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Magazine Chapter 10: Q4P (page 292)
Taking the relative reactivities of 1掳, 2掳, and 3掳 hydrogen atoms into account, what product(s) would you expect to obtain from monochlorination of 2-methylbutane? What would the approximate percentage of each product be? (Don鈥檛 forget to take into account the number of each kind of hydrogen.)
Short Answer
The products obtain from monochlorination of 2-methylbutane with their approximate percentage are as follows:
Step by step solution
Monochlorination
It means substitution of one hydrogen atom from alkane, with one chlorine atom to form alkyl halide.
Types of hydrogen in 2-methylbutane
The compound have four types of hydrogens, designated as a,b,c and d.
Monochlorination of 2-methylbutane
When chlorine is added to 2-methylbutane in presence of sunlight, following four products are formed:
Calculation of approximate percentage of products
Type of H | Number of H of each type | Relative reactivity | Number times reactivity = Number of H*Relative reactivity | Percentage chlorination =(Number of H*100/21) |
a | 6 | 1.0 | 6.0 | 29% |
b | 1 | 5.0 | 5.0 | 24% |
c | 2 | 3.5 | 7.0 | 33% |
d | 3 | 1.0 | 3.0 | 14% |
| Total number times reactivity | 21 |
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