Haloalkanes and Haloarenes

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.

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 : -

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: -

Here is the reasoning that you can learn more while practicing the NCERT Solutions for Haloalkanes And Haloarenes. 

 

This outcome is directly attributed to the free radical halogenation mechanism. Usually, in photochemical chlorination, a chlorine atom can replace any hydrogen atom in the alkane. The number of unique products, or isomeric monochlorides, depends on the number of distinct types of hydrogen atoms in the starting isomeric alkane. 

 

For instance, the high symmetry of 2,2-dimethylpropane (neopentane) means all its hydrogen atoms are equivalent. That itself yields only a single monochloride. But, in contrast, the different primary and secondary hydrogens in n-pentane and 2-methylbutane create multiple products.

In the presence of sulphuric acid (H₂SO₄), KI produces HI as follows : -

2KI + H₂SO₄ → 2KHSO₄ + 2KI

Since H₂SO₄ is an oxidising agent, it oxidises HI (produced in the reaction to I₂)

2HI + H₂SO_{4 →} I₂ + SO₂ +;

As a result, the reaction between an 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-oxidising acid such as H3PO4 is used in the reaction to get the desired product.

A few things you can remember here, while solving NCERT Solutions for Haloalkanes And Haloarenes.

• Sulphuric acid is a powerful oxidising agent. We know this because the sulphur atom in sulphuric acid is in its highest possible oxidation state (+6). It can only be reduced. And it can readily accept electrons from other substances.

• It oxidises the necessary intermediate, hydrogen iodide (HI), to iodine (I2). We know this because iodide ions are strong reducing agents, and they can simply give up electrons. So when sulphuric acid accepts electrons, it can convert the iodide to elemental iodine.

• The alcohol cannot be converted to alkyl iodide. The reason for this is that the side reaction consumes the iodide ions necessary to act as a nucleophile and replace the alcohol's -OH group. This is the desired alkyl iodide.

• A non-oxidising acid like phosphoric acid (H3PO4) is used instead to produce the desired alkyl iodide. Phosphoric acid is a weaker oxidising agent than sulphuric acid and does not react with the iodide ions. That allows the intended reaction with the alcohol to proceed.  
  
  
  

10.1 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:

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: -

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: -

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

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