The conversion of ethyl bromide to ethyl iodide using sodium iodide and dry acetone is known as:

The correct order for $E_2$ reaction with alc. $KOH$ will be

$A$ compound $A$ $(C_5H_{10}Cl_2)$ on hydrolysis gives $C_5H_{10}O$ which reacts with $NH_2OH$,forms iodoform,but does not give Fehling's test. $A$ is:

Allyl chloride is hydrolysed more readily than $n-$propyl chloride. Why?

Reaction of $(R)-2$-butanol with $p$-toluenesulphonyl chloride in pyridine followed by $LiBr$ gives:

Arrange the following compounds in order of decreasing rate of hydrolysis for $S_N1$ reaction:
$(i)$ $C_6H_5CH_2Br$
(ii) $p-CH_3-C_6H_4-CH_2Br$
(iii) $p-CH_3CH_2-C_6H_4-CH_2Br$
(iv) $p-(CH_3)_2CH-C_6H_4-CH_2Br$