See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Alkene stability increases with the degree of substitution at the double bond carbons. More alkyl substituents on the double bond carbons = greater hyperconjugation and inductive stabilization = more stable alkene. The order is: tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted. Step 1: Identify the substitution pattern for each option. (a) Option (a) shows CH2=C(Me)(Et) — this is a 1,1-disubstituted terminal alkene (two substituents on one carbon of the double bond, zero on the other). Total substituents on double bond = 2 (disubstituted). (b) Option (b) shows CH2= with a branched chain — this is also a terminal alkene (=CH2 end), making it at most disubstituted (1,1-disubstituted), with the terminal carbon bearing no substituents. (c) Option (c) shows CH2=CH- (vinyl type) with a branched substituent on the adjacent carbon — this is a monosubstituted terminal alkene (only one alkyl group on the double bond carbons total). This is the least substituted. (d) Option (d) shows an internal alkene (C=C with no terminal =CH2), with substituents on both carbons of the double bond. The structure appears to be a trisubstituted alkene: one carbon of the double bond bears one alkyl group (methyl, shown as the shorter branch going left/up) and the other carbon bears two alkyl groups (an ethyl and another group), giving three total alkyl substituents on the double bond. Trisubstituted > disubstituted. Step 2: Rank the options. - (c): monosubstituted — least stable - (a) and (b): disubstituted (1,1-disubstituted terminal) — intermediate stability - (d): trisubstituted internal alkene — most stable Step 3: Why other options fail. (a) and (b) are both disubstituted terminal alkenes; neither achieves the higher substitution of (d). (c) is monosubstituted, the least stable. (d) being trisubstituted (internal, with three alkyl groups on the double bond carbons) has the greatest hyperconjugative and inductive stabilization, making it the most stable. Therefore, the correct answer is D.