The monomer of natural rubber is:

C H_{2} = C (C H_{3}) C H = C H_{2}

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C H_{2} = C (C l) C H = C H_{2}

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C H_{2} = C H - C H = C H_{2}

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None of these

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Solution

The correct option is A $C H_{2} = C (C H_{3}) C H = C H_{2}$

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Similar questions

C H_{2} = C H - C H = C H - + C H_{2}

C H_{2} = C H - + C H - C H = C H_{2}

C H_{2} = C H - + C H - - C H - + C H_{2}

+ C H_{2} - C H = C H - - C H - + C H_{2}

C H_{2} = C H - C H_{2} C l + R M g X \to

C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{3} | C H - C H_{3}

C H_{3} - C H_{3}, C H_{2} = C H_{2}

C H \equiv C H

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

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