Q4P
Question
Taking the relative reactivities of 1°, 2°, and 3° 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’t forget to take into account the number of each kind of hydrogen.)
Step-by-Step Solution
VerifiedThe products obtain from monochlorination of 2-methylbutane with their approximate percentage are as follows:
It means substitution of one hydrogen atom from alkane, with one chlorine atom to form alkyl halide.
The compound have four types of hydrogens, designated as a,b,c and d.
When chlorine is added to 2-methylbutane in presence of sunlight, following four products are formed:
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 |