A) 1, 1, 7, 7-tetramethyl-2, 5-octadiene
B) 2, 8-dimethyl-3, 6-decadiene
C) 1, 5-diisopropyl-1, 4-hexadiene
D) 3, 9-dimethyl-4, 6-decadiene
E) 2, 8-dimethyl-4, 6-decadiene
Correct Answer: E
Solution :
\[\overset{1}{\mathop{C}}\,{{H}_{3}}^{2}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,\overset{3}{\mathop{C}}\,{{H}_{2}}\overset{4}{\mathop{C}}\,H=\overset{5}{\mathop{C}}\,H\overset{6}{\mathop{C}}\,H=\overset{7}{\mathop{C}}\,H\underset{\begin{align} & | \\ & ^{9}C{{H}_{2}} \\ & | \\ & ^{10}C{{H}_{3}} \\ \end{align}}{\mathop{\overset{8}{\mathop{C}}\,H}}\,C{{H}_{3}}\]
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