Archive for the "Organic Chemistry General" Category

This is the second part of a multi-part module on Free Radical Halogenation.

View the first part here: Part 1: The Mechanism

 

Radical bromination will always replace the C-H bond on the MOST subsituted carbon center because the stability of the radical intermediate is higher with increasing substituents on the carbon center.

(click on image to view larger)

This selectivity is the same, but a weaker consideration, for radical chlorination which obeys Hammond’s Postulate in that stability of the radical center is outweighed by the extreme exothermicity of radical chlorination (compared to bromination), thus a mixture of chlorinated products is observed.

 

There are multiple examples of this reaction to review in the StudyOrgo.com Study Mode. When ready, test your knowledge in the StudyOrgo.com Quiz Mode.

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Check out Part 3: Stereoslectivity – Determining stereochemistry of carbon centers

This is the first part of a multi-part module on Free Radical Halogenation.

 

There are THREE critical steps to free radical reactions – Memorize!

1) Initiation: The Br2 single bond is broken by high energy ligh (hv) to form radicals placing one electron on each atom.

2) Propagation: (Hint: One radical reacts with a sigle bond to form another radical, thus propigating the radical species to drive the reaction forward.

a) Radical Br abstracts one hydrogen from a C-H bond in propane to form radical propane and HBr.

b) Radical propane asbracts one Br from Br2 to form the bromoalkane and radical Br, thus restoring the reactants for another round as shown in step 2a.

3) Termination: Any two radicals combine to form a single bond.  These species will be in low abundance. Hint: Radicals are destroyed by combining two radicals to form a single bond.  This eliminates the radical necessary for radical alkane formation (green boxes) as shown in step 2a and ends the reaction.

 

There are multiple examples of this reaction to review in the StudyOrgo.com Study Mode. When ready, test your knowledge in the StudyOrgo.com Quiz Mode.

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This is the final part of a four part series in the Energy Diagram Module.

Click on the following link to see earlier parts:

Part 1

Part 2

Part 3

 

For each of the following questions- choose the best answer among the four answer choices.

Scroll to the bottom to see the answer key.

 

Question1

The net amount of heat released from the formation of products in a reaction is represented by _______________.

A) Temperature

B) Activation Energy

C) Heat of Reaction

D) Entropy

 

Question 2

A transition state represents _______________.

A) The highest potential energy molecule of the mechanism

B) The lowest potential energy molecule of the mechanism

C) The a molecule that does not occur in the reaction

D) A new reactant introduced to the reaction.

 

Question 3

A high energy, sometimes isolatable, chemical species during the course of reactants to products is referred to as a _______________.

A) Product

B) Midpoint

C) Transition State

D) Intermediate

 

Question 4

Refer to the Gibbs Free Energy Equation

ΔG⁰ = ΔH⁰ – TΔS

Under which circumstance will DG⁰ _­NEVER be spontaneous. (Hint: spontaneous reactions have a negative ΔG⁰)

A) ΔH⁰ < 0 and ΔS < 0

 ΔH⁰ > 0 and ΔS < 0

C) ΔH⁰ > 0 and ΔS > 0

D) ΔH⁰ < 0 and ΔS > 0

 

 

 

ANSWER KEY:

1. C

2. A

3. C

4. B

This is part 3 of a four part series in the Energy Diagram Module. Stay tuned for Part 4!

 Click on the following links to see earlier parts:

Part 1

Part 2

Sometimes reactions are more complex than simply a transition state (Graph 3), which would represent a single step in the reaction mechanism.  You will soon see most reactions proceed in a multistep fashion.  In this case, reaction mechanisms often form lower energy and sometimes isolatable intermediates.   The reaction intermediate occurs between two transition states however its energy is still higher than either products or reactants.  Note that the activation energy between reactant and the intermediate (step 1, ΔG^‡₁) is greater than the activation energy between the intermediate and the products (step 2, ΔG^‡₂). Thus it can be said that step 1 is the rate-limiting step of the reaction, which is the highest energy barrier that must be overcome.

 

 

 

Graph 3

(click on image to enlarge)

 

This is part 2 of a four part series in the Energy Diagram Module. Stay tuned for the other parts!

To see part 1 click here.

In order to talk about energy of the reaction, a few key concepts are needed.

• If the products have less potential energy than the reactants, the reaction will release a net amount of energy (an exothermic reaction).
• If the products are higher in energy than the reactions, the reaction will consume a net amount of energy (an endothermic reaction).

But, in order to predict how well a reaction will progress, or how spontaneous the reaction will be, enthalpy is insufficient to make this estimation.  Therefore, we rely on the thermodynamic calculation of Gibbs Free Energy (ΔG⁰) which is represented by the equation;

ΔG⁰ = ΔH⁰ – TΔS

The components of Gibbs Free Energy are:

• Enthalpy, ΔH⁰ – The heat consumed or released by the reaction.
• Temperature, T – Temperature of the system.
• Entropy, ΔS.  The change in the degree of disorder.

The sign of the reaction indicates the release (negative) or absorption (positive) of heat during the reaction. Therefore, if ΔG⁰ is negative the reaction is always spontaneous.  Similarly, if ΔG⁰ is positive the reaction is never spontaneous.  When ΔG⁰ equals zero, the reaction is at equilibrium. In accounting for these additional thermodynamic properties using Gibbs Free Energy, different terms are used.

• If a reaction has a negative  ΔG⁰ , it is therefore spontaneous and is said to be exergonic.
• Conversely, a reaction that has a  positive  ΔG⁰  is not spontaneous is considered endergonic.

 

Graph 2

(click on photo to enlarge)

With the addition of the temperature variable in the Gibbs Free Energy equation, it is easy to see that an endergonic reactions can be driven forward simply by increasing the temperature of the reaction so that the term TΔS is more negative than ΔH⁰, thus making ΔG⁰ negative and making the reaction spontaneous!