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Hydrocarbons — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryHydrocarbons2 Marks
Question
Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.
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Answer

In general chlorination of alkanes proceed by free radical mechanism. 2-methylpropane has both primary (1°) and tertiary (3°) carbon atoms. It will form two free radicals intermediates upon chlorination.$\ \ \ \ 1^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \circ\\\text{CH}_3-\text{CH}-\text{CH}_3\xrightarrow{\text{HOMOLYSIS}}\text{CH}_3-\text{CH}-\text{CH}_2+\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\3^\circ\ \text{free radical is more stable}\ \ \ \ \ \ 1^\circ\text{free radical(I)}\ \ \ \ \ \ \ 3^\circ\text{free radical (II)}\\1^\circ\text{ free radical due to}\\\text{Hyperconjugation}$

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