In the alkaline hydrolysis of a tertiary alkyl halide by aqueous alkali,if the concentration of the alkali is doubled,then the reaction rate:

Observe the following reactions:
$1$. $C_6H_5-CH=CH_2 + HBr \rightarrow X$
$2$. $C_6H_5-C(CH_3)=CH_2 + HBr \rightarrow Y$
$3$. $C_6H_5-CH=CH_2 + HBr \xrightarrow{(C_6H_5COO)_2} Z$
The correct order of reactivity of $X, Y, Z$ towards $S_N1$ reaction is:

Which is the correct order of the following alkyl halides for an $S_{N}2$ reaction?

The major product in the reaction is:

Consider the following four reactions: $(I) C_6H_5-CHCl-CH_3 \xrightarrow{95\% \text{ acetone}, 5\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(II) C_6H_5-CHCl-CH_3 \xrightarrow{90\% \text{ acetone}, 10\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(III) C_6H_5-CHCl-CH_3 \xrightarrow{80\% \text{ acetone}, 20\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(IV) C_6H_5-CHCl-CH_3 \xrightarrow{100\% \text{ water}} C_6H_5-CH(OH)-CH_3$ Arrange these reactions in decreasing order of the proportion of inverted product and select the correct answer from the codes given below:

Explain the factors favouring $S_{N}2$ reaction.