Haloalkanes And Haloarenes Class 12th Chemistry Part Ii CBSE Solution

Class 12th Chemistry Part Ii CBSE Solution
Intext Questions Pg-285
  1. 2-Chloro-3-methylpentane Write structures of the following compounds:…
  2. 1-Chloro-4-ethylcyclohexane Write structures of the following compounds:…
  3. 4-tert. Butyl-3-iodoheptane Write structures of the following compounds:…
  4. 1,4-Dibromobut-2-ene Write structures of the following compounds:…
  5. 1-Bromo-4-sec. butyl-2-methylbenzene. Write structures of the following compounds:…
Intext Questions Pg-289
  1. Why is sulphuric acid not used during the reaction of alcohols with KI?…
  2. Write structures of different halogen derivatives of propane.
  3. A single monochloride. Among the isomeric alkanes of molecular formula C5H12, identify the…
  4. Three isomeric monochlorides. Among the isomeric alkanes of molecular formula C5H12,…
  5. Four isomeric monochlorides. Among the isomeric alkanes of molecular formula C5H12,…
  6. Draw the structures of major monohalo products in each of the following reactions:…
  7. Draw the structures of major monohalo products in each of the following reactions:…
  8. Draw the structures of major monohalo products in each of the following reactions:…
  9. Draw the structures of major monohalo products in each of the following reactions:…
  10. CH3CH2Br + NaI → Draw the structures of major monohalo products in each of the following…
  11. Draw the structures of major monohalo products in each of the following reactions:…
Intext Questions Pg-307
  1. Which alkyl halide from the following pairs would you expect to react more rapidly by an…
  2. In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?…
  3. Identify A, B, C, D, E, R and R1 in the following:
Exercises
  1. Name the following halides according to IUPAC system and classify them as alkyl, allyl,…
  2. But-1-ene to but-2-ene How will you bring about the following conversions?…
  3. CH3CH(Cl)CH(Br)CH3:2-Bromo-3-chlorobutane Give the IUPAC names of the following compounds:…
  4. CHF2CBrClF :1-bromo-1-chloro-1,2,2-trifluoroethane Give the IUPAC names of the following…
  5. ClCH2C≡CCH2Br: 1-bromo-4-chlorobut-2-yne Give the IUPAC names of the following compounds:…
  6. (CCl3)3CCl Give the IUPAC names of the following compounds:
  7. CH3C(p-ClC6H4)2CH(Br)CH3 Give the IUPAC names of the following compounds:…
  8. (CH3)3CCH=CClC6H4I-p Give the IUPAC names of the following compounds:…
  9. 2-Chloro-3-methylpentane Write the structures of the following organic halogen compounds.…
  10. p-Bromochlorobenzene Write the structures of the following organic halogen compounds.…
  11. 1-Chloro-4-ethylcyclohexane Write the structures of the following organic halogen…
  12. 2-(2-Chlorophenyl)-1-iodooctane Write the structures of the following organic halogen…
  13. 2-Bromobutane Write the structures of the following organic halogen compounds.…
  14. 4-tert-Butyl-3-iodoheptane Write the structures of the following organic halogen…
  15. 1-Bromo-4-sec-butyl-2-methylbenzene Write the structures of the following organic halogen…
  16. 1,4-Dibromobut-2-ene Write the structures of the following organic halogen compounds.…
  17. Which one of the following has the highest dipole moment? (i) CH2Cl2 (ii) CHCl3 (iii) CCl4…
  18. A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro…
  19. Write the isomers of the compound having formula C4H9Br.
  20. 1-butanol Write the equations for the preparation of 1-iodobutane from…
  21. 1-chlorobutane Write the equations for the preparation of 1-iodobutane from…
  22. but-1-ene. Write the equations for the preparation of 1-iodobutane from…
  23. What are ambident nucleophiles? Explain with an example.
  24. Which compound in each of the following pairs will react faster in SN2 reaction with -OH?…
  25. 1-Bromo-1-methylcyclohexane Predict all the alkenes that would be formed by…
  26. 2-Chloro-2-methylbutane Predict all the alkenes that would be formed by…
  27. 2,2,3-Trimethyl-3-bromopentane. Predict all the alkenes that would be formed by…
  28. Ethanol to but-1-yne How will you bring about the following conversions?…
  29. Ethane to bromoethene How will you bring about the following conversions?…
  30. Propene to 1-nitropropane How will you bring about the following conversions?…
  31. Toluene to benzyl alcohol How will you bring about the following conversions?…
  32. Propene to propyne How will you bring about the following conversions?…
  33. Ethanol to ethyl fluoride How will you bring about the following conversions?…
  34. Bromomethane to propanone How will you bring about the following conversions?…
  35. 1-Chlorobutane to n-octane How will you bring about the following conversions?…
  36. Benzene to biphenyl. How will you bring about the following conversions?…
  37. The dipole moment of chlorobenzene is lower than that of cyclohexylchloride? Explain why…
  38. Alkyl halides, though polar, are immiscible with water? Explain why…
  39. Grignard reagents should be prepared under anhydrous conditions? Explain why…
  40. Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.…
  41. CH3CH2CH2Cl + NaI Write the structure of the major organic product in each of the…
  42. (CH3)3CBr + KOH Write the structure of the major organic product in each of the following…
  43. CH3CH(Br)CH2CH3 + NaOH Write the structure of the major organic product in each of the…
  44. CH3CH2Br + KCN Write the structure of the major organic product in each of the following…
  45. C6H5ONa + C2H5Cl Write the structure of the major organic product in each of the following…
  46. CH3CH2CH2OH + SOCl2 Write the structure of the major organic product in each of the…
  47. CH3CH2CH = CH2+ HBr Write the structure of the major organic product in each of the…
  48. CH3CH = C(CH3)2 + HBr Write the structure of the major organic product in each of the…
  49. Write the mechanism of the following reaction: nBuBr + KCN nBuCN
  50. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane Arrange the compounds of each set…
  51. 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane Arrange the…
  52. 1-Bromobutane, 1-Bromo-2,2 dimethylpropane, 1-Bromo-2-methylbutane,…
  53. Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.…
  54. p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.…
  55. Propene to propan-1-ol How the following conversions can be carried out?…
  56. Ethanol to but-1-yne How the following conversions can be carried out?…
  57. 1-Bromopropane to 2-bromopropane How the following conversions can be carried out?…
  58. Toluene to benzyl alcohol How the following conversions can be carried out?…
  59. Benzene to 4-bromonitrobenzene How the following conversions can be carried out?…
  60. Benzyl alcohol to 2-phenylethanoic acid How the following conversions can be carried out?…
  61. Ethanol to propanenitrile How the following conversions can be carried out?…
  62. Aniline to chlorobenzene How the following conversions can be carried out?…
  63. 2-Chlorobutane to 3, 4-dimethylhexane How the following conversions can be carried out?…
  64. 2-Methyl-1-propene to 2-chloro-2-methylpropane How the following conversions can be…
  65. Ethyl chloride to propanoic acid How the following conversions can be carried out?…
  66. But-1-ene to n-butyliodide How the following conversions can be carried out?…
  67. 2-Chloropropane to 1-propanol How the following conversions can be carried out?…
  68. Isopropyl alcohol to iodoform How the following conversions can be carried out?…
  69. Chlorobenzene to p-nitrophenol How the following conversions can be carried out?…
  70. 2-Bromopropane to 1-bromopropane How the following conversions can be carried out?…
  71. Chloroethane to butane How the following conversions can be carried out?…
  72. Benzene to diphenyl How the following conversions can be carried out?…
  73. tert-Butyl bromide to isobutyl bromide How the following conversions can be carried out?…
  74. Aniline to phenylisocyanide How the following conversions can be carried out?…
  75. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but…
  76. Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound…
  77. n-butyl chloride is treated with alcoholic KOH, What happens when…
  78. bromobenzene is treated with Mg in the presence of dry ether, What happens when…
  79. chlorobenzene is subjected to hydrolysis, What happens when
  80. ethyl chloride is treated with aqueous KOH, What happens when
  81. methyl bromide is treated with sodium in the presence of dry ether, What happens when…
  82. methyl chloride is treated with KCN? What happens when
Intext Questions Pg-291
  1. Arrange each set of compounds in order of increasing boiling points.(i) Bromomethane,…

Intext Questions Pg-285
Question 1.

Write structures of the following compounds:

2-Chloro-3-methylpentane


Answer:

From the name of the compound it is clear that the parent ring is pentane and chloro and methyl groups are attached in the straight chain at 2nd and 5th position respectively. Hence, the structure is as follows: -



Question 2.

Write structures of the following compounds:

1-Chloro-4-ethylcyclohexane


Answer:

From the name of the compound given it is clear that the parent group is hexane with 2 attachments namely chloro and ethyl groups at 1 and 4 positions respectively. Hence, the structure is as follows: -



Question 3.

Write structures of the following compounds:

4-tert. Butyl-3-iodoheptane


Answer:

From the name of the compound given it is clear that the parent group is heptane with tertiary butyl and iodine groups attached at 4 and 3 positions respectively. Hence, the structure is as follows: -



Question 4.

Write structures of the following compounds:

1,4-Dibromobut-2-ene


Answer:

From the name of the compound given it is clear that the parent group is an alkene with the unsaturation present in between 2nd and 3rd carbon and 2 bromine atoms attached at 1st and 4th carbon. Hence, the structure is as follows : -



Question 5.

Write structures of the following compounds:

1-Bromo-4-sec. butyl-2-methylbenzene.


Answer:

From the name of the compound given it is clear that the parent group is benzene with bromine group attached at 1st carbon, secondary butyl group attached at 4th carbon and a methyl group attached at 2nd carbon. Hence, the structure is a follows : -




Intext Questions Pg-289
Question 1.

Why is sulphuric acid not used during the reaction of alcohols with KI?


Answer:

In the presence of sulphuric acid (H2SO4), KI produces HI

2KI + H2SO→ 2KHSO4 + 2HI

Since H2SO4is an oxidizing agent, it oxidizes HI (produced in the reaction to I2).

2HI + H2SO4 → I2 + SO2 + H2O

As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. 
So, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as H3PO4is used.


Question 2.

Write structures of different halogen derivatives of propane.


Answer:

There are four different dihalogen derivatives of propane. The structures of these derivatives are as shown below: -

(i) 


(ii) 


(iii) 


(iv) 



Question 3.

Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields

A single monochloride.


Answer:

To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C5H12. This is because of the fact that replacement of any H-atom leads to the formation of the same product. Therefore the isomer is 2,2-dimethylpropane.



Question 4.

Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields

Three isomeric monochlorides.


Answer:

To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain three different types of H-atoms. Therefore, the isomer is n-pentane. It can be observed that there are three types of H atoms labelled as a, b and c in n-pentane as shown below : -



Question 5.

Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields

Four isomeric monochlorides.


Answer:

To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain four different types of H-atoms in it. Therefore, the required isomer is 2-methylbutane. It can be observed that there are four different types of H-atoms labelled as a, b, c, and d in 2-methylbutane as shown below: -



Question 6.

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



Answer:

Cyclohexanol will react with thionyl chloride to form Chlorocyclohexane with the evolution of sulphur dioxide and hydrogen chloride gas as shown below : -



Question 7.

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



Answer:

4-Ethylnitrobenzene will undergo benzylic bromination in presence of heat or light. Now, as benzylic radicals are more stable therefore benzylic hydrogen is abstracted. Hence, the reaction yields 4-(1-Bromoethyl)nitrobenzene as a product.



Question 8.

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



Answer:

4-Hydroxymethylphenol will react with hydrochloric acid under thermal conditions to yield 4-Chloromethylphenol a shown in the reaction below: -



Question 9.

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



Answer:

1-Methylcyclohexene will react with hydrogen iodide via markovnikov’s addition mechanism to give 1-Iodo1-methylcyclohexane as shown in the reaction below: -



Question 10.

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

CH3CH2Br + NaI →


Answer:

Bromoethane will react with sodium iodide in a Finkelstein reaction manner to yield Iodoethane in presence of dry acetone and heat as shown in the reaction below : -



Question 11.

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



Answer:

Cyclohexene reacts with bromine in presence of heat and light to undergo allylic bromination to yield 3-bromocyclohex-1-ene as the major product as shown in the reaction below : -




Intext Questions Pg-307
Question 1.

Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer.

(i) 

(ii) 

(iii) 


Answer:

In SN2 Reaction, the product will be formed in a single step with no intermediates. Nucleophile attack will be easier for simple halides than hindered haloalkanes. Therefore,


Because of this in SN2


i) Bromobutane (1°)reacts faster than 2-Bromobutane(2°)


ii) 2-Bromobutane(2°) reacts faster than 2-Bromo-2-methylpropane or tert-Butyl bromide (3°)


iii) 1-Bromo-3-methyl butane reacts faster than the 1-Bromo-2-methyl butane as the former is less hindered with respect to leaving halide than the later which is more hindered comparatively.



Question 2.

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

(i) 

(ii) 


Answer:


Reason: SN1 occurs in two steps. In the first step, carbocation forms. The rate of reaction in SN1 depends upon the stability of the carbocation, greater the stability faster the reaction. Tertiary (3°) carbocation is more stable than secondary (2°), which is further stable than primary (1°)


By using this


i) Tert-Butyl Chloride (3°) reacts faster than 3-Chloropentane (2°)


ii) 2-Chloro heptane (2°) is more reactive than 1-Chloro hexane (1°)



Question 3.

Identify A, B, C, D, E, R and R1 in the following:



Answer:

i) Bromocyclohexane is reacting with the metal Mg to give organo-metallic compound/Grignard reagent i.e. Cyclohexyl Magnesium Bromide (A).


Grignard reagents are highly reactive with hydrocarbons to give alkanes. Cyclohexyl Magnesium Bromide reacts with water to cyclohexane (B) and Mg(OH)Br.



ii) As the above reaction, haloalkane with metal Mg react to form the Grignard reagent, alkyl magnesium halide, RMgX, and further with hydrocarbon gives alkane.


∴ C is CH3CH(MgBr)CH3, a Grignard Reagent, R is CH3CH(Br)CH3 and D ≅ H



iii) In the Wurtz Reaction, alkyl halide in the presence of sodium in dry ether gives the double number of carbons.


Now, in the reaction 2,2,3,3-tetramethyl butane is formed after the alkyl halide reacting in the presence of sodium in dry ether.


∴ Number of carbons will be 3 in R1


R1 –X is CH3C(X)(CH3)CH3


And the remaining reaction is same as the above i.e. alkyl halide reacting with Magnesium to give Grignard Reagent (D) and further with hydrocarbon H2O to give alkane (E)


∴ D is CH3C(CH3)(CH3)MgBr


E is CH3CH(CH3)CH3





Exercises
Question 1.

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

(i) (CH3)2CHCH(Cl)CH3

(ii) CH3CH2CH(CH3)CH(C2H5)Cl

(iii) CH3CH2C(CH3)2CH2I

(iv) (CH3)3CCH2CH(Br)C6H5

(v) CH3CH(CH3)CH(Br)CH3

(vi) CH3C(C2H5)2CH2Br

(vii) CH3C(Cl)(C2H5)CH2CH3

(viii) CH3CH=C(Cl)CH2CH(CH3)2

(ix) CH3CH=CHC(Br)(CH3)2

(x) p-ClC6H4CH2CH(CH3)2

(xi) m-ClCH2C6H4CH2C(CH3)3

(xii) o-Br-C6H4CH(CH3)CH2CH3CH3


Answer:

(i) 2-chloro-3-methylbutane, (2 alkyl halide)


Explanation: The Cl atom is attached with two alkyl groups therefore 2 alkyl halide.


(ii) 3-chloro-4-methylhexane (2 alkyl halide)


Explanation: The Cl atom is attached with two alkyl groups therefore 2 alkyl halide.


(iii) 1-iodo-2,2-dimethylbutane (1 alkyl halide)


Explanation: The atom I is attached with one alkyl group therefore 1 alkyl halide.


(iv) 1-bromo-3,3-dimethyl-1-phenylbutane (2 benzylic halide)


Explanation: The atom Br is attached with two alkyl groups with the benzene group therefore 2 benzylic halide.


(v) 2-bromo-3-methylbutane (2 alkyl halide)


Explanation: The atom Br is attached with two alkyl groups therefore 2 alkyl halide.


(vi) 1-bromo-2-ethyl-2-methylbutane (1 alkyl halide)


Explanation: The atom Br is attached with one alkyl group therefore 1 alkyl halide.)


(vii) 3-chloro-3-methylpentane (3 alkyl halide)


Explanation: The atom Cl is attached with three alkyl groups therefore 3 alkyl halide.


(viii) 3-chloro-5-methylhex-2-ene (vinylic halide)


Explanation: The atom Cl is attached with the double bond i.e. at vinyllic position ∴ vinylic halide.)


(ix) 4-bromo-4-methylpent-2-ene (allylic halide)


Explanation: The Br atom is attached with the carbon next to the double bond i.e. at allylic position therefore allylic halide.


(x) 1-chloro-4-(2-methylpropyl)benzene (aryl halide)


Explanation: The Cl atom is attached with the benzene i.e. at aryl position therefore aryl halide.


(xi) 1-chloro-3-(2,2-dimethylpropyl)benzene (1 benzylic halide)


Explanation: The Cl is attached with one alkyl group at the benzene carbon therefore 1 benzylic halide.


(xii) 1-bromo-2-(1-methylpropyl)benzene (aryl halide)


Explanation: The Br is attached with the benzene i.e. at aryl position therefore aryl halide.



Question 2.

How will you bring about the following conversions?

But-1-ene to but-2-ene


Answer:

CH3-CH2-CH=CH2 +HBr---> CH3-CH2-CH(Br)-CH3

This reaction happens according to Markonikoffs rule

Now reacting this with alcoholic KOH gives alkenes

CH3-CH2-CH(Br)-CH3 + KOH (alc)——-> CH3CH=CHCH3 + KBr + H2O



Question 3.

Give the IUPAC names of the following compounds:
CH3CH(Cl)CH(Br)CH3


Answer:

The IUPAC name of the give compound is: :2-Bromo-3-chlorobutane



Question 4.

Give the IUPAC names of the following compounds:

CHF2CBrClF :1-bromo-1-chloro-1,2,2-trifluoroethane


Answer:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical order. Eg: bromo is written before chloro and fluoro as B comes before C and F.


2. Always use a hyphen {-} between a number and a alphabet. Since Br is attached to first carbon and Cl Is also attached to first carbon i.e. it is written as 1-Bromo and 1-chloro.


3. Since there are 3 F present so it is written as trifluoro.


4. And at the last write the name of the longest hydrocarbon chain i.e. ethane of 2 carbons.



Question 5.

Give the IUPAC names of the following compounds:

ClCH2C≡CCH2Br: 1-bromo-4-chlorobut-2-yne


Answer:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical order. That is why bromo is written before chloro as B comes before C.


2. Always use a hyphen {-} between a number and a alphabet. Since Br is attached to 1 position and Cl isattached to fourthposition i.e. it is written as 1-Bromo and 4-chloro.


3. Also there is a presence of triple bond at the second carbon so the suffix used for triple bond is “yne” along with the position of the bond i.e. but-2-yne )



Question 6.

Give the IUPAC names of the following compounds:

(CCl3)3CCl


Answer:

2-(trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane


Explanation:


1. While writing the IUPAC name, write the name of side substituent according to the alphabetical order.


2. Always use a hyphen {-} between a number and a alphabet.


3. Because of the presence of 3 CCl3 group tri is used as a prefix.


4. Since 7 Cl atoms are present in all therefore hepta is used a prefix with the position of chlorine. And at the last write the name of the longest hydrocarbon chain i.e. propane)



Question 7.

Give the IUPAC names of the following compounds:

CH3C(p-ClC6H4)2CH(Br)CH3


Answer:

2-bromo-3,3-bis-(4-chlorophenyl)butane


Explanation:


1. While writing the IUPAC name, write the name of side substituent according to the alphabetical order.


2. Always use a hyphen {-} between a number and a alphabet.


3. Because of the presence of 2 (p-ClC6H2) group bis is used as a prefix instead of di because C6H4 contains itself contains numbers 6 & 4.


4. And at the last write the name of the longest hydrocarbon chain i.e. butane)



Question 8.

Give the IUPAC names of the following compounds:

(CH3)3CCH=CClC6H4I-p


Answer:

1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene


Explanation:


1. While writing the IUPAC name, write the name of side substituent according to the alphabetical order.


2. Always use a hyphen {-} between a number and a alphabet.


3. Because of the presence of 2 methyl group di is used as a prefix.


4. And the presence of double bond has the ending ‘ene’ and the presence of double bond at the first position is written as but-1-ene.)



Question 9.

Write the structures of the following organic halogen compounds.

2-Chloro-3-methylpentane


Answer:


Explanation:


1. Write the longest chain first i.e. in this case it is ‘pentane’ consisting of 5 carbons.


2. Now, place the substituents at their respective place. For eg. Chlorine(Cl) at second carbon and methyl(CH3) at third carbon.



Question 10.

Write the structures of the following organic halogen compounds.

p-Bromochlorobenzene


Answer:


Explanation:


1. Write the longest chain first i.e. in this case it is ‘benzene’ a cyclic structure consisting of 6 carbons with 3 double bonds.


2. Now, place the substituent Bromine(Br) at the para position i.e. at the fourth carbon.



Question 11.

Write the structures of the following organic halogen compounds.

1-Chloro-4-ethylcyclohexane


Answer:

(1-Chloro-4-ethyl cyclohecane)


Explanation:


1. Write the longest chain first i.e. in this case it is ‘cyclohexane’ a cyclic structure consisting of 6 carbons with 0 double bonds.


2. Now, place the substituent Chlorine(Cl) at the first carbon and ethyl(C2H5) at the fourth carbon.



Question 12.

Write the structures of the following organic halogen compounds.

2-(2-Chlorophenyl)-1-iodooctane


Answer:


Explanation:


1. Write the longest chain first i.e. in this case it is ‘octane’ i.e. straight chain of 8 carbons.


2. Now, place the substituent Iodine(I) at the first carbon and chlorophenyl at second carbon.



Question 13.

Write the structures of the following organic halogen compounds.

2-Bromobutane


Answer:


Explanation:


1. Write the longest chain first i.e. in this case it is ‘butane’ i.e. straight chain of 4 carbons.


2. Now, place the substituent Bromine(Br) at the second carbon.



Question 14.

Write the structures of the following organic halogen compounds.

4-tert-Butyl-3-iodoheptane


Answer:


Explanation:


1. Write the longest chain first i.e. in this case it is ‘heptane’ i.e. straight chain of 7 carbons.


2. Now, place the substituent Iodine(I) at the third carbon and tert butyl at fourth carbon.



Question 15.

Write the structures of the following organic halogen compounds.

1-Bromo-4-sec-butyl-2-methylbenzene


Answer:


Explanation:


1. Write the longest chain first i.e. in this case it is ‘benzene’ a cyclic structure consisting of 6 carbons with 3 double bonds.


2. Now, place the substituent Bromine(Br) at the first carbon,methyl(CH3) at second carbon and sec butyl at fourth carbon.



Question 16.

Write the structures of the following organic halogen compounds.

1,4-Dibromobut-2-ene


Answer:


Explanation:


1. Write the longest chain first i.e. in this case it is ‘butane’ i.e. straight chain of 4 carbons and a double and second carbon.


2. Now, place the substituents Bromine(Br) at the first and fourth carbon.



Question 17.

Which one of the following has the highest dipole moment?

(i) CH2Cl2 (ii) CHCl3 (iii) CCl4


Answer:

The three dimensional structures of the three compounds along with the direction of dipole moment in each of their bonds are given below:-



CCl4 being symmetrical has zero dipole moment. In CHCl3, the resultant of the two C-Cl dipole moments is opposed by the resultant of C-H and C-Cl bonds. Since the dipole Moment of latter resultant is expected to be smaller than the former,CHCl3 has a finite dipole moment(1.03D)


In CH2Cl2, the resultant of two C-Cl dipole moments is reinforced by resultant of two C-H dipoles. Therefore, CH2Cl2(1.62D) has a dipole moment higher than that of CHCl3. Thus ,CH2Cl2 has the highest dipole moment.



Question 18.

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


Answer:

The hydrocarbon with molecular formula C5H10 can either form a cycloalkane or an alkene. Since the compound does not react with Cl2in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since the cycloalkane reacts with Cl2 in the presence of bright sunlight to give a single monochloro compound, C5H9Cl, therefore, all the ten hydrogen atoms of the cycloalkanes must be equivalent. Thus, the cycloalkane is cyclopentane.




Question 19.

Write the isomers of the compound having formula C4H9Br.


Answer:

Double bond equivalent is used to find the level of unsaturations present in an organic molecule.


DBE = 


Where C= number of carbon atoms present


H=number of hydrogen atoms present


N=number of nitrogen atoms present


X=number of halogen atoms present


Double bond equivalent (DBE) for C4H9Br


=


=0


So none of the isomers has a ring or unsaturation, so the isomers are positions or chain isomers as shown below in the table:




Question 20.

Write the equations for the preparation of 1-iodobutane from

1-butanol


Answer:

1-butanol is treated with KI in the presence of H3PO4 where the OH is being replaced by the iodine and gives H2O and KH2PO4 as the by-product with 1-iodobutane as the final product.




Question 21.

Write the equations for the preparation of 1-iodobutane from

1-chlorobutane


Answer:

The conversion of 1-chlorobutane to 1-iodobutane simply takes place by treating the reactant with KI in the presence of acetone. The iodine from KI normally replaces the chlorine from the reactant and gives 1-iodobutane as the final product with KCl as the by product.




Question 22.

Write the equations for the preparation of 1-iodobutane from

but-1-ene.


Answer:

The conversion of but-1-ene to 1-iodobutane takes place according to anti-markovnikoff (positive charged ion H+ goes to the carbon which has less number of hydrogens and negative part Br- goes to the carbon which has more number of hydrogens.) i.e. the attack of HBr on but-1-ene in the presence of peroxide gives 1-bromobutane as the intermediate which on treatment with NaI+Acetone gives 1-iodobutane as the final product and NaBr as the by-product.




Question 23.

What are ambident nucleophiles? Explain with an example.


Answer:

Ambident Nucleophiles are those nucleophilies which can attack through different sites.


For example:-cyanide ions are a resonance hybrid of the following two structures:



It can attack through carbon to form cyanide and through N to form is O cyanide.


Example: NO2, NO3- etc.



Question 24.

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) Since I- ion is a better leaving group than Br- ion, therefore, CH3I reacts faster CH3Br in SN2 reaction with OH- ion.


Note: Better the leaving group, faster is the SN2 reaction.


(ii) On steric grounds, 1 alkyl halides are more reactive than tert-alkyl halides in SN2 reactions. Therefore, CH3Cl will react at a faster rate than (CH3)3CCl in a SN2 reaction with OH- ion.(bulkier the g8roup, slower is the SN2 reaction.)



Question 25.

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

1-Bromo-1-methylcyclohexane


Answer:

The reaction of 1-bromo-1-methylcyclohexane with sodium ethoxide in ethanol gives 2 products: - major and minor as shown below:



And,



The 2 products are obtained because of the presence of two types of beta (β) hydrogen.


According to saytzeff rule, the alkene formed in greatest amount (major) is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents.


When the 1-bromo-1-methylcyclohexane reacts with sodium ethoxide in ethanol, removal of HBr takes place giving two products one major and the other minor.


(In the major product number of alpha hydrogen=5 and in the minor product no. of alpha product=4)



Question 26.

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

2-Chloro-2-methylbutane


Answer:

The reaction of 2-chloro-2-methylbutane with sodium ethoxide in ethanol gives 2 products: - major and minor.


The 2 products are obtained because of the presence of two types of beta (β) hydrogen.


According to saytzeff rule, the alkene formed in greatest amount (major) is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents.


When the 2-chloro-2-methylbutane reacts with sodium ethoxide in ethanol, removal of HBr takes place giving two products one major and the other minor.


(In the major product number of alpha hydrogen=6 and in the minor product no. of alpha product=5)



And,




Question 27.

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

2,2,3-Trimethyl-3-bromopentane.


Answer:

The reaction of 2,2,3-trimethyl-3-bromopentane with sodium ethoxide in ethanol gives 2 products: - major and minor.


The 2 products are obtained because of the presence of two types of beta (β) hydrogen.


According to saytzeff rule, the alkene formed in greatest amount (major) is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents.


When the 2,2,3-trimethyl-3-bromopentane reacts with sodium ethoxide in ethanol, removal of HBr takes place giving two products one major and the other minor.


(In the major product number of alpha hydrogen=6 and in the minor product no. of alpha product=2)


and,




Question 28.

How will you bring about the following conversions?

Ethanol to but-1-yne


Answer:

Since the conversion of ethanol to but-1-yne involves the addition the two extra carbons that is why the conversion is carried out in 3 steps.


1. In the first step ethanol is treated with thionyl chloride (SOCl2) in the presence of pyridine to give chloroethane.



2. In the second step the acetylene is treated with sodamide(NaNH2) in the presence of liquid ammonia to give sodium acetylide.



3. The last step involves the reaction between chloroethane and sodium acetylide to give the final product as the But-1-yne and NaCl as the by-product.




Question 29.

How will you bring about the following conversions?

Ethane to bromoethene


Answer:

The bromination of ethane (Br2 in the presence of light at 520-670K) gives bromoethane which on further treatment with alcoholic KOH results in the alkene called ethane which again on bromination with Br2/CCl4 gives dibromide(vicinal bromide) called 1,2-dibromoethane which on treatment with alcoholic KOH gives bromoethene as the final product.




Question 30.

How will you bring about the following conversions?

Propene to 1-nitropropane


Answer:

Propene on treatment with HBr in the presence of peroxide (anti-markovnikoff’s rule which states the negative part i.e. Br- goes to the carbon which has more number of hydrogens and positive part i.e. H+ goes to the carbon which has lesser number of hydrogens) gives 1-bromopropane which on treatment with silver nitrite in the presence of alcohol or water gives 1-nitropropane as the final product.




Question 31.

How will you bring about the following conversions?

Toluene to benzyl alcohol


Answer:

Chlorination of toluene (Cl2 at 773K) gives benzyl chloride with the removal of HCl. Benzyl Chloride on further treatment with dilute KOH in the presence of heat gives benzyl alcohol as the final product.(dilute KOH removes Cl from the benzyl chloride and replaces it by OH)




Question 32.

How will you bring about the following conversions?

Propene to propyne


Answer:

Propene is treated with Br2/CCl4 i.e. bromination of alkene takes place to give dibromides, in this case 1,2-dibromopropane which on further treatment with alcoholic KOH gives propyne.(two times removal of KBr)




Question 33.

How will you bring about the following conversions?

Ethanol to ethyl fluoride


Answer:

Ethanol in treatment with thionyl chloride(SOCl2) in the presence of pyridine gives ethyl chloride with the evolution of SO2 gas. The ethyl chloride on treatment with mercury fluoride gives ethyl fluoride (the Cl of ethyl chloride is being replaced by F from Hg2F2).




Question 34.

How will you bring about the following conversions?

Bromomethane to propanone


Answer:

Bromoethane on treatment with alcoholic KCN forms Acetonitrile with the removal of KBr. CN is added in the case where there is a need to increase the number of carbons. Further the acetonitrile is treated with CH3MgBr(Grignard reagent) in the presence of ether, the intermediate so formed is hydrolysed, and the product formation takes place i.e. Propanone and NH3 is obtained as the by-product.




Question 35.

How will you bring about the following conversions?

1-Chlorobutane to n-octane


Answer:

The conversion of 1-chlorobutane to n-octane in the presence of Na metal and dry ether is called Wurtz reaction. Two equivalent Chlorobutane reacts to give n-octane as a final product and NaCl as the by-product.




Question 36.

How will you bring about the following conversions?

Benzene to biphenyl.


Answer:

The conversion of benzene to biphenyl using Na and dry ether is called FITTIG reaction.


Firstly, the benzene is treated with Br2/FeBr3 results in the formation of bromobenzene. It is an electrophilic substitution. Further the bromobenzene is treated with Na metal in presence of dry ether to form biphenyl as the final product and NaBr as the by-product.



If instead of Na metal, bromobenzene is treated with Cu metal then the reaction is called Ullmann’s reaction. The final product obtained in this reaction will also be biphenyl.



Question 37.

Explain why

The dipole moment of chlorobenzene is lower than that of cyclohexylchloride?


Answer:

sp2 hybrid carbon(s-character=33.33%) in chlorobenzene is more electronegative than a sp3-hybrid carbon(s-character=25%) in cyclohexylchloride, due to greater s-character. Thus, Carbon atom of chlorobenzene has lesstendency to release electrons to Cl than carbon atom of cyclohexylchloride.


As a result,C-Cl bond in chlorobenzene is less polar than in cyclohexylchloride. Further due to delocalisation of lone pairs of electrons of the Cl atom over the benzene ring, C-Cl bond in cholorobenzene acquires some double bond character while the C-Cl in cyclohexylchloride is a pure single bond. In other words, C-Cl bond in chlorobenzene is shorter than in cyclohexylchloride. Since the dipole moment is a product of charge and distance, therefore, chlorobenzene has lower dipole moment than cyclohexylchloride due to lower magnitude of negative charge on the Cl atom and shorter C-Cl distance.(double bond is of shorter length than single bond)





Question 38.

Explain why

Alkyl halides, though polar, are immiscible with water?


Answer:

Alkyl halides are polar molecules, therefore their molecules are held together by dipole-dipole interaction. The molecules of H2O are held together by hydrogen bonds (H-bonds). Since the new force of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halide-alkyl halide molecule and water-water molecules, therefore alkyl halides are immiscible (not soluble) in water. Alkyl halide are neither able to form H-bonds with water nor are able to break the H-bonding network of water.



Question 39.

Explain why

Grignard reagents should be prepared under anhydrous conditions?


Answer:

Grignard reagents are very reactive. They react with moisture present in the apparatus to form alkanes.


R-Mg-X + H-OH → R-H + Mg(OH)X


Thus Grignard reagents must be prepared under anhydrous conditions.



Question 40.

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


Answer:

(i) Freon 12:- the chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They are extremely stable,unreactive, non-toxic and non-corrosive. Example of Freon 12 is CCl2F2. It is mainly used in refrigeration and eventually makes its way into the atmosphere where it diffuses unchanged into the stratosphere. In stratosphere, Freon is able to initiate the radical chain reactions that can upset the natural ozone balance.


(ii) DDT: - it stands for p, p’-Dichlorodiphenyltrichloroethane (DDT). It is mainly used as an insecticide.


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


(iv) Carbon tetrachloride (CCl4):- it is used as an industrial solvent for oils, fats, resins etc. and also in dry cleaning. Used as a cleaning fluid both in industries, as a degreasing agent and in the home, as a spot remover and as a fire extinguisher.



Question 41.

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

CH3CH2CH2Cl + NaI


Answer:

It is a simple substitution reaction with Cl being replaced by iodide ion.




Question 42.

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

(CH3)3CBr + KOH 


Answer: 



(2-Bromo 2-Methyl Propane) (2-Methyl Propene)


Question 43.

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

CH3CH(Br)CH2CH3 + NaOH 


Answer:

It is also a simple substitution reaction in which under the presence of aqueous NaOH, bromide ion is replaced by hydroxide ion as it is a better leaving group than hydroxide ion.




Question 44.

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

CH3CH2Br + KCN 


Answer:

It is a simple substitution reaction in which a better leaving group leaves and is being substituted by an ion.


CH3CH2Br + kCN  CH3CH2CN + kBr



Question 45.

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

C6H5ONa + C2H5Cl 


Answer:

It is a simple substitution reaction in which a better leaving group leaves and is being substituted by an ion.


C6H5ONa + C2H5Cl → C6H5 – O – C2H5 + NaCl



Question 46.

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

CH3CH2CH2OH + SOCl2


Answer:

This reaction is called as wurtz reaction in which in which clubbing together of a alkyl halide group with a sodium or potassium salt of an ether group.




Question 47.

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

CH3CH2CH = CH2+ HBr 


Answer:

It is an antimarkovnikoff reaction in which under presence of a peroxide an alkene undergoes substitution, wherein the halo group is attached to that carbon which has least no. Of alkyl groups attached to it.




Question 48.

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

CH3CH = C(CH3)2 + HBr 


Answer:

It is a markovnikoff reaction in which an alkene undergoes substitution reaction with hydrohalide, wherein a halo group is attached to that carbon atom which has largest no. Of alkyl groups attached to it.




Question 49.

Write the mechanism of the following reaction:

nBuBr + KCN  nBuCN


Answer:

The given reaction is a nucleophillic reaction:


nBuBr + KCN  nBuCN


The given reaction is an SN2 reaction. In this reaction, CN acts as the stronger nucleophile and attacks the carbon atom to which Br is attached in nBuBr.


And, CN- ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.


The mechanism is shown below:




Question 50.

Arrange the compounds of each set in order of reactivity towards SN2 displacement:

2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane


Answer:

The reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered(does not have any free space or very crowded), then the reactivity towards SN2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.


1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane.


The structures are shown below:





Hence, the increasing order of reactivity towards SN2 displacement is:


2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane



Question 51.

Arrange the compounds of each set in order of reactivity towards SN2 displacement:

1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane


Answer:

The stearic hinderance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is given as-


3° < 2° < 1°.


The structures are given below:



Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN2 displacement as:


2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane



Question 52.

Arrange the compounds of each set in order of reactivity towards SN2 displacement:

1-Bromobutane, 1-Bromo-2,2 dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.


Answer:

The stearic hinderance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the stearic hinderance increases with an increase in the number of substituents. Therefore, the


increasing order of steric hindrances in the given compounds is as below:


1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane < 1-Bromo-2, 2-dimethylpropane


The structures are given below:



Hence, the increasing order of reactivity of the given compounds towards SN2 displacement


1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane



Question 53.

Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.


Answer:

Hydrolysis by KOH results in the formation of the carbocation. Those compounds which leads to the formation of stable carbocation are easily hydrolysed.C6H5CH2Cl leads to formation of 1°-carbocation, while C6H5CHClC6H5 forms 2°-carbocation, which is more stable than 1°-carbocation. Hence C6H5CHClC6H5, is hydrolyzed more easily than C6H5CH2Cl by aqueous KOH.




Question 54.

p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.


Answer:

p-Dichlorobenzene is more symmetrical than o-and m-isomers. Because it fits more closely and easily in the crystal lattice than o-and m-isomers. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. Therefore, p-dichlorobenzene is symmetrical and has a higher melting point and lower solubility than o-and m-isomers due to strong force of attraction in crystal. Therefore more energy required to break lattice and will not easily be soluble.



Question 55.

How the following conversions can be carried out?

Propene to propan-1-ol


Answer:

It is an antimarkovnikoff reaction in which under the presence of a peroxide an alkene undergoes substitution, wherein the halo group is attached to that carbon which has least no. Of alkyl groups attached to it.



Question 56.

How can the following conversions be carried out?

  • Ethanol to But-1-yne


Answer:

1. Treat ethanol with thionyl chloride SOCl2 to get chloroethane
2. Chloroethane when reacted with sodium acetylide forms but- 1-yne.