Bihar Board - Class 12 Chemistry - Chapter 10: Haloalkanes and Haloarenes Short Answer Question

                                         Short Question Answer 

Question. 1. What happens when KI is added in the reaction chamber where the following reaction is taking place ?

CH3-Cl CH3-CN 

Answer ⇒ Since I- is stronger nucleophile than CN- and also I- is a better leaving group than Cl-. Addition of I will increase the rate for the given reaction.

Question. 2. Name the Grignard reagent required for converting formaldehyde to n-Butyl alcohol.

Answer ⇒ n-Propyl magnesium iodide.

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Question. 3. Write structures of the following compounds :

(i) 2-chloro-3-methyl pentane
(ii) 1-chloro-4-ethyl cyclohexane
(iii) 4-tert-Butyl-3-iodoheptane
(iv) 1 , 4-Dibromo but-2-ene
(v) 1-Bromo-4-sec butyl 1-2methyl benzene.

Answer ⇒ (i) CH3CH2CH(CH3)CHClCH3

Question. 4. Write IUPAC name of the following:

Answer ⇒ (i) 2-Bromopent-2-ene
(ii) 3-Bromo-2-methylbut-1-ene
(iii) 4-Bromo-3-methylpent-2-ene
(iv) 1-Bromo-2-methylbut-2-ene
(v) 1-Bromobut-2-ene

Question.5. What is meant by dehydrohalogenation ?

Answer ⇒ Dehydrohalogenation is the elimination reaction shown by a haloalkane which involves the removal of the halogen atom together with a hydrogen atom bonded to a C atom adjacent to C atom bearing the halogen. As a result of dehydrohalogenation an alkene is formed. For example :

Question.6. What is Grignard reagent and is it prepared ?

Answer ⇒ Grignard reagents : Alkyl magnesium halides (RMgX) are called Grignard reagents.

Preparation of Grignard reagents : When a solution of alkyl halide (R-X) in ether is allowed to stand over magnesium turning for some time then metal gradually dissolves. In this reaction alkyl magnesium halide is formed. This compound is called Grignard reagent.

R-X+MgR-Mg-X
Haloalkane                 Grignard reagent

Question. 7. In contrast to arenes, aliphatic hydrocarbons do not undergo nitration easily.

Answer ⇒ Nitration proceeds through the attack of the electrophile NO2+  at a point of high electron density. In arenes, due to resonance such electrophilic reaction is possible, while similar situation in aliphatic hydrocarbons is not possible under normal conditions.

Question.8.  Give a chemical test to distinguish between H5Br and C6H5Br.
Answer ⇒ Both are heated with aqueous NaOH. C6H5Br ves  ethanol and NaBr, which reacts with AgNO3, ves yellow precipitate of AgBr. C6H5Br does not respond to this test. 

Question .9.  Arrange the following in increasing order of Boiling point:
(i) CH3CH2CH2CH2Br
(ii) (CH3)3C.Br
(iii) (CH3)2.CH.CH2.Br
Answer ⇒ (CH3)3C.Br <(CH3)2.CH.CH2.Br <CH3CH2CH2CH2Br  

Question.10.  What happens when CH3-Br is treated with KCN? 

Answer ⇒

CH3-Br +KCN CH3CN+KBr

Methyl       Potassium   Methyl     Potassium 

bromide      cyanide       cyanide     bromide

Question.11.  A hydrocarbon C5H12 gives only one mono-chlorination product. Identify the hydrocarbon. 

Answer ⇒ Since there is only one monochloro derivative, the compound contains 12 equivalent hydrogen in four equivalent CH3. The compound is 2 , 2-dimethyl propane. 

Question 12.  Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same.
Answer ⇒The reasons are: 

1. Due to resonance/diagrammatic representation C-Cl bond acquires a partial double bond character. As a result, the C-Cl the 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 substitution. 

2. Due to repulsion between nucleophilic and electron rich arenes. 

Question 13.  (a) Why does p-dichlorobenzene have a higher m.p. than its o-and m-isomers? 

(b) Why is ()-Butan-2-ol optically inactive? 

Answer ⇒

(a) p-isomers are comparatively more symmetrical and fit closely in the crystal lattice, thus requiring more heat to break these strong forces of attraction. Therefore higher melting point than o-and m-isomers. 

(b) ()-Butan-2-ol is optically inactive because in a racemic mix one type of rotation is canceled by another. 

Question 14.  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 other the S or N attacking. 

Question 15.  Write the IUPAC names of the following compounds: 

(i) CH2=CHCH2Br 

(ii) (CCl3)3 CCl

Answer ⇒

(i) CH2=CHCH2Br  

IUPAC name: 3-Bromopropene : 

(ii) (CCl3)3 CCl

IUPAC name: 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptafluoropropane  

Question 16.  Which compound in each of the following pairs will react faster in SN2 reaction with -OH? 

(i) CH3Br or CH3I 

(ii) (CH3)3 CCl or CH3Cl 

Answer ⇒

(i) CH3I: Because Iodide is better at leaving a group than bromide. 

(ii) CH3Cl : Carbon atom leaving group is less hindered. 

Question 17. What is plane polarized light?
Answer ⇒ Light which vibrates in one specific plane is known as plane polarized light. On passing normal light through a Nicol prism, plane polarized light is obtained.

Question 18. What is the order of boiling points of alkyl halides for the same alkyl group?

Answer ⇒ Decreasing order of boiling points : RI > RBr > RCl > RF

Question 19. Write an example of 3 alkyl chloride.

Answer ⇒ Tertiary butyl chloride

Question 20. Write the equation of Swart’s reaction.
Answer ⇒

CH3-Br/Cl =AgFCH3F+AgBr

Question 21. What are Enantiomers?
Answer ⇒ The stereoisomers related to each other as non – superimposable mirror images are called Enantiomers.

Question 22. What is Lucas reagent? Give its application.

Answer ⇒Solution of ZnCl2 in cone. HCl is known as Lucas reagent.

Application: It is used to differentiate primary, secondary and tertiary alcohols.

On adding the alcohol to Lucas reagent, a tertiary alcohol reacts immediately forming a ppt. of alkyl chloride. If the ppt. appears after a few minutes, then the alcohol is secondary. If no ppt. is obtained in cold the alcohol is primary.

Question 23: What are ambident nucleophiles? Explain with an example.

Answer ⇒When there are two nucleophilic sites in a nucleophile, they are called ambident nucleophiles. These nucleophilic sites are sites through which they can attack.

For ex- Nitrite ion

Alkyl nitrites are formed when nitrite ion can attack through oxygen. And, nitroalkanes are formed when it can attack through nitrogen.

Question 24 : What will be the mechanism for the following reaction?

Answer ⇒The given reaction is

CN- acts as the nucleophile and attacks the carbon atom on which the Br  is attached. CN- ion is an ambident nucleophile and can attack through both C  and N  positions. It attacks through C  atom, in this case.

Question 25.  Write the IUPAC name of the following: 

Answer ⇒

IUPAC name : 3-Methylpent-2-ene.