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    (Original post by alow)
    What do you mean?
    I wish I knew lol hang on let me find the question Name:  ImageUploadedByStudent Room1465078992.152548.jpg
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    So when I'm drawing the spectrum, why is the CH2 singlet peak between 3.3-4.3ppm instead of ~2-3ppm?




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    (Original post by Saywhatyoumean)
    I wish I knew lol hang on let me find the question Name:  ImageUploadedByStudent Room1465078992.152548.jpg
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    Name:  ImageUploadedByStudent Room1465079079.845987.jpg
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    So when I'm drawing the spectrum, why is the CH2 singlet peak between 3.3-4.3ppm instead of ~2-3ppm?




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    The CH2 is deshielded by virtue of its proximity to two electron withdrawing groups.
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    (Original post by alow)
    The CH2 is deshielded by virtue of its proximity to two electron withdrawing groups.
    surely it's because the carbon atom attached to the 2Hs is also attached to an oxygen.Therefore the corresponding chemical shift would be 3.3-4.3,which represents the HC-O proton signal.
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    (Original post by suibster)
    surely it's because the carbon atom attached to the 2Hs is also attached to an oxygen.Therefore the corresponding chemical shift would be 3.3-4.3,which represents the HC-O proton signal.
    The chemical shift of an environment isn't only determined by one of the groups it is attached to...
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    (Original post by suibster)
    No it's determined by the functional group the c is attached to.
    I don't think you understood what I said.
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    (Original post by Hunnybeebee)
    The recommended adult dose of chloral hydrate as a sedative is 250 mg, threetimes a day. Calculate the mass of trichloroethanal you would need to react with water tomake one week’s supply of chloral hydrate for an adult, assuming a 60% yield.Mr: chloral hydrate, 165.5; trichloroethanal, 147.5




    mass of trichloroethanal = g

    Anyone else getting 7.41g or?
    I got 7.8g, think I did it wrong though..
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    (Original post by AqsaMx)
    I got 7.8g, think I did it wrong though..
    I got 7.79 but I used a really old calculater so I couldn't store the values.

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    Hi, can somebody clarify how you make a diazonium ion (with a balanced equation) please?
    In the textbook, pg. 39, it says you need 2HCl but I think this is incorrect?
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    (Original post by chuckster111)
    Hi, can somebody clarify how you make a diazonium ion (with a balanced equation) please?
    In the textbook, pg. 39, it says you need 2HCl but I think this is incorrect?
    Hey you need 1 HCl to make HNO2 and another to make the ion
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    (Original post by chuckster111)
    Hi, can somebody clarify how you make a diazonium ion (with a balanced equation) please?
    In the textbook, pg. 39, it says you need 2HCl but I think this is incorrect?
    Yes, that equation is wrong, it's only 1 HCl. If you wanted to write a full equation, its:

    phenylamine + NaNO2 + 2HCl ----> diazonium ion + NaCl + 2H2O
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    (Original post by rory58824)
    Yes, that equation is wrong, it's only 1 HCl. If you wanted to write a full equation, its:

    phenylamine + NaNO2 + 2HCl ----> diazonium ion + NaCl + 2H2O
    Ok, so if you want the overall equation, it's the above? (Wouldn't it be the diazonium salt rather than ion btw?).

    And if you do it in two steps, it would be:
    1. NaNO2 + HCl ----> NaCl + HNO2
    2. phenylamine + HNO2 + HCl ---> diazonium salt + 2H2O

    ?
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    (Original post by chuckster111)
    Ok, so if you want the overall equation, it's the above? (Wouldn't it be the diazonium salt rather than ion btw?).

    And if you do it in two steps, it would be:
    1. NaNO2 + HCl ----> NaCl + HNO2
    2. phenylamine + HNO2 + HCl ---> diazonium salt + 2H2O

    ?
    Isn't my attachment loading?
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    (Original post by chuckster111)
    Ok, so if you want the overall equation, it's the above? (Wouldn't it be the diazonium salt rather than ion btw?).

    And if you do it in two steps, it would be:
    1. NaNO2 + HCl ----> NaCl + HNO2
    2. phenylamine + HNO2 + HCl ---> diazonium salt + 2H2O

    ?
    I've noticed the textbook changes between ion and salt (maybe someone who knows more could clarify this lol). Probably best to use salt, my bad.

    And yep, that looks fine.
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    (Original post by Serine Soul)
    Isn't my attachment loading?
    Hi, sorry I'm on my phone I didn't realise there was an attachment!
    Thanks for that
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    (Original post by rory58824)
    I've noticed the textbook changes between ion and salt (maybe someone who knows more could clarify this lol)

    And yep, that looks fine.
    It is a salt if another negative ion is with it, like Chlorine. It then becomes Diazonium chloride which is a salt


    If it is just the first part by itself, like below, then it is a diazonium ion
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    (Original post by rory58824)
    I've noticed the textbook changes between ion and salt (maybe someone who knows more could clarify this lol)

    And yep, that looks fine.
    I haven't looked at the textbook for this topic (too simplified) but remember that the diazonium salt is essentially the diazonium ion with a Cl-

    In synthetic routes etc on a question paper, it often says to ignore the presence of the Cl- ion next to the diazonium ion.

    When writing the full reaction in itself though, I'd go for the HNO2 + HCl route and draw the salt as the product rather than just the ion
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    Can anyone explain the specification point - Explain how Condensation polymers may be photodegradable as the C=O bond absorbs radiation.

    Is it just that the C=O bond absorbs IR and becomes weak and brittle, and the amide/ester bonds break?
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    (Original post by AqsaMx)
    Can anyone explain the specification point - Explain how Condensation polymers may be photodegradable as the C=O bond absorbs radiation.

    Is it just that the C=O bond absorbs IR and becomes weak and brittle, and the amide/ester bonds break?
    Please never use the word brittle to describe a chemical bond.
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    (Original post by alow)
    Please never use the word brittle to describe a chemical bond.
    I meant the polymer becomes brittle
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    Can anyone explain what a labile proton actually means? A proton that can be exchanged?
 
 
 
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