Dehydrogenation of $CH_3-CH(OH)-CH_3$ gives

The product obtained in the reaction is:

$CH_2=CH-CH(OH)-CH_2-CH_2OH \xrightarrow{MnO_2} A$. $A$ is:

Iodoform is formed on warming $I_2$ and $NaOH$ with:

In the following sequence of reactions,
$C_{3}H_{6}$ $\xrightarrow{H^{+}/H_{2}O} A$ $\xrightarrow[dil. KOH]{KIO} B + C$
The compounds $B$ and $C$ respectively are :

Arrange the following alcohols in decreasing order of the ease of ionization under acidic conditions.