arrange the following alkyl halides in order from least reactive to most reactiv

Question

arrange the following alkyl halides in order from least reactive to most reactive in an SN1 reaction: 2-bromo-2-methylpentane, 2-chloro-methylpentane, 3-chloropentane, and 2-iodo-2-methylpentane

Explanation / Answer

Reactivity of alkyl halides in an SN1 reaction depends on the stability of the carbocation being formed: 3rd degree carbocation > 2nd degree carbocation > 1st degree carbocation > methyl 1) 2-bromo-2-methylpentane forms a tertiary carbocation. 2) 2-chloro-methylpentane. You haven't indicated the position of the methyl group. If it's on carbon 2 (carbon bonded to chlorine), a secondary carbocation will be formed. If it's on carbon 3, a secondary carbocation will be formed. If it's on carbon 4, secondary carbocation will be formed. I am going to assume the methyl group is on carbon 2 in which case a tertiary carbocation wil be formed. 3) 3-chloropentane forms a secondary carbocation. 4) 2-iodo-2-methylpentane forms a tertiary carbocation. Here are the steps to an SN1 mechanism: 1. Leaving group leaves, forming carbocation. 2. Carbocation undergoes nucleophilic attack, forming product. There is a tie between 1, 2, and 4 (both form tertiary carbocations). To narrow it down, see which one of the three has the best leaving group. Leaving group ranking: I > Br > Cl > F The better of the leaving groups will be more reactive. Here's the order from least to most reactive: 3-chloropentane < 2-chloro-2-methylpentane < 2-bromo-2-methylpentane < 2-iodo-2-methylpentane Hope this helps! :)

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