JEE Main Practice — Chemistry Set 4

JEE Main · 2026 · with answers and explanations

questions
25
Total marks
100
Duration
60 minutes
Negative marking
−1

Subjects: Chemistry

Question 1

Chemistry · 4 marks · −1

A buffer is 0.2 M in acetic acid and 0.1 M in sodium acetate. If pKa\mathrm{p}K_a of acetic acid is 4.74, the pH of the buffer is (log⁡2=0.30\log2=0.30)

  1. A.5.04
  2. B.4.44
  3. C.4.74
  4. D.4.14

Correct answer: B

Explanation: Henderson equation: pH =pKa+log⁡[salt][acid]=4.74+log⁡0.10.2=4.74−0.30=4.44=\mathrm pK_a+\log\frac{[\text{salt}]}{[\text{acid}]}=4.74+\log\frac{0.1}{0.2}=4.74-0.30=4.44.

Question 2

Chemistry · 4 marks · −1

The solubility product of CaF2\mathrm{CaF_2} is 3.2×10−113.2\times10^{-11}. Its molar solubility in pure water is

  1. A.5.7×10−65.7\times10^{-6} M
  2. B.2×10−42\times10^{-4} M
  3. C.3.2×10−43.2\times10^{-4} M
  4. D.4×10−44\times10^{-4} M

Correct answer: B

Explanation: Ksp=s(2s)2=4s3⇒s3=8×10−12⇒s=2×10−4K_{sp}=s(2s)^2=4s^3\Rightarrow s^3=8\times10^{-12}\Rightarrow s=2\times10^{-4} M.

Question 3

Chemistry · 4 marks · −1

At a certain temperature, PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)} is 50% dissociated when the total equilibrium pressure is 3 atm. At the same temperature, the total pressure at which PCl5\mathrm{PCl_5} would be one-third dissociated is

  1. A.4.5 atm
  2. B.6 atm
  3. C.12 atm
  4. D.8 atm

Correct answer: D

Explanation: For degree of dissociation α\alpha at total pressure PP: Kp=α2P1−α2K_p=\frac{\alpha^2P}{1-\alpha^2}. With α=12\alpha=\frac12, P=3P=3: Kp=0.25×30.75=1K_p=\frac{0.25\times3}{0.75}=1 atm. With α=13\alpha=\frac13: 1=P/98/9=P8⇒P=81=\frac{P/9}{8/9}=\frac P8\Rightarrow P=8 atm.

Question 4

Chemistry · 4 marks · −1

One mole each of A and B are placed in a 1 L vessel and allowed to reach equilibrium A+B⇌C+D\mathrm{A+B\rightleftharpoons C+D}. At equilibrium, 0.8 mol of C is present. The equilibrium constant KcK_c is ______.

Correct answer: 16

Explanation: At equilibrium: [C] = [D] = 0.8, [A] = [B] = 0.2. Kc=0.8×0.80.2×0.2=16K_c=\frac{0.8\times0.8}{0.2\times0.2}=16.

Question 5

Chemistry · 4 marks · −1

The solubility of AgCl (Ksp=1.8×10−10K_{sp}=1.8\times10^{-10}) in 0.01 M NaCl solution is x×10−9x\times10^{-9} mol L−1^{-1}. The value of xx is ______.

Correct answer: 18

Explanation: Common-ion effect: s(s+0.01)=1.8×10−10s(s+0.01)=1.8\times10^{-10} with s≪0.01s\ll0.01: s=1.8×10−100.01=1.8×10−8=18×10−9s=\frac{1.8\times10^{-10}}{0.01}=1.8\times10^{-8}=18\times10^{-9} M.

Question 6

Chemistry · 4 marks · −1

The major product of the addition of HBr to propene (in the absence of peroxide) is

  1. A.1-bromopropane
  2. B.2-bromopropane
  3. C.1,2-dibromopropane
  4. D.propane

Correct answer: B

Explanation: Markovnikov addition: H+\mathrm{H^+} adds to the terminal carbon forming the more stable secondary carbocation, which captures Br−\mathrm{Br^-}: CH3CHBrCH3\mathrm{CH_3CHBrCH_3}.

Question 7

Chemistry · 4 marks · −1

Ozonolysis of but-2-ene followed by reductive work-up (Zn/H2O\mathrm{H_2O}) gives

  1. A.two molecules of ethanal
  2. B.one molecule each of methanal and propanal
  3. C.two molecules of propanone
  4. D.one molecule of butanal

Correct answer: A

Explanation: CH3CH=CHCH3\mathrm{CH_3CH{=}CHCH_3} is cleaved at the C=C, each half becoming C=O: 2 CH3CHO2\,\mathrm{CH_3CHO} (ethanal).

Question 8

Chemistry · 4 marks · −1

1-Bromopropane is obtained as the major product when propene is treated with

  1. A.HBr in the dark, without peroxide
  2. B.HBr in the presence of benzoyl peroxide
  3. C.HCl in the presence of benzoyl peroxide
  4. D.Br2\mathrm{Br_2} in CCl4\mathrm{CCl_4}

Correct answer: B

Explanation: Peroxide effect (Kharasch): HBr adds by a free-radical chain; Br∙\mathrm{Br^\bullet} adds to the terminal carbon giving the more stable secondary radical, so Br ends up on C1 (anti-Markovnikov). HCl does not show this effect; Br2\mathrm{Br_2} gives 1,2-dibromopropane.

Question 9

Chemistry · 4 marks · −1

The correct decreasing order of acidic character of ethyne, ethene and ethane is

  1. A.ethane > ethene > ethyne
  2. B.ethene > ethyne > ethane
  3. C.ethyne > ethene > ethane
  4. D.ethyne > ethane > ethene

Correct answer: C

Explanation: Acidity follows the s-character of the C–H carbon: spsp (50%) > sp2sp^2 (33%) > sp3sp^3 (25%). The more s-character, the better the carbanion is stabilised.

Question 10

Chemistry · 4 marks · −1

Which of the following species is aromatic?

  1. A.Cyclopentadienyl cation
  2. B.Cyclooctatetraene
  3. C.Cycloheptatrienyl anion
  4. D.Cyclopentadienyl anion

Correct answer: D

Explanation: Hückel: cyclic, planar, fully conjugated with (4n+2)π(4n+2)\pi electrons. Cyclopentadienyl anion has 6π6\pi (aromatic). The cation (4π4\pi) and cycloheptatrienyl anion (8π8\pi) are antiaromatic; cyclooctatetraene (8π8\pi) is non-planar, tub-shaped and non-aromatic.

Question 11

Chemistry · 4 marks · −1

A hydrocarbon X (C6H10\mathrm{C_6H_{10}}) takes up two moles of H2\mathrm{H_2} on catalytic hydrogenation and gives a white precipitate with ammoniacal silver nitrate. On hydration with dilute H2SO4/HgSO4\mathrm{H_2SO_4}/\mathrm{HgSO_4}, X gives 3,3-dimethylbutan-2-one. X is

  1. A.3,3-dimethylbut-1-yne
  2. B.hex-1-yne
  3. C.4-methylpent-1-yne
  4. D.4-methylpent-2-yne

Correct answer: A

Explanation: Two moles of H2\mathrm{H_2} and a silver acetylide precipitate indicate a terminal alkyne. Markovnikov hydration converts R−C≡CH\mathrm{R{-}C{\equiv}CH} to R−CO−CH3\mathrm{R{-}CO{-}CH_3}; the product (CH3)3C−CO−CH3\mathrm{(CH_3)_3C{-}CO{-}CH_3} requires R=(CH3)3C\mathrm{R=(CH_3)_3C}, i.e. X =(CH3)3C−C≡CH=\mathrm{(CH_3)_3C{-}C{\equiv}CH} (C6H10\mathrm{C_6H_{10}}). Hex-1-yne would give hexan-2-one and 4-methylpent-1-yne gives 4-methylpentan-2-one; 4-methylpent-2-yne is an internal alkyne and gives no precipitate with ammoniacal AgNO3\mathrm{AgNO_3}.

Question 12

Chemistry · 4 marks · −1

The number of structural isomers of the alkane C6H14\mathrm{C_6H_{14}} is ______.

Correct answer: 5

Explanation: Hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane: 5.

Question 13

Chemistry · 4 marks · −1

The volume of oxygen at STP (in litre; 1 mol of gas = 22.4 L) required for the complete combustion of 11 g of propane is ______.

Correct answer: 28

Explanation: C3H8+5O2→3CO2+4H2O\mathrm{C_3H_8+5O_2\to3CO_2+4H_2O}. 11 g propane =0.25=0.25 mol, needs 1.251.25 mol O2\mathrm{O_2} =1.25×22.4=28=1.25\times22.4=28 L.

Question 14

Chemistry · 4 marks · −1

Which of the following undergoes an SN2\mathrm{S_N2} reaction with hydroxide ion fastest?

  1. A.CH3Br\mathrm{CH_3Br}
  2. B.CH3CH2Br\mathrm{CH_3CH_2Br}
  3. C.(CH3)2CHBr\mathrm{(CH_3)_2CHBr}
  4. D.(CH3)3CBr\mathrm{(CH_3)_3CBr}

Correct answer: A

Explanation: SN2\mathrm{S_N2} proceeds through backside attack; steric hindrance at the carbon slows it. Rate order: methyl > 1° > 2° > 3°.

Question 15

Chemistry · 4 marks · −1

Chloroform stored in the presence of air and light slowly forms a poisonous gas. The gas is

  1. A.chlorine
  2. B.carbon monoxide
  3. C.phosgene (COCl2\mathrm{COCl_2})
  4. D.hydrogen chloride only

Correct answer: C

Explanation: 2CHCl3+O2→hν2COCl2+2HCl\mathrm{2CHCl_3+O_2\xrightarrow{h\nu}2COCl_2+2HCl}. Phosgene is highly toxic; chloroform is therefore stored in dark, completely filled bottles.

Question 16

Chemistry · 4 marks · −1

In the Finkelstein reaction, CH3CH2Cl\mathrm{CH_3CH_2Cl} is converted into CH3CH2I\mathrm{CH_3CH_2I} using NaI in dry acetone. The reaction goes to completion because

  1. A.iodide is a weaker nucleophile than chloride
  2. B.NaCl formed is insoluble in dry acetone and precipitates
  3. C.the C–I bond is stronger than the C–Cl bond
  4. D.acetone reacts with the chloride ion

Correct answer: B

Explanation: NaI is soluble in acetone but NaCl is not; precipitation of NaCl removes a product and drives the equilibrium forward (Le Chatelier). The C–I bond is actually weaker than C–Cl.

Question 17

Chemistry · 4 marks · −1

(R)-2-Bromobutane is treated with aqueous NaOH under conditions where only the SN2\mathrm{S_N2} mechanism operates. The product is

  1. A.(R)-butan-2-ol
  2. B.a racemic mixture of butan-2-ol
  3. C.but-2-ene
  4. D.(S)-butan-2-ol

Correct answer: D

Explanation: SN2\mathrm{S_N2} occurs with Walden inversion. OH replaces Br (both are priority 1 at that carbon; the other groups keep their order), so inversion of configuration changes R to S: (S)-butan-2-ol.

Question 18

Chemistry · 4 marks · −1

2-Bromobutane on heating with alcoholic KOH gives as the major product

  1. A.but-1-ene
  2. B.but-2-ene
  3. C.butan-2-ol
  4. D.2-ethoxybutane

Correct answer: B

Explanation: Dehydrohalogenation follows Saytzeff's rule: the more substituted alkene (but-2-ene, mainly trans) predominates over but-1-ene.

Question 19

Chemistry · 4 marks · −1

The correct decreasing order of reactivity towards nucleophilic substitution by aqueous hydroxide of (I) chlorobenzene, (II) 1-chloro-4-nitrobenzene, (III) 1-chloro-2,4-dinitrobenzene and (IV) 2-chloro-1,3,5-trinitrobenzene is

  1. A.I > II > III > IV
  2. B.IV > III > II > I
  3. C.II > III > IV > I
  4. D.IV > II > III > I

Correct answer: B

Explanation: Nucleophilic aromatic substitution goes via a Meisenheimer (carbanion) intermediate that is stabilised by −M-M groups at ortho/para positions. More ortho/para NO2\mathrm{NO_2} groups ⇒\Rightarrow faster: trinitro (reacts with warm water) > dinitro > mononitro > chlorobenzene (needs ~623 K, high pressure).

Question 20

Chemistry · 4 marks · −1

The number of structurally distinct monochloro derivatives obtained by free-radical chlorination of 2-methylbutane is ______.

Correct answer: 4

Explanation: CH3−CH(CH3)−CH2−CH3\mathrm{CH_3{-}CH(CH_3){-}CH_2{-}CH_3} has four kinds of H: the two equivalent CH3\mathrm{CH_3} on C2 (→ 1-chloro-2-methylbutane), the C2–H (→ 2-chloro-2-methylbutane), CH2\mathrm{CH_2} at C3 (→ 2-chloro-3-methylbutane) and the terminal CH3\mathrm{CH_3} C4 (→ 1-chloro-3-methylbutane). Total 4 structural isomers.

Question 21

Chemistry · 4 marks · −1

A mixture of bromomethane and bromoethane is treated with sodium in dry ether (Wurtz reaction). The number of different alkanes formed is ______.

Correct answer: 3

Explanation: Coupling possibilities: CH3−CH3\mathrm{CH_3{-}CH_3} (ethane), CH3−C2H5\mathrm{CH_3{-}C_2H_5} (propane), C2H5−C2H5\mathrm{C_2H_5{-}C_2H_5} (butane): 3.

Question 22

Chemistry · 4 marks · −1

Which of the following is the strongest base in aqueous solution?

  1. A.NH3\mathrm{NH_3}
  2. B.CH3NH2\mathrm{CH_3NH_2}
  3. C.(CH3)2NH\mathrm{(CH_3)_2NH}
  4. D.(CH3)3N\mathrm{(CH_3)_3N}

Correct answer: C

Explanation: In water, basicity depends on the +I effect, solvation of the conjugate acid (H-bonding) and steric factors. For methylamines the order is (CH3)2NH>CH3NH2>(CH3)3N>NH3\mathrm{(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3}; the tertiary amine's cation is poorly solvated.

Question 23

Chemistry · 4 marks · −1

Which amine gives a foul-smelling isocyanide when warmed with chloroform and alcoholic KOH?

  1. A.N-Methylaniline
  2. B.Aniline
  3. C.N,N-Dimethylaniline
  4. D.Triethylamine

Correct answer: B

Explanation: The carbylamine reaction is given only by primary amines: RNH2+CHCl3+3KOH→RNC+3KCl+3H2O\mathrm{RNH_2+CHCl_3+3KOH\to RNC+3KCl+3H_2O}. Aniline is primary; the others are secondary or tertiary.

Question 24

Chemistry · 4 marks · −1

An amine reacts with benzenesulfonyl chloride to give a product that is insoluble in aqueous KOH. The amine is

  1. A.ethanamine
  2. B.N-ethylethanamine (diethylamine)
  3. C.N,N-diethylethanamine (triethylamine)
  4. D.aniline

Correct answer: B

Explanation: Hinsberg test: a secondary amine gives an N,N-disubstituted sulfonamide with no acidic N–H, so it is insoluble in alkali. Primary amines (ethanamine, aniline) give alkali-soluble sulfonamides (acidic N–H); tertiary amines do not react.

Question 25

Chemistry · 4 marks · −1

Benzenediazonium chloride reacts with phenol in mildly alkaline solution to give

  1. A.4-hydroxyazobenzene (an orange dye)
  2. B.chlorobenzene
  3. C.benzene
  4. D.4-aminophenol

Correct answer: A

Explanation: Azo coupling: the electrophilic diazonium ion attacks the para position of the activated phenoxide ion, giving pp-hydroxyazobenzene, C6H5−N=N−C6H4−OH\mathrm{C_6H_5{-}N{=}N{-}C_6H_4{-}OH} (orange).

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