Question

Why are there multiple organic or hydrocarbon products produced from the halogenation of alkanes via ultraviolet light? I am grading pretty big on the why part, and it is not only about the number of H's present.....10 pts.) 5) Using the following monomer, draw the steps and/or mechanism for the formation of the polymer, stopping after attaching three units together. DETAIL!! (20 pt.) 6) Br H

Answer 1

5)During halogenation of alkanes via UV light, the reaction mixture undergoes a chain reaction which follow free radical mechanism.Depending on the reactivity of the different type of H-atoms and the reactivity of the halogen there are mixture of products formed.

Order of reactivity of H-atoms in a branched alkane, tertiary H>secondary H> primary H atom

Order of reactivity of halogens: F₂>> Cl₂> Br₂> I₂

Reactivity of F₂ is too violent and that of I₂ is too slow.Preferably, Cl₂ and Br₂ are chosen for direct halogenation.

With I₂, the reaction will be reversible.To improve the yield the reaction can be carried out using oxidising agents

Ex: Chlorination of methane

CH₄ + Cl₂ --UV-----> CH₃Cl+ HCl

CH₃Cl+ Cl₂ -----> CH₂Cl₂ + HCl

CH₂Cl₂ + Cl₂ -----> CHCl₃ + HCl

CHCl₃ + Cl₂ -------> CCl₄ + HCl

Free Radical Mechanism:

Step1: Chain initiation: Cl₂ -----> 2Cl^.

Step2: Chain Propagation: CH₄ + Cl^. -----> CH₃^. + HCl

CH₃^. + Cl₂ -----> CH₃Cl + Cl^.

Step3: Termination: CH₃^. + CH₃^. ---------> C₂H₆

Cl^. + Cl^. --------> Cl₂

CH₃^. + Cl^. --------> CH₃Cl

6) Alkenes undergo polymerisation in general due to the presence of the double bond. They undergo addition polymerisation reaction.

n CFCl=CHBr --------> ----[CFCl-CHBr]_n---- where n= is the repeating units

n CFCl=CHBr --------> ----CFCl-CHBr-CFCl-CHBr-CFCl-CHBr----

The reaction follows free radical mechanism at the double bond.Steps are similar to the one discussed in Q5.