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Equation showing formation of a compound

  1. Mar 21, 2006 #1
    Write balanced equations to show the formation of each of the following compunds:
    a) butyl propanote
    b) propyl methanote


    I am doing a horribly constructed independent learning course, they offered absolutly no information on this topic, and then asked me to solve this question..
    I really don't even know how to start it. :grumpy:
    any help would be much appreciated!
     
  2. jcsd
  3. Mar 21, 2006 #2
    2-butene and 1-butene reactions with HCl

    This is another question from my independent learning course:

    When 2-butene reacts with hydrogen chloride gas, only one product is detected, whereas when 1-butene reacts similarily, two products are usually found. Explain this.




    I don't understand why they would say, "usually" found.

    Does the answer have something to do with the fact that the second carbon in 1-butene is chiral?

    I was not given very much background information on this topic, only a very brief introduction. The only reason I even know what 'chiral' means, is because my brother, told me that might be one of the reasons.
     
  4. Mar 21, 2006 #3

    siddharth

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    Hi and welcome to PF christinaa! You've come to the right place. The folks here will be happy to assist you with the difficulties you have in your course.

    Both the above compounds are esters. One of the simplest ways to prepare esters is by the esterfication reaction. Can you see how you can get each of the products now?

    For your second question, have you learnt about the rearrangement of Carbocations? If so, can you see how to apply it here?

    BTW, what level is your independent learning course and what is the text you are following?
     
    Last edited: Mar 21, 2006
  5. Mar 21, 2006 #4
    thanx

    My independent learning course is at an academic 12 level, and the textbook is Foundations of Chemistry, Second Addition, (Harcourt)

    I have not been introduced to the rearrangement of Carbocations in this course. As well we have not been introduced to esterfication, but i will look both these topics up, in order to better understand these two questions.

    Thank-you for your help :smile:
     
  6. Mar 22, 2006 #5

    siddharth

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    If you have access to a local library, you might want to look at Organic Chemistry (6th Edition) by Morrison and Boyd. The book is an excellent introduction to Organic Chemistry.
     
  7. Mar 23, 2006 #6
    :)

    thank-you for the suggestion i will definitely look for that textbook.

    For one of the questions I have to draw structural formulas for all the alcohols with the molecular formula of C5H11OH, and give two uses for each type of alcohol, as well as write how they were prepared.

    When writing how they were prepared do i just put,
    '_____ was prepared by reacting with sulphuric acid solution'
    '_____ was prepared by oxidation'
    " by hydration...

    Also are there only 8 alcohols with the structual formula of (C5H11OH) ?
    and, how can I find the uses for the different types of alcohol, i looked then up but i didn't find anything about them. :uhh:
     
  8. Mar 23, 2006 #7
    a half-cell balancing question

    i really need help with this question. There were about 7 equations that i have to balance by using the half-cell method. I have already finished most of them fairly easily, but these two, i am stuck on. I have been rewriting these 2 equations for the last 7 hours :eek: , but i cannnot succesfully balance them. I don't know what I'm doing wrong..

    a) SO3(-2) + MnO4(-) + H(+) <--> Mn(+2) + SO4(-2) + H2O(l)
    so far i got this:

    its an acid reaction,

    =SO3(-2) --> SO4(-2) (was oxidized as 2 electrons were removed)
    =SO3(-2) + H2O --> SO4(-2) + 2H(+)
    = SO3(-2) + H2O --> S(-2) + 2H(+) + 2e-

    = MnO4(-) --> Mn(2+) ( was reduced as 5 electrons were added)
    = 5e(-) + MnO4(-) + 8H(+) -->Mn(+2) + 4H2O

    b) Cl2(g) + OH(-) <--> Cl(-) + ClO3(-) + H2O(l)

    its a basic reaction:

    Cl2 --> Cl(-) + ClO(-3) -- (reduced, 2 electrons were added)
    OH(-) --> H2O -- (oxidized, 1 electron was removed )
     
  9. Mar 25, 2006 #8

    siddharth

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    It's better if you write the complete reaction, and if possible, the mechanism as well.

    From

    5e(-) + MnO4(-) + 8H(+) -->Mn(+2) + 4H2O (-I)
    and
    SO3(-2) + H2O --> SO4(-2) + 2H(+) + 2e(-) (-II)

    You have to multiply the equations so that the same number of electrons are involved in both and then add them up

    These links should help
    http://www.chemguide.co.uk/inorganic/redox/equations.html
    http://www.chemguide.co.uk/inorganic/redox/equations2.html#top
     
  10. Mar 26, 2006 #9
    thanx

    Thank-you for your help. The website was ver y helpful. These are the two answers I got. I am not sure about the second one, could you check to see if its right.


    a) SO3(-2) + MnO4(-) + H(+) <--> Mn(+2) + SO4(-2) + H2O(l)

    =5H2O + 5SO3-2 + 2MnO4(-) + 16H+ <-> 8H2O + 2Mn(+2) + 10H+ + 5S-2


    b) Cl2(g) + OH(-) <--> Cl(-) + ClO3(-) + H2O(l)

    =2Cl + 4OH(-) + 6HOH(-) --> 6H2O + 2Cl(-) + 2ClO(-3) + 2HOH(-)
     
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