$n_{H_2}= \frac{0,46}{22,4}= 0,02 mol$
Ancol có dạng $C_nH_{2n+1}OH$
$2C_nH_{2n+1}OH+ 2Na \to 2C_nH_{2n+1}ONa+ H_2$
=> $n_X= 0,04 mol$
=> $M_X= \frac{3,7}{0,04}= 92,5= 14n+18$
$\Leftrightarrow n= 5$
=> CTPT: $C_5H_{12}O$
CTCT:
(1) $CH_2OH-CH_2-CH_2-CH_2-CH_3$ (pentan-1-ol)
(2) $CH_3-CHOH-CH_2-CH_2-CH_3$ (pentan-2-ol)
(3) $CH_3-CH_2-CHOH-CH_2-CH_3$ (pentan-3-ol)
(4) $CH_2OH-CH(CH_3)-CH_2-CH_3$ (2-metylbutan-1-ol)
(5) $CH_3-COH(CH_3)-CH_2-CH_3$ (2-metylbutan-2-ol)
(6) $CH_3-CH(CH_3)-CHOH-CH_3$ (3-metylbutan-2-ol)
(7) $CH_3-CH(CH_3)-CH_2-CH_2OH$ (3-metylbutan-1-ol)
(8) $CH_2OH-C(CH_3)_3$ (2,2-đimetylpropan-1-ol)
