Relative stability of the contributing structures is : CH2=CH–C–H O (I) CH2–CH=C–H O (II) – + CH2–CH=C–H O (III) + – (A) (I) (III) (II) (B) (I) (II) (III) (C) (II) (I) (III) (D) (III) (I

Question

Relative stability of the contributing structures is : CH2=CH–C–H O (I) CH2–CH=C–H O (II) – + CH2–CH=C–H O (III) + –

(A) (I) > (III) > (II)
(B) (I) > (II) > (III)
(C) (II) > (I) > (III)
(D) (III) > (II) > (I)

Prashant Kotian · Accepted Answer

Correct answer: B. (1) Neutral structures are more stable than charged ones. Therefore I is more stable than II and III. (2) +ve charge on less electronegative atom is more stable i.e., Cis more stable than O   Order is I > II > III

💡 Solution & Explanation

Practice 22,000+ questions like this