What is the order of reactivity of halogen acids $(HX)$ with $C_2H_5OH$?

$CH_3 CH_2 OH \underset{\text{step-}1}{\stackrel{Cl_2}{\longrightarrow}} CH_3 CHO$ $\underset{\text{step-}2}{\stackrel{3 Cl_2}{\longrightarrow}} Cl_3 CCHO$
In the above reactions,the role of $Cl_2$ in step-$1$ and step-$2$ respectively is:

Select the best method for the preparation of the following compounds:
(Image shows a mixture of trans$-2-$methylcyclohexanol enantiomers)
($MCPBA$ = meta-chloroperoxybenzoic acid)

$ROH + HX \to RX + H_2O$
In the above reaction,the reactivity order of hydrogen halides is

An optically active alcohol of formula $C_9H_{12}O_2$ produced the following compound when refluxed with $KMnO_4$. What is the structure of $(A)$? The original compound showed these properties also:
$\mathop {C_9H_{12}O_2}\limits_{(A)} \xrightarrow{Na} H_2 \text{ liberated}$
$\mathop {C_9H_{12}O_2}\limits_{(A)} \xrightarrow{Br_2} \text{ no rapid reaction}$
$\mathop {C_9H_{12}O_2}\limits_{(A)} \xrightarrow[{cool}]{CrO_3 / H^{+}} C_9H_8O_3$

In $CH_3CH_2OH \xrightarrow{X} CH_2 = CH_2 + H_2O$ at $350 \ ^oC$; $'X'$ is