Question

show accurately how reaction of 2-methlypropane with chlorine at high temperature, could lead to the formation of 2 -chloro-2-methyl propane in a free radicle process?

Answer 1

Answer on Question #42107 - Chemistry - Organic Chemistry

Question:

Show accurately how reaction of 2-methlypropane with chlorine at high temperature, could lead to the formation of 2-chloro-2-methylpropane in a free radical process?

Answer:

At high temperature chlorine molecule splits into two chlorine free radicals

$$\begin{array}{c} \text{Cl} - \text{Cl} \xrightarrow{\text{T.}^\circ\text{C}} \text{Cl}^+ + \text{Cl}^+ \end{array}$$

chlorine

The chlorine radical attacks 2-methlypropane and detaches H resulting in 2-methylprop-2-yl free radical and hydrogen chloride

$$\begin{array}{c} \text{H}_3\text{C} \\ \text{H}_3\text{C} \xrightarrow{\text{H}_3\text{C}} \text{H} + \text{Cl}^+ \longrightarrow \text{H}_3\text{C} \xrightarrow{\text{H}_3\text{C}} \text{C}^+ + \text{HCl} \\ \text{H}_3\text{C} \\ \end{array}$$

2-methylpropane

The question, which might arise is "why exactly the hydrogen in position 2 is detached, not any of other 9 hydrogens?" It is because in such case most stable free radical is formed.

The 2-methylprop-2-yl free radical reacts with another chlorine molecule resulting in 2-chloro-2-methylpropane and chlorine free radical formation

$$\begin{array}{c} \text{H}_3\text{C} \\ \text{H}_3\text{C} \xrightarrow{\text{C}^*} + \text{Cl} - \text{Cl} \longrightarrow \text{H}_3\text{C} \xrightarrow{\text{H}_3\text{C}} \text{Cl} + \text{Cl}^+ \\ \text{H}_3\text{C} \\ \end{array}$$

2-chloro-2-methylpropane

Overall reaction is as follows

$$\begin{array}{c} \text{H}_3\text{C} \\ \text{H}_3\text{C} \xrightarrow{\text{H}_3\text{C}} \text{H} + \text{Cl} - \text{Cl} \longrightarrow \text{H}_3\text{C} \xrightarrow{\text{H}_3\text{C}} \text{Cl} + \text{H} - \text{Cl} \\ \text{H}_3\text{C} \\ \end{array}$$

2-methylpropane

2-chloro-2-methylpropane

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