$HBr$ reacts with $CH_2=CH-OCH_3$ under anhydrous conditions at room temperature to give:

The reaction of carbon tetrachloride with water vapor at $500\,^oC$ gives .................

Arrange the following compounds in order of decreasing rate of hydrolysis for $S_{N}1$ reaction:
$(I) \ C_6H_5-CH_2-Br$
$(II) \ p-CH_3-C_6H_4-CH_2-Br$
$(III) \ p-CH_3CH_2-C_6H_4-CH_2-Br$
$(IV) \ p-(CH_3)_2CH-C_6H_4-CH_2-Br$

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A):$ $R-I$ undergoes $S_N2$ reaction faster than $R-Cl$.
Reason $(R):$ Iodine is a better leaving group because of its large size.
In the light of the above statements,choose the correct answer from the options given below:

Assertion $(A)$: $S_N1$ hydrolysis of optically active $2-$bromooctane results in the formation of $(\pm)-$octan$-2-$ol.
Reason $(R)$: The reaction proceeds through a planar carbocation which can be attacked by the nucleophile from either side.
The correct answer is

$A$ solution of $(\pm)-2$-chloro-$2$-phenylethane in toluene is mixed slowly with a small amount of $SbCl_5$. The reaction proceeds due to the formation of: