Which of the following compounds would undergo $S_N 1$ reaction faster and why?
Compound (B) will give $S_N 1$ reaction faster than compound $(A)$ because $S_N 1$ reaction depends upon the stability of carbocation. Benzyl chloride on ionisation gives $\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{H}_2$ carbocation which is resonance srabilised while the carbocation obtained from compound $(A)$ is not stabilised by resonance.
Allyl chloride is hydrolysed more readily than n-propyl chloride. Why?
As we know that $\mathrm{S}_{\mathrm{N}} 1$ mechanism depends upon the stability of carbocation. Allyl chloride on hydrolysis gives resonance stabilised carbocation while no resonance is observed in the carbocation formed by $n$-propyl chloride.
Hence, allyl chloride undergoes hydrolysis much faster than $n$-propyl chloride.
Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?
Grignard reagents are highly reactive and react with water to give corresponding hydrocarbons.
How do polar solvents help in the first step in $\mathrm{S}_{\mathrm{N}}{ }^1$ mechanism?
Polar solvents help in the first step in $\mathrm{S}_{\mathrm{N}} 1$ mechanism because leaving group and carbocation both are stabilised by polar solvent. Polarity of a solvent depends upon the value of dielectric constant. Higher the value of dielectric constant, higher will be the polarity of the solvent, faster will be the rate of $S_N 1$ mechanism. These polar solvents can work as a nucleophile and stabilise the carbocation as follows
Write a test to detect the presence of double bond in a molecule.
Presence of double bond in a molecule is detected by following two methods:
(i) $\mathrm{Br}_2$ in $\mathrm{CCl}_4$ test ${\mathrm{When} \mathrm{Br}_2 / \mathrm{CC}_4}^2$ is added unsaturated compound then orange colour of bromine disappears and dibromoderivative is formed. (colourless).
(ii) Bayer's test When alkaline solution of $\mathrm{KMnO}_4$ is added to the solution of unsaturated compound then its pink colour disappears due to the formation of dihydroxy derivative.
