1.5. Circle The oxidation state of the carbon atom in the molecule chloroform (C

Question

1.5. Circle The oxidation state of the carbon atom in the molecule chloroform (CHCl3) 1.6. Circle the order of halogen reactivity for the reaction below 1.7. Circle the conditions that would give the polymer polyvinylchloride (PVC) shown below 1.8. The reaction but that was an example of....
1.5) Circle the oxidation state of the carbon atom in the molecule chloroform (CHCl3) a) -2 b)-1 c) 0 e) +3 1.6) Circle the order of halogen reactivity for the reaction belovw 1.7) Circle the conditions that would give the polymer, polyvinylchloride(PVC), shown below CI CI CI CI Cl2 Cl2 a) b) hv Cl peroxide peroxide c) d) 1.8) The reaction below is an example of a) addition b) oxidation c) substitution d) reduction

Explanation / Answer

Ans 1.5 : d) +2

The oxidation number of C can be determined as :

C + 1 + 3(-1) = 0

C = +2

Ans 1.6 : b) I2 > Br2 > Cl2 > F2

Due to the largest size of iodine molecule here amongst the others , it has least bond dissociation enthalpy , hence it is most reactive and fluorine being the smallest is least.

Ans 1.7 : c)

PVC refers to the polymer named - Polyvinyl Chloride. It is formed as the polymerization of the monomer ethyl chloride in the presence of benzoyl peroxide.

Ands 1.8 : b) Oxidation

The reactant molecule is alcohol , whereas the product is a carboxylic acid. Alcohol undergoes strong oxidation ( in the presence of strong oxidising agent) to produce carboxylic acids.

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