Which of the following compound does not de-colourise the Baeyer's reagent?

Question

Which of the following compound does not de-colourise the Baeyer's reagent? (A) CH3-C≡ C-CH3 (B) CH2=CH-CH=CH-CH=CH2 (C) <img class="img-fluid question-image" alt="Option 3" src="https://cdn.tardigrade.in/q/nta/c-mcf4dtnhwexujroq.jpg" /> (D) H2C=C=CH2

Tardigrade Admin · Accepted Answer

Benzene is an aromatic cyclic compound.
Benzene is not a pure unsaturated compound due to its de-localised double bonds.
It is reasonance stabilised and it cannot be oxidised by

K M n O_{4}

,
hence benzene does not de-colourise the Baeyer's reagent.

Q. Which of the following compound does not de-colourise the Baeyer's reagent?

$C H_{3} - C \equiv C - C H_{3}$

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

$H_{2} C = C = C H_{2}$

Benzene is an aromatic cyclic compound.
Benzene is not a pure unsaturated compound due to its de-localised double bonds.
It is reasonance stabilised and it cannot be oxidised by $K M n O_{4}$ ,
hence benzene does not de-colourise the Baeyer's reagent.

Q. Which of the following compound does not de-colourise the Baeyer's reagent?

CH3​−C≡C−CH3​

CH2​=CH−CH=CH−CH=CH2​

H2​C=C=CH2​

Benzene is an aromatic cyclic compound. Benzene is not a pure unsaturated compound due to its de-localised double bonds. It is reasonance stabilised and it cannot be oxidised by KMnO4​ , hence benzene does not de-colourise the Baeyer's reagent.

$C H_{3} - C \equiv C - C H_{3}$

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

$H_{2} C = C = C H_{2}$

Benzene is an aromatic cyclic compound.
Benzene is not a pure unsaturated compound due to its de-localised double bonds.
It is reasonance stabilised and it cannot be oxidised by $K M n O_{4}$ ,
hence benzene does not de-colourise the Baeyer's reagent.