4, Na2Cr2O7 or chromic acid). Secondary alcohols give ketones (CrO3). Tertiary alcohols resist oxidation; under harsh conditions (hot KMnO4) C-C bonds break to give acids with fewer carbons.
Special oxidising agents: PCC in CH2Cl2 stops at the aldehyde stage and does not touch C=C bonds (it does not oxidise 3° alcohols). Chromic acid and Jones reagent (H2CrO4 in acetone at about 35 °C) oxidise 1° and 2° alcohols without attacking C=C, and Jones reagent is mainly used for 2° alcohols. MnO2 oxidises only allylic and benzylic 1° or 2° alcohols, giving aldehydes or ketones. Cycloalkanols with 50 percent HNO3 at 55 °C are cleaved to dicarboxylic acids.

8. Dehydrogenation over hot copper (300 °C or 573 K): 1° alcohols give aldehydes, 2° give ketones and 3° give alkenes (by dehydration).

Distinguishing Tests for 1°, 2° and 3° Alcohols

Lucas test: Lucas reagent is concentrated HCl with anhydrous ZnCl2. The zinc chloride is a Lewis acid that converts the OH into a better leaving group because chloride is only a weak nucleophile. The alkyl chloride is insoluble, so the solution turns cloudy.
Victor Meyer test: the alcohol is converted to the alkyl iodide (P/I2 or HI), then to the nitroalkane (AgNO2), and then treated with HNO2 followed by NaOH.

Allyl and benzyl alcohols react with Lucas reagent as fast as tertiary alcohols because they form stable carbocations.

Worked Problems

Worked Problem 1. A compound of formula \( C_7H_8O \) does not dissolve in water or dilute \( Na_2CO_3 \) but dissolves in dilute NaOH. With bromine water it readily gives a precipitate of \( C_7H_5OBr_3 \). Identify it.
Answer:
Benzyl alcohol and anisole do not dissolve in NaOH, so the compound must be a cresol (a phenol).
OH is strongly activating, so bromine replaces three ring hydrogens. Only m-cresol can take three bromines at positions activated by both OH and CH3 (2, 4 and 6 relative to OH).
The compound is m-cresol (3-methylphenol).

Worked Problem 2. A compound X (\( C_5H_8O \)) does not react appreciably with Lucas reagent at room temperature but gives a precipitate with ammoniacal \( AgNO_3 \). With excess MeMgBr, 0.42 g of X gives 224 mL of methane at STP. Hydrogenation over Pt, then boiling with excess HI, gives n-pentane. Find X.
Answer:
Precipitate with ammoniacal AgNO3 shows a terminal alkyne, \( -C\equiv C-H \).
Moles of X = 0.42 / 84 = 0.005; moles of CH4 = 224 / 22400 = 0.01. So one mole of X releases 2 mol CH4, i.e. it has two acidic hydrogens (the alkyne H and the OH).
No Lucas reaction means a primary alcohol, and the straight chain follows from n-pentane. X is pent-4-yn-1-ol, \( HC\equiv C-CH_2CH_2CH_2OH \).

Worked Problem 3. What are the main products when (a) ethanol is heated with propan-1-ol and conc. H2SO4 at about 140 °C, (b) ethanol is heated with tert-butyl alcohol and conc. H2SO4 at about 140 °C?
Answer:
(a) Both alcohols are primary, so no stable carbocation forms. By the SN2 path a mixture of three ethers is obtained: diethyl ether, dipropyl ether and ethyl propyl ether.
(b) tert-Butyl alcohol forms a stable carbocation (SN1), which is trapped by ethanol, so the main product is ethyl tert-butyl ether. The carbocation also loses a proton to give isobutene as a minor product (and this becomes the major product at 160 to 170 °C).

Worked Problem 4. 3,3-Dimethylbutan-2-ol is heated with hot conc. H2SO4. What is the major product and why?
Answer:
Protonation and loss of water give a secondary carbocation next to a quaternary carbon.
A methyl group shifts to the cationic carbon, giving the more stable tertiary carbocation.
Loss of a proton from this cation follows Saytzeff's rule and gives the most substituted alkene, 2,3-dimethylbut-2-ene (tetramethylethylene), as the major product. The minor product is 2,3-dimethylbut-1-ene.

Worked Problem 5. What is the stereochemical outcome when an optically active alcohol with a chiral carbon (a) is converted to an ether with Na then ethyl bromide, (b) is treated with TsCl and then HS-, (c) is treated with SOCl2?
Answer:
(a) Na only breaks the O-H bond and the ether forms by SN2 attack of the alkoxide on ethyl bromide, so the chiral carbon is untouched: configuration retained.
(b) TsCl converts OH to the tosylate (configuration retained at this step). HS- then attacks the chiral carbon from the back by SN2, so the configuration is inverted.
(c) SOCl2 without a base reacts by SNi, so configuration is retained; with pyridine it reacts by SN2 and the configuration is inverted.