Under catalytic hydrogenation with H2 at 1 atm over Pd/C, the carbon-carbon double bond of cyclohex-2-en-1-one is reduced while the ketone carbonyl remains unchanged. Product A is therefore cyclohexanone.
CN– attacks the electrophilic carbonyl carbon of acetone to give a tetrahedral alkoxide intermediate which upon protonation yields the cyanohydrin. This is nucleophilic addition to a carbonyl.
Answer: (c) Hydrazine in slightly acidic solution. Hydrazine adds to the carbonyl (nucleophilic addition) and then elimination of water gives a hydrazone.
Answer: (b) Victor Meyer test. Hydration of acetylene under these conditions gives acetaldehyde (CH3CHO), which gives Tollen's, Fehling's and iodoform tests, but the Victor Meyer test is for distinguishing types of alcohols, not for aldehydes.
Answer: c. Ozonolysis of ethene gives two molecules of formaldehyde (HCHO). Formaldehyde reacts with ammonia to give hexamethylenetetramine (urotropine): 4 NH3 + 6 HCHO → (CH2)6N4.
CH3COOH → CH3COCl (PCl5). Friedel–Crafts acylation on benzene gives acetophenone (C6H5COCH3). Addition of CH3MgBr to the ketone followed by acid workup gives the tertiary carbinol C6H5C(OH)(CH3)2, i.e. (CH3)2C(OH)C6H5.
2,2-Dimethylpropanoic acid (pivalic acid, (CH3)3C–COOH) has no α-hydrogen on the carbon adjacent to the carboxyl group; HVZ requires α-hydrogens to form the enol/enolate intermediate. Thus the assertion and reason are both true and the reason correctly explains the assertion.
Electron withdrawing power increases acidity. For haloacetic acids: F > Cl > Br; all are stronger acids than unsubstituted acetic acid. So FCH2COOH > ClCH2COOH > BrCH2COOH > CH3COOH.
On reduction of an appropriate derivative to aniline, treatment with NaNO2/HCl diazotizes the aromatic amine to give the benzene diazonium chloride (Ar–N2+Cl–). Thus C is benzene diazonium chloride.
The α-halogenation of carboxylic acids using P and Br2 (or PBr3/Br2) to give α-bromo acids is the Hell–Volhard–Zelinsky (HVZ) reaction.
CH3Br → CH3CN (KCN). Hydrolysis of CH3CN (acidic) gives CH3COOH (acetic acid). Treatment with PCl5 converts CH3COOH → CH3COCl (acetyl chloride). So (C) after PCl5 is acetyl chloride.
Formic acid (HCOOH) is easily oxidized (or acts as a reducing agent) and reduces Tollens' reagent (Ag(NH3)2+) to metallic silver, giving a silver mirror.
Grignard reagent RMgBr reacts with CO2 to give the carboxylate salt, which on acidification (H3O+) yields the corresponding carboxylic acid (R–COOH).
Number the chain from the carboxylic-acid carbon (C-1). In CH2=CH-CH2-COOH the carbons are C1(COOH)-C2H2-C3H=C4H2, so the C=C lies between C-3 and C-4: the name is but-3-ene-1-oic acid.
Hydrazine followed by a strong base carries out the Wolff-Kishner reduction. It converts the carbonyl group of acetophenone, C6H5COCH3, into a methylene group, giving C6H5CH2CH3 (ethylbenzene).
Option (a) is a symmetrical ketone. After HCN addition, the former carbonyl carbon is bonded to OH, CN and two identical ethyl groups, so it is not chiral. The other carbonyl compounds have two different carbon groups at the carbonyl carbon, so cyanohydrin formation generates a chiral center.
Answer: (b) Both assertion and reason are true, but the reason is not the correct explanation. The aldehyde can undergo benzoin condensation (cyanide-catalysed coupling). The directing influence of ‑CHO in electrophilic substitution (meta directing) is true but irrelevant to the benzoin condensation mechanism.
The Cannizzaro reaction is a disproportionation in which one molecule of an aldehyde (without α‑H) is reduced to the corresponding alcohol while another is oxidized to the carboxylic acid.
Benzaldehyde (phenylmethanal) lacks α‑hydrogen and undergoes the Cannizzaro reaction with concentrated base to give benzyl alcohol (reduction product) and benzoic acid (oxidation product).
Answer: b. Fehling's solution oxidises aliphatic aldehydes (e.g. acetaldehyde gives a red precipitate) but does not oxidise aromatic aldehydes like benzaldehyde, so it distinguishes between them.
Phenyl methanal = benzaldehyde (PhCHO). Under concentrated NaOH (no α‑H) benzaldehyde undergoes the Cannizzaro reaction to give benzyl alcohol (PhCH2OH) and sodium benzoate (PhCO2– Na+). The alcohol (X = PhCH2OH) reacts with metallic Na to give H2. Thus X = phenyl methanol (benzyl alcohol) and Y = sodium benzoate.
Aldol, Friedel–Crafts (acylation/alkylation) and Kolbe electrolysis (coupling of radicals) all form new C–C bonds. Wolff–Kishner reduction converts a carbonyl to a methylene (C=O → CH2) and does not form a new C–C bond.
Alkene A that on ozonolysis gives acetone (CH3COCH3) and acetaldehyde (CH3CHO) is CH3–CH= C(CH3)2. Addition of HCl (Markovnikov) places Cl on the more substituted carbon (tertiary carbon). The major product is CH3–CH2–C(Cl)(CH3)2, i.e. H3C–CH2–C(Cl)(CH3)2, which corresponds to option (c).
Carboxylic acids form strong intermolecular hydrogen‑bonded dimers (O–H···O) which greatly increases their effective molecular association and hence their boiling points compared with comparable alcohols, aldehydes or ketones.
(a) From an alcohol: Oxidation of 1‑propanol with KMnO4 or K2Cr2O7/H+: \(CH3CH2CH2OH \xrightarrow{[O]} CH3CH2COOH\).
(b) From an alkyl halide: Convert 1‑bromopropane to Grignard and carboxylate with CO2, then acidify: \(CH3CH2CH2Br \xrightarrow{Mg/ether} CH3CH2CH2MgBr \xrightarrow{CO2} CH3CH2CH2CO2^- \xrightarrow{H3O^+} CH3CH2COOH\). Alternatively: \(CH3CH2CH2Br \xrightarrow{NaCN} CH3CH2CH2CN \xrightarrow{H3O^+, heat} CH3CH2COOH\).
(c) From an alkene: Hydroboration–oxidation to give 1‑propanol then oxidation: \(CH3CH=CH2 \xrightarrow{1) BH3/THF 2) H2O2/OH^-} CH3CH2CH2OH \xrightarrow{[O]} CH3CH2COOH\).
See solution
A = benzonitrile (C6H5CN). Hydrolysis: \(C6H5CN + 2H2O + H^+ \to C6H5COOH\) (B = benzoic acid). Thionyl chloride: \(C6H5COOH + SOCl2 \to C6H5COCl\) (C = benzoyl chloride). Friedel–Crafts acylation: \(C6H6 + C6H5COCl \xrightarrow{AlCl3} C6H5COC6H5\) (D = benzophenone). Clemmensen reduction: \(C6H5COC6H5 \xrightarrow{Zn/Hg, HCl} C6H5CH2C6H5\) (E = diphenylmethane). (Equations written above.)
See solution
X: the alkoxide salt, e.g. CH3C(OMgBr)(CH3)CH2CH2COOEt (magnesium alkoxide). Y (after H3O+): CH3C(OH)(CH3)CH2CH2COOEt (the alcohol: 3‑hydroxy‑3‑methylbutanoate ester). (Identity: X = alkoxide intermediate; Y = the corresponding tertiary alcohol on the former ketone carbon.)
Starting compound is a β‑keto ester (ethyl 4‑oxobutanoate, shown as CH3COCH2CH2COOEt). Addition of methyl Grignard (1 equiv.) attacks the ketone carbonyl to give the magnesium alkoxide intermediate X; acidic workup gives the corresponding tertiary/secondary alcohol Y.
benzoic acid --(PCl5)--> A = benzoyl chloride (C6H5COCl). Benzene + benzoyl chloride/AlCl3 --> B = benzophenone (C6H5COC6H5) (Friedel–Crafts acylation). B + H^+ / C2H5OH (acidic ethanol) converts the ketone to its acetal (diethyl acetal): C = benzophenone diethyl acetal, (C6H5)2C(OC2H5)2. (Alternatively, protonation/acetalization of the ketone with ethanol produces the corresponding ketal/acetal.)
One convenient sequence (common textbook transformations): (1) D = CH3COCl (acetyl chloride) obtained from ethanoic acid by SOCl2: \(CH3COOH + SOCl2 \to CH3COCl + SO2 + HCl\). (2) Acetyl chloride (D) on Rosenmund reduction (H2/Pd–BaSO4) gives acetaldehyde (A): \(CH3COCl \xrightarrow{H2, Pd/BaSO4} CH3CHO\). (3) Acetaldehyde (A) on treatment with NaOH undergoes aldol condensation to give the aldol product (B) = 3‑hydroxybutanal which on heating dehydrates to crotonaldehyde (C). Thus identifications: A = acetaldehyde, B = 3‑hydroxybutanal (or its dehydrated product crotonaldehyde after heating), C = crotonaldehyde, D = acetyl chloride. (Any equivalent sequence that uses SOCl2 then Rosenmund and NaOH aldol is acceptable.)
A plausible set (standard textbook scheme): Take A = CH3–CH=C(CH3)2 (an alkene which ozonolyses to acetone and an aldehyde). If B is isobutyraldehyde (2‑methylpropanal), oxidation gives C = 2‑methylpropanoic acid (isobutyric acid). HVZ α‑bromination (Br2/P) on C gives D = α‑bromo‑2‑methylpropanoic acid; hydrolysis of the C–Br gives E = 2‑hydroxy‑2‑methylpropanoic acid (α‑hydroxy‑isobutyric acid). Also acetone on treatment with HCN (to give acetone cyanohydrin) followed by hydrolysis of –CN gives the same α‑hydroxy acid E: (CH3)2C(OH)COOH. Thus: A = the alkene that yields acetone + 2‑methylpropanal on ozonolysis; B = 2‑methylpropanal; C = 2‑methylpropanoic acid; D = α‑bromo‑2‑methylpropanoic acid; E = 2‑hydroxy‑2‑methylpropanoic acid. (Note: the exact structural formula of A is the alkene whose cleavage gives those carbonyls — e.g. appropriate branched pentene.)
See solution
(ii) benzaldehyde → benzoic acid: Oxidation (KMnO4 or K2Cr2O7/H+): \(PhCHO \xrightarrow{[O]} PhCOOH\).
(iii) benzaldehyde → α‑hydroxyphenylacetic acid: Addition of HCN to give benzaldehyde cyanohydrin, then hydrolysis of –CN to –COOH: \(PhCHO + HCN \to PhCH(OH)CN \xrightarrow{H3O^+, heat} PhCH(OH)COOH\). (This is mandelic acid, α‑hydroxyphenylacetic acid.)
(i) benzaldehyde → benzophenone: Oxidize benzaldehyde to benzoic acid, convert to benzoyl chloride, then Friedel–Crafts acylation with benzene: \(PhCHO \xrightarrow{[O]} PhCOOH \xrightarrow{SOCl2} PhCOCl;\; PhCOCl + PhH \xrightarrow{AlCl3} PhCOPh\).
HCN adds to carbonyl compounds to give cyanohydrins. (i) Propanone (acetone): \((CH3)2CO + HCN \to (CH3)2C(OH)CN\) (acetone cyanohydrin, 2‑hydroxy‑2‑methylpropanenitrile). (ii) 2,4‑Dichlorobenzaldehyde: \(Cl2C6H3CHO + HCN \to Cl2C6H3CH(OH)CN\) (the corresponding aromatic cyanohydrin). (iii) Ethanal (acetaldehyde): \(CH3CHO + HCN \to CH3CH(OH)CN\) (acetaldehyde cyanohydrin, 2‑hydroxypropanenitrile). Cyanohydrins can be further hydrolysed to α‑hydroxy acids on acidic hydrolysis.
Positive iodoform and no Fehling reduction identifies a methyl ketone (CH3CO–R). A C5H10O methyl ketone is pentan‑2‑one (CH3COCH2CH2CH3) or pentan‑2‑one (also written 2‑pentanone). 2‑pentanone forms a bisulphite adduct (bisulphite soluble) and gives iodoform. Therefore A = pentan‑2‑one (2‑pentanone).
Acetone (enolate donor) condenses with two equivalents of benzaldehyde to give dibenzalacetone (major product), structure: PhCH=CH–CO–CH=CHPh (trans,trans‑dibenzalacetone). (This is the crossed aldol condensation product where both α‑positions of acetone are condensed with benzaldehyde.)
- (a) propanal into butanone
- (b) Hex-3-yne into hexan-3-one.
- (c) phenylmethanal into benzoic acid
- (d) phenylmethanal into benzoin
(a) propanal → butanone (CH3COCH2CH3): Reduce propanal to propanol (NaBH4), convert to alkyl halide (PBr3), make Grignard (Mg/ether) and acylate with acetyl chloride: CH3CH2CHO \xrightarrow{NaBH4} CH3CH2CH2OH \xrightarrow{PBr3} CH3CH2CH2Br \xrightarrow{Mg} CH3CH2CH2MgBr; \; CH3CH2CH2MgBr + CH3COCl → CH3COCH2CH2CH3 (butanone). (Alternatively prepare propyl Grignard and react with acetyl chloride to give butanone.)
(b) Hex‑3‑yne → hexan‑3‑one: hydration of internal alkyne (HgSO4/H2SO4) or hydroboration–oxidation (for anti‑Markovnikov) but for internal symmetrical give hexan‑3‑one: \(R–C≡C–R \xrightarrow{H2O, HgSO4/H2SO4} R–CO–CR\).
(c) phenylmethanal (benzaldehyde) → benzoic acid: oxidation with KMnO4 or Ag2O/Tollen's: \(PhCHO \xrightarrow{[O]} PhCOOH\).
(d) benzaldehyde → benzoin: benzoin condensation (cyanide catalysed): \(2 PhCHO \xrightarrow{CN^-} PhCH(OH)COPh\) (benzoin).
See solution
Under dry HCl, pentan-2-one reacts with propane-1,3-diol to form the cyclic ketal 2-methyl-2-propyl-1,3-dioxane and water.
HOCH2CH2CH2OH + CH3CH2CH2COCH3 --(dry HCl)--> 2-methyl-2-propyl-1,3-dioxane + H2O.
Benzyl bromide (C6H5CH2Br) + NaCN → A = benzyl cyanide (phenylacetonitrile, C6H5CH2CN). Hydrolysis of A (acidic hydrolysis) → B = phenylacetic acid (C6H5CH2COOH). Alternatively, benzyl bromide + Mg/ether → benzylmagnesium bromide (C6H5CH2MgBr) (C); quenching with CO2 then H3O+ gives phenylacetic acid (same B). So: A = C6H5CH2CN, B = C6H5CH2COOH, C = C6H5CH2MgBr (Grignard reagent). (Equations: C6H5CH2Br + NaCN → C6H5CH2CN; C6H5CH2CN + 2H2O/H+ → C6H5CH2COOH; C6H5CH2Br + Mg → C6H5CH2MgBr; C6H5CH2MgBr + CO2 → (after H3O+) C6H5CH2COOH.)
See solution
Structure of 2,5‑dimethylhexan‑3‑one (numbered C1–C6): CH3–CH(CH3)–CO–CH2–CH(CH3)–CH3. Oxidative cleavage at the carbonyl (R–CO–R') gives two carboxylic acids corresponding to the two alkyl groups R and R'. Left fragment R = CH3–CH(CH3) → 2‑methylpropanoic acid (isobutyric acid), i.e. CH3CH(CH3)COOH (2‑methylpropanoic acid). Right fragment R' = CH2–CH(CH3)–CH3 → 3‑methylbutanoic acid (isovaleric acid), i.e. CH3CH(CH3)CH2COOH. So the oxidation products are 2‑methylpropanoic acid and 3‑methylbutanoic acid.
ii) Ethyl acetate from methyl acetate: Transesterification: CH3COOCH3 + C2H5OH \xrightarrow{H^+} CH3COOC2H5 + CH3OH.
iii) Acetamide from methyl cyanide (acetonitrile): Partial hydrolysis of CH3CN: CH3CN + H2O \xrightarrow{acid, controlled} CH3CONH2.
iv) Lactic acid from ethanal: Ethanal + HCN → cyanohydrin CH3CH(OH)CN; hydrolysis of –CN → CH3CH(OH)COOH (lactic acid).
v) Acetophenone from acetyl chloride: Friedel–Crafts acylation: C6H6 + CH3COCl \xrightarrow{AlCl3} C6H5COCH3 (acetophenone).
vi) Ethane from sodium acetate: Decarboxylation with soda lime: CH3COONa + NaOH (CaO), heat → CH4? (Actually sodium acetate on heating with soda lime gives methane). For ethane from sodium acetate, use Kolbe electrolysis of sodium propionate? To obtain ethane: decarboxylation of sodium acetate gives methane. To get ethane, use decarboxylation of sodium propionate (CH3CH2COONa) gives ethane; or Kolbe coupling of acetate gives ethane: 2 CH3COO^- → CH3–CH3 + 2 CO2 + 2 e^- (Kolbe electrolysis). Thus Kolbe electrolysis of sodium acetate yields ethane.
vii) Benzoic acid from toluene: Side‑chain oxidation with KMnO4 (hot, acidic): C6H5CH3 \xrightarrow{KMnO4} C6H5COOH.
viii) Malachite green from benzaldehyde: Condensation of benzaldehyde with dimethylaniline derivatives in acid followed by oxidation yields triarylmethane dye (malachite green). (Briefly: benzaldehyde + dimethylaniline (2 mol) in presence of HCl gives leuco dye which on oxidation gives malachite green.)
ix) Cinnamic acid from benzaldehyde: Perkin condensation of benzaldehyde with acetic anhydride (or sodium acetate + acetic anhydride) yields cinnamic acid (PhCH=CHCOOH).
x) Acetaldehyde from ethyne: Hydration of ethyne (HgSO4/H2SO4) (oxy‑mercuration) gives acetaldehyde: HC≡CH + H2O \xrightarrow{HgSO4/H2SO4} CH3CHO.
i) Acetic anhydride: Dehydrate acetic acid, e.g. 2 CH3COOH \xrightarrow{P2O5, heat} (CH3CO)2O + H2O. Or CH3COOH + CH3COCl → (CH3CO)2O + HCl.