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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread Watch

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    (Original post by ayvaak)
    Hey guys! another question i cant seem to get my head around. Could use some help as to why the answers A. Thanks in advance
    cant zoom in on the image, what is the question
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    (Original post by samb1234)
    cant zoom in on the image, what is the question
    Q18 IAL paper January 2014
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    (Original post by ayvaak)
    Q18 IAL paper January 2014
    We know because it reacts with fehlings it has an aldehyde group, and we also know that it must either have or be able to form something with a methyl ketone group. The oh group in A can be oxidised to a ketone (as secondary alcohol) leaving a methyl ketone group hence why can undergo iodoform
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    (Original post by samb1234)
    We know because it reacts with fehlings it has an aldehyde group, and we also know that it must either have or be able to form something with a methyl ketone group. The oh group in A can be oxidised to a ketone (as secondary alcohol) leaving a methyl ketone group hence why can undergo iodoform
    I guess what i wasn't sure on was the fact that the question insists that P ( with the OH functional group) undergoes iodoform. I didn't realise it was oxidised as a result of the reaction with I2 and NaOH. Thank you for the help
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    (Original post by ayvaak)
    I guess what i wasn't sure on was the fact that the question insists that P ( with the OH functional group) undergoes iodoform. I didn't realise it was oxidised as a result of the reaction with I2 and NaOH. Thank you for the help
    there's some good info on iodoform and alcohols here http://www.chemguide.co.uk/organicpr.../iodoform.html , should give you a better idea of what is going on
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    Just to confirm, this is for the old spec right? How did people people find the Unit 1 retake today?
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    (Original post by narcotica)
    Just to confirm, this is for the old spec right? How did people people find the Unit 1 retake today?
    Yes this is old spec
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    Does anyone know if theres a IAL Jan 2016 Unit 4 and 5 paper?
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    (Original post by ayvaak)
    Does anyone know if theres a IAL Jan 2016 Unit 4 and 5 paper?
    Yes there are
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    (Original post by narcotica)
    Just to confirm, this is for the old spec right? How did people people find the Unit 1 retake today?
    Generally, I found it good wbu?
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    (Original post by samb1234)
    Yes there are
    do you have any pdf copies that can be shared with this forum? That would be greatly appreciated
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    (Original post by ayvaak)
    do you have any pdf copies that can be shared with this forum? That would be greatly appreciated
    I don't, best bet is to ask the PAMT guy if he will send it to you
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    (Original post by samb1234)
    I don't, best bet is to ask the PAMT guy if he will send it to you
    thanks. I'll try now. I'll keep you guys posted
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    CAN SOMEONE PLEASE OUTLINE THE METHOD (from unit 4) OF CALCULATING THE PH OF SOLUTION PART WAY THROUGH A REACTION: 1. WHEN THEY ARE ARE EQUIMOLAR 2. WHEN THEY HAVE A DIFFERENT MOLE RATION
    *when i refer to they i mean acid and base
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    (Original post by dpoojaraa)
    Generally, I found it good wbu?
    I wasn't expecting some of the multiple choice questions and the first mass spectrum question about naming 2 different ions. I had 20 mins extra to check over the paper but I had to waste that time trying to guess those multiple choice questions. Did you manage to find an unofficial mark scheme btw?
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    (Original post by ayvaak)
    Does anyone know if theres a IAL Jan 2016 Unit 4 and 5 paper?
    Yes, you'll find them here.
    https://56leomessiphotoshop.blogspot...materials.html
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    (Original post by tayloryeah)
    CAN SOMEONE PLEASE OUTLINE THE METHOD (from unit 4) OF CALCULATING THE PH OF SOLUTION PART WAY THROUGH A REACTION: 1. WHEN THEY ARE ARE EQUIMOLAR 2. WHEN THEY HAVE A DIFFERENT MOLE RATION
    *when i refer to they i mean acid and base
    Write a balanced chemical equation between them. Once you have done that, calculate the initial moles of the reactants and products. The initial concentration for both the acid and base will be given to you. Now, once you have gotten the initial moles find out the moles that have reacted. This is done by taking the difference in their moles. Now you'll get the moles of the products. Find the new concentration using your moles of the salt and the total volume of the solution. Then if it is between let's say a carboxylic acid and a strong base such as NaOH then use the Kw value to get your hydrogen ion concentration then substitute your value of the hydrogen concentration in the formula pH= -Log[H+]. Hope this helps.
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    I have a few questions: halogenoalkane nucleophilic substitution with OH- 1.Why do primary halogenoalkanes follow sn2 mechanism and not sn1?

    2. Why does sn1 occur in tertiary halogenoalkane and not sn2?

    3. What mechanism occurs in secondary halogenoalkanes and why?

    Thanks a lot. I am studying for edexcel and i couldnt find the exact info anywhere( perhaps because i didnot scan every page)
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    (Original post by thebrahmabull)
    I have a few questions: halogenoalkane nucleophilic substitution with OH- 1.Why do primary halogenoalkanes follow sn2 mechanism and not sn1?

    2. Why does sn1 occur in tertiary halogenoalkane and not sn2?

    3. What mechanism occurs in secondary halogenoalkanes and why?

    Thanks a lot. I am studying for edexcel and i couldnt find the exact info anywhere( perhaps because i didnot scan every page)
    Firstly SN1, SN2 stands for nucleophillic substitution and the numbers refer to the number of molecules in the RDS (rate determining step)

    Tertiary halogenoalkanes undergo SN1 because the halogen functional group is attached to three alkyl groups which feeds electrons to the carbonation, enabling it to be stable (more steroic hindrance)

    Primary halogenoalkanes undergo SN2 because the halogen functional group is only attached to one alkyl group, so no stable carbocation forms, and as a result, the nucleophille attacks to the electron deficient area immediately. (it can form a transition state)

    Secondary halogenoalkanes can go either route as there are two alkyl groups attached

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    (Original post by thebrahmabull)
    I have a few questions: halogenoalkane nucleophilic substitution with OH- 1.Why do primary halogenoalkanes follow sn2 mechanism and not sn1?

    2. Why does sn1 occur in tertiary halogenoalkane and not sn2?

    3. What mechanism occurs in secondary halogenoalkanes and why?

    Thanks a lot. I am studying for edexcel and i couldnt find the exact info anywhere( perhaps because i didnot scan every page)
    There are a few slight issues with the other answer so I will go over it quickly. Firstly, as stated before the SN stands for nucleophilic substitution, and the number tells us how many species are involved in the rate determining step.

    If we want to understand why the mechanisms are the way around that they are, you need to look at the reaction mechanisms themselves. Firstly, what is a nucleophile? A nucleophile is a species which is attracted to a +vely charged region, so we can imply from that that it itself is either negatively or partially negatively charged (the same mechanisms happen regardless of what the nucleophile is).
    We also know that in the sn2 mechanism, the slightly positive carbon attached to the slightly -ve halogen is attacked directly by the nucleophile. But which direction does this attack take place from? Well we know that the nucleophile is slightly or fully -ve, and the halogen is also slightly -ve, so as they would repel the attack must happen from the opposite side of the carbon to the halogen. So why can this happen for primary haloalkanes and not for tertiary ones?

    This is due to an effect called steric hindrance, which is essentially blocking. As an analogy, imagine that you see one of your friends at the other end of the corridor, and there are 3 year 7s between you and them. You can easily get through and get to your friend, which is the equivalent of a primary haloalkane, as at least 2 of the groups attached to the carbon are small hydrogen atoms. Now imagine that instead there are 3 large groups of year 13 students who are standing around waiting for their teacher to arrive. The corridor is now almost completely blocked, and it is now very difficult for you to get through to reach your friend. This is exactly what steric hindrance is in tertiary haloalkanes - methyl groups are bulky and as such they leave little if any room for a nucleophile to come through from that side, so as such the nucleophile can't directly attack the slightly positive carbon.

    So why is it that tertiary haloalkanes can react via sn1 but primary can't? If we look at the sn1 mechanism, we form a carbocation intermediate. In the case of a tertiary haloalkane, this is a tertiary carbocation intermediate. Methyl groups have an inductive effect, meaning that the electrons of the methyl group are pushed slightly towards the positive carbon, which delocalises the charge slightly across the molecule, and as a result the carbocation is just about stable (which means that not a huge amount of energy is needed to get it into that point). In comparison, in a primary haloalkane, the hydrogen atoms do not have such an inductive effect, meaning that the positive charge is highly localised on the carbon and as a result it is extremely unstable and requires an immense amount of energy to form, which is why primary haloalkanes go via sn2.

    As mentioned by the other answer, secondary do both
 
 
 
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