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    I would like to clarify please - usually the reactivity of halides decrease down Group 7 right? So why does the halide C-X (in a halogenoalkane) increase down the group? Also what is the number one factor/priority that is given when determining the reactivity of a compound?
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    In haloalkanes the reactivity of the halogen group is primarily affected by the size of the halogen-alkane bond. The forces holding the atoms together are all just electrostatic attractions. Electrostatic attraction strangths are dependent on the distance between the charges, for electrostatic forces: strength is inversely proportional to distance^2. So as distance increases the strength decreases.

    Likewise in haloalkanes, the bond strength is inversely proportional to the bond length squared. So as the bond length increases (as the halogen atoms get bigger) the bond strength decreases and therefore the reactivity increases.
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    (Original post by I <3 WORK)
    I would like to clarify please - usually the reactivity of halides decrease down Group 7 right? So why does the halide C-X (in a halogenoalkane) increase down the group? Also what is the number one factor/priority that is given when determining the reactivity of a compound?
    When we're reacting with a halogenoalkane, perhaps a nucleophilic substitution with a cyanide ion to form a nitrile, notice how we're breaking the C-X bond (where X is your halide)

    If we can break that C-X bond easily, then it's more reactive because the mechanism is more likely to proceed.

    Compare C - F and C - Br.
    Bromine is less reactive than fluorine. That's because it has a large atomic radius and lots of shielding. This makes it far less electronegative.

    C - F is very polar. The fluorine atom is very small so the bond is very short, and the electrons are going to be strongly attracted to the fluorine atom. This forms a strong bond.

    Contrast that with the C - Br bond which is much longer, less polar and more weaker; this makes it easier to break. That's why it makes it more reactive.
    Spoiler:
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    You might think the fact that the C - F bond is so polar, it should be far more reactive. Surely because the carbon atom has a high delta charge, a nucleophile will readily attack it, and fluorine would want to accept those electrons in the bond? Well that's not the case, because if fluorine get's dejected from the carbon, it's intense electric charge will just cause it to bond back again.

    In the reactivity of haloalkanes for organic mechanisms, bond enthalpy is the considering factor. It's the energy to break that C - X bond.
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    (Original post by Peroxidation)
    In haloalkanes the reactivity of the halogen group is primarily affected by the size of the halogen-alkane bond. The forces holding the atoms together are all just electrostatic attractions. Electrostatic attraction strangths are dependent on the distance between the charges, for electrostatic forces: strength is inversely proportional to distance^2. So as distance increases the strength decreases.

    Likewise in haloalkanes, the bond strength is inversely proportional to the bond length squared. So as the bond length increases (as the halogen atoms get bigger) the bond strength decreases and therefore the reactivity increases.
    Thank you very much , so in this case for haloalkanes the bond length is the main factor in determining reactivity. So why does this rule differ with other halides that decrease in reactivity down the group? And how are you able to determine which rule/factor you apply?
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    (Original post by RMNDK)
    When we're reacting with a halogenoalkane, perhaps a nucleophilic substitution with a cyanide ion to form a nitrile, notice how we're breaking the C-X bond (where X is your halide)

    If we can break that C-X bond easily, then it's more reactive because the mechanism is more likely to proceed.

    Compare C - F and C - Br.
    Bromine is less reactive than fluorine. That's because it has a large atomic radius and lots of shielding. This makes it far less electronegative.

    C - F is very polar. The fluorine atom is very small so the bond is very short, and the electrons are going to be strongly attracted to the fluorine atom. This forms a strong bond.

    Contrast that with the C - Br bond which is much longer, less polar and more weaker; this makes it easier to break. That's why it makes it more reactive.
    Spoiler:
    Show
    You might think the fact that the C - F bond is so polar, it should be far more reactive. Surely because the carbon atom has a high delta charge, a nucleophile will readily attack it, and fluorine would want to accept those electrons in the bond? Well that's not the case, because if fluorine get's dejected from the carbon, it's intense electric charge will just cause it to bond back again.

    In the reactivity of haloalkanes for organic mechanisms, bond enthalpy is the considering factor. It's the energy to break that C - X bond.
    Ahh I see thank you so much! So would you say that the only exception to the usual trend is with this carbon bond? As in are there any other cases in which the rule you explained above applies?
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    (Original post by I <3 WORK)
    Ahh I see thank you so much! So would you say that the only exception to the usual trend is with this carbon bond? As in are there any other cases in which the rule you explained above applies?
    I can't really think of any other cases off the top of my head, but I do want to illustrate the fact that there's a whole host of factors that come into play when determining the reactivity of a compound...

    For instance, we've just said the bond enthalpy for the halogenoalkanes is the driving factor.

    The other user, peroxidation, highlighted bond length, which to be honest is inseperable from bond enthalpy. Longer bond lengths are weaker due to the weaker electrostatic attraction, so the enthalpy change to break it is lower.


    Then you've got bond polarity which is the driving factor for acids and bases. Because the HCl bond is more polar than HBr, it makes it more ready to ionise and donate a proton as a BL acid.
    If you haven't covered this, then whoops sorry.

    I guess this is just one of them things you got to learn.
    Trust, when you start learning more in chemistry and you come across concepts like charge density and polarisation, you'll realise how you can't say there's this one rule you apply to all compounds that determines it's reactivity
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    (Original post by RMNDK)
    I can't really think of any other cases off the top of my head, but I do want to illustrate the fact that there's a whole host of factors that come into play when determining the reactivity of a compound...

    For instance, we've just the bond enthalpy for the halogenoalkanes is the driving factor.

    The other user, I <3 work, highlighted bond length, which to be honest is inseperable from bond enthalpy. Longer bond lengths are weaker due to the weaker electrostatic attraction, so the enthalpy change to break it is lower.


    Then you've got bond polarity which is the driving factor for acids and bases. Because the HCl bond is more polar than HBr, it makes it more ready to ionise and donate a proton as a BL acid.
    If you haven't covered this, then whoops sorry.

    I guess this is just one of them things you got to learn.
    Trust, when you start learning more in chemistry and you come across concepts like charge density and polarisation, you'll realise how you can't say there's this one rule you apply to all compounds that determines it's reactivity
    I really appreciate this, you have explained the concept very well! I'm only at AS stage, but my problem is I can never proceed with my revision when I don't understand something along the way. That's why I result to finding out information beyond my level. But thank you once again, I understand now how reactivity can be grouped into different categories when determining it for a certain compound.
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    (Original post by I <3 WORK)
    I really appreciate this, you have explained the concept very well! I'm only at AS stage, but my problem is I can never proceed with my revision when I don't understand something along the way. That's why I result to finding out information beyond my level. But thank you once again, I understand now how reactivity can be grouped into different categories when determining it for a certain compound.
    And that's perfectly fine! And I really really encourage you to keep doing it! Where they grossly simplified things at GCSE, they still leave out many elements at A-Level that provokes us to research and question more. But its can be an enjoyable and thoroughly rewarding process.

    Good luck with your revision,
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    (Original post by I <3 WORK)
    Thank you very much , so in this case for haloalkanes the bond length is the main factor in determining reactivity. So why does this rule differ with other halides that decrease in reactivity down the group? And how are you able to determine which rule/factor you apply?
    Sorry I've been a bit distracted. Looks like RMNDK's covered it all though
 
 
 

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