In the following reaction,$P$ gives two products $Q$ and $R$ each in $40 \,\%$ yield. If the reaction is carried out with $420 \,mg$ of $P$,the reaction yields $108.8 \,mg$ of $Q$. The amount of $R$ produced in the reaction is closest to $....\, mg$.

Consider the following reactions:
$X + HCl \xrightarrow{\text{Anhydrous } AlCl_3} C_2H_5Cl$ (Addition)
$Y \xrightarrow{\text{Anhydrous } ZnCl_2 / HCl} C_2H_5Cl$ (Substitution)
$Y$ can be converted to $X$ on heating with $...$ at $...$ temperature:

Alkenes react with water in the presence of acid as a catalyst to form alcohols: $CH_3-CH=CH_2 + H_2O \xrightarrow{H^{+}} CH_3-CH(OH)-CH_3$. The reaction takes place in accordance with:

Identify the type of the given reactions:
$(a) \ CH_3CH_2OH \xrightarrow[\Delta]{\text{Conc. } H_2SO_4} CH_2 = CH_2 + H_2O$
$(b) \ CH_2BrCH_2Br + Zn \xrightarrow{\Delta} CH_2 = CH_2 + ZnBr_2$
$(c) \ CH_3CHBr - CH_2Br + Zn \xrightarrow{\Delta} CH_3CH = CH_2 + ZnBr_2$
$(d) \ RC \equiv CR' + H_2 \xrightarrow{Na, \text{liquid } NH_3} RCH = CHR'$
$(e) \ CH_3CH_2Cl + KOH \xrightarrow{\text{ethanol, } KOH} CH_2 = CH_2 + KCl + H_2O$

Explain the Anti-Markovnikov rule or peroxide effect (Kharasch effect) or free radical addition reaction with an example.

Identify $Z$ in the following series: $C_2H_5I$ $\xrightarrow{Alco. KOH} X$ $\xrightarrow{Br_2} Y$ $\xrightarrow{KCN} Z$