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Write hydrocarbon radicalsthat can be formed as intermediates during monochlorination of 2-methylpropane ? Which of them is more stable ? Give reasons . |
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Answer» Solution :2-methylpropane has two sets of equivalent hydrogens marked as 'a' and 'B' and hence GIVES two radicals, I and II. `undersetI(oversetalphaCH_3-underset*oversetoversetoversetalpha(CH_3)|C-oversetalphaCH_3) larr underset"2-Methylpropane"(oversetalphaCH_3-undersetboversetoversetoverseta(CH_3)|C-oversetaCH_3) to underset"II"(CH_3-oversetoverset(CH_3)|(.^alphaCH)-oversetdotCH_2)` Radical (I) is more STABLE than radical (II) because of the following two reasons : (i)Radicals (I) is `3^@` while radical (II) is `1^@` (ii)Radical (I) has nine `ALPHA`-hydrogen and hence is STABILIZED by nine hyperconjugation structues while radical (II) has only one `alpha`-hydrogens and is stabilized by one hyperconjugation structures as shown below : 
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