What is the most abundant product in the bromination of 2-methylbutane?

[Michael Ross]

This is question 41, of passage VII, chapter 1 in the Berkley Ochem book.

Options are...
A) 1-bromo-2-methylbutane
B) 2-bromo-2-methylbutane
C) 2-bromo-3-methylbutane
D) 1-bromo-3-methylbutane

The answer key said B, but I am 99% sure the answer is A.

The passage states that "the reactivity preference for carbon substitution through free radicals follows the trend tertiary>secondary>primary by a factor of 4:2.5:1 at room temp"

BUT THERE ARE SIX TIMES AS MANY PRIMARY H's THAN TERTIARY H's. Therefore, the quantity of six primary hydrogens overcomes the quality of tertiary being selected at a 4:1 ratio. Am I wrong??

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[pepocho]

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[pepocho]

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[Michael Ross]

Oh I found it. Its on Ch 1 Passage I Question 4 for me. Correct me if I am wrong, but I think you misinterpret the information. There is no preference for tertiary hydrogens. The preference is tertiary carbons. The second carbon would have tertiary carbocation while the first carbon would create a primary carbocation. Carbocations are most stable if there are tertiary, which is why choice B is correct.

I meant preference to replace tertiary hydrogens because it is a halogenation reaction. Carbocations are 4 times more stable on the tertiary carbon than primary. However, I believe I am right because the fact that there are 6 times as many primary hydrogens available for replacement than tertiary hydrogens available for replacement, that should outweigh the tertiary preference.

 

[m.z]

Bromine is one of the lesser reactive halogens, and one of the more selective ones. According to EK, Bromine would replace the tertiary hydrogen because it is more selective of where it replaces hydrogens. Chlorine is less selective but more reactive so the secondary product would predominate. Fluorine is the least selective and most reactive of the halogens and the primary product would predominate for Fluorine. According to EK the reactivity of halogens is F, Cl, Br, I (they say iodine wouldn't react at all) ; and the selectivity is the opposite which would be I, Br, Cl, F. I guess you have to take into account selectivity as well. Hope that helps.

 

[syoung]

Br2 is more selective, and prefers a tertiary carbon radical. This is more stable than a primary carbon radical, regardless of the number of primary H's. Simple matter of stability > quantity of available H's.

 

[DrknoSDN]

free radicals follows the trend tertiary>secondary>primary by a factor of 4:2.5:1 at room temp"

BUT THERE ARE SIX TIMES AS MANY PRIMARY H's THAN TERTIARY H's. Therefore, the quantity of six primary hydrogens overcomes the quality of tertiary being selected at a 4:1 ratio. Am I wrong??

That is the ratio for chlorine.
Bromine is much much more selective.

http://depts.washington.edu/chemcrs/bulkdisk/chem237A_aut01/info_Ch_10_Overview.pdf

statistical factor (selectivity factors):
relative rate factors for chlorination 3° : 2° : 1° = 5 : 3.8 : 1
Halogenation with Other Halogens
selectivity of bromination
relative rate factors for bromination 3° : 2° : 1° = 1600 : 82 : 1