Hex$-4-$en$-2-$ol on treatment with $PCC$ gives $'A'.$ $'A'$ on reaction with sodium hypoiodite gives $'B',$ which on further heating with soda lime gives $'C'.$ The compound $'C'$ is $....... .$

The product $(B)$ is:

The above compounds can be differentiated by the following reagent:

Match List-$I$ with List-$II$.

List-$I$	List-$II$
$1.$ Ammoniacal $AgNO_3$	$a.$ Primary amine
$2.$ $HIO_4$	$b.$ Aldehyde
$3.$ Alkaline $KMnO_4$	$c.$ Vicinal-$OH$
$4.$ Chloroform $NaOH$	$d.$ Double bond

Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$ : $Image-I$ can be easily reduced using $Zn-Hg / HCl$ to $Image-II$.
Reason $R : Zn-Hg / HCl$ is used to reduce carbonyl group to $-CH_2-$ group.
In the light of the above statements,choose the correct answer from the options given below:

Identify the correct statements about $Z$ in the following reaction sequence:
$C_2H_5NH_2$ $\xrightarrow[NaNO_2 / HCl]{} X$ $\xrightarrow{H_2O} Y$ $\xrightarrow{Cu / 573 \ K} Z$
$I$. $Z$ is an aldehyde
$II$. $Z$ undergoes Cannizzaro reaction
$III$. $Z$ gives iodoform test
$IV$. $Z$ does not give test with Tollens' reagent