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    hi
    i have a confusion can alchols react with NaHCo3 ,naoh,na2Co3
    i think no ....cuz like iv solved a couple of past papers and the answer is always that alcohols dont react i just wanted to make sure ..that they dont and the reasoning why they dont react with any of these sodium compound?
    thank u
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    (Original post by alyoan tariq)
    hi
    i have a confusion can alchols react with NaHCo3 ,naoh,na2Co3
    i think no ....cuz like iv solved a couple of past papers and the answer is always that alcohols dont react i just wanted to make sure ..that they dont and the reasoning why they dont react with any of these sodium compound?
    thank u
    No they don't. They react with Na metal though.
    Carboxylic acids react with all of those.
    The reason being that...well these are just not reactions which alcohols can undergo. Just learn the ones i the syllabus (oxidation, combustion, nucleophilic subs etc) and you should be fine.

    If a compound contains both acid and alcohol functional groups it will react bt only the acid part.
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    (Original post by InkRed)
    No they don't. They react with Na metal though.
    Carboxylic acids react with all of those.
    The reason being that...well these are just not reactions which alcohols can undergo. Just learn the ones i the syllabus (oxidation, combustion, nucleophilic subs etc) and you should be fine.

    If a compound contains both acid and alcohol functional groups it will react bt only the acid part.


    hey thank u so much !!!!
    that totally cleared up my confusion!!
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    (Original post by alyoan tariq)
    hi
    i have a confusion can alchols react with NaHCo3 ,naoh,na2Co3
    i think no ....cuz like iv solved a couple of past papers and the answer is always that alcohols dont react i just wanted to make sure ..that they dont and the reasoning why they dont react with any of these sodium compound?
    thank u
    These reagents are bases.

    Alcohols are not acidic enough to react with these bases, although they do enter into an equilibrium with NaOH even if there is no visible reaction. This is because NaOH is a strong enough base to abstract the proton from the alcohol.

    ROH + NaOH <==> RO- + H2O + Na+
 
 
 
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