NCERT Solutions for Class 12 Chemistry Chapter 10 - Haloalkanes and Haloarenes

Question 1:

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

Answer:

Eight isomers are possible.

Question 2:

Write IUPAC names of the following compounds :

Answer:

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

Question 3:

Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH3)2CHCH2CH3.

Answer:

There are four different types of hydrogen atoms. Replacement of these hydrogen atoms will give the following isomers.

Question 4:

Write the products of the following reactions :

Answer:

Question 5:

Haloalkanes react with KCN to form alkyl cyanides as the major product while AgCN forms isocyanides as the chief product. Explain.

Answer:

The cyanide ion is a resonance hybrid of the following structures :

It is clear from the resonating structures that cyanide ion can attack the nucleophilic site through C as well as through N. Such a nucleophile which is capable of attacking through more than one site is called ambident nucleophile. Thus, CN behaves as an ambident nucleophile.

KCN is predominantly ionic and therefore, both C and N atoms are free to donate electron pair. Since, C—C bond is relatively more stronger than C—N bond, therefore, the attack occurs mostly through the carbon atom of the cyanide group forming alkyl cyanides as the major product.

On the other hand, AgCN is predominantly covalent. Therefore, in the case only N atom is free to donate electron pair and the attack mostly occurs through the N atom of the cyanide group forming alkyl isocyanides as the major product.

Question 6:

Answer:

Question 7:

Predict the order of reactivity of the following compounds in SN1 and SN2 reactions :

(a) The four isomeric bromobutanes

Answer:

(a) The four isomeric bromobutanes are :

Question 8:

Identify chiral and achiral molecules in each of the following pairs of compounds :

Answer:
  1. In structure (i), the central carbon atom is bonded to four different substituents (H, OH, Br and CH3) and hence (i) is chiral. Structure (ii) has two identical Br atoms attached to central carbon atom and hence it is achiral molecule.
  2. (i) is chiral and (ii) is achiral. 
  3. (i) is chiral and (ii) is achiral.
Question 9:

Although chlorine is an electron withdrawing group, yet it is ortho-, para- directing in electrophilic aromatic substitution reactions. Why?

Answer:

Chlorine is an electron withdrawing group and has –I (inductive) effect. Therefore, it withdraws electrons from the benzene ring and tends to destabilize the intermediate carbocation formed during the electrophilic substitution reaction. The intermediate carbocations for ortho- and para- attacks are shown below :

Since –I effect of Cl is stronger than its +R effect, therefore, causes electron withdrawal and this causes net deactivation. The resonance effect tends to oppose the inductive effect for attack at ortho and para position and hence makes deactivation less for ortho- and para- attack. Thus, we can say that the reactivity is controlled by the stronger inductive effect and orientation is controlled by the resonance effect. Thus, although chlorobenzene is less reactive than benzene but it is ortho, para directing in electrophilic substitution reaction.

Question 10:

Write structures of the following compounds :

  1. 2-Chloro-3-methylpentane
  2. 1-Chloro-4-ethylcyclohexane
  3. 4-tert-Butyl-3-iodoheptane
  4. 1-4-Dibromobut-2-ene
  5. 1-Bromo-4-sec-butyl-2-methylbenzene
Answer:

Question 11:

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

Answer:

Sulphuric acid is an oxidising agent. It will oxidise HI produced during the reaction to I2 and therefore, will prevent the reaction between an alcohol and HI to form alkyl iodide.

Question 12:

Write structures of different dihalogen derivatives of propane.

Answer:

Four isomeric dihalogen derivatives of propane are formed. For example,

Question 13:

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

  1. A single monochloride
  2. Three isomeric monochlorides
  3. Four isomeric monochlorides.
Answer:

Question 14:

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

Answer:

Question 15:

Arrange each set of compounds in order of increasing boiling points.

  1. Bromomethane, Bromoform, Chloromethane, Dibromomethane
  2. 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane
Answer:
  1. For the same alkyl group, boiling point increases with the increase in size of the halogen atom. Therefore, boiling point of bromomethane is more than that of chloromethane. Further, the boiling point increases with increase in number of halogen atoms. Therefore, boiling point of bromoform is higher than that of dibromomethane, which is higher than that of bromomethane. Thus, the boiling point increases as :
    Chloromethane < Bromomethane < Dibromomethane < Bromoform.
  2. For the same halogen, boiling point increases with increase in size of the alkyl group due to increase in van der Waals forces of attraction. Therefore, boiling point of 1-chlorobutane is more than that of 1-chloropropane. Further, the boiling point decreases as branching increases so that the boiling point of 1-chloropropane is higher than that of isopropyl chloride. Thus, the boiling point increases as :
    Isopropyl chloride < 1-chloropropane < 1-chlorobutane
Question 16:

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

Answer:

Question 17:

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

Answer:

The reactivity of alkyl halide towards SN1 reaction depends upon the stability of the carbocation formed as 3° > 2° > 1°.

Question 18:

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

Answer:

Question 19:

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

Answer:

Question 20:

Give the IUPAC names of the following compounds :

Answer:

Question 21:

Write the structure of the following organic halogen compounds :

  1. 2-Chloro-3-methylpentane (ii) p-Bromochlorobenzene
  2. 1-Chloro-4-ethylcyclohexane (iv) 2-(2-Chlorophenyl)-1-iodooctane
  3. 2-Bromobutane (vi) 4-tert-Butyl-3-iodoheptane
  4. 1-Bromo-4-sec-butyl-2-methylbenzene (viii) 1,4-Dibromobut-2-ene
Answer:

Question 22:

Which one of the following has highest dipole moment ?

  1. CH2Cl2
  2. CHCl3
  3. CCl4
Answer:

Question 23:

Answer:

Question 24:

Write the isomers of the compound having formula C4H9Br.

Answer:

The compound is saturated C4H4×2+1Br. It has following four isomers :

Question 25:

What are ambident nucleophiles ? Explain with an example.

Answer:

The nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide is an ambident nucleophile because it can attack through C or N because of the following resonance structures :
–:CN: :C==N:–

Question 26:

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

Answer:

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 :

  1. 1-Bromo-1-methylcyclohexane
  2. 2-Chloro-2-methylbutane
  3. 2,2,3-Trimethyl-3-bromopentane
Answer:

Question 28:

How will you bring the following conversions?

  1. Ethanol to but-1-yne
  2. Ethane to bromoethene
  3. Propene to 1-nitropropane
  4. Toluene to benzyl alcohol
  5. Propene to propyne
  6. Ethanol to ethyl fluoride
  7. Bromomethane to propanone
  8. But-1-ene to but-2-ene
  9. 1-Chlorobutane to n-octane
  10. Benzene to biphenyl.
Answer:

Question 29:

Explain why

  1. the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride,
  2. alkyl halides, though polar, are immiscible with water,
  3. Grignard reagents should be prepared under anhydrous conditions ?
Answer:
  1. In chlorobenzene, the C of C—Cl bond is sp2 hybridised while the C of C—Cl bond in cyclohexyl chloride is sp3-hybridised.

  2. Alkyl halides are polar molecules and therefore, their molecules are held together by dipole-dipole forces. On the other hand, the molecules of H2O are held together by hydrogen bonds. When alkyl halides are added to water, the new forces of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halide-alkyl halide molecules and water-water molecules. Hence, alkyl halides are immiscible in water.
  3. Grignard reagents are very reactive. They react with the moisture present in the apparatus or the starting materials (RX or Mg).
    R MgX + HOH R—H + Mg(OH)X
    Therefore, Grignard reagents must be prepared in anhydrous conditions.
Question 30:

Give the uses of Freon 12, DDT, Carbon tetrachloride and Iodoform.

Answer:

Uses of Freon-12

  1. Freons are used as refrigerants in refrigerators and air conditioners.
  2. These have also been used extensively as propellants for aerosols and foams to spray out deodorants, cleansers, shaving creams, hair sprays and insecticides.
    However, the use of freon is restricted or banned in many countries because they are responsible for ozone depletion.

Uses of DDT

  1. D.D.T. is almost insoluble in water but it is moderately soluble in polar solvents.
  2. D.D.T. is a powerful insecticide. It is widely used as an insecticide for killing mosquitoes and other insects.

Uses of CCl4

  1. It is used as a solvent for oils, fats, waxes, etc.
  2. It is used as a fire extinguisher under the name pyrene.
  3. It is used in dry cleaning.
  4. It is used for the manufacture of freon.

Uses of Iodoform

  1. It is used as an antiseptic and this nature is due to iodine that it liberates. However, because of its very unpleasant smell, it has now been replaced by better antiseptics.
  2. It is used in the manufacture of pharmaceuticals.
Question 31:

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

Answer:

Question 32:

Explain the following reaction :

Answer:

KCN has nucleophile CN– ion which is ambident nucleophile because of the following two contributing structures:
–:CN: :C===N:–
Therefore, it can attack the carbon atom of C—Br bond in n-BuBr either through C or N. Since C—C bond is stronger than C—N bond, therefore, the attack occurs through C to form n-butyl cyanide as given below :

Question 33:

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

  1. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
  2. 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane
  3. 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane
Answer:

The reactivity in SN2 reaction depends upon steric hindrance. Lesser the steric hindrance more the reactivity. Therefore, the reactivity of different alkyl halides towards SN2 reactions is 1° > 2° > 3°.

Thus, reactivity decreases with steric hindrance as :
1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2, 2-dimethylpropane.

Question 34:

Answer:

Question 35:

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

Answer:

The melting point of p-isomer of dichlorobenzene is higher than that of o- and m-isomers. This is because, p-isomer has symmetrical structure and therefore, its molecules can easily pack closely in crystal lattice. Hence, it has stronger intermolecular forces of attraction than o- and m-isomer. Therefore, greater energy is required to break the intermolecular forces to melt or dissolve the p-isomer than the corresponding o- and m-isomers. In other words, the melting point of p-isomer is higher than the corresponding o- and m-isomers.

Question 36:

How the following conversions can be carried out ?

  1. Propene to propan-1-ol
  2. Ethanol to but-1-yne
  3. 1-Bromopropane to 2-bromopropane
  4. Toluene to benzyl alcohol
  5. Benzene to 4-bromonitrobenzene
  6. Benzyl alcohol to 2-phenylethanoic acid
  7. Ethanol to propanenitrile
  8. Aniline to chlorobenzene
  9. 2-Chlorobutane to
  10. 2-Methyl-1-propene to 2-chloro-3, 4-dimethylhexane 2-methylpropane
  11. Ethyl chloride to propanoic acid
  12. But-1-ene to n-butyliodide
  13. 2-Chloropropane to 1-propanol
  14. Isopropyl alcohol to iodoform
  15. Chlorobenzene to p-nitrophenol
  16. 2-Bromopropane to 1-bromopropane
  17. Chloroethane to butane
  18. Benzene to diphenyl
  19. tert-Butyl bromide to isobutyl bromide
  20. Aniline to phenylisocyanide
Answer:

Question 37:

The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in presence of alcoholic KOH, alkenes are major products. Explain.

Answer:

In aqueous KOH, KOH is almost completely ionized to give OH ions. These being strong nucleophiles result into substitution reaction on alkyl chlorides to form alcohols. Moreover, in aqueous solution, the OH ions are highly hydrated (solvated). The hydration reduces the basic character of OH– ions which therefore, fails to abstract a hydrogen from the β-carbon of the alkyl chloride to form an alkene. On the other hand, an alcoholic solution of KOH contains alkoxide (RO) ions which being stronger base than OH ions preferentially eliminates a molecule of HCl from an alkyl halide to form alkenes.

Question 38:

Primary alkyl halide (a) C4H9Br reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) was reacted with sodium metal it gave a compound (d) C8H18 that was different than the compound when n-butyl bromide was reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Answer:

(i) Two primary alkyl halides having the molecular formula C4H9Br are possible. These are :

Question 39:

What happens when

  1. n-butyl chloride is treated with alcoholic KOH,
  2. bromobenzene is treated with Mg in the presence of dry ether,
  3. chlorobenzene is subjected to hydrolysis,
  4. ethyl chloride is treated with (aq) KOH,
  5. methyl bromide is treated with sodium in the presence of dry ether,
  6. methyl chloride is treated with KCN ?
Answer:

Helpline