Reactant A converts to product D through the given mechanism (with the net evolution of heat) :
$\mathrm{A} \rightarrow \mathrm{B} \quad$ slow $; \Delta \mathrm{H}=+\mathrm{ve}$
$\mathrm{B} \rightarrow \mathrm{C}$ fast; $\Delta \mathrm{H}=-\mathrm{ve}$
$\mathrm{C} \rightarrow \mathrm{D} \quad$ fast ; $\Delta \mathrm{H}=-\mathrm{ve}$
Which of the following represents the above reaction mechanism?

• Correct answer
  A

  
  
• B

  
  
• C

  
  
• D

  
  

Solution

$\begin{array}{ll}\mathrm{A} \rightarrow \mathrm{B} Slow & \\ \Delta \mathrm{H}=+\mathrm{ve} & \mathrm{E}_{\mathrm{a}_1} \rightarrow \text { High } \\ \mathrm{B} \rightarrow \mathrm{C} fast& \\ \Delta \mathrm{H}=-\mathrm{ve} & \mathrm{E}_{\mathrm{a}_2} \rightarrow \text { Low } \\ \mathrm{C} \rightarrow \mathrm{D} & \\ \Delta \mathrm{D}=-\mathrm{ve} & \mathrm{E}_{\mathrm{a}_3} \rightarrow \text { Low }\end{array}$

Hence, the correct answer is option: 'A'.