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Edexcel - Chemistry Unit 2 - 4 June 2013 Watch

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    this may sound stupid but u know how after all reactions they say carry out distiliation

    what is distilation ? is just purifying the product with distilled water?
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    why are tertiary alcohols more reactive?
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    (Original post by edex123)
    why are tertiary alcohols more reactive?
    It's down to the positive inductive effect of surrounding alkyl groups.
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    (Original post by edex123)
    why are tertiary alcohols more reactive?
    has 3 R groups and R groups break the c-x bond quickly cos 3 is more than 2 R groups (secondary) or 1 R group (primary)

    R group = alkyl groups
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    (Original post by HarryMWilliams)
    It's down to the positive inductive effect of surrounding alkyl groups.
    but why does this make it more reactive?
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    (Original post by Priya08)
    has 3 R groups and R groups break the c-x bond quickly cos 3 is more than 2 r groups (secondary) or 1 r group (primary)
    I am really sorry if this is a dumb question but why does the c-x bond break more quickly if there are more r groups?
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    (Original post by HarryMWilliams)
    A bit of reaction mechanism practice; would anyone like to explain the procedure of producing a carboxylic acid from ethane? (Hint: there are multiple steps.. )
    ethane + (uv) + cl2 -> CH2CHCl + NaOH(aq) -> CH2CHOH + KCr2O7 + H2SO4 (reflux). Distil final mixture and dry with anhydrous sodium sulfate.
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    The grade boundaries are scary for this exam..
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    (Original post by edex123)
    I am really sorry if this is a dumb question but why does the c-x bond break more quickly?
    Baiscally, Alkyl groups are e- donating groups so

    primary - 1 R group present
    secondary - 2 R groups present
    tertiary - 3 R groups present

    so because tertiary has more R' groups (electron donating groups) than secondary or primary, it breaks the c-x bond quicker
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    (Original post by edex123)
    but why does this make it more reactive?
    The 3 alkyl groups feed their electrons to the carbocation, hence, it forms a much more stable intermediate that primary and secondary alcohols, it's therefore more 'available' to react - the fact theres 3 alkyl groups also means the halogen group can leave easier. That's my understanding at least.

    (Original post by Goods)
    ethane + (uv) + cl2 -> CH2CHCl + NaOH(aq) -> CH2CHOH + KCr2O7 + H2SO4 (reflux). Distil final mixture and dry with anhydrous sodium sulfate.
    Very good.

    But! You're missing a hydrogen in your haloalkane.
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    (Original post by scientific222)
    The grade boundaries are scary for this exam..
    whyy what are they like roughly??
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    (Original post by Priya08)
    whyy what are they like roughly??
    Its been slowly creeping up, it was 63/80 in June 2012 and now 65/80 for Jan 2013's paper (which was real hard in my opinion). That is really insane considering its a paper where mistakes can be made so easily
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    Hello I'm confused on question 17fii on the May 2011 paper. What apparatus is the mark scheme looking for? (If you have a page number in the edexcel textbook would be very helpful so I can see it) thank you
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    (Original post by HarryMWilliams)
    The 3 alkyl groups feed their electrons to the carbocation, hence, it forms a much more stable intermediate that primary and secondary alcohols, it's therefore more 'available' to react - the fact theres 3 alkyl groups also means the halogen group can leave easier. That's my understanding at least.



    Very good.

    But! You're missing a hydrogen in your haloalkane.
    Thank you very much i understand now!
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    (Original post by scientific222)
    Its been slowly creeping up, it was 63/80 in June 2012 and now 65/80 for Jan 2013's paper (which was real hard in my opinion). That is really insane considering its a paper where mistakes can be made so easily
    I'd like the grade boundaries from January 2010:

    A - 51/80
    B - 45/80
    C - 39/80 etc.
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    (Original post by Priya08)
    Baiscally, Alkyl groups are e- donating groups so

    primary - 1 R group present
    secondary - 2 R groups present
    tertiary - 3 R groups present

    so because tertiary has more R' groups (electron donating groups) than secondary or primary, it breaks the c-x bond quicker
    oh ok thanks, i understand now
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    A quick question, in unit2 chem jan 12 question 21 b (ii) , the reaction between KOH(aq) and a primary and tertiary halogenoalkane(they are isomers of eac hother) is different in terms of the mechanism and the intermediate formed...i think it has something to do with the 1 step and 2 step mechanism but im not too sure, also why does the structure of the halogenoalkane effect the mechanism? If someone could look at the question it would really help thank you.
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    (Original post by Kenzo-k)
    A quick question, in unit2 chem jan 12 question 21 b (ii) , the reaction between KOH(aq) and a primary and tertiary halogenoalkane(they are isomers of eac hother) is different in terms of the mechanism and the intermediate formed...i think it has something to do with the 1 step and 2 step mechanism but im not too sure, also why does the structure of the halogenoalkane effect the mechanism? If someone could look at the question it would really help thank you.

    maybe this helps:Name:  Chem.JPG
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    How much unit 1 do we need?
    I'm resitting so barely know anything in the unit.

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    (Original post by HarryMWilliams)
    A bit of reaction mechanism practice; would anyone like to explain the procedure of producing a carboxylic acid from ethane? (Hint: there are multiple steps.. )
    Alkane + Halogen (with UV rays) ----> Halogenoalkane

    Halogenoalkane + NaOH -----> Alcohol + NaX

    Alcohol + potassium dichromate -----> Carboxylic acid + H2O

    [conditions being: dilute H2SO4... heat under reflux]

    ?????
 
 
 
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