Organic Chemistry Intro. to Radical Halogenation, Thermodynamics, Assignment

Organic Chemistry Intro. to Radical Halogenation, Thermodynamics, Kinetics, Radical Halogenation Cont., Alkyl halides - Assignment Example At this state, injection of energy is needed to shift the reaction. The shifting takes place from a stable state to a state where it converts and reacts to conform to other products. In contrary, thermodynamic reactions are more stable in a product’s state. This is as a result of instant occurrence of the reaction that causes shifts in the reaction without injection of any energy. Consequently, a substance whose stability is kinetic would crave to remain in the reactant form. Thermodynamically stable substances need energy for conversion from products to reactants unlike kinetic which needs the energy to shift a reaction forward. It can be therefore concluded that thermodynamic and kinetic reactions uses energy for different purposes. For instance, thermodynamic reactions need energy to move from an opposite state compared to kinetic reactions that required energy to shift forward. 2. Draw the mechanism and all possible products for A simple format for the mechanism Applying t he above concept will help you to draw the mechanism as below. Termination overall reaction 3. What type of reaction is this? Radical reaction. 4. What is the major organic product for the reaction, please explain why in details? 2-bromo-2-methylbutane The reason behind this is that in the radical halogenations, bromine is more selective but less reactive than chlorine. Consequently, in case there is a formation of a radical at one chiral centre, then the product will be racemic. The process shows that the radical is planar having equal forces. Bromine sticks itself to a tertiary carbon of degree 3 to structure into a main product. A radical of bromine adds a more stable carbon of degree three to alkanes. 5. Using the bond energies posted or in the textbook, calculates the ?HÂ ° for the reactions providing each product above. Do these values support your answer in question? What other factors must be taken into account? Initiation Br-Br 2Br ?HÂ °=192 k j/mol From the above reactio ns, 1) ?HÂ °= ? products- ? reactants = (368+343)- 343 = 368 Kj/mol 2) ?HÂ °= ? products- ? reactants = 272 -(343 + 192) = -263Kj/mol Overall enthalpy = 368- 263 =+105Kj/mol Second product 3. ?HÂ °? products- ? reactants (343+368) – (343) = 368 kj/mol 4. ?HÂ °= ? products- ? reactants = (272) – (343+192) = -263 Kj/mol Overall= 368 – 263= 105 Kj/mol The other product in this level of propagation reaction repeats itself. This confirms that the ?HÂ °= ? products- ? reactants remains constant showing that the major organic product is that of question 4 above. Since bromine experience an endothermic reaction, weakly bounded hydrogens are only removed by bromine from carbon atoms they are bonded to. The reaction between 2 – methylbutane and bromine mostly gives 2 – bromo – 2 – methylbutane and a few secondary bromides without primary bromides. The other factors that must be taken into account are pressure, temperature and catalyst. Tem perature affects this reaction if it is either decreased or increased. If temperature of this reaction is increased, the rate of this reaction increases and if it is lowered the rate of reaction decreases. A catalyst can either increase or decrease reaction depending on the type. There are those catalysts that slow down reaction rate and there are those that increases reaction rate. Lastly, for enthalpy to occur there should be constant pressure. There