Chemical Reactions of Haloalkanes

What is the order of reactivity of following compounds towards  $S_N{}^1$ substitutions reaction.
(i) 

(ii) 

(iii) 

What is the order of reactivity of following compounds towards  ${\mathrm{S}}_{\mathrm{N}}1$ substitution reaction.

Which one of the following two substances undergoes ${\mathrm{S}}_{\mathrm{N}}1$ reaction faster?

(i)    or   (ii) 

(CH₃)₃CMgCl on reaction with D₂O produces –

An S_N2 reaction at an asymmetric carbon of a compound always gives:

A solution of (+) -2-chloro-2-phenylethane in toluene racemises slowly in the presence of small amount of  $SbC{l}_5$ , due to the formation of –

A solution of (+) -2-chloro-1-phenylethane in toluene racemises slowly in the presence of small amount of  ${\mathrm{SbCl}}_5$, due to the formation of:

${\left({\mathrm{CH}}_3\right)}_3\mathrm{CMgCl}$ on treatment with ${\mathrm{D}}_2\mathrm{O}$ produces:

n–propyl bromide on treatment with ethanolic potassium hydroxide produces

The major product of the following reaction is

The product of following reaction is –

${\text{C}}_2{\text{H}}_5{\text{O}}^{-}+\text{C}{\text{H}}_3\text{C}{\text{H}}_2\text{I}\to$

The reaction of $4$-methyloct-$1$-ene $(\mathrm{P},2.52 \mathrm{g})$ with $\mathrm{HBr}$ in the presence of ${\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CO}\right)}_2{\mathrm{O}}_2$ gives two isomeric bromides in a $9: 1$ ratio, with a combined yield of $50\%$. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with aq. ${\mathrm{K}}_2{\mathrm{CO}}_3$ to give a non-ionic product $\mathrm{S}$ in $100\%$ Yield. The mass (in $\mathrm{mg}$ ) of $\mathrm{S}$ obtained is

[Use molar mass (in ${\mathrm{gmol}}^{-1}$ ): $\mathrm{H}=1,\mathrm{C}=12, \mathrm{N}=14,\mathrm{Br}=80$ ]

Which of the following is/are the property(ies) of ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}$?

What is the configuration of product $\left(\mathrm{P}\right)$?

Product (B) is:

In the reaction, ${\mathrm{CH}}_3\mathrm{Br} \stackrel{\mathrm{KCN}}{⟶} \mathrm{A} \stackrel{{\mathrm{H}}^{+}}{\to } \mathrm{B} \stackrel{{\mathrm{LiAlH}}_4}{⟶} \mathrm{C}$

The product C is.