CBSE Class 12 Chemistry Haloalkanes and Haloarenes MCQs Set H

Question. Identify X and Y in the following sequence
(a) X = KCN, Y = LiAlH₄
(b) X = KCN, Y = H₃O+
(c) X = CH₃Cl, Y = AlCl₃ HCl
(d) X = CH₃NH₂, Y = HNO₂
Answer: a

Question. Chloropicrin is formed by the reaction of
(a) steam on carbon tetrachloride.
(b) nitric acid on chlorobenzene.
(c) chlorine on picric acid.
(d) nitric acid on chloroform.
Answer: d

Question. S_N1 reaction of alkyl halides leads to
(a) Retention of configuration
(b) Racemisation
(c) Inversion of configuration
(d) None of the above
Answer: b

 Question. p-dichlorobenzene has higher melting point than its o- and m- isomers because
(a) p-dichlorobenzene is more polar than o- and m- isomer.
(b) p-isomer has a symmetrical crystalline structure.
(c) boiling point of p-isomer is more than o- and m-isomer.
(d) All of these are correct reasons.
Answer: b

Question. Fitting reaction can be used to prepare
(a) Toluene
(b) Acetophenon
(c) Diphenyl
(d) Chlorobenzene
Answer: c

Assertion-Reason Questions

DIRECTION: Mark the option which is most suitable:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Question. Assertion: Nucleophilic substitution reaction on an optically active alkyl halide gives a mixture of enantiomers.
Reason: The reaction occurs by SN2 mechanism.
Answer: c

Question. Assertion: n-Butyl bromide has higher boiling point than isobutyl bromide.
Reason: The branching of the chain makes the molecule compact and therefore decreases the surface area. 
Answer: a

Question. Assertion: Tertiary haloakanes are more reactive than primary haloakanes towards elimination reactions.
Reason: The +I-effect of the alkyl groups weakens the C—X bond.
Answer: a

Question. Assertion: Vinyl chloride is less reactive than alkyl chloride.
Reason: Stability of alkyl halide decreases as the strength of C-X bond decreases. 
Answer: b

Question. Assertion: Boiling point of alkyl halides increases with increase in molecular weight.
Reason: Boiling point of alkyl halides is in the order Rl > RBr > RCl > RF. 
Answer: b

Question. Assertion: Boiling point of RCI is greater than RF.
Reason: R-CI is more stable than R-F. 
Answer: a

Question. Assertion: Lower members of alkyl halides are colourless gases.
Reason: Alkyl iodides in general turn black on exposure to air and light.
Answer: c

Question. Assertion: p-Dichlorchlorobenzene is less soluble in organic solvents than the corresponding o-isomer.
Reason: o-Dichlorobenzene is polar while p-dichlorobenzene is not.
Answer: b

Question. Assertion: p-Dichlorobenzene has higher melting point than o-dichlorobenzene.
Reason: Stronger the van der Waals’ forces of attraction, higher is the melting point. 
Answer: b

Question. Assertion: Benzyl chloride is more reactive than p-chlorotoluene towards aqueous NaOH.
Reason: The C—Cl bond in benzyl chloride is more polar than C—Cl bond in p-chlorotoluene. 
Answer: a

Case Based Questions 

A chlorocompound (A) on reduction with Zn- Cu ethanol gives the hydrocarbon (B) with five carbon atoms. When (A) is dissolved in dry ether and treated with sodium metal it gave 2,2,5,5-tetramethylhexane. The treatment of (A) with alcoholic KCN gives compound (C).

Answer the following questions by choosing the most appropriate options:

Question. The reaction of (C) with Na, C₂H₅OH gives
(a) (CH₃)₃C CH₂CONH₂
(b) (CH₃)₃C NH₂
(c) (CH₃)₃C CH₂ CH₂NH₂
(d) (CH₃)₂CHCH₂NH₂  
Answer: c

Question. The compound (A) is
(a) 1-chloro-2,2-dimethylpropane
(b) 1-chloro-2,2-dimethyl butane
(c) 1-chloro-2 methyl butane
(d) 2-chloro-2-methyl butane.  
Answer: a

Question. The reaction of (C) with Na, C₂H₅OH is called
(a) Gilman reaction
(b) Mendius reaction
(c) Grooves process
(d) Swart’s reaction  
Answer: b

Question. Compound (B) is
(a) n-pentane
(b) 2, 2-dimethylpropane
(c) 2-methylbutane
(d) none of above
Answer: b

Question. The reaction of (A) with aq. KOH will preferably favour
(a) S_N1 mechanism
(b) S_N2 mechanism
(c) E₁ mechanism
(d) E₂ mechanism Ans
Answer: a

Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (S_N1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards SN1 and S_N2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. S_N2 reaction mechanism is favoured mostly by primary alkyl halide followed by secondary and tertiary. This order is reversed in case of S_N1 reactions. Answer the following questions by choosing the most appropriate options:

Question. Isopropyl chloride undergoes hydrolysis by
(a) S_N1 mechanism
(b) S_N2 mechanism
(c) S_N1 and S_N2 mechanism
(d) neither S_N1 nor S_N2 mechanism 
Answer: c

Question. Which of the following is most reactive towards nucleophilic substitution reactions?
(a) C₆H₅Cl
(b) CH₂ = CHCl
(c) ClCH₂CH = CH₂
(d) CH₃CH = CHCl
Answer: c

Question. The most reactive nucleophile among the following is
(a) CH₃O^–
(b) C₆H₅O^–
(c) (CH₃)₂CHO^–
(d) (CH₃)₂CO^–  
Answer: a

Question. Which of the following is the correct order of decreasing SN2 reactivity?
(a) RCH₂X > R₂CHX > R₃CX
(b) R₃CX > R₂CHX > RCH₂X
(c) R₂CHX > R₂CX > RCH₂X
(d) RCH₂X > R₃CX > RCHX
Answer: b

Question. Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of
(a) insolubility
(b) instability
(c) inductive effect
(d) steric hindrance
Answer: d

When haloalkanes with b-hydrogen atom are boiled with alocholic solution of KOH, they undergo elimination of hydrogen halide resulting in the formation of alkenes. These reactions are called b-elimination reactions or dehydro halo genation reactions. These reactions follow Saytzeff’s rule. Substitution and elimination reactions often compete with each other. Mostly bases behave as nucleophiles and therefore can engage in substitution or elimination reactions depending upon the alkyl halide and the reaction conditions.

Answer the following questions by choosing the most appropriate options:
                                            aq. OH
Question. The general reaction, R—X     →    ROH + X–, is expected to follow decreasing order of reactivity as in
(a) t-BuI > t-BuBr > t-BuCl > t-BuF
(b) t-BuF > t-BuCl > t-BuBr > t-BuI
(c) t-BuBr > t-BuCl > t-BuI > t-BuF
(d) t-BuF > t-BuCl > t-BuI > t-BuBr 
Answer: a

Question. The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is
(a) 3º < 2º < 1º
(b) 3º > 2º > 1º
(c) 3º < 2º > 1º
(d) 3º > 2º < 1º 
Answer: b

Question. Reaction of t-butyl bromide with sodium methoxide produces
(a) sodium t-butoxide
(b) t-butyl methyl ether
(c) iso-butane
(d) iso-butylene
Answer: d

Question.Among the following the most reactive towards alcoholic KOH is
(a) CH₂ = CHBr
(b) CH₃COCH₂CH₂Br
(c) CH₃CH₂Br
(d) CH₃CH₂CH₂Br 
Answer: d

Question. In the elimination reactions the reactivity of alkyl halides follows the sequence
(a) R – F > R – Cl > R – Br > R – I
(b) R – I > R – Br > R – Cl > R – F
(c) R – I > R – F > R – Br > R – Cl
(d) R – F > R – I > R – Br > R – Cl 
Answer: b

Nucleophilic substitution reaction of haloalkane can be conducted according to both S_N1 and S_N2 mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent.
Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leaves the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I⁻ < Br⁻ < Cl⁻ < F⁻ and the order of their leaving tendency should be I⁻ > Br⁻ > Cl⁻ > F⁻ . Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI> RBr > RCl > RF . In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on S_N1 mechanism. If the leaving group is not easy to leave, the reaction is based on S_N2 mechanism.
Influences of solvent polarity: In S_N1 reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In S_N2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate ( S_N1) of tertiary chlorobutane in 25ºC water (dielectric constant 79) is 3,00,000 times faster than in ethanol (dielectric constant 24). The reaction rate (S_N2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both SN1 and S_N2 reactions, but with different results. Generally speaking, weak polar solvent is favourable for S_N2 reaction, while strong polar solvent is favourable for S_N1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favourable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on S_N1 mechanism in solvents with a strong polarity (for example, ethanol containing water).

(Source: Ding, Y. (2013). A Brief Discussion on Nucleophilic Substitution Reaction on Saturated Carbon Atom. In Applied Mechanics and Materials (Vol. 312, pp. 433-437). Trans Tech Publications Ltd.)

Answer the following questions by choosing the most appropriate options:

Question. Nucleophilic substitution will be fastest in case of:
(a) 1-Chloro-2,2-dimethyl propane
(b) 1-Iodo-2,2-dimethyl propane
(c) 1-Bromo-2,2-dimethyl propane
(d) 1-Fluoro-2,2-dimethyl propane 
Answer: b

Question. S_N1 mechanism is favoured in which of the following solvents:
(a) benzene
(b) carbon tetrachloride
(c) acetic acid
(d) carbon disulphide
Answer: c

Question. S_N1 reaction will be fastest in which of the following solvents?
(a) Acetone (dielectric constant 21)
(b) Ethanol (dielectric constant 24)
(c) Methanol (dielectric constant 32)
(d) Chloroform (dielectric constant 5) 
Answer: c

Question. S_N1 reaction will be fastest in case of:
(a) 1-Chloro-2-methyl propane
(b) 1-Iodo-2-methyl propane
(c) 1-Chlorobutane
(d) 1-Iodobutane 
Answer: b

Question. Polar solvents make the reaction faster as they:
(a) destabilize transition state and decrease the activation energy
(b) destabilize transition state and increase the activation energy
(c) stabilize transition state and increase the activation energy
(d) stabilize transition state and decrease the activation energy 
Answer: c

Haloarenes are less reactive than haloalkanes.The low reactivity of haloarenes can be attributed to
— resonance effect
— sp² hybridisation of C—X bond
— polarity of C—X bond
— instability of phenyl cation (formed by self-ionisation of haloarene)
— repulsion between the electron rich attacking nucleophiles and electron rich arenes.
Reactivity of haloarenes can be increased or decreased by the presence of certain groups at certain positions. For example, nitro (–NO₂) group at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.

Answer the following questions by choosing the most appropriate options:

Question. Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to
(a) the formation of less stable carbonium ion
(b) resonance stabilisation
(c) larger carbon-halogen bond
(d) inductive effect
Answer: b

Question. The reactivity of the compounds (i) MeBr,
(ii) PhCH₂Br, (iii) MeCl, (iv) p-MeOC₆H₄Br decreases as
(a) (i) > (ii) > (iii) > (iv)
(b) (iv) > (ii) > (i) > (iii)
(c) (iv) > (iii) > (i) > (ii)
(d) (ii) > (i) > (iii) > (iv) 
Answer: d

Short Answer Type Questions-I 

Question. Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same.
Answer. The reasons are:
(i) Due to resonance/diagrammatic repre-sentation, C – Cl bond acquires a partial double bond character. As a result, the C – Cl bond in chlorobenzene is shorter and hence stronger. Thus, cleavage of C – Cl bond in benzene becomes difficult which makes it less reactive towards nucleophilic substition.
(ii) Due to repulsion between nucleophile and electron rich arenes.

Question. What are ambident nucleophiles? Explain with an example.
Answer. Ambident nucleophile : A nucleophile that can form new bonds at two or more spots in its structure, usually due to resonance contributors.
Example : S = C = N– can act as a nucleophile with either the S or N attacking.

Question. Which compound in each of the following pairs will react faster in S_N2 reaction with —OH?
(i) CH₃Br or CH₃I (ii) (CH₃)₃ CCl or CH₃Cl
Answer. (i) CH₃I : Because Iodide is better leaving group than bromide.
(ii) CH₃Cl : Carbon atom leaving group is less hindered.

Question. Give reasons :
(a) n-Butyl bromide has higher boiling point than t-butyl bromide.
(b) Racemic mixture is optically inactive.
(c) The presence of nitro group (–NO₂) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer. (a) n-Butyl bromide has higher boiling point than t-butyl bromide because it has larger surface area hence have more Van der Waals’ forces.
(b) Rotation due to one enantiomer is cancelled by another enantiomer.
(c) The presence of nitro group (–NO₂) at ortho and para positions withdraws the electron density from benzene ring and thus facilitating the attack of nucleophile.

Short Answer Type Questions-II 

Question. Rearrange the compounds of each of the following sets in order of reactivity towards S_N2 displacement :
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methyl-butane, 3-Bromo-2-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2, 2-dimethyl-propane, 1-Bromo-2-methylbutane 
Answer.  (i) 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane
( i i ) 1-Bromo-3-methylbutane > 3-Bromo-2 methylbutane > 2-Bromo-2-methylbutane
(iii) 1-Bromobutane > 1-Bromo-2-methylbutane > 1-Bromo-2, 2-dimethylpropane.

Question. Following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane
(i) Write the compound which is most reactive towards S_N2 reaction.
(ii) Write the compound which is optically active.
(iii) Write the compound which is most reactive towards β-elimination reaction. 
Answer.  (i) 1-Bromopentane is most reactive towards S_N2 reaction.
(ii) 2-Bromopentane is optically active.
(iii) 2-Bromo-2-methylbutane is most reactive towards β-elimination reaction.

Question. Answer the following :
(i) Haloalkanes easily dissolve in organic solvents, why?
(ii) What is known as a racemic mixture? Give an example.
(iii) Of the two bromoderivatives, C₆H₅CH (CH₃)Br and C₆H₅CH(C₆H₅)Br, which one is more reactive in SN1 substitution reaction and why? (Delhi)
Answer. (i) Because the new forces of attraction set up between haloalkanes and solvent molecules are of the same strength as the forces of attraction being broken.
(ii) A mixture which contains the equal proportions of two enantiomers of a compound in equal proportions is called racemic mixture
Example : (±) butan-2-ol
(iii) Since the reactivity of S_N1 reactions increases as the stability of intermediate carbocation increases. Of the two 2° bromides, the carbocation intermediate derived from C₆H₅CH(C₆H₅)Br i.e. C₆H₅C+HC₆H₅ is more stable as compared to the carbocation C₆H₅C + H CH3 obtained from C₆H₅CH(CH₃)Br because it is stabilized by two phenyl groups due to resonance.

Question. Give reasons for the following :
(i) Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH₃—X. 
Answer. (i) I, is better leaving group/C—I bond is weaker than C—Br bond.
(iii) In halobenzene, halogen atom is attached to the sp² hybrid carbon atom while in
CH₃-X halogen atom is attached to sp³ hybrid carbon atom. Hence C-X bond length in halo benzene is smaller than CH₃-X.

Question. Account for the following :
(a) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(b) Alkyl halides, though polar, are immiscible with water.
(c) Grignard’s reagents should be prepared under anhydrous conditions. 
Answer. (a) Electron pairs of Cl atom are in conjugation with π electrons of the benzene ring so C-Cl bond in chlorobenzene acquires some double bond character while C-Cl bond in cyclohexyl chloride is a pure single bond.C – Cl bond in chlorobenzene is shorter than in cyclohexyl chloride. Since dipole moment is a product of charge and distance, so chlorobenzene has lower dipole moment than cyclohexyl chloride.
(b) Alkyl halides are polar molecules, therefore, their molecules are held together by dipole-dipole attraction.
The molecules of H₂O are held together by H-bonds. Since the new forces of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halidealkyl halide molecules and water- water molecules, therefore, alkyl halides are immiscible with water.
(c) Grignard’s reagents are very reactive. They react with alcohol, water, amines etc. to form corresponding hydrocarbon.
R–MgX + HOH → RH + Mg(OH)X
Therefore, Grignard’s reagents must be prepared under anhydrous conditions.

 Question. Write the IUPAC names of the following compounds :
(i) CH₂ = CHCH₂Br
(ii) (CCl₃)₃ CCl
Answer. (i) CH₂ = CHCH₂ – Br
IUPAC name : 3-Bromopropene
(ii) (CCl₃)₃—C—Cl :
IUPAC name : 2-(Trichloromethyl) – 1, 1, 1, 2,3, 3, 3-heptachloropropane