A compound (C5 H8) reacts with ammonical AgNO3 to give a white precipitate and reacts with excess of KMnO4 solution to give (C H3)2CH-COOH . The compound is -

Question

A compound (C5 H8) reacts with ammonical AgNO3 to give a white precipitate and reacts with excess of KMnO4 solution to give (C H3)2CH-COOH . The compound is - (A) CH2=CH-CH=CH-CH3 (B) (C H3)2CHC≡ CH (C) CH3(C H2)2C≡ CH (D) (C H3)2C=C=CH2

Tardigrade Admin · Accepted Answer

Q. A compound $(C_{5} H_{8})$ reacts with ammonical $A g N O_{3}$ to give a white precipitate and reacts with excess of $K M n O_{4}$ solution to give $(C H_{3})_{2} C H - COO H$ . The compound is -

$C H_{2} = C H - C H = C H - C H_{3}$

$(C H_{3})_{2} C H C \equiv C H$

$C H_{3} (C H_{2})_{2} C \equiv C H$

$(C H_{3})_{2} C = C = C H_{2}$

Q. A compound (C5​H8​) reacts with ammonical AgNO3​ to give a white precipitate and reacts with excess of KMnO4​ solution to give (CH3​)2​CH−COOH . The compound is -

CH2​=CH−CH=CH−CH3​

(CH3​)2​CHC≡CH

CH3​(CH2​)2​C≡CH

(CH3​)2​C=C=CH2​

Q. A compound $(C_{5} H_{8})$ reacts with ammonical $A g N O_{3}$ to give a white precipitate and reacts with excess of $K M n O_{4}$ solution to give $(C H_{3})_{2} C H - COO H$ . The compound is -

$C H_{2} = C H - C H = C H - C H_{3}$

$(C H_{3})_{2} C H C \equiv C H$

$C H_{3} (C H_{2})_{2} C \equiv C H$

$(C H_{3})_{2} C = C = C H_{2}$