Question

For the following reactions the major product are shown :
$H_2=CH-CH=C{H}_2\underset{0^{\circ }C}{\overset{HBr}{\to }}{H}_{2}C=CH-\underset{Br}{\underset{|}{CH}}-C{H}_3\stackrel{+{25}^{\circ }C}{\to }\underset{Br}{\underset{|}{C{H}_{2}CH}}=CHC{H}_3$
These provide an example of $\underset{–}{1}$ control at low temperatue and $\underset{–}{2}$ control at higher temperature.

• A. 1- Kinetic 2- Thermodynamic
• B. 1-Thermodynamic 2- Kinetic
• C. 1- Kinetic 2- Kinetic
• D. 1-Thermodynamic 2-Thermodynamic

Answer 1

Answer: (A)

1- Kinetic 2- Thermodynamic

At low temperature the lowest energy transition state determine the major product and this is kinetic control.At high temperature,the more substituted alkene will give most stable product,this is thermodynamic control.So,answer will be $1-$thermodynamo and $2$ kinetic.