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): \(C_6H_5CH_2OH \xrightarrow{[O]} C_6H_5COOH\).
The provided reaction sequences are corrupted/ambiguous in the OCR text (missing subscripts, reagents and arrow placements). To give correct completions I need the original clear sequences (showing starting structures, reagents and intended steps). Please repost these four sequences clearly and I will complete them.
Unable to unambiguously complete the sequences because the OCR/text is garbled. Please provide a clearer, correctly formatted set of reaction sequences.
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.
The OCR for this item is unclear; two reasonable reconstructions:
(i) If the intended sequence is: Ph–OH + PhCOCl → (A) A + HNO3/H2SO4 (nitration) → (B)
Then: Ph–OH (phenol) on benzoylation with benzoyl chloride gives phenyl benzoate (Ph–O–CO–Ph) (A). Nitration of phenyl benzoate occurs on the phenyl ring bonded to oxygen; the –O– group (though its lone pair is tied up in the ester linkage) still directs ortho/para, so the major nitration product is the para-nitro derivative: p-nitrophenyl benzoate (B).
(ii) For the second line the OCR appears garbled; if it represents dehydration of a secondary alcohol R–CH(OH)–CH(R')–R'' in presence of conc. H2SO4 (or conversion of a β‑hydroxy compound under acidic conditions), the typical result is the corresponding alkene (Zaitsev's rule gives the more substituted alkene as major product).
Please supply the original clear text for a precise, stepwise completion.
Question text is ambiguous (OCR unclear). Please confirm the exact reactants and arrow directions. Likely interpretation: (i) Phenol + benzoyl chloride → phenyl benzoate (A). Nitration of phenyl benzoate gives nitro substitution mainly on the phenyl ring that is not benzoyl-substituted (para relative to the –O– group) — major product: p-nitrophenyl benzoate (B). (ii) If the second line is dehydration of a secondary alcohol in presence of conc. H2SO4, the product will be the corresponding alkene (Zaitsev product).
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.
The provided text is garbled. A common textbook sequence involving CH3MgBr and formaldehyde is: 1) Preparation: CH3Br + Mg (ether) → CH3MgBr (methylmagnesium bromide) (A). 2) Nucleophilic addition to formaldehyde: CH3MgBr + H–CHO → CH3–CH2O–MgBr (alkoxide intermediate) 3) Acid workup (H3O+) gives primary alcohol CH3CH2OH (ethanol) (D). If the problem intended another specific sequence (for example CH3MgBr reacting with an alkyl halide or with an acid chloride), please supply the clear original steps so the intermediates A,B,C,D can be identified exactly.
OCR for this item is unclear. Typical useful sequences are: - Formation of Grignard: R–Br + Mg/ether → R–MgBr (A). - Reaction of R–MgBr with HCHO followed by acid workup → primary alcohol (D). Provide the exact original text for a precise answer.
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).
Interpreting the sequence: 1) butan‑2‑ol (CH3–CH(OH)–CH2–CH3) + SOCl2 → 2‑chlorobutane (A) (inversion of configuration at C if chiral). 2) 2‑Chlorobutane + Mg/ether → sec‑butylmagnesium chloride (B, CH3–CH(MgCl)–CH2–CH3). 3) Treatment of an organomagnesium with copper salts gives organocuprate species which on heating can undergo coupling; coupling of two sec‑butyl units (Wurtz‑type or via organocuprate) leads to the symmetric C8 hydrocarbon (Y). The expected symmetrical coupling product from sec‑butyl units is the branched C8 alkane often written as 3,4‑dimethylhexane (one of the major coupling products). (Notes: such couplings give mixtures; the important intermediates are A = 2‑chlorobutane, B = sec‑butyl‑MgCl, X = formation of organocuprate, Y = C8 coupling product.)
A = 2‑chlorobutane. B = sec‑butylmagnesium chloride (the Grignard reagent). X = Cu (conversion to organocuprate) and Y = coupling product (symmetrical coupling of sec‑butyl groups) — major alkane: 3,4‑dimethylhexane (symmetrical C8 coupling product).
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).