Chapter 6 Haloalkanes and Halorenes Intext Questions NCERT Solutions Class 12 Chemistry- PDF Download
Page No. 163
6.1. Write structures of the following compounds:
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert-Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. butyl-2-methylbenzene
Solution
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert-Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. butyl-2-methylbenzene
Page No. 167
6.2. Why is sulphuric acid not used during the reaction of alcohols with KI?
Solution
In the presence of sulphuric acid (H2SO4), KI produces HI
2KI + H2SO4 → 2KHSO4 + 2HI
Since H2SO4 is 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. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as H3PO4 is used.
6.3. Write structures of different dihalogen derivatives of propane.
Solution
There are four different dihalogen derivatives of propane. The structures of these derivatives are shown below.
(i) 1, 1-Dibromopropane
(ii) 2, 2-Dibromopropane
(iii) 1, 2-Dibromopropane
(iv) 1, 3-Dibromopropane
Br-CH2-CH2-CH2-Br
6.4. 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.
Solution
(i) 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, replacement of any H-atom leads to the formation of the same product. The isomer is neopentane.
Neopentane
(ii) To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12should 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.
(iii) To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain four different types of H-atoms. Therefore, the isomer is 2-methylbutane. It can be observed that there are four types of H-atoms labelled as a, b, c, and d in 2-methylbutane.
6.5. Draw the structures of major monohalo products in each of the following reaction:
Solution
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Page No. 169
6.6. Arrange each set of compounds in order of increasing boiling points
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Solution
(i) Chloromethane < Bromomethane < Dibromomethane < Bromoform
The reason is:
- For same alkyl group, B.Pt increases with size of halogen atom.
- Boiling point increases as number of halogen atoms increase.
(ii) For alkyl halides containing the same halide, the boiling point increases with an increase in the size of the alkyl group. Thus, the boiling point of 1-chlorobutane is higher than that of isopropyl chloride and 1-chloropropane.
Further, the boiling point decreases with an increase in branching in the chain. Thus, the boiling point of isopropyl alcohol is lower than that of 1-chloropropane.
Hence, the given set of compounds can be arranged in the increasing order of their boiling points as:
Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane
Page No. 186
6.7. Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
Solution
(i)
2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an SN2 mechanism.
(ii)
2-Bromobutane is 2° alkylhalide whereas 2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers of substituents are present in 3o alkyl halide than in 2° alkyl halide to hinder the approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism.
(iii)
Both the alkyl halides are primary. However, the substituent -CH3 is at a greater distance to the carbon atom linked to Br in 1-bromo-3-methylbutane than in 1-bromo- 2methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by SN2 mechanism.
6.8. In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
Solution
(i)
(ii)
The alkyl halide (I) is 2° while (II) is 1°. 2° carbocation is more stable than 1° carbocation. Therefore, (I), 2−chloroheptane, undergoes faster SN1 reaction than (II), 1-chlorohexane.
6.9. Identify A, B, C, D, E, R and R1 in the following:
Solution
Since, D of D2O gets attached to the carbon atom to which MgBr is attached, C is
Therefore, the compound R − Br is
When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz reaction. Therefore, the halide, R1−X, is
Therefore, compound D is
And, compound E is
