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    In the formation of 2-bromoethanol there is a stage where the water molecule has just attached to the bromoethane compound.

    In this state the oxygen atom from the water moelcule has a positive charge, why ?
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    (Original post by Ari Ben Canaan)
    In the formation of 2-bromoethanol there is a stage where the water molecule has just attached to the bromoethane compound.

    In this state the oxygen atom from the water moelcule has a positive charge, why ?
    What is your starting point? ethene or 1,2-dibromoethane?
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    (Original post by Plato's Trousers)
    What is your starting point? ethene or 1,2-dibromoethane?
    ethene.
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    (Original post by Ari Ben Canaan)
    ethene.
    ooh... no wonder i couldn't work it out.
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    (Original post by Princess_perfect786)
    ooh... no wonder i couldn't work it out.
    Basically I'm asking when ethene is reacted with bromine water you will get 2-bromoethanol.

    In the formation of this there is a stage where the water molecule has just attached to to the bromoethane molecule.

    At this point the Oxygen atom has a positive charge, why ?
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    (Original post by Ari Ben Canaan)
    Basically I'm asking when ethene is reacted with bromine water you will get 2-bromoethanol.

    In the formation of this there is a stage where the water molecule has just attached to to the bromoethane molecule.

    At this point the Oxygen atom has a positive charge, why ?
    Well, first you will get 1,2-dibromoethane as the Br2 molecules react with ethene. Then water can react with this molecule via Sn2 to produce an intermediate that looks like bromoethane with H2O stuck on with a positive charge. This is because, the H2O lone pair (2e-) attack the C-Br bond in the dibromoethane. The Br- is substituted for H2O. This means that the 2e- in the lone pair are now being used to form a bond with the C that the Br- was attached to. A bond requires 2 e-, so the O now has a ~50% share of the 2e- in the bond where it previously had a 100% share of the 2e- (since they were both in the lone pair on the O). The charge is balanced overall due to the loss of the Br-. Make sense?
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    (Original post by Ari Ben Canaan)
    Basically I'm asking when ethene is reacted with bromine water you will get 2-bromoethanol.

    In the formation of this there is a stage where the water molecule has just attached to to the bromoethane molecule.

    At this point the Oxygen atom has a positive charge, why ?
    CH2=CH2 + Br-OH ----> CH2Br-CH2+ + OH-

    CH2Br-CH2+ + OH- --> CH2Br-CH2OH
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    (Original post by Toneh)
    Well, first you will get 1,2-dibromoethane as the Br2 molecules react with ethene. Then water can react with this molecule via Sn2 to produce an intermediate that looks like bromoethane with H2O stuck on with a positive charge. This is because, the H2O lone pair (2e-) attack the C-Br bond in the dibromoethane. The Br- is substituted for H2O. This means that the 2e- in the lone pair are now being used to form a bond with the C that the Br- was attached to. A bond requires 2 e-, so the O now has a ~50% share of the 2e- in the bond where it previously had a 100% share of the 2e- (since they were both in the lone pair on the O). The charge is balanced overall due to the loss of the Br-. Make sense?
    No, wait, look at the picture below.

    Explain what's happend in the second step.
    Attached Images
     
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    (Original post by Plato's Trousers)
    what does he mean by cool story, bromoethanol?
    One other question for you.

    The acid hydrolysis of HYDROXY-nitriles produces a HYDROXY-carboxylic acid and what ?

    What would the alkaline hydrolysis of HYDROXY-nitriles produce ?
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    (Original post by Ari Ben Canaan)
    No, wait, look at the picture below.

    Explain what's happend in the second step.
    O, ok. I probably should've drawn out the mechanism. So the water attacks the carbocation intermediate instead of the 1,2-dibromoethane. The principle for the positive oxygen is essentially the same though, the 2e- in the lone pair attack into an empty C+ p orbital to produce a C-O sigma bond with 2e-, so the O now shares ~50% (>50% due to the electronegativity, but that's not important) of the 2e- now in the C-O bond, so becomes +1 (because prior to attacking it had 100% of the 2e- in the lone pair and was neutral, so it loses the charge of 1 e-). The C which was +1 now gains 50% of 2e- (i.e. a charge of -1) and is hence neutrally charged.

    Does that make sense given the mechanism shown?
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    (Original post by Ari Ben Canaan)
    No, wait, look at the picture below.

    Explain what's happend in the second step.
    OK, I wrote a simplified form showng Br2(aq) as BrOH. This is perfectly acceptable but doesnpt answer your question based on the mechanism given.

    When water attacks the positive carbon it uses two of its own electrons and after the bond is formed it has an effective share of only one of them. It hence has a positive charge (the one less electron of its 'own')

    Logically, when a plus adds to a neutral you must end up with a plus. As the oxygen has three bonds it is clear that the plus resides there.
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    (Original post by Toneh)
    O, ok. I probably should've drawn out the mechanism. So the water attacks the carbocation intermediate instead of the 1,2-dibromoethane. The principle for the positive oxygen is essentially the same though, the 2e- in the lone pair attack into an empty C+ p orbital to produce a C-O sigma bond with 2e-, so the O now shares ~50% (>50% due to the electronegativity, but that's not important) of the 2e- now in the C-O bond, so becomes +1 (because prior to attacking it had 100% of the 2e- in the lone pair and was neutral, so it loses the charge of 1 e-). The C which was +1 now gains 50% of 2e- (i.e. a charge of -1) and is hence neutrally charged.

    Does that make sense given the mechanism shown?
    So, in short, the positive carbon attracts an electron from the oxygen atom such that it neutralises its previously positive charge, however, the oxygen atom is now left with one less electron such that it gains a positive charge.
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    (Original post by charco)
    OK, I wrote a simplified form showng Br2(aq) as BrOH. This is perfectly acceptable but doesnpt answer your question based on the mechanism given.

    When water attacks the positive carbon it uses two of its own electrons and after the bond is formed it has an effective share of only one of them. It hence has a positive charge (the one less electron of its 'own')

    Logically, when a plus adds to a neutral you must end up with a plus. As the oxygen has three bonds it is clear that the plus resides there.
    Not to worry, Toneh explained it well. Thanks.

    Could you answer this other query of mine ?


    (Original post by Ari Ben Canaan)
    One other question for you.

    The acid hydrolysis of HYDROXY-nitriles produces a HYDROXY-carboxylic acid and what ?

    What would the alkaline hydrolysis of HYDROXY-nitriles produce ?
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    Think of it this way. In your picture, you have a carbocation and water molecule, you then combine them together to make a single molecule.

    If you've started with a +ve charge and a neutral charge and combined them together, you must have a positive charge leftover. Since you have just one molecule now it must be localised somewhere in this molecule. Is it on the carbon as it was before? No, since the water has just moved in and donated its lone pair to forming a bond with this carbon. So the +ve charge must be written on the water. Loss of a proton to some base will then remove this charge and give you your neutral alcohol.

    As a rule, when you see oxygen with three bonds it is going to be +ve charged, when you see it with only one bond it will be -ve. However its better to understand why than to learn that by rote!

    EDIT: as i wrote that three other people answered the question :P

    However to help with your other problem:

    http://www.chemguide.co.uk/organicpr...ydrolysis.html
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    (Original post by Ari Ben Canaan)
    So, in short, the positive carbon attracts an electron from the oxygen atom such that it neutralises its previously positive charge, however, the oxygen atom is now left with one less electron such that it gains a positive charge.
    Well, really it attracts a pair of electrons in order to form a bond, and in doing so "gains" an electron whilst the oyxgen "loses" an electron. Not really strictly true, but it will do to understand it. So pretty much, yer.
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    (Original post by Blocker)
    However to help with your other problem:

    http://www.chemguide.co.uk/organicpr...ydrolysis.html
    I've alrady been there.

    My question is specifically about HYDROY-nitriles NOT good old nitriles.
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    (Original post by Ari Ben Canaan)
    One other question for you.

    The acid hydrolysis of HYDROXY-nitriles produces a HYDROXY-carboxylic acid and what ?

    What would the alkaline hydrolysis of HYDROXY-nitriles produce ?
    ok, from having a look at potential mechanisms for that, the only difference between an hydroxy nitrile and a nitrile is that the hydroxy nitrile can undergo intramolecular attack to form a ring structure. The size of the ring depends on the position of the hydoxyl group, and the size of the ring will determine how likely that reaction is to happen (e.g. if the OH is on the adjacent C, a 3 membered ring would form, which would be very stressed and probably quite unstable and liable to reform the structure before intramolecular attack and then undergo the normal acid hydolysis of a nitrile. In comparison if it was on say the 4th C of some chain (with the 1st being the CN) a 5m ring would form which would be much more stable and thus more likely to form.)
    So in other words, the reaction mechanism is broadly the same and would produce either a hydroxy carboxylic acid as you suggested, or an ester ring structure. (i.e. in the second example I said, a 5 membered ring with one oxygen member and a C=O would form). Both mechanisms would also produce ammonia.

    That's what I reckon would happen, anyway. If you want I could draw out the mechanisms for you.

    As for alkaline hydrolysis, I'm currently unsure. Let me get back to you on that.
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    (Original post by Toneh)
    ...
    So you're sure that NH3 would be produced in alkaline and acid hydrolysis ?
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    (Original post by Ari Ben Canaan)
    So you're sure that NH3 would be produced in alkaline and acid hydrolysis ?
    Pretty sure, yer.
 
 
 

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