Important Questions Class 12 Chemistry Chapter 10

Important Questions for CBSE Class 12 Chemistry Chapter 10 – Haloalkanes and Haloarenes

Important Questions Class 12 Chemistry Chapter 10 prepared by Extramarks help students prepare for board exams effectively. This set contains the marks distribution of Important Questions, numerical, terminologies and concepts related to Haloalkanes and Haloarenes. Chapter 10 Class 12 Chemistry Important Questions prepared by Extramarks subject matter experts will help students gain in-depth understanding of the above mentioned topics. 

These questions are curated by subject matter experts in accordance with the CBSE Syllabus. Chemistry Class 12 Chapter 10 Important Questions are presented with step-by-step solutions. Students can refer to this set of important questions from the Extramarks website. 

CBSE Class 12 Chemistry Chapter-10 Important Questions

Important Questions For Class 12 Chemistry Chapter 10

Q1.]  Arrange the following in the increasing order of properly indicated:

i.] bromomethane, chloromethane, dichloromethane. (Increasing order of boiling points).

Ans. All the above-mentioned compounds are haloalkanes. The order will be:

Chloromethane < Bromomethane < Dichloromethane

The reason behind this is that as the halogen size increases the boiling point will increase. Similarly, the boiling point will increase as the number of halogen atoms rises in the similar chain.

ii.] 1-chloropropane, isopropyl chloride, 1-chlorobutane  (Increasing order of boiling point)

Ans. A chlorine atom is existent in the compounds and there are distinct sizes of the alkyl chain. The order will be:

Isopropyl chloride < 1- Chloropropane < 1 – Chlorobutane 

This happens as the branching of the chain expands the boiling point will fall and as the chain’s size increases the boiling point will inflate.

iii.] o,m,p-dichlorobenzenes (Increasing order of melting points)

Ans. p-dichlorobenzene has the highest melting point due to its symmetry and structure.  The melting point of a compound is connected to its symmetry. Similarly, the symmetry of the compound goes after the same pattern as the melting point. The order is given below:

m-Dichlorobenzene < o-Dichlorobenzene < p-Dichlorobenzene

Q2] How do the below-stated conversions occur?

  • Ethanol to but-1-yne

Ans. Chloroethane Is formed when ethanol reacts with SOCl2 and pyridine. Acetylene reacts with NaNH2 which forms sodium acetylide. But-1-yne is formed when Chloroethane and Sodium acetylide react. The reactions are :


CH3CH2OH     ⟶    CH3CH2Cl


CHCH+NaNH2      HCCNa+

CH3CH2Cl+HCCNa+ ⟶ CH3CH2C ≡ CH + NaCl

  1.   1-bromopropane to 2-bromopropane

Ans. Propene is formed when 1-Bromopropane reacts with alcoholic KOH. HBr reacts with Propene which produces 2-Bromopropane. The reaction is as follows:

                                                    Alc. KOH                                                    HBr    

               CH3CH2CH2Br        CH3CH=CH2CH3CHBr-CH3 

    iii.   Ethyl chloride to propanoic acid

Ans. KCN reacts with ethyl chloride to give propanenitrile. Propanoic acid is produced when propanenitrile goes through hydrolysis. The reaction is as follows: 

                                 KCN                              H+/H2O

                    CH3CH2Cl ⟶  CH3CH2CN  ⟶   CH3CH2COOH

  1.     But-1-ene to n-butyl iodide

Ans: When but-1-ene reacts with HBr in the existence of peroxide, 1-Bromobutane is produced. NaI reacts with 1-Bromobutane in the presence of Acetone to form  n-butyl iodide. The reaction is as follows:                                                            


                       CH3CH2CH=CH2       CH3CH2CH2CH2Br



                        CH3CH2CH2CH2Br      CH3CH2CH2CH2I  


Q3.] What is the difference between following 

  1. i) Electrophilic and Nucleophilic substitution reactions.


Electrophilic substitution reactions Nucleophilic substitution reactions
1. A chemical reaction where the functional group is connected to a compound is exchanged by an electrophile is called an electrophilic substitution reaction.

2. The two primary types of electrophilic substitution reaction are electrophilic aromatic substitution reactions and electrophilic aliphatic substitution reactions.

1. When a nucleophile strikes a haloalkane with a partial positive charge atom or group, a Nucleophilic substitution reaction takes place. 

2. The two types of nucleophilic substitution reactions are substitution nucleophilic unimolecular and Substitution Nucleophilic Bimolecular. 


  1. ii) Retention and Inversion of configuration. 


Retention of configuration Inversion of configuration
1. The conservation of integrity of the spatial adjustment of bonds to an asymmetric centre during a chemical reaction or transformation is called 

Retention of configuration.

2. The retention in configuration transforms the R-configuration of the compound into R and S-configuration of the compound converts into S.

1. A process in which the absolute and relative configurations of atoms or molecules are not changed are called Inversion of configuration.

2. The inversion in configuration converts the R-configuration of the compound into S and S-configuration of the compound converts  into R.

Q4.) Identify which compound in the pairs will react faster in Sn2 reaction with OH ?

  1. a) CH3Br or CH3I

 Ans. The iodide ion is a bigger atom than bromide ion still both compounds are considered to be alkyl halide. So, I ion is better leaving group than Br ion. Thus, CH3I will react quicker than CH3Br with regards Sn2 reaction in the presence of hydroxyl ion.

  1. b) CH33CCl or CH3Cl

Ans. The steric obstacles should be reduced in Sn2 reaction . CH33CCl has extreme steric hindrance and CH3Cl has limited steric hindrance. Therefore, CH3Cl will react rapidly to the Sn2 reaction in the presence of hydroxyl ion.

Q5.) In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

  1. i) CH33C-Cl and C6H5CH2Cl
  2. ii) CH2=CH-Cl and CH2=CH-CH2Cl

iii) C6H5CH2Cl and C6H5CClC6H5

Ans. For the SN1 reaction the order of reactivity is 3°>2°>1°,

  1. i) The former compound is a tertiary compound and the latter compound is primary compound. Therefore, the former compound will go through a quicker SN1
  2. ii) The former compound is the vinylic primary compound whereas the second compound is the primary compound. Therefore, the second compound will go through rapid SN1 reaction as there will be resonance in the former compound.

iii) The former compound is the primary compound and the latter compound is the secondary compound. Hence, the second compound will go through quicker SN1 reaction in comparison to the first compound.

Q6.) Give one use of each of the following:

(i) Freon-12

Ans: Freon-12 CCl2F2, is used by the industry which is the most prevalent form of the refrigerant or air-conditioning components, aerosol propellants.

(ii) Iodoform

Ans: The Iodoform was used earlier as an antiseptic due to the liberation of free iodine from it, and not due to the substance itself. As it leaves an peculiar odour it has been replaced by other iodine-containing formulations.

(iii) DDT

Ans: DDT was first discovered to be used as chlorinated organic insecticides in 1939. After World War II, DDT was used as it due to its effectiveness against mosquitoes that led to disease like malaria and insects harming the crops leading to rise in its demand.

Q7.) An organic compound (A) having molecular formula C3H7Cl reaction with alcoholic solution of KCN gives compound B. The compound B on hydrolysis with dilute HCl gives compound C. C on reduction with H2/Ni gives 1-aminobutane. Identify A, B and C.

Ans. The formula C3H7Cl depicts that compound (A) is an alkyl halide. Compound (B) is formed when there is a reaction with KCN. When compound (C) is reduced with hydrogen and nickel it produces 1-aminobutane that concludes that all the compounds mentioned above in the question are straight-chain compounds. Therefore, compound (A) is a 1-Chloropropane, compound (B) is a Propionitrile, and compound (C) is a Butanamide. The reactions are mentioned below:

CH3-CH2-CH2-Cl+KCN    ⟶   CH3-CH2-CH2-CN+KCl


CH3-CH2-CH2-CN      CH3-CH2-CH2-CONH2



Q8. Elimination reactions (especially B-elimination) are as common as the nucleophilic substitution reaction in case of alkyl halides. State the reagents applied for use in both the  cases.

Ans. The reactions like nucleophilic substitution and elimination (beta-elimination) reactions are more likely to be possible with alkyl halides. The elimination reaction is well-suited to strong, huge bases and high temperatures. Whereas, the substitution reaction works well at lower temperatures for smaller and weaker bases.

Q9. Diphenyls are a potential threat to the environment. How are aryl halides used to form these?

Ans. When coal and mineral oil are not burned entirely, the diphenyl is produced in the environment. These are usually present in car exhaust gases, exhaust air of home and industrial heating systems. Several case studies have shown that ingestion of  diphenyl has led to several  health problems in humans as well as animals such as well as harmful effects on the liver, kidneys, eye and skin irritation, as well as damage to the central/peripheral nervous system. In the presence of dry ether, when aryl halides are in contact with sodium, diphenyl is produced. This reaction is termed a fitting reaction.

Q10. Describe the reactions mentioned below and give a example:

(A) Swarts reaction.

(B) Finkelstein reaction.

(C) Wurtz reaction.

Ans. (A) Swarts reaction.

Alkyl fluorides from alkyl chlorides or alkyl bromides are formed by using the Swarts reaction. This can be achieved by heating the alkyl chloride/ bromide in the existence of fluoride in specific heavy metals. The reaction is as follows:

                         CH3Br+AgF → CH3F+AgBr

(B) Finkelstein reaction.

When an alkyl bromide or alkyl chloride is changed into an alkyl iodide, which is further treated with a sodium iodide solution present in an acetone. The reaction is as follows:

                      CH3CH2Br+NaI → CH3CH2I+NaBr

(C) Wurtz reaction.

When alkyl halides come in contact with sodium metal in a dry ethereal (moisture-free) solution form  higher alkanes. It can also be applied to produce higher alkanes with an equal number of carbon atoms. The reaction is as follows: 

                        2RX+2Na ⟶RR+2NaX

Q11. Why is it mandatory to eliminate water, alcohols, and amines while using  a Grignard reagent?

Ans. Grignard reagents are considered as extremely reactive substances. Liquids such as water, alcohols, and amines are adequately acidic to transform the grignard reagents into the corresponding hydrocarbon. Therefore, it is mandatory to eliminate water, alcohols and amines while using a Grignard reagent.

FAQs (Frequently Asked Questions)

1. Why should students of Class 12 refer to extra questions for Class 12 Chemistry Chapter 10 Solutions?

Class 12 is an important academic year for every student. Therefore, students should extensively practise the papers of past years as well as extra questions for their board exams. Class 12 Chemistry Chapter 10 Important Questions will be beneficial for CBSE Class 12 students to score excellent marks in their Class 12 board exams. These notes consist of the important topics from Chapter 10 of Chemistry which will aid the students to get a thorough understanding of the chapter and help them clear their doubts. Class 12 Chemistry Chapter 10 Important Questions are made in accordance with the CBSE Syllabus.

2. What are organo-metallic compounds?

Compounds consisting of carbon-metal bonds are derived when some specific metals react with organic chlorides, bromides, and iodides. These compounds are called organo-metallic compounds.