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

Long Question Answer

Question. 1. Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.

Answer ⇒ (i) Freon-12 :

Preparation : The chlorofluorocarbon compounds of methane and ethane are collectively known as freons. Dichloro difluoro methane (Freon 12) is the most common freons. It is manufactured from tetrachloromethane by the action of antimony trifluoride in the presence of antimony pentafluoride. This is known as swarts reaction.

3CCl₄ +                2SbF₃               3CCl₂F₂+2SbCl₃
Carbon                   Antimony                 (Freon)
tetrachloride         trifluoride.

Uses : (a) It is used as a refrigerant (cooling agent) in refrigerators and air conditioners.

(b) It is also used as a propellant in aerosols and foams to spray out deodorants, cleaners, hair sprays, and shaving creams.

(c) It is also used as an insecticide.

(ii) DDT (p, p’-Dichloro trichloro ethane) :

Preparation : It is manufactured by the condensation of chlorobenzene with trichloro acetal-dehyde (chloral) in the presence of sulphuric acid.

uses: (a) It is a powerful insecticide. It is highly stable and not easily decomposed.

(b) It is used for killing insects and mosquitoes.

(iii) Carbon tetrachloride (CCl4) :

Preparation : It is prepared industrially by chlorination of methane and by the action of chlorine
on carbon disulphide in the presence of aluminum chloride as catalyst.

(i) CH₄+4Cl₂ CCl₄+4HCl

(ii) CS₂+3Cl₂ CCl₄+           S₂Cl₂
Carbon                            Carbon             Sulphur
disulphide               tetrachloride         monochloride

Uses : (a) It is used as a solvent for oils, fats and waxes.

 (b) It is used as a fire extinguisher under the name pyrene.
(c) It is used for dry cleaning.
(d) It is used for the manufacture of freon.

(iv) Iodoform (CHI₃) :

Preparation : It is prepared by using ethanol or acetone with sodium hydroxide and iodine or sodium carbonate and iodine in water.

(a) CH₃CH₂OH+6NaOH+4I₂

.                                                   CHI₃+HCOONa+5H₂O+5NaI (Iodoform)

(b) CH3COCH₃+4NaOH+₃I₂
.                                            CHI₃CH₃COONa+3H₂O+3NaI  (Iodoform)

Uses : (a) It is used as an antiseptic and this nature is due to iodine that it liberates. However, because of its very unpleasant smell, it has now been replaced by better antiseptics.

(b) It is used in the manufacture of pharmaceuticals.

Question. 2. How will you convert ethanal into the following compounds ?

(i) Butane-1,3-diol
(ii) But-2-enal
(iii) But-2-enoic acid.

Answer ⇒ (i)

Where pcc stands for ‘Pyridinium chloro chromate’.

Question. 3. Draw the structures of major monohalo products in each of the following reactions :

Answer : 

Question. 4. Draw the structure of all eight structural isomers that have the molecular formula C₅H₁1Br . Name each isomer according to the IUPAC system and classify them as primary, secondary or tertiary bromide.

Answer ⇒ CH₂CH₂CH₂CH₂CH₂Br..........1-Bromopentane (1)

CH₂CH₂CH₂CHBrCH₃.............2-Bromopentane (2)

CH₃CH₂CH(Br)CH₂CH₂..........3-Bromopentane (2)
(CH₃)₂CHCH₂CH₂Br.......1-Bromo-3-methyl butane (1)
(CH₃)₂CHCHBrCH₃.......2-Bromo-3-methyl butane (2)
(CH₃)₂CBrCH₂CH₃.......2-Bromo-2-methyl butane (3)
CH₃CH₂CH(CH)₃CH₂Br.......1-Bromo-2-methyl butane (1)
(CH₃)₃CCH₂Brl......-Bromo-2,2-dimethylpropane (1)

Question. 5. Write the chemical equations for the following conversions :

(i) Chloroform to acetic acid.

(ii) Acetaldehyde to chloroform.

(iii) Chloroform to ethyne (acetylene).

(iv) Ethanol to chloroform.

Answer ⇒

(i) CHCl₃ CH₃CNCH₃COOH

Chloroform                Methyl                      Acetic Acid

                                     Cyanide

(ii) CH₃CHOCHCl₃

  Acetaldehyde                                           Chloroform

(iii) 2CHCl₃ CH CH

  Chloroform                               Ethyne (acetylene)

(iv) CH₃CH₂OHCH₃CHO

     Ethanol                                                      Acetaldehyde

CCl₃CHOCHCl₃

Chloral                            Chloroform

Question . 6. Write the structure of isomers of C₅H1₁Br.

Answer ⇒ (i) CH₃-CH₂-CH₂-CH₂-CH₂Br

                1-bromopentane

                  2,2-methyl-1-bromo pentane

Question. 7. Write the structure of the major organic product in each of the following reactions :

(a) CH₂CH₂CH₂Cl+NaI
(b) (CH₃)₃CBr+KOH

(c) CH₂CH(Br)CH₂CH₃+NaOH aq. ethanol

(d) CH₃CH₂Br+KCN

(e) C₆H₅ONa+C₂H₅Cl

(f) CH₃CH₂CH₂OH+SOCl₂
(g) CH₃CH₂CH=CH+HBr

(h) CH₃CH=C(CH₃)₂+HBr

Answer ⇒ (a)

CH₂CH₂CH₂Cl+NaI CH₃CH₂CH₂I+NaCl
(Propyl chloride)                    (Major product)

(b) (CH₃)₃CBr+KOH → -C=CH₂+KBr
Tertiary butyl bromide                Isobutane
.                                                    (Major product)

(c) CH₂CH(Br)CH₂CH₃+NaOH
(2- bromobutane)

CK₃CH=CHCH₃+CH₃CH₂OH=CH₂
(but-2 -ene)           (but-1 -ene)
(Major product)        (Minor product)

(d) CH₃CH₂Br+KCN CH₃CH₂C ≅ KBr
(e) C₆H₅ONa+C₂H₅Cl C₆H₅OC₂H₅+NaCl
.                                                         (Ethoxy benzyl ethane)

(f) CH₃CH₂CH₂OH+SOCl₂CH₃CH₂CH₂Cl+SO₂+HCl 

(Propan- l -ol)                           Propyl chloride

.                                                                (Major product)

.                                                                Peroxide

(g) CH₃CH₂CH₂CH=CH+HBr CH₃CH₂CH₂CH₂CH₂Br

              (2-pentene)                                       (1-bromopentane)

                                                                                     (Major product)

(h)

Question.8. Give the IUPAC name of the following:

(a) (CH₃)₂CHCH₂I
(b) (CH₃)₂CHClCH₃
(c) CH₃CHBr(CH₃)₂CH₂CH
(d) (CH₃)2CClCBr(CH₃)₂

Answer : 

              2-chloro  -2-methylpropanz

(c) CH₃CHBrC (CH₃)₂ CH₂ CH₃                   

       4-Bromo, 3-dimethyl pentane        

(d) (CH₃)₂ CClCBr (CH₃)₂   

 1-Bromo dimethyl  2- chloro  2 methyl propane

Question . 9. Explain why ortho-nitrophenol is more acidic than ortho methoxy benzene ?

Answer ⇒

NO₂ The group is electron withdrawing due to resonance and inductive effect. The OCH₃ group has +R effect and increases the electron density around H of O-H group. Therefore, removal of Has H+ ion is easier in O-nitrophenol than in O-methoxy benzene. Hence, o-nitrophenol is more acidic than O-methoxy benzene.

Question. 10. What do you understand about the nucleophilic substitution reaction ? Why do haloalkanes undergo a nucleophilic substitution reaction ?

Answer ⇒ Nucleophilic Substitution Reaction : A chemical reaction in which a stronger nucleophilic (electron rich group which loves +ve center) substitutes a weaker nucleophile called the nucleophilic substitution reaction. For example,

CH₂Br +OH- CH₃OH+Br-

is a nucleophilic substitution reaction.

Nucleophilic Substitution reactions of haloalkanes : In haloalkanes (R-X) the halogen is more electronegative than carbon. Thus,C-X bond is polarized as C₅+-X₅-. Therefore, a nucleophilic (electron rich species) attacks the positively charged carbon atom and replaces the negatively charged halogen from R-X. Thus a new product is formed.

     |                                 |
-C₅+ – X₅- + : Z → C-Z + X–
    |                                 |
Nucleophile       Product    Halide ion

Question. 11. Write the equation for the following reactions :

(i) Alkyl halide reacts with the alcoholic solution of potassium cyanide.
(ii) Bromoethane reacts with alcoholic ammonia.
(iii) Haloalkane reacts with silver cyanide.
(iv) Ethyl chloride reacts with sodium lead alloy.
(v) 1-chloropropane is heated at about 573 K in the presence of AlCl₃
(vi) Haloalkane is heated with benzene in the presence of anhydrous aluminum chloride.

Answer ⇒

(i) R-X+                 KCN                R-C=N+KX
Alkyl                           Potassium          Alkyl cyanide Potassium
halide                         cyanide              (alkyl nitrile) iodide

(ii) C₂H₅Br+NH₃(alc.)C₂H₂NH₂+HBr
Bromoethane                                       Ethanamine

(iii) R-X+            AgCn               R-N=C+AgX
      Haloalkane           Silver                     Aminoalkane
.                                   cyanide

(iv) 4C₂H₅Cl+4Na+4Na-Pb(C₂H₅)4Pb+4NaCl+3Pb
Ethyl chloride  Sodium            Tetraethyl
                          lead alloy              lead

(v)

CH₃-CH₂-CH₂-Cl+ C₆H₅Br+HX
Haloalkane                 Benzene           Alkyl Benzene   
                          (Friedel crafts reaction)

Question 12. Write Iodoform reaction.

Iodoform gives yellow ppt. with an AgNO3 solution but chloroform doesn’t Why?

What happens when ethyl bromide is heated with alcoholic KOH?

Answer: 1. Iodoform reaction:
When ethyl alcohol or acetone is heated with iodine and NaOH yellow crystals of iodoform are formed.

C₂H₅OH+4I₂+6NaOH₅NaF+HCOONa+5H₂O+CHI₃

2. When iodoform is heated with AgNO3  solution a yellow ppt. (Agl)  is obtained but chloroform doesn’t give this reaction because in chloroform C-Cl bond is more stable than C-I bond in iodoform.

3. On boiling Ethyl bromide with alcoholic KOH  ethylene is formed.

C₂H₅Br+KOH (alc.)C₂H₄+KBr+H₂O 

                                     Ethylene

Question 13. Explain the Sandmeyer reaction with an example.

Answer ⇒ Decomposition of diazonium salts (Sandmeyer reaction):

When a diazonium salt solution is added to a solution of cuprous halide dissolved in the corresponding halogen acid, the diazo group is replaced by a halogen atom.

Question 14. Give chemical reaction between chlorobenzene and chloral in presence of cone. H₂SO₄.

Or, How is D.D.T. formed? Write its one application.

Answer ⇒ DDT (Dichlorodiphenyltrichloroethane) is formed by the condensation of one molecule of chloral with two molecules of chlorobenzene in presence of cone, H₂SO₄.

Application : It is an important pesticide.

Question 15. What product is formed by the reduction of chloroform? Give the chemical equation when it reacts to nitric acid and acetone.
Answer ⇒ Reduction:
1. On heating with Zn and HCl, it reduces to form methylene dichloride.

CHCl₃+2HCH₂Cl₂                           + HCl

                                               Methylene dichloride

2. On heating with zinc dust and water, methane is formed.

CHCl₃+6H CH₄+3HCl

                                          Methane

Reactions of Chloroform :
1. With Conc.HNO₃ :
On treating chloroform with concentrated nitric acid, the hydrogen atom of chloroform is replaced by nitro group and nitro chloroform (or chloropicrin) is formed. It is a liquid (b.p. 112C) which is used in war as a poisonous gas.

2. With Acetone:
Chloroform condenses with acetone in presence of sodium hydroxide to form chloretone which is a hypnotic (sleep inducing drug) of high grade.

Question 16. What is 666 (lindane)? Explain its preparation and use in agriculture.

Answer ⇒ It is 1,2,3,4,5,6-HexachLorocyclohexane. It is obtained by heating benzene with chlorine in the presence of sunlight.

Preparation : It is prepared by the chlnncriUiv benzene in the presence of ultraviolet light.

1,2,3,4,5,6- Hexachlorocyclohexane (B.H.C.)

Uses:

Benzene hexachloride is an additional compound and its - isomer is called gammexane. It is an important pesticide used in agriculture. It is also called lindane or 666.

Question 17. Explain the following reaction of chlorobenzene:

1. Ration with chlorine in the presence of FeCl₃ in dark

2. Fittig reaction.

Answer ⇒
1. When Chlorobenzene reacts with Cl₂ hi the presence of FeCl₃ in the dark. o- dichlorobenzene and p- dichlorobenzene is obtained.

2. Fittig reaction : When Chlorobenzene is heated at 200C with Cu powder in a sealed tube Diphenyl is formed.

When two molecules of aryl halide react with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as the Fittig reaction.

Question 18. Write method of preparation, properties and uses of Freon.

Answer ⇒ Freon : Dichloro, Difluoro methane.

it is formed by the action of SbF₃ with CCl₃ in presence of SbCl₅.

3CCl₄+2SbF₃2SbCl₃+3CCl₂F₂

It has a very low boiling point due to which by increasing the pressure at room temperature it can be liquefied. It is a non – poisonous, non – combustible and inactive substance which is used as a cooling agent in the refrigerator. It is used in aerosol and foam.

Question 19. 1. B.P. of ethyl iodide is higher than b.p. of ethyl bromide. Give a reason.

2. Explain why the m.p. of para dichlorobenzene is higher than its ortho arid meta derivatives.

Answer ⇒

1. In alkyl halides containing the same alkyl group boiling point increases with increase in atomic weights of halogen atoms. Molecular weight of ethyl iodide is more than ethyl bromide and therefore boiling point of ethyl iodide is also high.

2. Para derivatives of dichlorobenzene are more symmetrical than its ortho and meta derivatives therefore its m.p. is higher.

Question 20. Explain Friedel – Craft’s reaction with chemical equations?

Answer ⇒ Friedel – Crafts reaction (alkylation) : Alkyl halides react with benzene in presence of anhydrous aluminum chloride to give alkyl benzene.

Acetylation:
Acetyl chloride reacts with benzene in presence of anhydrous aluminum chloride to give acetophenone.

Question 21. Write the equation of following reactions of chlorobenzene : 

Halogenation

Nitration

Sulphonation

Alkylation.

Answer ⇒
1. Halogenation : Haloarene reacts with halogen in presence of halogen carrier like FeCl₃ to form ortho and para substituted di haloarene.

2. Nitration : Haloarenes react with nitrating mixture to form o- nitro and p-nitro substituted haloarenes.

3. Sulphonation : On heating with a cone. H₂SO₂ , 2- Chlorobenzene sulfonic acid and 4-Chlorobenzene sulfonic acid is formed.

4. Alkylation : Alkylation takes place with alkyl halide in presence of anhydrous AlCl₃.

Question 22. Explain the nucleophilic substitution reaction in alkyl halide by SN₁ and SN₂  medianism.
Answer ⇒ Nucleophilic substitution reaction:

In the carbon halogen bond of haloalkane. The halogen atom is more electronegative as compared to the carbon atom hence, the shared pair of electrons between carbon and halogen is more attracted by the halogen atom. As a result a small negative charge and an equivalent positive charge develops on the halogen atom and carbon atom respectively.

Nucleophile attacks the electron deficient carbon due to the presence of partial positive charge on it and replaces the weaker nucleophilic ion i.e. the halide ion. Thus, the reaction is known as nucleophilic substitution reaction.

The order of reactivity of different alkyl halide towards nucleophilic substitution reaction is:

RI > RBr > RCl>RF 

Mechanism of Nucleophilic substitution reactions :
Nucleophilic substitution reaction occurs through two different mechanism :

(1) SN₁  Mechanism (Unimolecular nucleophilic substitution) : In this mechanism following steps are involved :
(a) Formation of carbocation by dissociation of substrate i.e., reactant molecule.

(b) Attack of nucleophile on carbocation forming the product.

(2) SN₂  mechanism (Bimolecular nucleophilic substitution):
Reactions of this type occur in one step i.e. they are concerted reactions. This reaction nucleophilic attack results in a transition state in which both the reactant molecules are partially bonded to each other and then the halide ion escapes out forming the product.

ROH-+CH₃XCH₃OH+RX


Rate of reaction =K[(RX)OH-]
Order of reactivity of alkyl halide is : Primary > Secondary > Tertiary.

Question 23. Draw labeled diagram of laboratory method for preparation of iodoform from alcohol. Write related chemical equations.
Answer ⇒

Chemical reaction:

CH₃-CH₂-OH+I₂ CH₃CHO+2HI

CH₃-CHO+3I₂CI₃-CHO+3HI

CI₃CHO +NaOH CHI₃+HCOONa

5HI +5NaOH 5NaI+5H₂O

CH₃-CH₂OH+4I₂+6NaOHCHI₃+5NaI+HCOONa+5H₂O

Question 24. Give the laboratory method for the preparation of chloroform. Describe the formation of chloroform by ethanol with labeled diagram, equation and principle.
Answer ⇒ Laboratory method : Chloroform is prepared in the laboratory by the action of water and bleaching powder on ethyl alcohol or acetone.

Method: About 100 gm of bleaching powder made into a paste by adding about 200 ml of water and taken in a flask fitted with a condenser. Now, 25 ml If alcohol or acetone is added and the mixture is distilled, chloroform collects as a heavy liquid under water.

It is washed with dilute NaOH solution then with water, dried over fused calcium chloride and redistilled. The available chlorine of bleaching powder acts as oxidizing as well as chlorinating agent during the preparation of chloroform from alcohol and acetone.

CaOCl₂+H₂OCa(OH)₂+Cl₂


The chemistry involved in the conversion of alcohol and acetone into chloroform is as shown below:

(A) From alcohol: The steps involved are :
(i) Ethyl alcohol is oxidized by chlorine to acetaldehyde.

CH₃CH₂OH+Cl₂CH₃CHO+2HCl

Ethyl alcohol             Acetaldehyde

(ii) Acetaldehyde reacts with chlorine to give chloral, i.e. trichloroacetaldehyde

CH₃CHO+       3Cl2 CCl₃CHO+3HCl

Acetaldehyde               Chloral 

(iii) Two moles of chloral react with one mole of calcium hydroxide to produce chloroform.

Calcium hydroxide Chloral Chloroform Calcium formate.

Question 25. HaloaLkanes are more reactive than haloarenes. Give a reason.
or,
Why, aryl halides less reactive than alkyl halides?
Answer ⇒
In aryl halides, the halogen atom is attached more strongly to the nucleus therefore the nucleophilic substitution takes slower than alkyl halides. There are two reasons for the less reactivity of aryl halides.
(i) In aryl halides sp₃ hybridization takes place whereas in alkyl halides sp₂ hybridization is present due to sp2 hybridization in haloarenes the halogen are attached to the nucleus more strongly.

(ii) Due to the presence of resonance in aryl halides there is some double bond character in C-Cl bond. Thus, the bond length of C-Cl bond is lesser than C-Cl bond in haloalkanes. Therefore, it is difficult to replace the halogen of haloarenes.

हिंदी के सभी अध्याय के महत्वपूर्ण प्रशन उत्तर के लिए अभी Download करें Vidyakul App - Free Download Click Here