Question

Show that the following two mechanisms give the same net overall reaction. Mechanism 1 \(\mathrm{OCl}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HOCl}+\mathrm{OH}^{-}\) \(\mathrm{HOCl}+\mathrm{I}^{-} \longrightarrow \mathrm{HOI}+\mathrm{Cl}^{-}\) \(\mathrm{HOI}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{OI}^{-}\) Mechanism 2 \(\begin{aligned} \mathrm{OCl}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow & \mathrm{HOCl}+\mathrm{OH}^{-} \\ \mathrm{I}^{-}+\mathrm{HOCl} & \longrightarrow \mathrm{ICl}+\mathrm{OH}^{-} \\ \mathrm{ICl}+2 \mathrm{OH}^{-} & \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-}+\mathrm{H}_{2} \mathrm{O} \end{aligned}\)

Short answer

Both Mechanism 1 and Mechanism 2 lead to the net overall reaction \(\text{OCl}^- + \text{I}^- \longrightarrow \text{Cl}^- + \text{OI}^-\).

Step-by-step solution

Write out the steps of Mechanism 1

For Mechanism 1, the reactions are as follows:1) \(\text{OCl}^- + \text{H}_2\text{O} \longrightarrow \text{HOCl} + \text{OH}^-\)2) \(\text{HOCl} + \text{I}^- \longrightarrow \text{HOI} + \text{Cl}^-\)3) \(\text{HOI} + \text{OH}^- \longrightarrow \text{H}_2\text{O} + \text{OI}^-\)These reactions occur sequentially. Now, let's identify the reactants and products that appear on both sides of the equation and cancel them out to find the net reaction.

Determine the net reaction for Mechanism 1

By adding up all the steps in Mechanism 1 we get:\(\text{OCl}^- + \text{HOCl} + \text{HOI} + \text{H}_2\text{O} + \text{OH}^- \longrightarrow \text{HOCl} + \text{OH}^- + \text{HOI} + \text{Cl}^- + \text{OH}^- + \text{H}_2\text{O} + \text{OI}^-\).Cancel out species that appear on both sides to find the net reaction:Net reaction: \(\text{OCl}^- + \text{I}^- \longrightarrow \text{Cl}^- + \text{OI}^-\).

Write out the steps of Mechanism 2

For Mechanism 2, the reactions are as follows:1) \(\text{OCl}^- + \text{H}_2\text{O} \longrightarrow \text{HOCl} + \text{OH}^-\)2) \(\text{I}^- + \text{HOCl} \longrightarrow \text{ICl} + \text{OH}^-\)3) \(\text{ICl} + 2\text{OH}^- \longrightarrow \text{OI}^- + \text{Cl}^- + \text{H}_2\text{O}\).Just like before, we will add these steps together and then cancel out species that appear on both sides of the reactions to determine the net reaction.

Determine the net reaction for Mechanism 2

By adding up all the steps in Mechanism 2 we get:\(\text{OCl}^- + \text{HOCl} + \text{I}^- + 2\text{OH}^- + \text{H}_2\text{O} \longrightarrow \text{HOCl} + \text{OH}^- + \text{ICl} + \text{OH}^- + \text{OI}^- + \text{Cl}^- + \text{H}_2\text{O}\).Cancel out species that appear on both sides to find the net reaction:Net reaction: \(\text{OCl}^- + \text{I}^- \longrightarrow \text{Cl}^- + \text{OI}^-\).

Compare the net reactions

Now that we have the net reactions for both mechanisms, we can compare them:Mechanism 1 net reaction: \(\text{OCl}^- + \text{I}^- \longrightarrow \text{Cl}^- + \text{OI}^-\)Mechanism 2 net reaction: \(\text{OCl}^- + \text{I}^- \longrightarrow \text{Cl}^- + \text{OI}^-\)Since the net reactions of Mechanism 1 and Mechanism 2 are identical, this shows that both mechanisms result in the same net overall reaction.

Key concepts

Net Reaction Determination

Understanding the net reaction determination is crucial in learning how chemical processes unfold. In essence, the goal is to decipher the overall change brought about by a chemical reaction, without getting tangled in the complexities of the individual steps. To do this, as our exercise illustrates, one writes out all of the steps of a given mechanism and systematically cancels out species that are present on both the reactant and product sides.

These 'cancelable' species typically represent reaction intermediates or catalysts that are not present in the final net chemical equation, as they are regenerated or used up during the reaction process. The remaining species then form the net reaction - which reflects the ultimate result of the entire chemical process. This simplification highlights the transformation from initial reactants to final products, providing a macroscopic view that is easier to analyze and understand.

Chemical Reaction Steps

When we zoom in on the individual chemical reaction steps, we gain insight into the intricate dance of atoms and molecules that result in the net reaction. Each step is a mini-reaction in itself, involving bond breakage, electron shifting, and the reformation of new bonds. These steps are usually classified as either elementary reactions, which happen in a single event, or complex reactions, which are multistep processes.

In our example, each mechanism is broken down into its steps and analyzed separately. By considering the order of these steps, and the stability of the intermediates formed, chemists can infer the speed of the reaction and potential bottlenecks. It’s these insights that allow chemists to modify conditions to optimize reaction rates and yields - essential knowledge for applications ranging from synthesis to industry.

Reaction Intermediates

Diving deeper into the concept of reaction intermediates, they are the transient species which are formed during the reaction process but are not seen in the net reaction because they are consumed just as quickly as they are generated. These species play the role of a molecular 'middleman', holding on to atoms or energy before passing these on to the next step.

In both mechanisms of our exercise, intermediates such as HOCl and HOI appear in the early steps only to be consumed in subsequent steps. Understanding these intermediates is key to mastering reaction mechanisms, as their stability and reactivity can greatly influence the path and rate of a reaction. By examining intermediates, chemists can devise strategies to stabilize them if necessary, or to harness their reactive nature to drive a reaction forward.