Chapter 10, Problem 122QRT

(a)

Interpretation Introduction

Interpretation:

The chemical equation of addition reaction of $\text{3-Hexene}$ with $\text{HCl}$ has to be stated with help of Lewis structures.

Concept Introduction:

The addition of one or more than one halogen atom in alkene is known as halogenation reaction.  The pathway of the halogenation reaction depends on the structure of the substrate.  In this the halogen is added to $\text{C}=\text{C}$ bond of the alkene to form haloalkane.

Explanation of Solution

The halogenation reaction of $\text{3-Hexene}$ gives an alkyl halide, $\text{3-chlorohexene}$.  The alkene has a $\pi$ bond in it.  The $\pi$ bond of $\text{3-Hexene}$ gets brake to form $\text{3-chlorohexene}$. The chloro-group is added to the third carbon atom and the hydrogen atom is added to the fourth carbon atom.  The corresponding chemical reaction is shown below.

Figure 1

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy change for the overall addition reaction of $\text{3-Hexene}$ with $\text{HCl}$ has to be calculated.

Concept Introduction:

The amount of energy which is produced or absorbed during the reaction is the enthalpy change of a chemical reaction.  When bonds gets break, some amount of energy is absorbed by the system.  When bonds are formed, some amount of energy is produced by the system.

Explanation of Solution

The overall addition reaction of $\text{3-Hexene}$ with $\text{HCl}$  is shown below.

Figure 1

The reactant $\text{3-Hexene}$ has $12$ $\text{C}-\text{H}$ bonds, $4$ $\text{C}-\text{C}$ bonds and a $\text{C}=\text{C}$ bond.

The reactant $\text{HCl}$ has a $\text{H}-\text{Cl}$ bond.

The product octane has $13$ $\text{C}-\text{H}$ bonds, a $\text{C}-\text{Cl}$ bond and $5$ $\text{C}-\text{C}$ bonds.

The bond enthalpy for $\text{C}-\text{H}$ is $416\text{kJ}/\text{mol}$.

The bond enthalpy for $\text{C}=\text{C}$ is $598\text{kJ}/\text{mol}$.

The bond enthalpy for $\text{C}-\text{C}$ is $356\text{kJ}/\text{mol}$.

The bond enthalpy for $\text{C}-\text{Cl}$ is $327\text{kJ}/\text{mol}$.

The bond enthalpy for $\text{H}-\text{Cl}$ is $431\text{kJ}/\text{mol}$.

The standard enthalpy change for the overall reaction is calculated by the formula as shown below.

  $\Delta H°=\left(\begin{array}{l}{\displaystyle \sum \left[\left(\text{number}\text{of}\text{bonds}\right)\times D\left(\text{bonds}\text{broken}\right)\right]}\\ -{\displaystyle \sum \left[\left(\text{number}\text{of}\text{bonds}\right)\times D\left(\text{bonds}\text{formed}\right)\right]}\end{array}\right)$

Where,

• $\Delta H°$ is the standard enthalpy change for the overall reaction.
• $D$ is the bond enthalpy.

The above equation can rewritten for addition reaction of $\text{3-Hexene}$ with $\text{HCl}$ as shown below.

  $\Delta H°=\left[\begin{array}{l}\left(12\times D\left(\text{C}-\text{H}\right)+4\times D\left(\text{C}-\text{C}\right)+D\left(\text{C}=\text{C}\right)\right)\\ +D\left(\text{H}-\text{Cl}\right)-\left(13\times D\left(\text{C}-\text{H}\right)+D\left(\text{C}-\text{Cl}\right)+5\times D\left(\text{C}-\text{C}\right)\right)\end{array}\right]$

Substitute the value of bond enthalpy for $\text{C}-\text{H}$, bond enthalpy for $\text{C}-\text{Cl}$, bond enthalpy for $\text{C}=\text{C}$, bond enthalpy for $\text{C}-\text{C}$, and bond enthalpy for $\text{H}-\text{Cl}$ in the above equation.

  $\begin{array}{c}\Delta H°=\left[\begin{array}{l}\left(12\times \left(416\text{kJ}/\text{mol}\right)+4\times \left(356\text{kJ}/\text{mol}\right)+\left(598\text{kJ}/\text{mol}\right)\right)\\ +\left(431\text{kJ}/\text{mol}\right)-\left(13\times \left(416\text{kJ}/\text{mol}\right)+\left(327\text{kJ}/\text{mol}\right)+5\times \left(356\text{kJ}/\text{mol}\right)\right)\end{array}\right]\\ =\left[4992+1424+598+431-5408-327-1780\right]\text{kJ}/\text{mol}\\ =\left[7445-7515\right]\text{kJ}/\text{mol}\\ =\underset{\_}{-70kJ/mol}\end{array}$

The standard enthalpy change for the overall addition reaction of $\text{3-Hexene}$ with $\text{HCl}$ is $\underset{\_}{-70kJ/mol}$.