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    Can you use NaBH4 and LiAlH4 for reducing all unsaturated compounds? If not, which can and can't you? Also, are there any other conditions necessary when you use these two agents. Also, can you just use H2 gas and Nickel catalyst for the same effect, are there any other conditions necessary here?

    If anyone is doing the AQA A2 Chem tomorrow, do you think putting H2/Ni is the best option..?

    Ta
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    Im doing it!

    I've decided if its nitrobenzene -----> phenylamine do Sn/HCl

    If its anything else other than alcohols etc put H2 / Ni

    and only NaBH4 for the alcohols/aldehydes/acids
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    I'm doing the exam aswell. Good luck to you!
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    (Original post by Casshern1456)
    I'm doing the exam aswell. Good luck to you!
    Agreed -- good luck yourself!
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    (Original post by Lou Reed)
    Can you use NaBH4 and LiAlH4 for reducing all unsaturated compounds? If not, which can and can't you? Also, are there any other conditions necessary when you use these two agents. Also, can you just use H2 gas and Nickel catalyst for the same effect, are there any other conditions necessary here?

    If anyone is doing the AQA A2 Chem tomorrow, do you think putting H2/Ni is the best option..?

    Ta
    No, this is often the question of reactivity.
    ie NaBH4 will reduce aldehydes to alcohol, which LiALH4 will do too, but for carboxylic acids, you'd need a more powerful reducing agent(ie LIAlH4) to reduce it, NaBH4 won't work to reduce carboxylic acid.

    Conditions for these are mostly in inert solvent like tetrahydrofuran(THF) as any presence of water in aqueous solution would be reduced too(maybe to H2 gas).

    H2 and Ni cataylst are "ok" choices for hydrogenation of alkenes, though it won't work for benzene. you can also use C/Pd catalyst with H2 for the same process.
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    (Original post by raceforthefishman)
    Im doing it!

    I've decided if its nitrobenzene -----> phenylamine do Sn/HCl

    If its anything else other than alcohols etc put H2 / Ni

    and only NaBH4 for the alcohols/aldehydes/acids
    NaBH4 doesn't reduce alcohols.
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    (Original post by shengoc)
    No, this is often the question of reactivity.
    ie NaBH4 will reduce aldehydes to alcohol, which LiALH4 will do too, but for carboxylic acids, you'd need a more powerful reducing agent(ie LIAlH4) to reduce it, NaBH4 won't work to reduce carboxylic acid.

    Conditions for these are mostly in inert solvent like tetrahydrofuran(THF) as any presence of water in aqueous solution would be reduced too(maybe to H2 gas).

    H2 and Ni cataylst are "ok" choices for hydrogenation of alkenes, though it won't work for benzene. you can also use C/Pd catalyst with H2 for the same process.
    Reductions with NaBH4 are in fact almost always done in a protic solvent (for example, ethanol), the reason is that a proton is involved in the mechanism, and most organic substrates are not soluble in water.

    LiALH4 cannot be used as a reducing agent with water as the solvent, because it reacts with water very violently releasing H2 gas.

    H2/nickel is an excellent combo for reducing double bonds and is probably the most used reaction for this purpose.
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    (Original post by StadtJunky)
    NaBH4 doesn't reduce alcohols.
    I am, in fact, aware of this...
    Don't really know why I put it -- just because its those kind of things it all comes under
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    (Original post by raceforthefishman)
    Im doing it!

    I've decided if its nitrobenzene -----> phenylamine do Sn/HCl

    If its anything else other than alcohols etc put H2 / Ni

    and only NaBH4 for the alcohols/aldehydes/acids
    i actually love you!!! where have you been all my life??

    Do you know why the Ka for weak acids is [H+]2 divided by reactants rathr than just the [h+] ???
    Last minute revision is hurting my head
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    (Original post by beckyburch)
    i actually love you!!! where have you been all my life??

    Do you know why the Ka for weak acids is [H+]2 divided by reactants rathr than just the [h+] ???
    Last minute revision is hurting my head

    Right Ka is normally = ([H+][A-])/[HA]

    but
    if you don't mess about with the solution beforehand the acid molecules will just dissociate as much as they like. but if you think about it, for everyone one H+ ion that dissociates one A- ion will be present. This means theyre the same quantities and so the eqn can be simplified to Ka = [H+]2/[HX]

    This doesn't apply in buffer solutions as [H+] is not = [A-]

    I think thats what you were asking!
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    so [h+]x[a-] is the same as [h+]2 ?? thankyouu x
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    (Original post by beckyburch)
    so [h+]x[a-] is the same as [h+]2 ?? thankyouu x
    When they are the same number, essentially yes!

    If they were both 0.150M then itd be
    0.150 x 0.150
    Do you see the square now?
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    excellent.. thankyou

    The other thing was, does anyone know how to work out the order of reactions from the tables?? i can do it if there is a standard to compare against.. but its thte uncomparable ones i get stuck on
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    (Original post by beckyburch)
    excellent.. thankyou

    The other thing was, does anyone know how to work out the order of reactions from the tables?? i can do it if there is a standard to compare against.. but its thte uncomparable ones i get stuck on
    Right...

    They might have given you an order...or you may already have worked one out that changes...

    All you do is see how that one changes..so how it would effect the rate of reaction, then do that to your initial rate...then look at the one you're puzzling over and use the rate you will have worked out.

    so If you know A is second order
    [A] is doubled
    [B] is doubled

    Initial rate 1) 2
    Initial rate 2) 16

    So A is second order so multiply rate 1) by 4 *giving 8*

    Then, what would you have to do to 8 to get 16
    double it

    so b would be 1st order
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    (Original post by StadtJunky)
    Reductions with NaBH4 are in fact almost always done in a protic solvent (for example, ethanol), the reason is that a proton is involved in the mechanism, and most organic substrates are not soluble in water.

    LiALH4 cannot be used as a reducing agent with water as the solvent, because it reacts with water very violently releasing H2 gas.

    H2/nickel is an excellent combo for reducing double bonds and is probably the most used reaction for this purpose.
    Alright, agreed, it is indeed in protic solvent for borohydride.
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    what happens if thre is 3?
    or i dnt no any of thm?
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    Do you have an example of a question?
    I'll take you through it
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    its okay thanx,, a friend mailed me a sheet which has helped.. :-)
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    Okay then! Well good luck for tomorrow -- You'll be fine
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    thanks, good luck to you toooo xx
 
 
 
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