A) n-pentane > 2, 2-dimethylpropane > 2-methylbutane
B) n-pentane > 2-methylbutane > 2, 2-dimethylpropane
C) 2, 2-dimethylpropane > 2-methylbutane > n-pentane
D) 2-methylbutane > n-pentane > 2, 2-dimethylpropane
E) 2-methylbutane > 2, 2-dimethylpropane > n-pentane
Correct Answer: B
Solution :
Amongst isomeric alkanes, the branched chain isomer has the lower boiling point than the corresponding n-alkane. This is due to the reason that with branching the shape of the molecule tends to approach that of a sphere. As a result, the surface area of the branched isomer decreases. Due to lesser surface area of these molecules, the van der Waals forces of attraction operating between their molecules become comparatively weaker and hence lesser amount of energy is required to overcome then. Thus the order of boiling points of chain isomers of pentane: \[\underset{\underset{(b.p.=309.1\text{ }K)}{\mathop{n-pentane}}\,}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,>\underset{\begin{smallmatrix} 2-methylbutane \\ (b.p.\text{ }=\text{ }301\text{ }K) \end{smallmatrix}}{\mathop{C{{H}_{3}}-\overset{\underset{|}{\mathop{C{{H}_{3}}}}\,}{\mathop{CH}}\,-C{{H}_{2}}C{{H}_{3}}}}\,\] \[\underset{\begin{smallmatrix} 2,\text{ }2-\dimethylpropane \\ (b.p.\text{ }=\text{ }282.5\text{ }K) \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}}}\,\]
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