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Explain why chlorination of n-butane in ...

Explain why chlorination of n-butane in
presence of light at 298 K gives a mixture of 72% of
2-chlorobutane and 28% of 1-chlorobutane.

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To explain why chlorination of n-butane in the presence of light at 298 K results in a mixture of 72% of 2-chlorobutane and 28% of 1-chlorobutane, we can follow these steps: ### Step 1: Understand the Structure of n-Butane n-Butane (C4H10) has the following structure: - CH3-CH2-CH2-CH3 This molecule has two types of hydrogen atoms: those attached to the terminal carbon atoms (1° hydrogens) and those attached to the middle carbon atoms (2° hydrogens). ### Step 2: Identify the Types of Chlorination Products ...
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If relative rates of substitution of 1^(@) and 2^(@) H are in the ratio 1 : 3.8, show that in the presence of light at 298 K, the chlorination of n-butane gives a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane.

Statement-I: Free radical chlorination of n-butane gives 72% of 2-chlorobutane and 28% of 1-chlorobutane though it has six primary and four secondary hydrogens. Because Statement-II: A secondary hydrogen is abstracted more easily than the primary hydrogen.

Which of the following organic compounds will give a maxiture of 1-chlorobutane and 2-chlorobutance on chlorintaion

2-chlorobutane obtained by chlorination of butane will be .

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Propane reacts with chlorine in sunlight to give two products. 1-chloropropane is obtained in 44% yield and 2-chloropropane is obtained in 56% yield of the total product. What will be the percent yield of the major product obtained when butane is treated with Cl_(2) in similar conditions