Compound $X$ on heating with $Zn$ dust gives compound $Y$,which on treatment with $O_3$ followed by reaction with $Zn$ dust and $H_2O$ gives propionaldehyde. The structure of $X$ is:

$A$ hydrocarbon $X$ adds one mole of hydrogen to give another hydrocarbon and decolourises bromine water. $X$ reacts with $KMnO_4$ in the presence of acid to give two moles of the same carboxylic acid. The structure of $X$ is:

Identify the product of the following reaction: $CH_3-CH=CH-CH_3 \xrightarrow[(ii) H_2O, Zn]{(i) O_3} \text{Product}$

$but-2-yne$ is reacted separately with one mole of $H_2$ as shown below:
$\underline{ B } \xleftarrow[\text{liq } NH_3]{Na} CH_3-C \equiv C-CH_3 \xrightarrow[\Delta]{Pd/C} \underline{ A }$
Identify the incorrect statements from the options given below:
$A.$ $A$ is more soluble than $B$.
$B.$ The boiling point and melting point of $A$ are higher and lower than $B$ respectively.
$C.$ $A$ is more polar than $B$ because dipole moment of $A$ is zero.
$D.$ $Br_2$ adds easily to $B$ than $A$.

The compound $CH_3-C(CH_3)=CH-CH_3$ on reaction with $NaIO_4$ in the presence of $KMnO_4$ gives:

The reaction of an alkene $X$ with bromine produces a compound $Y$,which has $22.22 \% \ C$,$3.71 \% \ H$ and $74.07 \% \ Br$. The ozonolysis of alkene $X$ gives only one product. The alkene $X$ is,
[Given,atomic mass of $C = 12$; $H = 1$; $Br = 80$ ]