JEE Main Previous year questions (2016-22): Haloalkanes & Haloarenes - Notes | Study Chemistry for JEE - JEE

Q.1. The number of isomeric tetraenes (NOT containing sp-hybridized carbon atoms) that can be formed from the following reaction sequence is ___________.    (JEE Advanced 2022)

Ans. 2

Q.2. Compound 'A' undergoes following sequence of reactions to give compound 'B'.
The correct structure and chirality of compound 'B' is           (JEE Main 2022)
[where Et is - C₂H₅]

(a)
(b)
(c)
(d)

Ans. c

Q.3. Identify the correct statement for the below given transformation.          (JEE Main 2022)

(a) A − CH₃CH₂CH = CH − CH₃, B − CH₃CH₂CH₂CH = CH₂, Saytzeff products
(b) A − CH₃CH₂CH = CH − CH₃, B − CH₃CH₂CH₂CH = CH₂, Hofmann products
(c) A − CH₃CH₂CH₂CH = CH₂, B − CH₃CH₂CH = CHCH₃, Hofmann products
(d) A − CH₃CH₂CH₂CH = CH₂, B − CH₃CH₂CH = CHCH₃, Saytzeff products

Ans. c

Q.4. Major product 'B' of the following reaction sequence is :          (JEE Main 2022)

(a)
(b)
(c)
(d)

Ans. b

Q.5. Major product of the following reaction is          (JEE Main 2022)

(a)
(b)
(c)
(d)

Ans. d

Q.6. Given below are two statements :          (JEE Main 2022)
Statement I : The presence of weaker π-bonds make alkenes less stable than alkanes.
Statement II : The strength of the double bond is greater than that of carbon-carbon single bond.
In the light of the above statements, choose the correct answer from the options given below:
(a) Both Statement I and Statement II are correct.
(b) Both Statement I and Statement II are incorrect.
(c) Statement I is correct but Statement II is incorrect.
(d) Statement I is incorrect but Statement II is correct.

Ans. b

Q.7. The major product in the reaction          (JEE Main 2022)

(a) t-Butyl ethyl ether
(b) 2,2-Dimethyl butane
(c) 2-Methyl pent-1-ene
(d) 2-Methyl prop-1-ene

Ans. d

Q.8. The product formed in the following reaction.          (JEE Main 2022)

(a)
(b)
(c)
(d)

Ans. b

Q.9. Product 'A' of following sequence of reactions is          (JEE Main 2022)

(a)
(b)
(c)
(d)

Ans. d

Q.10. What will be the major product of following sequence of reactions?          (JEE Main 2022)

(a)
(b)
(c)
(d)

Ans. c

Q.11. Which one of the following compounds is inactive towards S_N1 reaction?          (JEE Main 2022)

(a)
(b)
(c)
(d)

Ans. c

Q.12. The major product (P) in the reaction          (JEE Main 2022)
 is
(a)
(b)
(c)
(d)

Ans. c

Q.13.

The stable carbocation formed in the above reaction is          (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. c

Q.14. The stereoisomers that are formed by electrophilic addition of bromine to trans-but-2-ene is/are :          (JEE Main 2021)
(a) 2 enantiomers and 2 mesomers
(b) 2 identical mesomers
(c) 2 enantiomers
(d) 1 racemic and 2 enantiomers

Ans. b

Q.15. In the reaction given below, the total number of atoms having sp2 hybridization in the major product P is _________.          (JEE Advanced 2021)

Ans. 12

Alkene on treatment with ozone (O₃) followed by treatment with Zn/H₂O gives carbonyl compounds.

Carbonyl compounds react with NH₂OH to give oximes.

There are four sp²-carbon atoms, four sp²-nitrogen atoms and four sp²-oxygen atoms as shown in structures I and II.
Therefore, total number of atoms having sp2-hybridisation are twelve (12).

Q.16. The major product formed in the following reaction of           (JEE Advanced 2021)

(a)
(b)
(c)
(d)

Ans. b
It is a case of Birch reduction. Alkynes on reaction with alkali metal in liq. NH3 gives trans-alkene. But terminal alkynes do not get reduced.

Q.17. For the following sequence of reactions, the correct products are :          (JEE Main 2021)

(a)
(b)
(c)
(d)

Ans. a

Q.18. The major product (A) formed in the reaction given below is :          (JEE Main 2021)

(a)
(b)
(c)
(d)

Ans. b

Q.19. In the following sequence of reactions, the final product D is :          (JEE Main 2021)

(a)
(b) CH₃ − CH = CH − CH₂ − CH₂ − CH₂ − COOH
(c) H₃C − CH = CH − CH(OH) − CH₂ − CH₂ − CH₃
(d)

Ans. d

Q.20.  Excess of isobutane on reaction with Br₂ in presence of light at 125ºC gives which one of the following, as the major product?          (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. d

Bromination is highly selective, reactivity order –
3° >> 2° > 1

Q.21. Among the following compounds I-IV, which one forms a yellow precipitate on reacting sequentially with (i) NaOH (ii) dil. HNO₃ (iii) AgNO₃?          (JEE Main 2021)


(a) II
(b) IV
(c) I
(d) III

Ans. b

Other compounds halide can't be removed because corresponding C⁺ is highly unstable.

Q.22. The major product formed in the following reaction is :          (JEE Main 2021)
(a)
(b)
(c)

(d)

Ans. a

This addition is known as 1, 4-addition.
Mechanism This reaction can undergoes two pathways

Out of the two intermediates formed in two paths, the path B intermediate is more stable as it has more stable carbocation. So, a major product will be formed because of path B.

1, 2-addition product is formed at low temperature and will be less stable as the double bond is less substituted. 1, 4-addition is a thermodynamically stable product as the double bond is more substituted. As diene is in excess and HBr is limited in reaction, so diene cannot be formed. So option (b) is incorrect.

Q.23. Metallic sodium does not react normally with :          (JEE Main 2021)
(a) gaseous ammonia
(b) But-2-yne
(c) Ethyne
(d) tert-butyl alcohol

Ans. b
Metallic sodium does not react with 2-butyne because 2-butyne does not have acidic hydrogen.

Q.24. The correct pair(s) of the ambident nucleophiles is(are)          (JEE Main 2021)
(A) AgCN/KCN
(B) RCOOAg/RCOOK
(C) AgNO₂/KNO₂
(D) AgI/KI
(a) (B) and (C) only
(b) (B) only
(c) (A) and (C) only
(d) (A) only

Ans. c
Ambident ligands are those which have two donor sites but at once it has the ability to donate either site.
Hence in AgCN/KCN, AgNO₂/KNO₂, CN^-, NO₂^- have two donar sites act as ambident Nu^-.

Q.25. Choose the correct statement regarding the formation of carbocations A and B given.          (JEE Main 2021)

(a) Carbocation A is more stable and formed relatively at faster rate
(b) Carbocation B is more stable and formed relatively at slow rate
(c) Carbocation B is more stable and formed relatively at faster rate
(d) Carbocation A is more stable and formed relatively at slow rate

Ans. c
Carbocation B is more stable as it is secondary carbocation having more number of α-hydrogens and having a greater +I effect.
∴ Carbocation B formed at a faster rate than carbocation A.

Q.26. Which of the following reaction is an example of ammonolysis?          (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. d
Ammonolysis of alkyl halides is the reaction of an alkyl halide with NH3 which leads to the preparation of amines.
C₆H₅CH₂Cl + NH₃ → C₆H₅CH₂NH₂

Q.27. Product ''A'' in the chemical reaction is :          (JEE Main 2021)

(a)

(b)
(c)
(d)

Ans. b

Q.28.

The above reaction requires which of the following reaction conditions?          (JEE Main 2021)
(a) 623 K, Cu, 300 atm
(b) 623 K, 300 atm
(c) 573 K, 300 atm
(d) 573 K, Cu, 300 atm

Ans. b
Chlorobenzene is fused with NaOH at 623 K and 300 atmospheric pressure to get sodium phenoxide.

Q.29.

Identify the reagent(s) 'A' and condition(s) for the reaction          (JEE Main 2021)
(a) A = HCl, ZnCl₂
(b) A = HCl; Anhydrous AlCl₃
(c) A = Cl₂; UV light
(d) A = Cl₂; dark, Anhydrous AlCl₃

Ans. c

Q.30.

The products ''A'' and ''B'' formed in above reactions are :          (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. b

Q.31.

Considering the above reaction, the major product among the following is :          (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. a

Q.32. Match List - I with List - II.

Choose the correct answer from the options given below :          (JEE Main 2021)
(a) (a) → (ii), (b) → (iv), (c) → (i), (d) → (iii)
(b) (a) → (iii), (b) → (iv), (c) → (i), (d) → (ii)
(c) (a) → (iii), (b) → (i), (c) → (iv), (d) → (ii)
(d) (a) → (ii), (b) → (i), (c) → (iv), (d) → (iii)

Ans. a

Q.33. For the given reaction :          (JEE Main 2021)

What is 'A'?

(a) 
(b)
(c)
(d) CH₃CH₂CH₂NH₂  

Ans. a

Q.34.

Which of the following reagent is suitable for the preparation of the product in the above reaction?          (JEE Main 2021)
(a) Red P + Cl₂
(b)
(c) NaBH₄
(d) Ni/H₂ 

Ans. b
To reduce the carbonyl groups into alkane, Wolff-Kishner reduction is used, without affecting the double bond.
Wolff-Kishner reagent It utilises hydrazine (NH₂ − NH₂) as the reducing agent in the presence of strong base in a high boiling protic solvent.
The driving force for the reaction is the conversion of hydrazine to nitrogen gas.

Q.35. The product formed in the first step of the reaction of          (JEE Main 2021)

(a)
(b)
(c)
(d)

Ans. a
Here, in first step only one mole of Mg/ Et₂O attacks on bromine and form two 2MgBr in the first step.
On further moving in the reaction, two MgBr are eliminated to form alkene in respective positions.

Q.36. What is the final product (major) 'A' in the given reaction ?          (JEE Main 2021)

(a)
(b)
(c)
(d)

Ans. a
Steps involved in this reaction are as follows

Step 2 : Elimination reaction always give minor product than substitution reaction.

Q.37. Identify products A and B.          (JEE Main 2021)

(a)
(b)
(c)
(d)

Ans. b
In first step, KMnO₄(dil.) help to break alkene in ketone and aldehyde and in second step, CrO₃ is used for selective over oxidation of aldehyde only. CrO₃ will form aldehyde to acid in product (B).

Q.38. The dipole moments of CCl₄, CHCl₃ and CH₄ are in the order    (2020)
(a) CHCl₃ < CH₄ = CCl₄
(b) CCl₄ < CH₄ < CHCl₃
(c) CH₄ < CCl₄ < CHCl₃
(d) CH₄ = CCl₄ < CHCl₃

Ans. d
Molecules having symmetrical geometry and identical side atoms have zero dipole moment, that is, the resultant dipole moment is zero. Whereas unsymmetrical molecules shows net dipole moment.

Q.39. For the following reactions

Where

k_s and k_e, are the rate constants for substitution and elimination respectively, and µ = k_s/k_e, the correct option is _______.    (2020)
(a)
(b)
(c)
(d)

Ans. b
The reaction involved is


Since, the nucleophile (A) is small so it will favor substitution rather than elimination product. Nucleophile (B) is bulkier, so it will favor elimination product. Thus,  and

Q.40. The decreasing order of reactivity towards dehydrohalogenation (E₁) reaction of the following compounds is    (2020)
(I)
(II)
(III)
(IV)
(a) (IV) > (II) > (III) > (I)
(b) (II) > (IV) > (I) > (III)
(c) (II) > (IV) > (III) > (I)
(d) (II) > (I) > (IV) > (III)

Ans. a
The strength of dehydrohalogenation depends upon the stability of alkene formed. More the stable alkene form, more the strength of dehydrohalogenation of the haloalkane.
(I)
(II)
(III)
(IV)
Since, the transition state of (IV) is more stable than (II), so, the alkene formed in (IV) is more stable than in (II). Thus, the correct decreasing order of dehydrohalogenation of the following compound is: (IV) > (II) > (III) > (I).

Q.41. Arrange the following bonds according to their average bond energies in descending order    (2020)
C – Cl, C – Br, C – F, C – I
(a) C – F > C – Cl > C – Br > C – I
(b) C – Br > C – I > C – Cl > C – F
(c) C – I > C – Br > C – Cl > C – F
(d) C – CI > C – Br > C – I > C – F

Ans. a
The bond enthalpy of C–X bond depends upon the size of the halogen X, as the size of the halogen increases, the bond energy decreases. Thus, the correct decreasing order of C–X, bond enthalpy is: C – F > C – Cl > C – Br > C – I.

Q.42. Which of these will produce the highest yield in Friedel Crafts reaction?   (2020)
(a)
(b)
(c)
(d)

Ans. c
The highest yield in Friedel Crafts reaction ins given by chlorobenzene. Since, phenol and aniline react with Lewis acid so reaction could not occur. –CONH₂ group is a strong deactivating groups towards electrophilic aromatic subsitution reactions.

Q.43. X melts at low temperature and is a bad conductor of electricity in both liquid and solid state. X is    (2020)
(a) zinc sulphide
(b) mercury
(c) silicon carbide
(d) carbon tetrachloride

Ans. d
Since, CCl₄ is a non-polar compound, so, it has low melting point and it is covalent compound, thus, it is bad conductor of electricity in both solid and liquid state.

Q.44. Which of the following reactions will not produce a racemic product?    (2020)
(a)
(b)
(c)
(d)

Ans. d
The reaction involved is

Q.45. The major product of the following reaction is:    (2019)

(a)
(b)
(c)
(d)

Ans. a
Mechanism involved for the given reaction is:

Q.46. Which amongst the following is the strongest acid?    (2019)
(a) CHBr₃
(b) CHI₃
(c) CH(CN)₃
(d) CHCl₃

Ans. c
Due to the resonance stabilisation of the conjugate base, CH(CN)₃ is the strongest acid amongst the given compounds.


The conjugate bases of CHBr₃ and CHl₃ are stabilised by inductive effect of halogens. This is why, they are less stable. Also, the conjugate base of CHCl₃ involves back-bonding between 2p and 3p orbitals

Q.47. The major product of the following reaction is:    (2019)

(a)
(b)
(c)
(d)

Ans. a
Dehydrohalogenation (β-elimination) occurs as:

Q.48. Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light?    (2019)

(a) α - hydrogen
(b) γ - hydrogen
(c) δ - hydrogen
(d) β - hydrogen

Ans. b
Allylic H is replaced due to the greater stability of allylic free radical.

Q.49. The major product of the following reaction is:    (2019)

(a)
(b)
(c)
(d)

Ans. b

Q.50. The major product of the following reaction is:    (2019)

(a)
 
(b)
 
(c)
 
(d) 

Ans. b
Heating of the given compound in presence of strong base is favoured for elimination reaction resulting in more stable alkene.

Q.51. The major product of the following reaction is:    (2019)

(a)
(b)
(c)
(d)

Ans. a

Q.52. The major product of the following reaction is:    (2019)

(a)
(b)
(c)
(d)

Ans. c

Q.53. The major product of the following reaction is:    (2019)

(a)
(b)
(c)
(d)

Ans. c

Q.54. Increasing order of reactivity of the following compounds for S_N1 substitution is:    (2019)

 
(a) (B) < (C) < (D) < (A)
(b) (B) < (C) < (A) < (D)
(c) (B) < (A) < (D) < (C)
(d) (A) < (B) < (D) < (C)

Ans. c
In S_N1 reaction carbocation acts as an intermediate.

Carbocation produced by (C) is more stable than carbocation produced by (D) due to + I effect of -OCH₃ group.
Further in (A) there is formation of tertiary carbocation after rearrangement while (B) is primary carbocation.

So, the correct order is (C) > (D) > (A) > (B).

Q.55. Which of the following potential energy (PE) diagrams represents the S_N1 reaction?    (2019)
(a)
(b)
(c)
(d)

Ans. a
The S_N1 reaction energy diagram illustrates the dominant part of the substrate with respect to the reaction rate.
The rate determining step is the formation of the intermediate carbocation.

Q.56. Increasing rate of S_N1 reaction in the following compounds is:    (2019)

 
(a) (A) < (B) < (C) < (4)
(b) (B) < (A) < (C) < (D)
(c) (B) < (A) < (D) < (3) 
(d) (A) < (B) < (D) < (C)

Ans. b
The rate of S_N1 is decided by the stability of carbocation formed in the rate determining step.

Carbocation(D) is most stable due to +R effect of CH₃O group, (C) is stabilised by +I and +H effects of the CH₃ group; (B) is least stable due to -I effect of MeO group. So increasing order of rate of S_N1 is (B) < (A) < (C) < (D)

Q.57. The major product of the following reaction is    (2019)

(a)
(b)
(c)
(d)

Ans. c

Q.58. The major product ‘Y’ in the following reaction is:    (2019)

(a)
(b)
(c)
(d)

Ans. c

Q.59. Which one of the following is likely to give a precipitate with AgNO₃ solution?    (2019)
(a) CH₂ = CH —Cl
(b) CCI₄
(c) CHCl₃
(d) (CH₃)₃CCl

Ans. d
Carbocation is formed on reaction with Ag⁺
 
is most stable carbocation compared to other carbocations i.e.

Q.60. Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product. Reaction of X with Hg (OAc)₂/H₂O followed by NaBH₄ gives Y as the major product. Y is:    (2019)
(a)
(b)
(c)
(d)

Ans. c

Q.61. An ‘Assertion’ and a ‘Reason’ are given below. Choose the correct answer from the following options :
Assertion (A): Vinyl halides do not undergo nucleophilic substitution easily.
Reason (R): Even though the intermediate carbocation is stabilized by loosely held π-electrons, the cleavage is difficult because of strong bonding.    (2019)
(a) Both (A) and (R) are wrong statements.
(b) Both (A) and (R) are correct statements and (R) is the correct explanation of (A).
(c) Both (A) and (R) are correct statements but (R) is not the correct explanation of (A).
(d) (A) is a correct statement but (R) is a wrong statement.

Ans. d

Due to partial double bond character of C-halogen bond, halogen leaves with great difficulty, if at all it does. Hence, vinyl halides do not undergo nucleophilic substitution easily. So, assertion is correct.

Intermediate carbocation is not stabilised by loosely held-π electrons because empty orbital, being at 90°, cannot overlap with p-orbitals of π bond. So, reason is wrong.

Q.62. The major product of the following reaction is:    (2018)

(a)
(b)
(c)
(d)

Ans. b

Q.63. Which of the following will most readily give the dehydrohalogenation product?    (2018)
(a)
(b)
(c)
(d)

Ans. b
Carbanion formed after removal of hydrogen is resonance stabilised as it is adjacent to phenyl ring.

Q.64. The major product of the following reaction is:    (2018)

(a)
(b)
(c)
(d)

Ans. c

Inversion takes place at the carbon containing bromine atom. 

Q.65. The increasing order of the reactivity of the following halides for the S_N1 reaction is    (2017) 

(a) (III) < (II) < (I) 
(b) (II) < (I) < (III) 
(c) (I) < (III) < (II) 
(d) (II) < (III) < (I) 

Ans. b
For any S_N1 reaction, reactivity is decided by ease of dissociation of alkyl halide
R - X ⇌ R⁺ + X^-
Higher the stability of R⁺ (carbocation) higher would be reactivity of S_N1 reaction. Since the stability of cation follows the order.
 

Q.66. The major product obtained in the following reaction is    (2017)

(a) (±)C₆H₅CH(OtBu)CH₂C₆H₅ 
(b) C₆H₅CH = CHC₆H₅ 
(c) (+)C₆H₅CH(OtBu)CH₂C₆H₅ 
(d) (-)C₆H₅CH(OtBu)CH₂C₆H₅ 

Ans. b

Q.67. The major product of the following reaction is:    (2017)

(a)
(b)
(c)
(d)

Ans. a

Q.68. The major product of the following reaction is:    (2017)

(a) CH₃CH = CH – CH = CHCH₃
(b) CH₂ = CHCH = CHCH₂CH₃
(c) CH₃CH = C = CHCH₂CH₃
(d) CH₂ = CHCH₂CH = CHCH₃

Ans. a

Q.69. 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:    (2016)
 

(1) All of these 
(2) (a) and (c)
(3) (c) only
(4) (a) and (b) 

Ans. (1)
Strong nucleophile (O^- Me) polar solvent (MeOH) gives elimination products over substitution products but all products are possible in different yields.

Q.70. Which one of the following reagents is not suitable for the elimination reaction?    (2016)

(a) NaI
(b) NaOH/H₂O-EtOH
(c) NaOH/H₂O
(d) NaOEt/EtOH

Ans. a
NaI gives Nucleophilic substitution reaction.