The reaction of $2-$methylbuta$-1,3-$diene with $O_3$ followed by $Zn-H_2O$ gives:

Rank the following alkenes in order of decreasing heats of hydrogenation (largest first):
$1$. $1-$butene
$2$. $2-$methyl$-2-$butene
$3$. trans$-2-$butene
$4$. cis$-2-$butene

In the reaction sequence $CH_3CH = CHCH_3$ $\xrightarrow{O_3} A$ $\xrightarrow{H_2O/Zn} B$,the alkene forms product $B$. What is product $B$?

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

Appropriate reducing agent for the following conversion is:
$CH_2=CH-CH_2-CHO \to CH_3-CH_2-CH_2-CH_2OH$

The correct order of reactivity of hydrogen halides with propene is