A) Head-to-head
B) Tail-to-tail
C) Head-to-tail
D) All of these
Correct Answer: C
Solution :
\[\overset{\text{Head}}{\mathop{C{{H}_{2}}}}\,=\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,}{\mathop{C-}}\,}}\,CH=\overset{\text{Tail}}{\mathop{C{{H}_{2}}}}\,+\overset{\text{Head}}{\mathop{C{{H}_{2}}}}\,=\underset{\,\,\,\,\,\,\,\,C{{H}_{3}}}{\mathop{\underset{\,\,\,\,\,\,|}{\mathop{C-CH}}\,}}\,=\overset{\text{Tail}}{\mathop{C{{H}_{2}}\,}}\,\to \] \[{{(-C{{H}_{2}}-\underset{C{{H}_{2}}\,\,\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,}{\mathop{C=}}\,CH}}\,-C{{H}_{2}}-C{{H}_{2}}-\underset{C{{H}_{3}}\,\,\,\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,}{\mathop{C=}}\,CH}}\,-C{{H}_{2}}-)}_{n}}\] From steric effects, the polymer formed has head to tail configuration.
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