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

Discussion in 'MCAT Study Question Q&A' started by Michael Ross, May 19, 2014.

  1. Michael Ross

    Michael Ross

    Feb 14, 2014
    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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  3. pepocho

    pepocho 2+ Year Member

    Mar 13, 2013
    Last edited: Apr 11, 2016
  4. pepocho

    pepocho 2+ Year Member

    Mar 13, 2013
    Last edited: Apr 11, 2016
    syoung likes this.
  5. Michael Ross

    Michael Ross

    Feb 14, 2014
    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.
  6. m.z


    Feb 3, 2014
    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.
    Dr Turkelton likes this.
  7. syoung

    syoung MS-3 5+ Year Member

    Jan 2, 2011
    Over the rainbow
    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.
  8. DrknoSDN


    Feb 21, 2014
    That is the ratio for chlorine.
    Bromine is much much more selective.

    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

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