Arrange the following alkyl halides in order from most reactive to least reactive in an sn1 reaction: 2-bromo-2-methylpentane, 2-chloro-2-methylpentane, 3-chloropentane, and 2-iodo-2-methylpentane.

2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.

Explanation:

At a substitution reaction by SN1, the alkyl halide must lose its halide, and then an intermediary will be formed: a carbonium, which is an alkyl group with a positive charge in the carbon. The halide lost will be formed the halide ion, which is also an intermediary of the reaction.

The reactivity depends on the stability of the intermediaries (first of the carbonium, and second of the halide ion). As more bonded with carbons is the carbonium, more stable it is. The order of stability of the halides ions is from their electronegativity: as lower is it, as stable is the ion. The order is then: I⁻ > Br⁻ > Cl⁻ > F⁻.

2-bromo-2-methylpentane, 2-chloro-2-metylpentane, and 2-iodo-2-methylpentane, will form a 3-degree intermediary, so they will be more reactive than 3-chloropentane, which form a 2-degree intermediary. So, for the order of the stability of the halide ions, the order of reactivity is:

2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.

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explanation:

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.