Question

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Answer 1

Answer:

11. Allylic bromination of an olefin is

(1) Nucleophilic substitution

(2) Electrophilic substitution

(3) Free radical substitution ☑️

(4) Electrophilic addition

: (3) Free radical substitution

12. Which of the following is a free radical reaction?

(1) C6H6 + NO2 ----> C6H5-NO2

(2) CH2=CH2 + HBr ----> CH3-CH2-Br

(3) CH3-CH=CH2 ----> CH3-CH2-CH2-Br

(4) CH3-CH=CH2 + HCl ----> CH3-CH(Cl)-CH3. ☑️

: (4) CH3-CH=CH2 + HCl ----> CH3-CH(Cl)-CH3

Explanation:

11.

»» Allylic bromination is the replacement of hydrogen on a carbon adjacent to a double bond or aromatic ring with bromine.

12.

»» Since, for a free radical addition there should be only the addition of free radicals (such as Cl ,Br, I) from backside and so as the the radical Cl get attached to the carbon atom of alkene which is next to double bond.

»» In propene, the double bond is formed in between 1° (primary) carbon atom and 2° (secondary) carbon atom. The secondary C-atom bears a free radical due double bond and hence, the chloride ions get attached to secondary C-atom & hydrogen (H) atom get added up to primary C-atom .

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Answer 2

Answer:

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