MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Haloalkanes and Haloarenes NCERT Intext Exercises

Question 1.
Write structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. ButyI-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. buty1-2-methylbenzene.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 1

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI ?
Answer:
H2SO4 is an oxidizing agent. It oxidizes HI produce during the reaction to I2 and thus prevents the reaction between alcohol and HI to form an alkyl iodide.
KI + H2SO4 → KHSO4 + HI
2HI + H2SO4 → I2 + 2H2O + SO2
To remove this difficulty a non-oxidizing agent such as H3PO4 is used in place of H2SO4.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 2

Question 4.
Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields:
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 3
All the hydrogen are same i.e. 1°. Therefore, replacement of only one of them will give the same product.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 4
Replacement of a, b and c hydrogen atoms give three isomeric monochlorides.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 5
Replacement of a, b, c and d hydrogen atoms give four isomeric monochlorides.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 6
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 7
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 8

MP Board Solutions

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
(a) For the same alkyl group boiling point increases with increase in the molar mass of halogen atom; increase in the number of halogen atoms.
(b) For the same halogen with the increase in branching boiling point decreases.
On the basis of this following order is predicted :
(i) Chloromethane < Bromomethane < Dichloro-methane < Bromoform.
(ii) Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism ? Explain your answer.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 9
Answer:
If the leaving group is same in different isomers of a particular formula, the reactivity of the isomers towards SN2 mechanism decreases with the steric hindrance.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 10
which is 2° alkyl halide having some steric hindrance.

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 11
which is 3° alkyl halide having much more steric hindrance (than 2°).
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 12
are 2° alkyl halides. But in (II) the CH3 group is at C2 atom which is closer to Br (exerts more steric hindrance) to the attacking nucleophile at C1 atom as compared to (I)
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 13

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 14
Answer:
The reactivity of alkyl halide towards SN1 reaction depends upon the stability of the intermediate carbocation formed as 3° > 2° > 1°
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 15

Question 9.
Identify A, B, C, D, E, R and R’ in the following :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 16
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 17
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 18
Since, D of D2O gets attached to same C-atom on which – MgBr or Br was present so that
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 19
tert-Alkyl halides do not undergo Wurtz reaction. Therefore, the question is not correct. They undergo dehydrohalogenation to give alkenes. Hence,
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 20

MP Board Solutions

Haloalkanes and Haloarenes NCERT Textbook Exercises

Question 1.
Name of the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides :

  1. (CH3)2CHCH(Cl)CH3
  2. CH3CH2CH(CH3)CH(C2H5)Cl
  3. CH3CH2C(CH3)2CH2I
  4. (CH3)3CCH2CH(Br)C6H5
  5. CH3CH(CH3)CH(Br)CH3
  6. CH3C(C2H5)2CH2Br
  7. CH3C(Cl)(C2H5)CH2CH3
  8. CH3CH=C(Cl)CH2CH(CH3)2
  9. CH3CH=CHC(Br)(CH3)2
  10. p-ClC6H4CH2CH(CH3)2
  11. m-ClCH2C6H4CH2C(CH3)3
  12. o-Br-C6H4CH(CH3)CH2CH3.

Answer:

  1. 2-Chloro-3-methylbutane (2° alkyl)
  2. 3-Chloro-4-methylhexane (2° alkyl)
  3. l-Iodo-2,2-dimethylbutane (1° alkyl)
  4. l-Bromo-3,3-dimethyl-l-phenylbutane (2° benzylic)
  5. 2-Bromo-3-methylbutane (2° alkyl)
  6. 3-Bromomethyl-3-methylpentane (1° alkyl)
  7. 3-Chloro-3-methylpentane (3° alkyl)
  8. 3-Chloro-5-methylhex-2-ene (vinyl)
  9. 4-Bromo-4-methylpent-2-ene (allylic)
  10. 1-Chloro-4-(2′-methylpropyl) benzene (aryl) or p-Chloro isobutyl benzene
  11. 1-Chloromethyl-3-(2’2′-diethylpropyl) benzene (benzylic) or /n-Neopentyl ben-zyl chloride
  12. 1-Bromo-2-(l’-methylpropyl) benzene (aryl).

Question 2.
Give the IUPAC names of the following compounds :

  1. CH3CH(Cl)CH(Br)CH3
  2. CHF2CBrCIF
  3. ClCH2C = CCH2Br
  4. (CCl3)3CCl
  5. CH3C(p-CIC6H4)2CH(Br)CH3
  6. (CH3)3CCH=ClC6H4I-p.

Answer:

  1. 2-Bromo-3-chlorobutane
  2. 1 -Bromo-1 -chloro-1,2,2-trifluoroethane
  3. l-Bromo-4-chlorobut-2-yne
  4. 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl) propane
  5. 3-Bromo-2,2,-bis (4′-chlorophenyl) butane
  6. 1-Chloro-l-(4′-iodophenyl)-3,3-dimethyl-but-1-ene.

Question 3.
Write the structures of the following organic halogen compounds :
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 21
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 22

Question 4.
Which one of the following has the highest dipole moment:
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 23
CCl4 (iii) is a symmetrical and has resultant zero dipole moment. In CHCl3 (ii), the resultant of two C – Cl dipoles is opposed by the resultant of C – H and C – Cl bonds. Since, the latter resultant is expected to be smaller than the former, therefore, CHCl3 has finite dipole (1.03 D) moment.
In CH2Cl2 (i), the resultant of two C-Cl dipole moments is reinforced by resultant of two C-H dipoles, therefore, CH2Cl2 (1.62 D) has a dipole moment higher than that of CHCl3.
Thus, CH2Cl2 has the highest dipole moment.

Question 5.
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight Identify the hydrocarbon.
Answer:

  1. The molecular formula suggests that it can either be a cycloalkane or alkene.
  2. Since, the hydrocarbon does not react with Cl2 in the dark, it cannot be alkene. Therefore, it must be a cycloalkane.
  3. The hydrocarbon reacts with Cl2 in the presence of bright sunlight to give a single mono-chloro compound C5H9Cl, therefore, all the ten H-atoms of cycloalkane must be equivalent.

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 24

Question 6.
Write the isomers of the compound having formula C4H9Br.
Answer:
The isomers of C4H9Br along with their common names are given below :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 25

MP Board Solutions

Question 7.
Write the equations for the preparation of 1-iodobutane from :
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 26

Question 8.
What are ambident nucleophiles ? Explain with an example.
Answer:
The nucleophile having two nucleophilic centres are called ambident nucleo-phile. Example is cyanide group because it can attack through C or N because of the following resonance structures :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 27

Question 9.
Which compound in each of the following pairs will react faster in SN2 reaction with OH-:
(i) CH3Br or CH3I
(ii) (CH3)3CCl or CH3Cl.
Answer:
(i) CH3I reacts faster than CH3Br in SN2 reaction with OH because I ion is a better leaving group than Br ion because of its large size.
(ii) CH3Cl reacts faster than (CH3)3CCl because of steric hindrance in (CH3)3CCl.

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene :
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.
Answer:
(i) In 1-Bromo-l-methylcyclohexane, the β hydrogens on either side of the Br atom are equivalent, therefore, only 1-alkene is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 28
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 29
Since, the alkene (A) is more substituted according to SaytzefTs rule, it is more stable and will be the major product.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 30

Question 11.
How will you bring about the following conversions :
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-l-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 31
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 32
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 33

MP Board Solutions

Question 12.
Explain why:
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride ?
(ii) Alkyl halides, though polar, are immiscible with water ?
(iii) Grignard reagents should be prepared under anhydrous conditions ?
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 34
Due to sp2 hybridization of C-atom in chlorobenzene, C-atom is more electronegative (greater s-character) whereas, cyclohexyl chloride, C-atom is sp3 hybridized i.e., less electronegative (lesser 5-character). So, polarity of C—Cl bond in chlorobenzene is less than C—Cl bond in cyclohexyl chloride. Further due to delocalisation of lone pair of electrons of Cl-atom over the benzene ring, C—Cl bond in chlorobenzene acquires partial double bond character while C—Cl bond in cyclohexyl chloride is a pure single bond. Thus, C—Cl bond in chlorobenzene is shorter than in cyclohexyl chloride. As dipole moment is a product of charge and distance, therefore, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

(ii) Water molecules have enough strong intermolecular H-bonding which is difficult to be broken by alkyl halides though they are polar in nature as well. Therefore, alkyl halide do not dissolve in water and form separate layers.

(iii) Grignard reagents (R-Mg-X) are readily decomposed by water to produce al-kanes. That is why, they should be prepared under anhydrous conditions. Instead, ether is used as a solvent during the preparation of Grignard reagent.

Question 13.
Give the uses of freon-12, D.D.T., Carbon tetrachloride and Iodoform.
Answer:
It is an important pesticide.

Freon-12 : Since freons have been found to be one of the factors responsible for the depletion of ozone layer, they are being replaced by other harmless compounds in many countries.

D.D.T. : DDT is highly toxic and has strong insecticidal properties and thus it was widely used as an insecticide and pesticide.

Carbon Tetrachloride : It is produced in large quantities for use in the manufacture of refrigerants and propellants for aerosol cans. It is used as feed stock in the synthesis of chlorofluorocarbons and other chemicals, pharmaceutical manufacturing and general sol¬vents use. Until the mid 1960’s, it was widely used as a cleaning fluid both in industry as a degreasing agent and in the home, as a spot remover and as fire extinguisher.

Iodoform : It was earlier used as an antiseptic but the antiseptic properties are due to the liberation of free iodine and not due to iodoform itself. Due to its objectionable smell, it has been replaced by other formulations containing iodine.

Question 14.
Write the structure of the major organic product in each of the following reactions :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 35
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 36
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 37
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 38

Question 15.
Write the mechanism of the following reaction :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 39
Answer:
KCN gives CN- ion as a nucleophile in an aqueous medium which is resonance hybrid of the following :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 40
Thus, cyanide ions is an ambident nucleophile. Therefore, it can attack the C-atom of C—Br bond in «-BuBr either through C-atom or through N-atom. Thus, two possible products are cyanides and isocyanides respectively.

But C—C bond is more stable than C—N bond, so attack occurs through C-atom and hence, cyanide is predominantly formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 41

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methylbutane, 1 Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methyIbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane
(iii) 1-Bromobutane, l-Bromo-2,2-dimethyl-propane, l-Bromo-2-methylbutane, l-Bromo-3-methylbutane.
Answer:
The reactivity of SN2 reaction depends upon steric hindrance. More the steric hindrance lesser will be the reactivity. Therefore, the reactivity of different alkyl halides towards SN2 reaction is 1° > 2° > 3°.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 42

Question 17.
Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH ?
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 43
However, under SN2 mechanism, the reactivity depends on steric hindrance, therefore, C6H5CH2Cl get hydrolysed more easily than C6H5CHClC6H5 under SN2 conditions.

Question 18.
p-Dichlorobenzene has higher m.p. and solubility than those of o- and m- isomers. Discuss.
Answer:
p-dichlorobenzene has higher melting point than its o-isomer due to symmetry of p-isomer that fits in the crystal lattice better than the o- or m- isomer. Therefore, it has stronger intermolecular forces of attraction than o- and m- isomers and thus greater energy is required to break crystal lattice to melt or dissolve the p-isomer than the corresponding o- and m- isomers. In other words melting point of p-isomer is higher and its solubility is lower than corresponding m- and o- isomers.

Question 19.
How the following conversions can be carried out:
(i) Propene to propan-1-ol
(ii) Ethanol to but-l-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3,4-dimethyIhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methyl-propane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) ttert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 44
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 45
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 46
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 47

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 48

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 49

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 50

MP Board Solutions

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
In the presence of water, KOH dissociates completely into OH ions which being a strong nucleophile brings about substitution on alkyl halides and produce alcohols from alkyl halide. Further in aqueous solution, OH ions are highly solvated (hydrated). This solvation reduces the basic character of OH ions which therefore, fails to abstract a proton from β-carbon of alkyl chloride to form alkenes. In alcoholic medium, (less polar than H2O) OH- is less highly hydrated therefore, acts as strong base and abstract the proton from β-carbon giving alkene as major product (dehydrohalogenation). Moreover alcoholic solution contains C2H5O- ethoxide ion in addition to OH ions. Being a stronger base than OH, they abstract proton gives alkene.

Qusetion 21.
Primary alkyl halide C4H9Br (a) is reacted with alcoholic KOH to give com-pound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d). C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural for-mula of (a) and write the equations for all the reactions.
Answer:
Two possible isomers of given 1° alkyl halide C4H9Br are :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 51
According to the question, compound (a) on reaction with sodium does not give the same product produced by n-butyl bromide. So (a) cannot be (I).
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 52
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 53

Question 22.
What happens when:
(i) n-butyl chloride is treated with alcoholic KOH.
(ii) bromobenzene is treated with Mg in the presence of dry ether.
(iii) chlorobenzene is subjected to hydrolysis.
(iv) ethyl chloride is treated with aqueous KOH.
(v) methyl bromide is treated with sodium in the presence of dry ether.
(vi) methyl chloride is treated with KCN ?
Answer:
(i) CH3CH2CH2CH2Cl + KOH (alc.)
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 54

MP Board Solutions

Haloalkanes and Haloarenes Other Important Questions and Answers

Haloalkanes and Haloarenes Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
Which of the following compound gives yellow precipitate with AgN03 solution :
(a) KIO3
(b) CHI3
(c) KI
(d) CH2I2.
Answer:
(c) KI

Question 2.
What is formed by the reaction of Ethyl bromide with lead sodium alloy :
(a) Tetraethyl lead
(b) Tetraethyl bromide
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Tetraethyl lead

Question 3.
Reaction CH3Br + OH → CH3 – OH + Br- is :
(a) Electrophilic substitution
(b) Electrophilic addition
(c) Nucleophilic addition
(d) Nucleophilic substitution.
Answer:
(d) Nucleophilic substitution.

Question 4.
When acetylene added to HC1, the product formed is :
(a) CH2 = CHCl
(b) CH3 – CH – Cl2
(c) Cl – CH = CHCl
(d) None of these.
Answer:
(b) CH3 – CH – Cl2

Question 5.
In Aryl halide, the carbon linked to halogen atom is :
(a) sp hybridized
(b) sp hybridized
(c) sp hybridized
(d) sp3d hybridized.
Answer:
(b) sp hybridized

Question 6.
In SN1 reaction in the first step is formed :
(a) Free radical
(b) Carbanion
(c) Carbcation
(d) Final product.
Answer:
(c) Carbcation

Question 7.
Chlorobenzene reacts with chloral and concentrated H2SO4 to form :
(a) PVC
(b) TNT
(c) B.H.C.
(d) DDT.
Answer:
(d) DDT.

Question 8.
Reaction CH3OH + OH → CH3OH + Br is :
(a) SN1
(b) SN2
(C) SE-1
(d)SE-2.
Answer:
(b) SN2

Question 9.
The following compound reacts with silver powder to form acetylene :
(a) CH2I2
(b) CH3I
(c) CHI3
(d) Cl4.
Answer:
(c) CHI3

Question 10.
Pyrene is used with any one of the following for extinguishing fire :
(a) CO2
(b) CH2Cl2
(c) CCl4
(d) CH2 = CHCl.
Answer:
(c) CCl4

Question 11.
Which of the following compound is known by the name freon :
(a) CHCl3
(b) CCl4
(c) CCl2F2
(d) CF4.
Answer:
(c) CCl2F2

Question 12.
On heating ethyl iodide with alcoholic KOH the product formed is :
(a) Ethanol
(b) Ethane
(c) Acetylene
(d) Ethylene.
Answer:
(d) Ethylene.

Question 13.
On heating C2H5OH with iodine and a base the product formed is :
(a) CH3I
(b) CHI3
(c) CH3CHO
(d) CHCI3.
Answer:
(b) CHI3

Question 14.
Which of the chemical formula is of chloropicrin :
(a) CCl3 – CHO
(b) C(NO2)Cl3
(c) CH3 – C(NO2)Cl2
(d) CCl3 – NH2.
Answer:
(b) C(NO2)Cl3

Question 15.
Order of polarity of CH3I, CH3Br and CH3Cl molecules is :
(a) CH3Br > CH3Cl > CH3I
(b) CH3I > CH3Br > CH3Cl
(c) CH3CI > CH3Br > CH3I
(d) CH3CI > CH3I > CH3Br.
Answer:
(c) CH3CI > CH3Br > CH3I

Question 16.
Correct order of reactivity of Alkyl halides is :
(a) Iodide > Bromide > Chloride
(b) Iodide < Bromide < Chloride
(c) Bromide > Iodide > Chloride
(d) Bromide < Chloride < Iodide.
Answer:
(a) Iodide > Bromide > Chloride

Question 17.
What is formed on heating iodoform with silver powder :
(a) Alkane
(b) Ethylene
(c) Acetylene
(d) Isocyanide.
Answer:
(c) Acetylene

Question 18.
Raschig method is used for the manufacture of which of the following :
(a) Chlorobenzene
(b) Benzene
(c) Toluene
(d) Nitro-benzene.
Answer:
(a) Chlorobenzene

Question 2.
Fill in the blanks :

  1. The formula of the harmful product formed on keeping chloroform open is ………………
  2. General formula of alkyl halide is ………………
  3. On heating aromatic primary amine with chloroform and alcoholic caustic potash a bad smelling gas ……………… is formed.
  4. B.H.C. is an insecticide whose commercial name is ………………
  5. Chloretone is a high grade ………………
  6. SN1 reaction occur in ……………… steps.
  7. In haloarene substitution reactions are mainly ………………
  8. Formula of refrigerant Freon is ………………

Answer:

    1. COCl2
  1. CnH2n+1X
  2. Phenyl isocyanide
  3. Gammaxene or lindane
  4. Hypnotic medicine
  5. Two
  6. Electrophilic
  7. CCl2F2.

Question 3.
Match the following :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 55
Answer:

  1. (d)
  2. (e)
  3. (f)
  4. (b)
  5. (c)
  6. (g)
  7. (a).

Question 4.
Answer in one word/sentence :

  1. On reacting Benzene with methyl chloride in presence of AlCl3, Toluene is formed, what is the name of the reaction ?
  2. Alkyl halide is polar in nature still it is insoluble in water.
  3. On heating alkyl halide with sodium metal, product formed is.
  4. On heating Benzene diazonium salt with cuprous halide and its corresponding acid haloarene is formed. Write name of the reaction.
  5. On heating iodobenzene with copper powder at 200°C, product obtained is.
  6. What is formed on treating benzene with Cl2 in the presence of sunlight ?
  7. Write laboratory method of preparation of chlorobenzene.
  8. Write the name of nucleophilic substitution reaction in primary alkyl halide.

Answer:

  1. Friedel-Crafts reaction
  2. Due to inability to form hydrogen bond
  3. Alkane
  4. Sandmeyer reaction
  5. Diphenyl
  6. B.H.C.
  7. Raschig method
  8. Bimolecular nucleophilic substitution reaction.

MP Board Solutions

Haloalkanes and Haloarenes Short Answer Type Questions

Question 1.
(i) Write Iodoform reaction.
(ii) Iodoform gives yellow ppt with AgNO3 solution but chloroform doesn’t Why ?
(iii) What happens when ethyl bromide is heated with alcoholic KOH ?
Answer:
(i) Iodoform reaction : When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C2H5OH +4I2 + 6NaOH → 5NaI + HCOONa + 5 H2O + CHI3

(ii) 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.

(iii) On boiling Ethyl bromide with alcoholic KOH ethylene is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 56
Question 2.
Explain the Sandmeyer reaction with 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.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 57

Question 3.
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 (Dichlorodiphenyl trichloroethane) is formed by the condensation of one molecule of chloral with two molecules of chlorobenzene in presence of cone. H2SO4.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 58
Application : It is an important pesticide.

Question 4.
What are Gem-dihalide and Vicinal-dihalide ?
Answer:
When both halogen atoms are linked to one carbon atom of hydrocarbon then it is known as Gem-dihalide. Gem means geminal i.e., same position.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 59
When both halogen atoms are connected to two different neighbouring carbon atoms, then it is known as vicinal dihalide. Vic means vicinal which means adjacent position.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 60

Question 5.
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 few minutes, then the alcohol is secondary. If no ppt. is obtained in cold the alcohol is primary.

Question 6.
Explain, Carbylamine reaction and give one application of this reaction.
Answer:
Carbylamine reaction: On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 61
Application : Chloroform and primary amine can be tested by this reaction.

Question 7.
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 presence of sunlight.
Preparation : It is prepared by the chlorination of benzene in the presence of ultra-violet light.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 62
1, 2, 3, 4, 5, 6-Hexachlorocyclohexane (B.H.C.)
Uses : Benzene hexachloride is an addition compound and its 7 -isomer is called gammexane. It is an important pesticide used in agriculture. It is also called lindane or 666.

Question 8.
Explain the following reaction of chlorobenzene :
(i) Reaction with chlorine in the presence of FeCI3 in dark
(ii) Fittig reaction.
Answer:
(i) When Chlorobenzene reacts with Cl2 in the presence of FeCl3 in dark o – dichlorobenzene and P – dichlorobenzene is obtained.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 63
(ii) Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 64
When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 65

Question 9.
How can you obtained following compounds from chloroform ? Give equations :
(a) Methane
(b) Acetylene
(c) Carbon tetrachloride.
Answer:
(a) Chloroform reduces into methane by Zn and H2O.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 66
(b) When CHCl3 heated with Ag powder it gives C2H2.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 67
(c) By the chlorination of CHCl3 in presence of sunlight CCl4 is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 68

Question 10.
Write short notes on :
(a) Hunsdiecker method and
(b) Raschig process.
Ans.
(a) Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl4, aryl bromide is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 69
This method is called Hunsdiecker method.

(b) Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl2(catalyst) at 230°C, chlorobenzene is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 70

Question 11.
Write method of preparation, properties and uses of Freon.
Answer:
Freon : Dichloro, Difluoro methane.
It is formed by the action of SbF3 with CCl4 in presence of SbCl5.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 71
It has 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 12.
(i) b.p. of ethyl iodide is higher than b.p. of ethyl bromide. Give reason
(ii) Explain why the m.p. of para dichlorobenzene is higher than its ortho and meta derivatives.
Answer:
(i) In alkyl halides containing same alkyl group boiling point increases with increases 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.
(ii) Para derivatives of dichlorobenzene is more symmetrical than its ortho and meta derivatives therefore its m.p. is higher.

Question 13.
Give main nucleophilic substitution reaction of alkyl halides.
Answer:
Nucleophilic substitution Reactions :

1. Substitution by —OH group (Hydrolysis): Alkyl halide on reacting with water or aqueous KOH hydrolyse to form alcohol.
C2H5Br + KOH → C2H5 — OH + KBr

2. Substitution by —OR group : Alkyl halide reacts with sodium alkoxide or Ag2O to form ether by substitution of halogen atom by —OR group.
C2H5Br + NaOC2H5 → C2H5 — OC2H5 + NaBr
2C2 H5I + Ag2O → (C2H5)2 O + 2AgI

3. Substitution by —CN group: Alkyl halide reacts with aqueous or alcoholic KCN to form alkyl cyanide.
C2H5Cl + KCN → C2H5CN + KCl

4. Substitution by ammonia (Hofmann method) : On heating alkyl halide with aqueous or alcoholic solution of NH3 in a sealed tube at 100°C, a mixture of various amine is obtained.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 72

Question 14.
Give the laboratory method for the preparation of chlorobenzene and explain its nitration and sulphonation reactions.
Answer:
Laboratory method: Chlorobenzene is prepared by direct halogenation of arene. Aryl chloride may be prepared from arenes by the action of chlorine or presence of halogen carrier like FeCl3, FeBr3 and AlCl3. Iodine and iron filings can also be used as halogen carrier.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 73
Nitration : Haloarenes react with nitrating mixture to form o-nitro and p-nitro sub-stituted haloarenes.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 74
Sulphonation : On heating with cone. H2SO4, 2-Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 75

Question 15.
Give the chemical reactions when ethyl halide reacts with the following :
(i) Alloy of Pb-Na
(ii) Mg metal
(iii) AgNO2
(iv) Na metal.
Answer:
(i) Reaction with lead-sodium alloy: Alkyl halides, form alkyl lead when treated with lead-sodium alloy.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 76
Tetraethyl lead (TEL) is an antiknock compound which is added in petrol.

(ii) Reaction with magnesium : Alkyl halide forms alkyl magnesium halide i.e., Grignard reagent with magnesium in presence of dry ether as a solvent.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 77

(iii) Reaction with AgNO2: Mainly nitroethane is formed.
C2H5I + AgNO2 → C2H5 NO2 + Agl

(iv) Reaction with sodium (Wurtz reaction) : When alkyl halide is heated with sodium, in the presence of dry ether, alkane is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 78

Question 16.
Give the preparation method, properties and applications of dichloroethane. Ans. Preparation of dichloroethane: It can be prepared by replacing two hydrogen of ethane by two chlorine atoms.
(i) By the passage of Cl2 into ethene :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 79
(ii) By the heating of mix of ethane diol and HCl in presence of anhydrous ZnCl2.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 80
Properties : (a) Reaction with aqueous KOH :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 81
(b) Reaction with alcoholic KOH :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 82
Vinyl chloride reacts with ale. KOH and form vinyl ethyl other.
CH2 = CHCl + HOC2H5 + KOH → CH2 = CH – OC2H5 + KCl + H2O
(c) Reaction with KCN :

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 83
(d) Reaction with Zn powder and methanol:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 84
Use : (i) As a solvent, (ii) As an antiknocking fuel, (iii) Removing paints.

MP Board Solutions

Question 17.
Write Frankland reaction.
Answer:
Reaction with zinc (Frankland’s reaction) : This reaction is similar to the Wurtz reaction but zinc is used in place of sodium.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 85

Question 18.
Explain Friedel-Craft’s reaction with chemical equation.
Answer:
Friedel-Craft’s reaction (alkylation) : Alkyl halides react with benzene in presence of anhydrous aluminium chloride to give alkyl benzene.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 86
Acetylation : Acetyl chloride reacts with benzene in presence of anhydrous aluminium chloride to give acetophenone.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 87

Question 19.
Identify ‘A’, ‘B’ ‘C’ and ‘D’
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 88
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 89

Question 20.
An alcohol ‘A’ on reaction with cone. H2SO4 gives an alkene ‘B’. ‘B’ after bromination with sodamide gives dehydrogenated compound ‘C’. ‘C’ on reaction of H2SO4 in presence of HgSO4 gives ‘D’ Identify ‘A’, ‘B% ‘C% and ‘D’.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 90

Haloalkanes and Haloarenes Long Answer Type Questions

Question 1.
Write short notes on :
(a) Hunsdiecker method
(b) Raschig process
(c) Wurtz’s reaction
(d) Westron
(e) Frankland reaction
(f) Carbylamine reaction
(g) Iodoform reaction
(h) Fittig reaction.
Answer:
(a) Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl4, aryl bromide is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 91
This method is called Hunsdiecker method.
(b) Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl2(catalyst) at 230°C, chlorobenzene is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 92
(c) Wurtz’s reaction : When alkyl halide is heated with sodium, in the presence of dry ether, alkane is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 93
(d) Westron : Symmetrical tetrachloromethane or acetylene tetrachloride are known as Westron. It can be prepared by the chlorination of acetylene.
Westron, it can be prepared by the chlorination of acetylene.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 94
It is poisonous, non- flammable liquid, which gives westrol when it is boiled with lime.
(e) Frankland reaction : When alkyl halide is heated with zinc dust, alkane is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 95
(f) Carbylamine reaction : On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has bad smell and is poisonous.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 96
(g) Iodoform reaction : When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C2H5OH +4I2 + 6NaOH → 5NaI + HCOONa + 5 H2O + CHI3

(h) Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 97
When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 98

Question 2.
Give the laboratory method for the preparation of chloroform. Describe the formation of chloroform by ethanol with labelled 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 of 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 oxidising as well as chlorinating agent during the preparation of chloroform from alcohol and acetone.

CaOCl2 + H2O →(OH)2 + Cl2
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 99
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.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 100
(ii) Acetaldehyde reacts with chlorine to give chloral, i.e. trichloro acetaldehyde.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 101
(iii) Two moles of chloral react with one mole of calcium hydroxide to produce chloroform.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 102

Question 3.
What product is formed by the reduction of chloroform ? Give the chemical equation when it reacts to nitric acid and acetone.
Answer:
Reduction: (i) On heating with Zn and HCl, it reduces to form methylene dichloride.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 103
(ii) On heating with zinc dust and water, methane is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 104
Nitration : 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. 112°C) which is used in war as a poisonous gas.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 105
Condensation : Chloroform condenses with acetone in presence of sodium hydroxide to form chloretone which is a hypnotic (sleep inducing drug) of high grade.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 106

MP Board Solutions

Question 4.
Give the Chemical reaction when chloroform reacts with following :
(a) Oxidation, (b) Carbylamine reaction, (c) Ag powder, (d) Nitration, (e) Reimer- Tiemann reaction.
Or,
How will you obtain the following from chloroform : (i) Carbonyl chloride, (ii) Acetylene, (iii) Chloropicrin, (iv) Phenyl isocyanide, (v) Chloretone, (vi) Salicylal- dehyde.
Or,
How trichloro methane reacts with : (a) Atmospheric air, (b) Aniline and ale. KOH, (c) Ag powder, (d) Cone. HNO3, (e) Phenol.
Answer:
(a) Action of air and light (Oxidation): Chloroform oxidizes in presence of sunlight and air and forms a poisonous gas, phosgene (carbonyl chloride).
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 107
Ordinary chloroform contains phosgene gas and is used as a solvent. Pure chloroform which is used as an anaesthetic does not contain even traces of phosgene. While preserving chloroform which is to be used as an anaesthetic, the following precautions are taken :

(i) The chloroform is filled in blue or brown coloured bottle up to the neck. After putting a stopper, the bottle is kept in dark. As there is no empty space in the bottle, it is also free from air.
(ii) One percent ethyl alcohol is added in the bottle. If phosgene gas is formed, alcohol reacts with it to form diethyl carbonate, a non-toxic substance, i.e. (C2H5)2CO3,
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 108

(b) Carbylamine reaction : On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 109

(c) Reaction with Ag powder or Dehalogenation : On heating chloroform with sil-ver powder, pure acetylene gas is formed.
CHCl3 + 6 Ag + Cl3CH → HC ≡ CH + 6 AgCl

(d) Nitration : 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. 112°C) which is used in war as a poisonous gas.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 110

(e) Reimer-Tiemann reaction : On heating chloroform with concentrated alkali and phenol at 60-70°C, o-hydroxy benzaldehyde (salicylaldehyde) is formed. Traces of p-hydro- xybenzaldehyde are also formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 111

Question 5.
Explain nucleophilic substitution reaction of chlorobenzene (Give only equations).
Answer:
Nucleophilic substitution reaction of Chlorobenzene : Halogen atom in haloarenes is very strongly linked directly to benzene ring due to which it cannot be easily substituted by nucleophilic reagents like -OH,-OR,-NH2,-CN etc. but at high pressure, temperature and in presence of suitable catalyst halogen can be substituted by these groups.

(i) Substitution by -OH group : On heating chlorobenzene with NaOH at 200 at-mospheric pressure and 300°C temperature phenol is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 112

(ii) Substitution by alkoxy (-OR) group :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 113
With sodium alkoxide mixed ether is formed.

(iii) Substitution by Amino group : On heating with aqueous ammonia in presence of Cu2O at 60°C atmospheric pressure and 200°C temperature aromatic amine is formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 114

(iv) Substitution by Cyano group:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 115

Question 6.
Explain the nucleophilic substitution reaction in alkyl halide by SN1 and SN2 mechanism.
Answer:
Nucleophilic substitution reaction : In the carbon halogen bond of haloalkane, 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 halogen atom and carbon atom respectively.

Nucleophile attacks the electron deficient carbon due to the presence of partial posi-tive charge on it and replaces the weaker nucleophilic ion i.e. the halide ion. Thus, the reaction is known as nucleophilic substitution reaction.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 116
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.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 117
(b) Attack of nucleophile on carbocation forming the product.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 118

(2) SN2 mechanism (Bimolecular nucleophilic substitution): Reactions of this type occur in one step i.e. they are concerted reactions. These 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.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 119
Rate of reaction = K[(RX)OH]
Order of reactivity of alkyl halide is : Primaiy > Secondary > Tertiary.

Question 7.
Draw labelled diagram of laboratory method for preparation at iodoform from alcohol. Write related chemical equation.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 120
Chemical reaction :
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 121

MP Board Solutions

Question 8.
Explain the following reactions of chlorobenzene :
(a) Reaction with chlorine in the absence of FeCl3 in dark.
(b) Ullmann reaction.
Answer:
(a) Chlorine reacts with chlorine in dark, in the presence of FeCl3 to form ortho -dichlorobenzene and p-dichlorobenzene.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 122
(b) On heating bromo or iodobenzene at 200°C temperature with Cu in a sealed tube, diphenyl is formed. This reaction is known as Ullmann reaction.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 123

Question 9.
Write the equation of following reactions of chlorobenzene :
(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Alkylation.
Answer:
(i) Halogenation : Haloarene reacts with halogen in presence of halogen carrier like FeCl3 to form ortho and para substituted dihaloarene.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 124

(ii) Nitration : Haloarenes react with nitrating mixture to form o-nitro and p-nitro substituted haloarenes.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 125

(iii) Sulphonation : On heating with cone. H2SO4, 2-Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 126

(iv) Alkylation: Alkylation takes place with alkyl halide in presence of anhydrous AlCl3.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 127

MP Board Solutions

Question 10.
Haloalkanes are more reactive than haloarenes. Give reason.
Or,
Why, aryl halides are less reactive than alkyl halides ?
Answer:
In aryl halides, halogen atom is attached more strongly to the nucleus therefore the nucleophilic substitution takes slowly than alkyl halides. There is 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 nucleus more strongly.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 128
(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.
MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes 129

MP Board Class 12th Chemistry Solutions