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Homework Help: How to produce C5H11OH

  1. Jan 14, 2008 #1
    1. The problem statement, all variables and given/known data
    I was asked to list all isomers and to describe the production and 2 uses for this type of alcohol. (C5H11OH)

    2. Relevant equations

    3. The attempt at a solution

    I got all the isomers and 2 use for the alcohol but cannot seem to figure out how it is produced. If someone can tell me what I should be looking for I would be grateful.
  2. jcsd
  3. Jan 15, 2008 #2


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    So, tell us what alcohols you came up with!
  4. Jan 15, 2008 #3
    Preperation depends from isomer to isomer. One very common method is hydration of alkenes. It shall also help you with the isomers as it is as simple as just moving the double bond. However, it shall not help in all the isomers.

    another method is reduction of carbonyl compounds. This shall also help with a few isomers by selecting the respective isomeric aldehydes/ketones/carboxylic acids/esters.
  5. Jan 15, 2008 #4
    For the isomers I have:
    3-methyl 1-butanol
    2-methyl 1-butanol

    so how can I determine how these are produced?
  6. Jan 15, 2008 #5


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    Do you not understand how these alcohols are produced or how alcohols are produced in general?
  7. Jan 15, 2008 #6
    Well, I'm not sure exactly but as far as I know you need an alkene, water and H2SO4 to make them react.

    So I guess for 1-pentanol the reaction would be

    CH3-CH2-CH-CH-CH3 + H2O ----------> CH2-CH2-CH2-CH2-CH3

    (I cant figure out how to space[H2SO4] [OH] text yet so look at the brackets and align H2SO4 and OH)

    Is this correct?
    Last edited: Jan 15, 2008
  8. Jan 15, 2008 #7
    actually.. nope.. remember i said hydration of alkenes. You have an alkane here...

    EDIT: i think u mean that the CH=CH is a double bond [u've written CH-CH]. Well, on the other hand side, an alcohol is to be formed. How do you think water [itex]H_2O[/itex] can provide an [itex]OH[/itex] group and how could it attach to the hydrocarbon.

    HINT: There is high electron density over the pi-bonds.
    Last edited: Jan 15, 2008
  9. Jan 15, 2008 #8
    CH3-CH2-CH=CH-CH3+H20 ----------------> CH2-CH2-CH2-CH2-CH3

    (and this would also apply to 2 and 3 pentanol by just switching the location of OH)

    I hope that's right :confused:
  10. Jan 16, 2008 #9

    CH3CH2CH=CHCH3 + H2O -------> CH3CH2CH(OH)CH2CH3

    depending on where the OH from water attaches, CH3CH2CH2CH(OH)CH3 can also be formed.

    the alcohol you want is propan-2-ol, and you take pent-2-ene for this. you can also get propan-3-ol as you said.

    if you were to produce 2,2 dimethyl propan-1-ol, which alkene you would use? think of where the -OH should attach.

    the key is to visualize the molecules.

    are you aware of the mechanism of this type of reaction? electrophilic addition reaction says something to you?
  11. Jan 16, 2008 #10
    Electrophilic addition reaction doesnt mean anything to me but i think that I understand what you are saying. The Oh will take the place of the double bond so depending on the alkene is what would affect the OH.
  12. Jan 17, 2008 #11
    H2O will add itself to the double bond. one of the carbon receives an H atom, and the other receives the OH group.
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