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    Basically I have one of the worst chemistry teachers ever. Doesn't explain anything and doesn't want anyone to succeed. We have been given an assignment to do and I don't have a clue how to do it and no one is willing to help. Listed Below is the assignment criteria.Predict the correct products with the use of equations fromthe following reactions with formula equations and FULL structural form.
    1) Pentanoic Acid + NaBH4
    2) Propanol + KCN
    3) Oxidation of butan-2-ol
    4) Esterification of butylpentanoate
    5) Production of nylon 6,6
    6) Hydrolysis of ethanoyl chloride
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    (Original post by NS_X1)
    Basically I have one of the worst chemistry teachers ever. Doesn't explain anything and doesn't want anyone to succeed. We have been given an assignment to do and I don't have a clue how to do it and no one is willing to help. Listed Below is the assignment criteria.Predict the correct products with the use of equations fromthe following reactions with formula equations and FULL structural form.
    1) Pentanoic Acid + NaBH4
    2) Propanol + KCN
    3) Oxidation of butan-2-ol
    4) Esterification of butylpentanoate
    5) Production of nylon 6,6
    6) Hydrolysis of ethanoyl chloride
    1. Nothing happens.
    2. Nothing happens.
    3. Get butanone (no number necessary - can you see why?)
    4. It's already an ester!
    5. http://en.wikipedia.org/wiki/Nylon_6...ring_Nylon_6-6
    6. Water + acid chloride --> acid + HCl

    3 of these questions seem pretty... wrong.
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    (Original post by illusionz)
    1. Nothing happens.
    2. Nothing happens.
    3. Get butanone (no number necessary - can you see why?)
    4. It's already an ester!
    5. http://en.wikipedia.org/wiki/Nylon_6...ring_Nylon_6-6
    6. Water + acid chloride --> acid + HCl

    3 of these questions seem pretty... wrong.
    Surely the NaBH4 would reduce pentanoic acid to a pentandiol, as NaBH4 (IIRC) reduces polar bonds, and the C=0 Bond is polar? For example, I was taught that NaBH4 could be used to reduce the CN bond in 2-hydroxypropan-2-nitrile to form 2-hydroxy-2-methylpropylamine, so surely it could reduce the (even more) polar C=O bond?
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    (Original post by AtomicMan)
    Surely the NaBH4 would reduce pentanoic acid to a pentandiol, as NaBH4 (IIRC) reduces polar bonds, and the C=0 Bond is polar? For example, I was taught that NaBH4 could be used to reduce the CN bond in 2-hydroxypropan-2-nitrile to form 2-hydroxy-2-methylpropylamine, so surely it could reduce the (even more) polar C=O bond?
    NaBH4 is a pretty weak reducing agent. It will only reduce ketones or more reactive carbonyls. It will not reduce esters (well, it will reduce esters but very slow), amides or carboxyllic acids. The reason for this is that the O/N in an ester/amide/acid donates electron density onto the carbon of the C=O and so it is less polar and much less reactive.
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    (Original post by illusionz)
    NaBH4 is a pretty weak reducing agent. It will only reduce ketones or more reactive carbonyls. It will not reduce esters (well, it will reduce esters but very slow), amides or carboxyllic acids. The reason for this is that the O/N in an ester/amide/acid donates electron density onto the carbon of the C=O and so it is less polar and much less reactive.
    Ah ok, thankyou. Does this same reason apply for why LiAlH4 cannot reduce carboxylic acids/amids/esters?
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    (Original post by AtomicMan)
    Ah ok, thankyou. Does this same reason apply for why LiAlH4 cannot reduce carboxylic acids/amids/esters?
    LiAlH4 will reduce esters and amides without issue, and acids slowly.
 
 
 
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