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Sa se scrie formulele de structura si denumirile alcoolilor monohidroxilici izomeri care contin 21.26% O.

Răspuns :

Mirceavaman
Alcool monohidroxilic: R-OH
Malcool=R+16+1=R+17 g/mol
R+17...........100%
16.............21,62%
R+17=1600/21,62=74g/mol
R=74-17=57g/mol
R=MCnH2n+1=12n+2n+1=14n+1=57
14n=56 => n=4
f.m: C4H10O
CH3-CH2-CH2-CH2-OH
CH3-CH(CH3)-CH2-OH
CH3-CH(OH)-CH2-CH3
CH3-C(CH3)(OH)-CH3

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