CBSE Class 12 Chemistry Chapter 10 Important Questions – Free PDF Download
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Important Questions Class 12 Chemistry Chapter 10 – Haloalkanes and Haloarenes
CBSE Class 12 Chemistry Important Questions Chapter 10 – Haloalkanes and Haloarenes
3 Mark Questions
1. 
Ans. X = 
Y = 
Z = 
2. An organic compound ‘A’ having molecular formula 
on treatment with aqueous 
given’ B’ which on treatment with Lucas reagent gives ‘C’. The compound ‘C’ on treatment with ethanolic KOH gives back on compound ‘A’. Identify A, B, & C.
Ans. 
The question are
3. An organic compound ‘A’ on heating with NH3 and cuprous oxide at high pressure gives compound ‘B’. The compound ‘B’ on treatment with ice cold solution of 
and HCl gives ‘C” , which on heating with copper turning and HCl gives ‘A’ again. Identify A, B & C. compound
Ans.
The question are
4. 
Ans.
5. 
Ans.
6. 
Ans.
7. 
Ans.
8. A compound ‘A’ contains carbon and hydrogen only and has molecular mass of 72. Its photo chlorination gives a mixture containing only one monochloro and two dichloro hydrocarbons. Deduce the structure of A and chlorinated products.
Ans. A is 
(mol. Wt. 72) Since its gives one mono chloro and two dichloro derivatives on photochemical chlorination, it is
The reactions are
9. Why is sulphuric acid not used during the reaction of alcohols with KI?
Ans. In the presence of sulphuric acid
, KI produces HI
Since 
is an oxidizing agent, it oxidizes HI (produced in the reaction to
).
As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as
is used.
10. A hydrocarbon
does not react with chlorine in dark but gives a single monochloro compound 
in bright sunlight. Identify the hydrocarbon.
Ans. A hydrocarbon with the molecular formula, belongs to the group with a general molecular formula
. Therefore, it may either be an alkene or a cycloalkane.
Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane.
Further, the hydrocarbon gives a single monochloro compound, by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H-atoms that are all equivalent. Also, as all H-atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.
Cyclopentane ()
The reactions involved in the question are:
11. Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Ans. (i)
(ii)
(iii)
12. What are ambident nucleophiles? Explain with an example.
Ans. Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.
For example, nitrite ion is an ambident nucleophile.
Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.
13. Which compound in each of the following pairs will react faster in 
reaction with
?
(i) 
(ii) 
Ans. (i) In the mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.
R-F << R-Cl < R-Br < R-I
Therefore, CH3I will react faster than 
in reactions with OH–.
(ii)
The mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of
, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in
. Hence, reacts faster than 
in reaction with.
14. Write the mechanism of the following reaction:
Ans.
The given reaction is an reaction. In this reaction, CN – acts as the nucleophile and attacks the carbon atom to which Br is attached. CN – ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.
15. Out of 
and
, which is more easily hydrolysed by aqueous KOH?
Ans.
Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now , forms 
carbocation, while forms 
carbocation, which is more stable than carbocation. Hence , is hydrolyzed more easily than by aqueous KOH.
16. p-Dichlorobenzene has higher m.p. and lower solubility than those of o– and
m-isomers. Discuss.
Ans.
p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers.
