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JEE • CHEMISTRY • HALOALKANES AND HALO ARENES

Haloalkanes and halo arenes practice · Mixed difficulty

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61 Medium
61 of 138
Analyzing +4 | -1
Wurtz reaction of methyl iodide yields an organic compound X\mathrm{X}. Which of the following reaction also yields X\mathrm{X}:
View answer and explanation
Correct option: A
C2H5Cl+LiAlH4→\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}+\mathrm{LiAlH}_4 \rightarrow
Option A (C2H5Cl + LiAlH4 →) is correct because LiAlH4 is a reducing agent that can reduce ethyl chloride to ethane, which is the product of the Wurtz reaction of methyl iodide.
62 Medium
62 of 138
Understanding +4 | -1
Aryl halides are less reactive towards electrophiles than alkyl halides due to:
View answer and explanation
Correct option: A
Resonance
Option A is correct because aryl halides are less reactive towards electrophiles than alkyl halides due to resonance, which stabilizes the aryl halide and reduces the electron density on the halogen.
63 Medium
63 of 138
Analyzing +4 | -1
pπ−dπp\pi-d \pi bonding is not exhibited by
View answer and explanation
Correct option: D
COCl2{COCl}_2
Option D (COCl2{COCl}_2) is correct because carbon does not have d-orbitals to participate in pπ−dπp\pi-d\pi bonding.
64 Medium
64 of 138
Analyzing +4 | -1
Out of following in which retention of configuration in product ?? will take
place?
View answer and explanation
Correct option: A
CH3−CH∣C2H5−OH→PySOCl2→AcetoneNaIPC{H_3} - \mathop {CH}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {{C_2}{H_5}} \end{array}} - OH\xrightarrow[{Py}]{{SOC{l_2}}}\xrightarrow[{{\text{Acetone}}}]{{NaI}}P
Option A is correct because the reaction with SOCl2SOCl_2 in the presence of pyridine leads to inversion, and the subsequent reaction with NaI in acetone leads to another inversion, resulting in overall retention of configuration.
65 Medium
65 of 138
Analyzing +4 | -1
Ethylene dichloride and ethylidene chloride are isomeric compounds. Identify the statement which is not applicable to both of them?
View answer and explanation
Correct option: B
React with aqueous potash and give the same products
Option B (React with aqueous potash and give the same products) is not applicable because ethylene dichloride and ethylidene chloride react with aqueous potash to yield different products (glycol and aldehyde, respectively).
66 Medium
66 of 138
Analyzing +4 | -1
Which process does not occur during formation of CHCl3  from    C2H5OH  and  bleaching  \mathrm{CHC}{\mathrm{l}_3}\mathrm{\;from\;\;}{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH\;and\;bleaching}\; powder?
View answer and explanation
Correct option: C
Elimination
Option C (Elimination) is correct because elimination reactions are not involved in the formation of chloroform from ethanol and bleaching powder.
67 Medium
67 of 138
Understanding +4 | -1
Although chlorobenzene does not give Ulmann’s reaction. However, presence of…group in chlorobenzene at o−,p−o - ,p -position enables it to give Ulmann’s reaction.
View answer and explanation
Correct option: A
NO2\mathrm{N}{\mathrm{O}_2}
Option A (NO2\mathrm{N}{\mathrm{O}_2}) is correct because the presence of electron-withdrawing groups like nitro (NO2\mathrm{N}{\mathrm{O}_2}) at ortho- and para- positions activates the chlorobenzene towards nucleophilic substitution in Ullmann's reaction.
68 Hard
68 of 138
Analyzing +4 | -1
Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reaction

i) Benaldehyde \quad ii) P-Tolualdehyde
iii) P-Nitro benzaldehyde \quad iv) Acetophenone
View answer and explanation
Correct option: C
iv < ii < i < iii
The correct answer is C (iv < ii < i < iii) because electron-withdrawing groups increase reactivity in nucleophilic addition, and electron-donating groups decrease it.
69 Easy
69 of 138
Remembering +4 | -1
The reaction between a primary amine, chloroform and
few drops of alcoholic KOH is known as:
View answer and explanation
Correct option: B
Carbylamine reaction
Option B (Carbylamine reaction) is correct because the reaction between a primary amine, chloroform, and alcoholic KOH is known as the Carbylamine reaction, which produces an isocyanide.
70 Medium
70 of 138
Analyzing +4 | -1
Which of these liquids would be unsuitable as an inert solvent for a Friedel-Crafts reaction?
View answer and explanation
Correct option: A
Chlorobenzene
Option A (Chlorobenzene) is correct because it is more likely to undergo Friedel-Crafts reaction itself compared to the other options.
71 Medium
71 of 138
Analyzing +4 | -1
Assertion : Chlorination of allylic hydrogen is difficult than vinylic hydrogen.

Reason : Allyl radical is stabilized by resonance.
View answer and explanation
Correct option: D
If the reason is true but the assertion is false
Option D (If the reason is true but the assertion is false) is correct because allylic chlorination is easier than vinylic chlorination, and the allyl radical is indeed stabilized by resonance.
72 Medium
72 of 138
Application +4 | -1
CH3CH2−CH=CH2→ NBS (X) One of the product → alc. KOH(Y)CH _3 CH _2- CH = CH _2 \xrightarrow{\text { NBS }} \underset{\text { One of the product }}{( X )} \xrightarrow{\text { alc. } KOH }( Y ) The compound (Y)( Y ) is
View answer and explanation
Correct option: C
1,3-Butadiene
Option C (1,3-Butadiene) is correct because NBS promotes allylic bromination followed by elimination with alc. KOH to form a conjugated diene.
73 Medium
73 of 138
Analyzing +4 | -1
Read the following statements
A. Gabriel phthalimide synthesis is used to prepare all 10{1^0} amines
B. C6H5ClC6H5−C6H5{C_6}{H_5}Cl{C_6}{H_5} - {C_6}{H_5} is known as Wurtz reaction.
C. In SN1S{N^1} mechanism there is more inversion than retention leading to partial racemization
D. All aldehydes give red precipitate with Fehling’s solution
E. Reactivity order of alkyl halides towards SN2S{N^2} reaction is 10>20>30{1^0} > {2^0} > {3^0}
Number of correct statements among these
View answer and explanation
Correct option: A
2
Option A (2) is correct because only two statements (C and E) are correct among the given statements.
74 Medium
74 of 138
Remembering +4 | -1
What is the common name of the given compound?
CH3−CH∣Cl −ClC{{H}_{3}}-\underset{\begin{matrix} | \\ Cl \\ \end{matrix}}{\mathop{CH}}\,-Cl
View answer and explanation
Correct option: C
Ethylidene dichloride
The correct answer is Ethylidene dichloride because the compound is a derivative of ethylidene with two chlorine atoms attached to the same carbon.
75 Easy
75 of 138
Understanding +4 | -1
Dehydrohalogenation in haloalkanes produces:
View answer and explanation
Correct option: B
A double bond
Option B (A double bond) is correct because dehydrohalogenation, which is the removal of a hydrogen and a halogen from a haloalkane, results in the formation of a double bond.
76 Medium
76 of 138
Remembering +4 | -1
Ethyl alcohol gives ethyl chloride on treatment with:
View answer and explanation
Correct option: B
SOCl2\mathrm{SOC}{\mathrm{l}_2}
Option B (SOCl2\mathrm{SOC}{\mathrm{l}_2}) is correct because thionyl chloride (SOCl2\mathrm{SOC}{\mathrm{l}_2}) is a reagent that converts alcohols to alkyl chlorides.
77 Easy
77 of 138
Remembering +4 | -1
For the carbylamine reaction we need hot alcoholic KOH and:
View answer and explanation
Correct option: D
Any monoalkyl amine and trichloro methane
Option D (Any monoalkyl amine and trichloro methane) is correct because the carbylamine reaction involves a primary amine, chloroform (trichloromethane), and a base like KOH.
78 Medium
78 of 138
Analyzing +4 | -1
Compare the rate of SN2\mathrm{S}{\mathrm{N}^2} reaction
CH3−CO∣∣−CH2−BrCH3−BrCH3−CO∣∣−CH2−ClCH3−Cl(1)(2)(3)(4)\begin{array}{cccccccccccccccccccc} {C{H_3} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - C{H_2} - Br}&{C{H_3} - Br}&{C{H_3} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - C{H_2} - Cl}&{C{H_3} - Cl} \\ {(1)}&{(2)}&{(3)}&{(4)} \end{array}
View answer and explanation
Correct option: C
1>3>2>41>3>2>4
Option D is correct because the rate of SN2 reaction depends on steric hindrance and the electronic effects of substituents, leading to the order 1 > 3 > 2 > 4.
79 Easy
79 of 138
Understanding +4 | -1
Ammonolysis of alkyl halides followed by the treatment with NaOHNaOH solution can be used to prepare primary, secondary and tertiary amines. The purpose of NaOHNaOH in the reaction is
View answer and explanation
Correct option: C
to remove acidic impurities
Option C is correct because NaOH is a base and is used to neutralize the acidic impurities (HX) formed during the ammonolysis of alkyl halides.
80 Medium
80 of 138
Application +4 | -1
In the reaction sequence,

C2H5Cl  +  KCN→C2H5OH    X→ΔC2H5OH  Y{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{Cl\;} + \mathrm{\;KCN}\xrightarrow{{{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH\;}}}\;X\xrightarrow[\Delta ]{{{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH\;}}}Y

What is the molecular formula of Y?
View answer and explanation
Correct option: A
C3H6O2{\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_2}
Option A (C3H6O2{\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_2}) is correct because the reaction sequence leads to the formation of propanoic acid.
81 Medium
81 of 138
Analyzing +4 | -1
The major product of the following reaction is:
CH3−CHCl∣−CH3→(ii) HBr, peroxide(iii)aq.KOH (i) Alc. KOHC{H_3} - \mathop {CH}\limits^{\begin{array}{cccccccccccccccccccc} {Cl} \\ | \end{array}} - C{H_3}\xrightarrow[{\begin{array}{cccccccccccccccccccc} {(ii){\text{ }}HBr,{\text{ peroxide}}} \\ {\left( {iii} \right){\kern 1pt} {\kern 1pt} {\text{aq}}.{\kern 1pt} {\kern 1pt} KOH} \end{array}}]{{{\text{ }}(i){\text{ Alc}}.{\text{ }}KOH}}
View answer and explanation
Correct option: B
CH3−CH2−CH2−OHC{H_3} - C{H_2} - C{H_2} - OH
The correct answer is B (CH3−CH2−CH2−OHC{H_3} - C{H_2} - C{H_2} - OH) because the reaction sequence involves anti-Markovnikov addition of HBr followed by substitution with KOH.
82 Medium
82 of 138
Understanding +4 | -1
Predict the order of reactivity of the following four isomers towards SN2{S}{{N}}^2 reaction.
(I) CH3CH2CH2CH2Cl{CH}_3 {CH}_2 {CH}_2 {CH}_2 {Cl}
(II) CH3CH2CH(Cl)CH3{CH}_3 {CH}_2 {CH}({Cl}) {CH}_3
(III) (CH3)2CHCH2Cl\left({CH}_3\right)_2 {CHCH}_2 {Cl}
(IV) (CH3)3CCl\left({CH}_3\right)_3 {CCl}
View answer and explanation
Correct option: C
(I) >> (III) >> (II) >> (IV)
Option D (I > II > III > IV) is correct because the reactivity of SN2S_N2 reactions decreases with increasing steric hindrance. Primary halides are more reactive than secondary, which are more reactive than tertiary.
83 Medium
83 of 138
Understanding +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: D
All of the above.
Option D (All of the above) is correct because all the statements regarding SN1 and SN2 reactions are accurate.
84 Medium
84 of 138
Analyzing +4 | -1
Which one is the least reactive towards SN1\mathrm{SN}^1 reaction?
View answer and explanation
Correct option: D
C6H5CH2Br{C}_6 {H}_5 {CH}_2 {Br}
Option D (C6H5CH2Br) is the least reactive towards SN1 because it forms a primary carbocation, which is the least stable among the given options.
85 Medium
85 of 138
Understanding +4 | -1
Which can be hydrolysed most easily?
View answer and explanation
Correct option: A
(C6H5)3CCl{\left( {{\mathrm{C}_6}{\mathrm{H}_5}} \right)_3}\mathrm{CCl}
Option A ((C6H5)3CCl{\left( {{\mathrm{C}_6}{\mathrm{H}_5}} \right)_3}\mathrm{CCl}) is correct because the carbocation formed after hydrolysis is highly stabilized by three phenyl groups.
86 Medium
86 of 138
Remembering +4 | -1
he catalyst used in the preparation of an alkyl chloride by the action of dry HCl on an alcohol is
View answer and explanation
Correct option: C
anhy.  ZnCl2\mathrm{anhy}.\mathrm{\;ZnC}{\mathrm{l}_2}
Option C (anhy.  ZnCl2\mathrm{anhy}.\mathrm{\;ZnC}{\mathrm{l}_2}) is correct because anhydrous zinc chloride is a common catalyst used in the reaction of alcohols with dry HCl to form alkyl chlorides (Lucas reagent).
87 Medium
87 of 138
Remembering +4 | -1
Wurtz’s reaction involves the reduction of alkyl halide with
View answer and explanation
Correct option: D
Na in ether
Option D (Na in ether) is correct because Wurtz reaction involves the reaction of alkyl halides with sodium metal in dry ether to form higher alkanes.
88 Easy
88 of 138
Remembering +4 | -1
Benzylidene chloride is
View answer and explanation
Correct option: C
C6H5CHCl2C_6H_5CHCl_2
Option C (C6H5CHCl2C_6H_5CHCl_2) is correct because Benzylidene chloride has the formula C6H5CHCl2C_6H_5CHCl_2.
89 Medium
89 of 138
Application +4 | -1
C6H5CH2Br→2. H3O+1. Mg,  EtherX{C_6}{H_5}C{H_2}Br\xrightarrow[{2.\,{H_3}{O^ + }}]{{1.\,Mg,\,\,{\text{Ether}}}}X
The product (X)({X}) is :-
View answer and explanation
Correct option: B
C6H5CH3{C}_6 {H}_5 {CH}_3
Option B (C6H5CH3{C}_6 {H}_5 {CH}_3) is correct because the Grignard reagent formed from C6H5CH2Br{C_6}{H_5}C{H_2}Br reacts with water to produce toluene (C6H5CH3{C}_6 {H}_5 {CH}_3).
90 Medium
90 of 138
Remembering +4 | -1
If chloroform is left open in air in the presence of sunlight, it gives
View answer and explanation
Correct option: B
carbonyl chloride
Option B (carbonyl chloride) is correct because chloroform reacts with oxygen in the presence of sunlight to produce phosgene (carbonyl chloride) and hydrochloric acid.
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