Victor Meyer test: blue colour indicates a secondary alcohol. Volume of H2 = 0.560 L at STP → n(H2)=0.560/22.4=0.025 mol. For 2 ROH + 2 Na → 2 RONa + H2, 1 mol alcohol produces 1/2 mol H2, so n(alcohol)=2×0.025=0.05 mol. Molar mass = 3.7 g / 0.05 mol = 74 g·mol−1. Secondary alcohol with molar mass 74 is butan-2-ol (CH3CH(OH)CH2CH3). Tert‑butyl alcohol is tertiary (would give different Victor Meyer colour).
Esters (RCOOR') react with excess Grignard reagent: two equivalents of RMgX add to the carbonyl carbon to give tertiary alcohols after hydrolysis. Methyl propanoate (an ester) with MeMgBr (excess) yields a tertiary alcohol. Aldehydes give secondary, acids destroy Grignard, esters give tertiary (with excess).
Hydroboration–oxidation adds H and OH across a C=C in anti‑Markovnikov fashion: the OH ends up on the less substituted (terminal) carbon. If none of the given options shows the correct terminal (primary) alcohol product, the correct choice is 'None of these.' (Based on the options as printed, the expected product is a primary (terminal) alcohol resulting from anti‑Markovnikov addition.)
Hydration of ethene (A) gives ethanol (by addition of H2O). Reaction of an alkene with HOCl gives a halohydrin (i.e. 2‑chloroethan‑1‑ol if starting from ethene oxide/diol intermediates), but the simplest identification for A is ethanol; the net sequence returns water/hydration steps. (Given the printed options and common elementary sequences, A = ethanol and X corresponds to H2O in the hydration context.)
Nitro is a strong –I and –R (electron withdrawing) group; para‑nitro stabilizes the phenoxide ion effectively by resonance without strong intramolecular hydrogen bonding that would stabilize the undissociated phenol (as in ortho). Hence 4‑nitrophenol is the strongest acid among the choices.
Concentrated H2SO4 at high temperature dehydratively converts ethanol mainly to ethene (elimination). At lower temperature (around 140 °C) ether (diethyl ether) can be formed, but 'predominantly' with conc. H2SO4 (dehydration) gives ethene.
Carbolic acid is the historical name for phenol (C6H5OH).
The Reimer–Tiémann reaction: phenol + chloroform (trichloromethane, CHCl3) in presence of base gives ortho‑formylation to salicylaldehyde after hydrolysis. CHCl3 is trichloromethane.
Acidic dehydration follows Zaitsev's rule: the more substituted (more stable) alkene predominates. From the given substrate the most substituted (and thus major) alkene corresponds to the option that places the double bond to give the most substituted C=C (option a as printed).
Number the chain so that the –OH carbon is C‑1. Substituents are methyls at C‑2 and C‑3 and a chloro at C‑4 → 4‑chloro‑2,3‑dimethylpentan‑1‑ol (option a). (Option b is the same name with different ordering of substituents but option a is the standard format given.)
Phenol is more acidic than ethanol because deprotonation gives phenoxide ion which is resonance stabilized over the aromatic ring; ethanol gives ethoxide which has no such resonance stabilization. Hence both statements are true and the reason correctly explains the assertion.
Ethanol —(PCl5)→ ethyl chloride. Ethyl chloride —(alc. KOH)→ ethene (elimination). Hydration of ethene (H2SO4/H2O, 298 K) regenerates ethanol. Thus Z is ethanol.
This is the Williamson ether synthesis: an alkoxide ion reacts with an alkyl halide (CH3I) in an SN2 step to give an ether (R–O–CH3).
Side‑chain oxidation of alkylbenzenes (when the alkyl side chain has at least one benzylic hydrogen) gives benzoic acid (C6H5COOH). Cumene (isopropylbenzene) oxidizes to benzoic acid under strong oxidative conditions.
The –OH group on phenol is an activating, ortho/para directing group: it donates electron density by resonance to the ring, stabilizing the σ‑complex (arenium ion) formed during electrophilic substitution, so phenol is more reactive than benzene. Both statements are true and the reason correctly explains the assertion.
Periodic acid cleaves vicinal diols to give carbonyl compounds. Ethylene glycol (HOCH2CH2OH) is cleaved to two molecules of formaldehyde (methanal, HCHO).
Ethane‑1,2‑diol (ethylene glycol) is commonly used as antifreeze because it lowers the freezing point and raises the boiling point of the coolant mixture.
Treatment of a diol with base forms a di‑alkoxide which can undergo intramolecular alkylation with a dihalomethane (CH2I2) to form a cyclic ether — this is an intramolecular variant of the Williamson ether synthesis.
A is C4H10O. Ethers cleaved by HI give alkyl iodides. If A is diethyl ether (ethoxyethane), HI gives ethyl iodide (Y) which on boiling with aqueous KOH yields ethanol (Z). Ethanol gives a positive iodoform test. So A = ethoxyethane (diethyl ether).
HI cleaves R–O–CH3 ethers to give R–I and CH3OH (methanol) when the methoxy group is present. Any ether of the form R–O–CH3 (e.g. ethyl methyl ether or isopropyl methyl ether) yields methanol; among the typical textbook options the ethyl‑methyl ether (R = ethyl) is a common choice. Cleavage is by SN2 on the less hindered centre when R is primary, so methyl alcohol is produced.
Williamson ether synthesis proceeds by nucleophilic attack of an alkoxide on an alkyl halide in a single concerted step (back-side attack) — an SN2 mechanism. Example: \(CH_3O^- + CH_3Br \xrightarrow{SN2} CH_3OCH_3 + Br^-\).
Phenol reacts with neutral FeCl_3 to give a violet coloured complex (iron–phenoxide complex).
1‑Methoxypropane (methyl propyl ether, \(CH_3OCH_2CH_2CH_3\)) is protonated on O; I^- does an SN2 attack on the less hindered alkyl (methyl) to give \(CH_3I\) and \(CH_3CH_2CH_2OH\). With excess HI and heat the formed 1‑propanol can be converted to 1‑iodopropane (SN2 after protonation).
Methyl iodide and 1-propanol (and with excess HI and heat, 1-iodopropane). Mechanism: protonation of ether followed by SN2 (and further SN2 on the alcohol with excess HI).
1‑Ethoxyprop‑1‑ene is a vinyl (enol) ether: \(CH(OEt)=CHCH_3\). Protonation and cleavage of the enol ether by I^- yields the carbonyl compound (propanal) and ethyl iodide: \(CH(OEt)=CHCH_3 + HI \rightarrow CH_3CH_2CHO + C_2H_5I\).
Propanal (CH3CH2CHO) and ethyl iodide (C2H5I) — major carbonyl product: propanal.
Addition of R–MgX to R–CHO gives a secondary alkoxide which on acidic hydrolysis yields the symmetric secondary alcohol \(RCH(OH)R\). Example: \(CH_3MgBr + CH_3CHO \xrightarrow{H_3O^+} CH_3CH(OH)CH_3\) (isopropanol).
React a Grignard reagent R–MgX with an aldehyde of the same R group, R–CHO, followed by acid workup: R–MgX + R–CHO → R–CH(OMgX)–R → (H^+/H2O) → R–CH(OH)–R.
Methyl benzoate (PhCOOCH_3) reacts with two equivalents of EtMgBr: first equivalent gives the ketone (PhCOEt) after methoxide departure; second equivalent adds to the ketone to give the tertiary alkoxide. Acid hydrolysis yields the tertiary alcohol \(PhC(OH)(Et)_2\).
Tertiary alcohol: phenyl(ethyl)2-carbinol i.e. Ph–C(OH)(Et)2 (a tertiary alcohol obtained by double addition).
For CH_3–C(CH_3)=CH–CH_3 (2‑methyl‑2‑butene): (i) acid hydration gives OH at more substituted C (C‑2) → 2‑methyl‑2‑butanol. (ii) hydroboration–oxidation gives OH at less substituted C (C‑3) → 2‑methyl‑butan‑3‑ol. (iii) cold dilute KMnO_4 (Baeyer) adds OH syn across the double bond → vicinal diol: 2‑methyl‑butane‑2,3‑diol.
(i) 2‑Methylbutan‑2‑ol (Markovnikov addition, OH at C‑2). (ii) 2‑Methylbutan‑3‑ol (anti‑Markovnikov addition, OH at the less substituted carbon C‑3). (iii) 2‑Methylbutane‑2,3‑diol (vicinal diol at C‑2 and C‑3).
Boiling points increase with ability to form intermolecular H‑bonds and molecular surface area. (i) Tertiary alcohol (2‑methylpropan‑2‑ol) is sterically hindered → weakest H‑bonding and lowest bp; secondary (butan‑2‑ol) intermediate; primary (butan‑1‑ol) forms strongest intermolecular H‑bonding (highest bp). (ii) Single OH secondary (propan‑2‑ol) has lower bp than primary propan‑1‑ol; diol (1,3‑diol) has two OH groups → higher bp; triol (glycerol) has three OH groups → highest bp.
(i) 2‑Methylpropan‑2‑ol < Butan‑2‑ol < Butan‑1‑ol. (ii) Propan‑2‑ol < Propan‑1‑ol < Propan‑1,3‑diol < Propan‑1,2,3‑triol.
Strong nucleophiles tend to deprotonate the alcohol (forming alkoxide) rather than displace OH. To achieve substitution, first convert the –OH into a good leaving group (R–OTs, R–Cl via SOCl_2/PCl_5) or use protonation (acidic conditions) for tertiary/benzylic systems (SN1) or use appropriate activated substrates for SN2.
Not directly. Alcohols are poor substrates for nucleophilic substitution because OH^– is a poor leaving group; direct attack by nucleophiles (NH2^–, CH3O^–) usually leads to deprotonation rather than substitution. Convert OH to a better leaving group (e.g., tosylate, halide) or use acidic conditions (protonation of OH) to enable substitution.
Oxidation of alcohols requires at least one hydrogen on the carbon with –OH. Tertiary alcohols have no such hydrogen and are resistant to oxidation to carbonyl compounds; strong conditions cause C–C bond cleavage rather than simple oxidation.
No. Tertiary alcohols (like t‑butyl alcohol) lack a hydrogen on the carbon bearing the OH and cannot be oxidized to a carbonyl by acidified dichromate under normal conditions.
Acidified KMnO_4 oxidizes secondary alcohols to ketones. For PhCH(OH)CH_3 the product is PhCOCH_3 (acetophenone).
1‑Phenylethanol (a secondary benzylic alcohol) is oxidized to acetophenone (phenyl methyl ketone, PhCOCH3).
1) \(CH_3CH_2OH + H^+ \rightarrow CH_3CH_2OH_2^+\). 2) \(CH_3CH_2OH_2^+ \rightarrow CH_3CH_2^+ + H_2O\) (loss of water). 3) \(CH_3CH_2^+ \xrightarrow{Base} CH_2=CH_2 + H^+\) (deprotonation). Net: ethanol → ethene + H_2O.
Stepwise acid‑catalysed mechanism: (1) Protonation of ethanol to give ethyl oxonium ion. (2) Loss of water to form ethyl carbocation. (3) Deprotonation of the carbocation to give ethene and H^+.
i) Dow process: \(C_6H_5Cl + 2NaOH \xrightarrow{623K, high\, pressure} C_6H_5ONa + NaCl;\) acidify to get phenol. ii) Cumene process: \(PhCH(CH_3)_2 \xrightarrow{O_2} PhC(CH_3)_2OOH\) (cumene hydroperoxide) then \(\xrightarrow{H^+}\) cleavage gives phenol + acetone.
i) From chlorobenzene by nucleophilic aromatic substitution under harsh conditions (Dow process: heating with NaOH at high T and pressure to give sodium phenoxide, then acidify). ii) From isopropylbenzene (cumene) by the cumene process: cumene → cumene hydroperoxide (oxidation) → acid cleavage → phenol + acetone.
Sodium phenoxide is treated with CO_2 at about 125 °C and high pressure to give sodium salicylate; acidification yields salicylic acid: \(C_6H_5ONa + CO_2 \xrightarrow{125°C, high\, P} o\!‑C_6H_4(OH)CO_2Na \xrightarrow{H^+} o\!‑C_6H_4(OH)CO_2H\). This is widely used industrially to make salicylic acid.
Kolbe–Schmitt (Kolbe) reaction: carboxylation of sodium phenoxide with CO2 under pressure and heat to give salicylate; acidification yields salicylic acid (ortho‑hydroxybenzoic acid).
Williamson ether synthesis requires an alkoxide and an alkyl halide. From ethanol prepare sodium ethoxide. Convert 2‑methylpentan‑2‑ol into the corresponding alkyl halide (e.g. R–Cl). React NaOEt with R–Cl to obtain 2‑ethoxy‑2‑methylpentane (R–OEt) plus NaCl. (Note: tertiary halides often undergo SN1/ elimination; this sequence outlines the textbook Williamson approach.)
Steps: (1) Convert ethanol to sodium ethoxide. (2) Convert 2‑methylpentan‑2‑ol to the corresponding alkyl halide. (3) Nucleophilic substitution to give the ether. Example equations: (1) \(CH_3CH_2OH + Na \to CH_3CH_2O^-Na^+ + \tfrac{1}{2}H_2\). (2) \( (CH_3CH_2)C(CH_3)(OH)CH_2CH_3 \xrightarrow{PCl_5\,or\,HCl} (CH_3CH_2)C(CH_3)Cl + H_2O\) (tertiary chloride). (3) \(CH_3CH_2O^-Na^+ + RCl \to R–OCH_2CH_3 + NaCl\) (R = 2‑methylpentan‑2‑yl).
Target alcohol: 4‑methylpent‑2‑en‑1‑ol: \(HOCH_2–CH=CH–CH(CH_3)–CH_3\). The corresponding carbonyl derivatives at C‑1 are: aldehyde: \(O=CH–CH=CH–CH(CH_3)–CH_3\) (4‑methylpent‑2‑enal); acid: \(HOOC–CH=CH–CH(CH_3)–CH_3\) (4‑methylpent‑2‑enoic acid); ester: e.g. methyl 4‑methylpent‑2‑enoate: \(CH_3OOC–CH=CH–CH(CH_3)–CH_3\).
Aldehyde: 4‑methylpent‑2‑enal (HOCH_2 precursor oxidized gives aldehyde at C‑1). Carboxylic acid: 4‑methylpent‑2‑enoic acid. Ester: methyl (or ethyl) 4‑methylpent‑2‑enoate. (Structures correspond to replacing –CH2OH by –CHO, –COOH, –COOR respectively.)
All three have formula C4H10O but differ in the groups attached to oxygen, demonstrating metamerism.
Metamerism is an isomerism in ethers where compounds have the same molecular formula but different alkyl groups on either side of the oxygen. Metamers of 2‑methoxypropane (methyl isopropyl ether) with formula C4H10O: (i) 2‑methoxypropane (methyl isopropyl ether) – CH3–O–CH(CH3)2 (IUPAC: 2‑methoxypropane); (ii) 1‑methoxypropane (methyl n‑propyl ether) – CH3–O–CH2CH2CH3 (IUPAC: 1‑methoxypropane); (iii) ethoxyethane (diethyl ether) – CH3CH2–O–CH2CH3 (IUPAC: ethoxyethane).
i) Nucleophilic substitution of benzyl chloride with aqueous base gives benzyl alcohol. ii) Oxidation of the benzylic primary alcohol with dichromate or permanganate converts it to benzoic acid.
i) Benzyl chloride → benzyl alcohol: hydrolysis with aqueous NaOH (or water) via SN1/SN2: \(C_6H_5CH_2Cl + NaOH_{aq} \to C_6H_5CH_2OH + NaCl\). ii) Benzyl alcohol → benzoic acid: oxidation with strong oxidant (KMnO_4 or K_2Cr_2O_7/H_2SO_4): \(C_6H_5CH_2OH \xrightarrow{[O]} C_6H_5COOH\).
- PBr3 replaces OH by Br; aqueous NaOH restores the alcohol, and sodium forms the alkoxide.
- Zn dust removes oxygen from phenol to give benzene. Friedel-Crafts methylation gives toluene, whose benzylic side chain is oxidized to COOH.
- tert-Butylation occurs mainly para to OCH3. Chlorination then occurs ortho to OCH3. HBr cleaves the methyl aryl ether at the methyl-O bond, giving the phenol and CH3Br.
- Acid-catalysed dehydration gives the terminal alkene. Ozonolysis cleaves its CH=CH2 group into an aldehyde and formaldehyde.
- A = CH3CH2Br (bromoethane); B = CH3CH2OH (ethanol); C = CH3CH2ONa (sodium ethoxide).
- A = C6H6 (benzene); B = C6H5CH3 (toluene); C = C6H5COOH (benzoic acid).
- A = 4-tert-butylanisole; B = 2-chloro-4-tert-butylanisole; C = 2-chloro-4-tert-butylphenol. The ether-cleavage coproduct is CH3Br.
- A = 1-ethenyl-1-methylcyclohexane; B = 1-methylcyclohexane-1-carbaldehyde + HCHO (with H2O2 formed during the aqueous ozonide work-up).
Amount of CH_4 liberated = 112 cm^3. Moles CH_4 = 112/22400 = 0.005 mol ⇒ moles alcohol = 0.005. Molar mass = 0.44 g / 0.005 mol = 88 g mol^{-1}. A monohydric alcohol of M = 88 is C_5H_{11}OH (C_5H_{12}O). PCC oxidation gives a carbonyl that gives Tollens (silver mirror) ⇒ an aldehyde, so the alcohol is primary. A suitable identification is 1‑pentanol (C_5H_{11}OH).
n‑Pentanol (a primary C5 alcohol, formula C5H11OH) — e.g., 1‑pentanol.
i) Phenoxide reacts with benzoyl chloride to form the ester phenyl benzoate. The -OCOC6H5 group directs nitration mainly to the para position on the phenoxy ring; the ortho isomer is minor.
ii) Protonation of OH followed by loss of water gives a carbocation. Direct beta-elimination gives the more substituted Saytzeff alkene as the major product. A methyl-shift pathway also gives rearranged alkenes.
i) A = phenyl benzoate, C6H5COOC6H5; B = 4-nitrophenyl benzoate, C6H5COO-C6H4-NO2 (para, major). The ortho nitro isomer is minor.
ii) Major product: C6H5CH2CH=C(CH3)2. Rearranged products shown in the textbook solution are C6H5CH2C(CH3)=CHCH3 and C6H5CH2CH(CH3)CH=CH2, the terminal alkene being minor.
Distillation of phenol with Zn reduces phenol to benzene (deoxygenation): Ph–OH + Zn (distn.) → Ph–H (benzene). Friedel–Crafts alkylation of benzene with propyl chloride gives propylbenzene (A): Ph–H + CH3CH2CH2Cl (AlCl3) → Ph–CH2CH2CH3 (propylbenzene). Oxidation of the side chain of propylbenzene with strong oxidants (eg. KMnO4 / heat) oxidises the alkyl side chain to –CO2H giving benzoic acid (B): Ph–CH2CH2CH3 → Ph–CO2H (benzoic acid). Thus A = propylbenzene; B = benzoic acid.
A = n‑propylbenzene (propylbenzene), B = benzoic acid.
- CH3MgBr adds CH3 to the carbonyl carbon of cyclohexanone; acid work-up protonates the magnesium alkoxide to give a tertiary alcohol.
- HBr replaces the tertiary OH group by Br.
- Mg in dry ether converts the bromide into the Grignard reagent.
- A Grignard reagent adds to HCHO and, after acid work-up, gives a primary CH2OH group containing one additional carbon.
A = 1-methylcyclohexan-1-ol; B = 1-bromo-1-methylcyclohexane; C = (1-methylcyclohexan-1-yl)magnesium bromide; D = (1-methylcyclohexyl)methanol.
Sequence: cyclohexanone + CH3MgBr -> 1-methylcyclohexan-1-OMgBr --(H3O+)--> A --(HBr)--> B --(Mg / ether)--> C --(HCHO)--> (1-methylcyclohexyl)CH2OMgBr --(H3O+)--> D.
Acetyl chloride (CH3COCl) reacts with excess methyl Grignard: two equivalents of CH3MgBr add to the acyl chloride carbonyl carbon to give, after acidic hydrolysis, the tertiary alcohol (neopentyl-type) — actually here it gives 2‑methyl‑2‑propanol (commonly written as tert‑butyl alcohol) (X): CH3COCl + 2 CH3MgBr → (CH3)3C–OMgBr + MgBrCl →(H3O+) (CH3)3C–OH (X). Tertiary alcohols are resistant to oxidation by dichromate under ordinary conditions, so treatment with acidified K2Cr2O7 does not oxidise X. Therefore A = no reaction (X remains the tertiary alcohol). (If only one equivalent of CH3MgBr is used and reaction is stopped at the ketone stage, the intermediate ketone would be acetone; but the usual outcome with Grignard on acyl chlorides is further addition to give tertiary alcohol.)
A convenient route: 1) HC≡CH + NaNH2 → HC≡C–Na+ (sodium acetylide). 2) HC≡C–Na+ + CH3CH2Br → CH3CH2–C≡CH (1‑butyne). 3) Partial hydrogenation (H2, poisoned Pd or Lindlar) or selective hydrogenation to alkene: CH3CH2–C≡CH + H2 (Lindlar) → CH3CH2–CH=CH2 (1‑butene). 4) Hydroboration–oxidation (BH3/THF then H2O2, OH–) of 1‑butene gives anti‑Markovnikov alcohol: CH3CH2–CH=CH2 → CH3CH2CH2CH2OH (1‑butanol, n‑butyl alcohol). Overall: acetylene → 1‑butyne → 1‑butene → 1‑butanol.
Sequence: (1) form sodium acetylide, alkylate with ethyl bromide → 1‑butyne; (2) hydrogenate to 1‑butene; (3) hydroboration–oxidation of 1‑butene → 1‑butanol (n‑butyl alcohol).
- A = 2-chlorobutane, CH3CH(Cl)CH2CH3.
- B = sec-butylmagnesium chloride, CH3CH(MgCl)CH2CH3.
- X = butan-2-one, CH3COCH2CH3, formed by dehydrogenation of butan-2-ol over Cu at 573 K.
- Y = 3,4-dimethylhexan-3-ol, CH3CH2C(OH)(CH3)CH(CH3)CH2CH3, formed by addition of B to X followed by hydrolysis.
Stepwise mechanism (concise): 1) Protonation: –OH is protonated by H2SO4 to form –OH2+. 2) Loss of water (rate‑determining) gives a secondary carbocation on C‑2. 3) A 1,2‑methyl shift from the adjacent C‑3 (which bears two methyl groups) to C‑2 converts the secondary carbocation into a much more stable tertiary carbocation. 4) Deprotonation (by HSO4– or solvent) from the carbon adjacent to the carbocation yields the most substituted alkene. The product is the tetra‑substituted alkene (tetramethyl ethylene), commonly represented as (CH3)2C=C(CH3)2. This sequence (protonation → loss of water → carbocation rearrangement by methyl shift → elimination) explains formation of the highly substituted (and therefore thermodynamically favored) alkene as the major product.
Mechanism: acid‑catalyzed dehydration (E1). Protonation of OH → loss of H2O to give a secondary carbocation; a methyl shift (1,2‑shift) from the gem‑dimethyl center produces a more stable tertiary carbocation; deprotonation of the tertiary carbocation gives the highly substituted alkene (tetramethyl ethylene, i.e. (CH3)2C=C(CH3)2).