a,
$n_{H_2}=\frac{1,12}{22,4}=0,05 mol$
$\Rightarrow n_X=0,05.2=0,1 mol$
$\Rightarrow M_X=\frac{6}{0,1}=60= 14n+18$
$\Leftrightarrow n=3$
Vậy CTPT A là $C_3H_8O$
b,
Oxi hoá X thu được anđehit nên X là ancol bậc I.
CTCT: $CH_3-CH_2-CH_2OH$ (propan-1-ol)
c,
Y có thể là propan-2-ol hoặc etylmetyl ete.
$CH_3-CH_2-CH_2OH+HCl \to CH_3CH_2CH_2Cl+ H_2O$
$CH_3-CH_2-CH_2Cl+ KOH \buildrel{{C_2H_5OH, t^o}}\over\to CH_3-CH=CH_2+ KCl+H_2O$
* Điều chế propan-2-ol:
$CH_3-CH=CH_2+H_2O\buildrel{{H^+, t^o}}\over\to CH_3-CHOH-CH_3$
* Điều chế etylmetyl ete:
$CH_3-CH=CH_2+H_2\buildrel{{Ni, t^o}}\over\to CH_3-CH_2-CH_3$
$CH_3-CH_2-CH_3 \buildrel{{\text{cracking}}}\over\to CH_4+C_2H_4$
$CH_4+Cl_2\buildrel{{t^o}}\over\to CH_3Cl+ HCl$
$CH_3Cl+NaOH \buildrel{{t^o}}\over\to CH_3OH+NaCl$
$C_2H_4+H_2O\buildrel{{t^o, H^+}}\over\to C_2H_5OH$
$CH_3OH+C_2H_5OH\buildrel{{140^oC, H_2SO_4}}\over\to CH_3-O-C_2H_5+H_2O$
