10.4 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

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 molecu

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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.

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<p>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.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690436100phpXwBrgB_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690436100phpXwBrgB.jpeg" alt="" width="218" height="125"></picture></div></div><p>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 : -</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690436120phpReKN6i_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690436120phpReKN6i.jpeg" alt="" width="223" height="85"></picture></div></div><p>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: -</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690436147phpVB3k7u_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690436147phpVB3k7u.jpeg" alt="" width="204" height="120"></picture></div><div>Here is the reasoning that you can learn more while practicing the &lt;a href="https://www.shiksha.com/preparation/chemistry-ncert-solutions-class-12th-haloalkanes-and-haloarenes-4511-ns" target="_blank" rel="noopener"&gt;NCERT Solutions for Haloalkanes And Haloarenes&lt;/a&gt;.&nbsp;</div><div>&nbsp;</div><div>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.&nbsp;</div><div>&nbsp;</div><div>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.<br><br></div></div>

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