Chapter 26
43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry · 70 exercises
Problem 3
An organic compound (A) (molecular formula \(\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\right)\) was hydrolysed with dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to give a carboxylic acid (B) and an alochol (C). 'C' gives white turbidity immediately when treated with anhydrous \(\mathrm{ZnCl}_{2}\) and conc. HCl. The organic compound (A) is :
5 step solution
Problem 4
The increasing order of the reactivity of the following compounds in nucleophilic addition reaction is: Propanal, Benzaldehyde, Propanone, Butanone [Main Sep. 03, 2020 (II)] (a) Benzaldehyde \(<\) Butanone \(<\) Propanone \(<\) Propanal (b) Butanone \(<\) Propanone \(<\) Benzaldehyde \(<\) Propanal (c) Propanal \(<\) Propanone \(<\) Butanone \(<\) Benzaldehyde (d) Benzaldehyde \(<\) Propanal \(<\) Propanone \(<\) Butanone
5 step solution
Problem 5
An organic compound [A], molecular formula \(\mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O}_{2}\) was hydrolyzed with dilute sulphuric acid to give a carboxylic acid [B] and an alcohol \([\mathrm{C}]\). Oxidation of \([\mathrm{C}]\) with \(\mathrm{CrO}_{3}-\mathrm{H}_{2} \mathrm{SO}_{4}\) produced \([\mathrm{B}]\). Which of the following structures are not possible for [A]?
4 step solution
Problem 9
An unsaturated hydrocarbon \(X\) absorbs two hydrogen molecules on catalytic hydrogenation, and also gives following reaction: \(X \frac{\mathrm{O}_{3}}{\mathrm{Zn} / \mathrm{H}_{2} \mathrm{O}} \mathrm{A} \stackrel{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}}{\longrightarrow}\) B(3-oxo-hexanedicarboxylic acid) \(\mathrm{X}\) will be :
5 step solution
Problem 10
What is the product of following reaction? Hex-3-ynal \(\frac{\text { (i) } \mathrm{NaBH}_{4}}{\text { (ii) } \mathrm{PBr}_{3} \text {, (iii) } \mathrm{Mg} / \text { ether, (iv) } \mathrm{CO}_{2} / \mathrm{H}_{3} \mathrm{O}^{\prime}} ? ?\)
4 step solution
Problem 14
In the following reaction [Main April 9, 2019 (II)] Carbonyl compound \(+\mathrm{MeOH} \stackrel{\mathrm{HCl}}{\rightleftharpoons}\) acetal Rate of the reaction is the highest for: (a) Acetone as substrate and methanol in excess. (b) Propanal as substrate and methanol in stoichiometric amount. (c) Propanal as substrate and methanol in excess. (d) Acetone as substrate and methanol in stoichiometric amount.
5 step solution
Problem 16
The correct decreasing order for acid strength is: (a) \(\mathrm{NO}_{2} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{FCH}_{2} \mathrm{COOH}>\) \(\mathrm{CNCH}_{2} \mathrm{COOH}>\mathrm{ClCH}_{2} \mathrm{COOH}\) (b) \(\mathrm{FCH}_{2} \mathrm{COOH}>\mathrm{CNCH}_{2} \mathrm{COOH}>\) \(\mathrm{NO}_{2} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{ClCH}_{2} \mathrm{COOH}\) (c) \(\mathrm{CNCH}_{2} \mathrm{COOH}>\mathrm{NO}_{2} \mathrm{CH}_{2} \mathrm{COOH}>\) \(\mathrm{FCH}_{2} \mathrm{COOH}>\mathrm{ClCH}_{2} \mathrm{COOH}\) (d) \(\mathrm{NO}_{2} \mathrm{CH}_{2} \mathrm{COOH}>\mathrm{CNCH}_{2} \mathrm{COOH}>\) \(\mathrm{FCH}_{2} \mathrm{COOH}>\mathrm{ClCH}_{2} \mathrm{COOH}\)
4 step solution
Problem 20
The correct statement about the synthesis of erythritol \(\left(\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{4}\right)\) used in the preparation of PETN is : [Main Online April 10, 2016] (a) Thy synthesis requires three aldol condensations and one Cannizzaro reaction. (b) Alpha hydrogens of ethanol and methanol are involved in this reaction. (c) The synthesis requires two aldol condensations and two Cannizzaro reactions. (d) The synthesis requires four aldol condensations between methanol and ethanol.
7 step solution
Problem 20
Sodium salt of an organic acid ' \(X\) ' produces effervescences with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\). \(' \mathrm{X}^{\prime}\) reacts with the acidified aqueous \(\mathrm{CaCl}_{2}\) solution to give a white precipitate which decolourises acidic solution of \(\mathrm{KMnO}_{4} .{ }^{\prime} \mathrm{X}^{\prime}\) is : (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa}\) (b) \(\mathrm{HCOONa}\) (c) \(\mathrm{CH}_{3} \mathrm{COONa}\) (d) \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)
4 step solution
Problem 22
Bouveault-Blanc reduction reaction involves: (a) Reduction of an acyl halide with \(\mathrm{H}_{2} / \mathrm{Pd}\) (b) Reduction of an anhydride with \(\mathrm{LiAlH}_{4}\) (c) Reduction of an ester with \(\mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (d) Reduction of a carbonyl compound with \(\mathrm{Na} / \mathrm{Hg}\) and \(\mathrm{HCl}\).
3 step solution
Problem 24
Which one of the following reactions will not result in the formation of carbon-carbon bond? [Main Online April 9, 2014] (a) Reimer-Tieman reaction (b) Friedel Craft's acylation (c) Wurtz reaction (d) Cannizzaro reaction
4 step solution
Problem 24
In the presence of a small amount of phosphorous, aliphatic carboxylic acids react with chlorine or bromine to yield a compound in which \(\alpha\)-hydrogen has been replaced by halogen. This reaction is known as : (a) Wolff - Kishner reaction (b) Rosenmund reaction (c) Etard reaction (d) Hell - Volhard - Zelinsky reaction
3 step solution
Problem 25
Tischenko reaction is a modification of: (a) Aldol condensation (b) Claisen condensation (c) Cannizzaro reaction (d) Pinacol-pinacolon reaction
4 step solution
Problem 26
An organic compound A upon reacting with \(\mathrm{NH}_{3}\) gives B. On heating B gives C. C in presence of \(\mathrm{KOH}\) reacts with \(\mathrm{Br}_{2}\) to give \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\). A is : (a) \(\mathrm{CH}_{3} \mathrm{COOH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\)
5 step solution
Problem 27
Formaldehyde can be distinguished from acetaldehyde by the use of : (a) Schiff's reagent (b) Tollen's reagent (c) \(\mathrm{I}_{2} / \mathrm{Alkal}\) (d) Fehling's solution
6 step solution
Problem 27
Monocarboxylic acids are functional isomers of: (a) Ethers (b) Amines (c) Esters (d) Alcohols
6 step solution
Problem 28
Clemmensen reduction of a ketone is carried out in the presence of : (a) \(\mathrm{LiAlH}_{4}\) (b) \(\mathrm{Zn}-\mathrm{Hg}\) with \(\mathrm{HCl}\) (c) Glycol with \(\mathrm{KOH}\) (d) \(\mathrm{H}_{2}\) with Pt as catalyst
6 step solution
Problem 28
The compound that does NOT liberate \(\mathrm{CO}_{2}\), on treatment with aqueous sodium bicarbonate solution, is (a) Benzoic acid (b) Benzenesulphonic acid (c) Salicylic acid (d) Carbolic acid (Phenol)
4 step solution
Problem 30
The carboxyl functional group (- COOH) is present in (a) picric acid (b) barbituric acid (c) ascorbic acid (d) aspirin
6 step solution
Problem 32
How will you convert butan-2-one to propanoic acid? (a) Tollen's reagent (b) Fehling's solution (c) \(\mathrm{NaOH} / \mathrm{I}_{2} / \mathrm{H}^{+}\) (d) \(\mathrm{NaOH} / \mathrm{NaI} / \mathrm{H}^{+}\)
4 step solution
Problem 36
36\. A mixture of benzaldehyde and formaldehyde on heating with aqueous \(\mathrm{NaOH}\) solution gives (a) benzyl alcohol and sodium formate (b) sodium benzoate and methyl alcohol (c) sodium benzoate and sodium formate (d) benzyl alcohol and methyl alcohol
4 step solution
Problem 39
Benzoyl chloride is prepared from benzoic acid by (a) \(\mathrm{Cl}_{2}, h v\) (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{SOCl}_{2}\) (d) \(\mathrm{Cl}_{2}, \mathrm{H}_{2} \mathrm{O}\)
4 step solution
Problem 40
In the Cannizzaro reaction given below, $$ 2 \mathrm{PhCHO} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{PhCH}_{2} \mathrm{OH}+\mathrm{PhCO}_{2}^{-}, $$ the slowest step is (a) the attack of \(-\mathrm{OH}\) at the carbonyl group, (b) the transfer of hydride to the carbonyl group, (c) the abstraction of proton from the carboxylic acid, (d) the deprotonation of \(\mathrm{PhCH}_{2} \mathrm{OH}\).
4 step solution
Problem 40
When propionic acid is treated with aqueous sodium bicarbonate, \(\mathrm{CO}_{2}\) is liberated. The ' \(\mathrm{C}\) ' of \(\mathrm{CO}_{2}\) comes from (a) methyl group (b) carboxylic acid group (c) methylene group (d) bicarbonate
4 step solution
Problem 41
\(m\)-Chlorobenzaldehyde on reaction with conc. \(\mathrm{KOH}\) at room temperature gives (a) potassium \(m\)-chlorobenzoate and \(m\)-hydroxybenzaldehyde (b) \(m\)-hydroxybenzaldehyde and \(m\)-chlorobenzyl alcohol (c) \(m\)-chlorobenzyl alcohol and \(m\)-hydroxybenzyl alcohol (d) potassium \(m\)-chlorobenzoate and \(m\)-chlorobenzyl alcohol.
4 step solution
Problem 42
The enolic form of acetone contains (a) 9 sigma bonds, 1 pi-bond and 2 lone pairs (b) 8 sigma bonds, 2 pi-bonds and 2 lone pairs (c) 10 sigma bonds, 1 pi-bond and 1 lone pair (d) 9 sigma bonds, 2 pi-bonds and 1 lone pair
6 step solution
Problem 44
The compound that will not give iodoform on treatment with alkali and iodine is : (a) acetone (b) ethanol (c) diethyl ketone (d) isopropyl alcohol
3 step solution
Problem 45
Formic acid when heated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) produces
4 step solution
Problem 46
When acetaldehyde is heated with Fehling's solution it gives a precipitate of (a) \(\mathrm{Cu}\) (b) \(\mathrm{CuO}\) (c) \(\mathrm{Cu}_{2} \mathrm{O}\) (d) \(\mathrm{Cu}+\mathrm{Cu}_{2} \mathrm{O}+\mathrm{CuO}\)
3 step solution
Problem 46
The boiling point of propionic acid is less than that of \(n\)-butyl alcohol, an alcohol of comparable molecular weight.
4 step solution
Problem 47
A compound that gives a positive iodoform test is (a) 1-pentanol (b) 2 -pentanone (c) 3-pentanone (d) pentanal
6 step solution
Problem 48
The reagent with which both acetaldehyde and acetone react easily is [1981 - 1 Mark] (a) Fehling's reagent (b) Grignard reagent (c) Schiff's reagent (d) Tollen's reagent
4 step solution
Problem 52
Consider the following reactions \(\mathrm{A} \frac{(\mathrm{i}) \mathrm{CH}_{3} \mathrm{MgBr}}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}} \mathrm{B} \frac{\mathrm{Cu}}{573 \mathrm{~K}} \longrightarrow 2\)-methyl-2-butene The mass percentage of carbon in \(\mathrm{A}\) is
4 step solution
Problem 53
The structure of the enol form of \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{2}-\mathrm{CO}-\mathrm{CH}_{3}\) with intramolecular hydrogen bonding is ......
4 step solution
Problem 54
Which of the following compounds will react with ethanolic \(\mathrm{KCN} ?\) (a) ethyl chloride (b) acetyl chloride (c) chlorobenzene (d) benzaldehyde
6 step solution
Problem 55
Which of the following compounds will give a yellow precipitate with iodine and alkali? (a) 2-Hydroxypropane (b) acetophenone (c) methyl acetate (d) acetamide
4 step solution
Problem 56
Each item from (i) to (x) given below indicates a reaction type, a process or a homologue. Match each of these items with the related phrase by writing the correct phrase in the corresponding vacant space given under each. The correct phrase must be picked only from those given below within brackets: (Bayer's process, Nucleophilic addition, Free radical substitution, Ostwald's process, Homologous pair, Cyanamide process, Electrophilic substitution, Homolytic addition, Thermite process, Nucleophilic substitution)
5 step solution
Problem 58
The yield of ketone when a secondary alcohol is oxidized is more than the yield of aldehyde when a primary alcohol is oxidized.
4 step solution
Problem 61
Compounds \(P\) and \(R\) upon ozonolysis produce \(Q\) and \(S\), respectively. The molecular formula of \(Q\) and \(S\) is \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O} . Q\) undergoes Cannizzaro reaction but not haloform reaction, whereas \(S\) undergoes haloform reaction but not Cannizzaro reaction
5 step solution
Problem 67
The smallest ketone and its next homologue are reacted with \(\mathrm{NH}_{2} \mathrm{OH}\) to form oxime $$ $$ (a) Two different oximes are formed (b) Three different oximes are formed (c) Two oximes formed are optically active (d) All oximes formed are optically active
5 step solution
Problem 69
Which of the following will undergo aldol condensation? (a) acetaldehyde (b) propanaldehyde (c) benzaldehyde (d) trideuteroacetaldehyde
3 step solution
Problem 70
Which of the following will react with water? (a) \(\mathrm{CHCl}_{3}\) (b) \(\mathrm{Cl}_{3} \mathrm{CCHO}\) (c) \(\mathrm{CCl}_{4}\) (d) \(\mathrm{CICH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\)
4 step solution
Problem 71
A new carbon-carbon bond formation is possible in (a) Cannizzaro reaction (b) Friedel-Craft alkylation (c) Clemmensen reduction (d) Reimer-Tiemann reaction
3 step solution
Problem 74
Base catalysed aldol condensation occurs with : (a) propionaldehyde (b) benzaldehyde (c) 2 -methylpropionaldehyde (d) 2,2 -dimethylpropionaldehyde
3 step solution
Problem 75
Statement-1 : Acetic acid does not undergo haloform reaction. Statement-2 : Acetic acid has no alpha hydrogens.
4 step solution
Problem 79
79\. Identify \((\mathrm{X}),(\mathrm{Y})\) and \((\mathrm{Z})\) in the following synthetic scheme and write their structures. \(\mathrm{Ba}^{*} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow(\mathrm{X})\) gas \(\left[\mathrm{C}^{*}\right.\) denotes \(\left.\mathrm{C}^{14}\right]\) $$ \begin{aligned} &\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Br} \frac{\text { (i) } \mathrm{Mg} / \text { ether }}{\text { (ii) } \mathrm{X} \text { (iii) } \mathrm{H}_{3} \mathrm{O}^{\prime}}(\mathrm{Y}) \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow}(\mathrm{Z}) \\ &\text { [2001-5 Marks] } \end{aligned} $$ Explain the formation of labelled formaldehyde \(\left(\mathrm{H}_{2} \mathrm{C}^{*} \mathrm{O}\right)\) as one of the products when compound (Z) is treated with HBr and subsequently ozonolysed. Mark the \(C^{*}\) carbon in the entire scheme.
4 step solution
Problem 80
An organic compound A, \(\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{3}\), in dry benzene in the presence of anhydrous \(\mathrm{AlCl}_{3}\) gives compound \(\mathbf{B}\). The compound \(\mathbf{B}\) on treatment with \(\mathrm{PCl}_{5}\), followed by reaction with \(\mathrm{H}_{2} / \mathrm{Pd}\left(\mathrm{BaSO}_{4}\right)\) gives compound \(\mathbf{C}\), which on reaction with hydrazine gives a cyclic compound \(\mathbf{D}\left(\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{~N}_{2}\right)\). Identify A, B, C and D. Explain the formation of \(\mathbf{D}\) from \(\mathbf{C}\).
5 step solution
Problem 83
An ester \(A\left(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\right)\), on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol \(B\) as the sole organic product. Alcohol \(B\), on oxidation with \(\mathrm{NaOCl}\) followed by acidification, gives acetic acid. Deduce the structures of \(A\) and \(B\). Show the reactions involved.
6 step solution
Problem 84
Give reasons for the following : (i) Although phenoxide ion has more number of resonating structures than benzoate ion, benzoic acid is a stronger acid than phenol. Why? (ii) Formic acid is a stronger acid than acetic acid; (iii) Acetic acid can be halogenated in the presence of red \(\mathrm{P}\) and \(\mathrm{Cl}_{2}\) but formic acid cannot be halogenated in the same way.
3 step solution
Problem 86
An organic compound \(A, \mathrm{C}_{8} \mathrm{H}_{6}\), on treatment with dilute sulphuric acid containing mercuric sulphate gives a compound \(B\), which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous aluminium chloride. The compound \(B\), when treated with iodine in aqueous \(\mathrm{KOH}\), yields \(C\) and a yellow compound \(D .\) Identify \(A, B, C\) and \(D\) with justification. Show how \(B\) is formed from \(A\).
6 step solution