$(A) (C_7H_{14}) \xrightarrow[Zn]{O_3} (B) + (C)$
Compound $(A)$ exists as geometrical isomers and $(B)$ gives Cannizzaro reaction. $(A)$ will be:

The product obtained by the reaction of propene with $HI$ in the presence of peroxide is.....

$CH_3-CH=CH-CH=CH_2 \xrightarrow[Zn]{O_3 / H_2O}$ Which will not form?

The major products '$A$' and '$B$',respectively,are

$but-2-yne$ 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 $A$.

Write the main product of the given reactions:
$(1)$ $CH_3CH = CHCH_3 + HBr \to$
$(2)$ $CH_3-C(CH_3)=CH_2 + HBr \to$