Sure, let's go through each of the given reactions step-by-step to identify their products:
### Reaction i: [tex]\(CH_4 + Cl_2 \xrightarrow{h \nu}\)[/tex]
This reaction is a photochemical halogenation of methane. When methane ([tex]\(CH_4\)[/tex]) is exposed to chlorine ([tex]\(Cl_2\)[/tex]) under UV light (denoted by [tex]\(h \nu\)[/tex]), a substitution reaction occurs where one hydrogen atom in methane is replaced by a chlorine atom, producing methyl chloride ([tex]\(CH_3Cl\)[/tex]) and hydrogen chloride ([tex]\(HCl\)[/tex]).
Product:
[tex]\[ CH_3Cl + HCl \][/tex]
### Reaction ii: [tex]\(CH_3-CH=CH_2 \xrightarrow{HCl}\)[/tex]
This reaction is an electrophilic addition of hydrogen chloride (HCl) to propene ([tex]\(CH_3-CH=CH_2\)[/tex]). According to Markovnikov's rule, the hydrogen atom from HCl will add to the carbon of the double bond that has the greater number of hydrogen atoms, and the chlorine will add to the carbon with fewer hydrogen atoms. This results in the formation of 2-chloropropane ([tex]\(CH_3-CHCl-CH_3\)[/tex]).
Product:
[tex]\[ CH_3-CHCl-CH_3 \][/tex]
### Reaction iii: [tex]\(CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}}\)[/tex]
In the presence of peroxide, the addition of hydrogen bromide (HBr) to propene ([tex]\(CH_3-CH=CH_2\)[/tex]) follows the anti-Markovnikov rule. This means that the bromine atom adds to the carbon of the double bond that has the greater number of hydrogen atoms, while the hydrogen adds to the carbon with fewer hydrogen atoms. This results in the formation of 1-bromopropane ([tex]\(CH_3-CH_2-CH_2Br\)[/tex]).
Product:
[tex]\[ CH_3-CH_2-CH_2Br \][/tex]
### Reaction iv: [tex]\(CH_2=CH-CH_3 + Br_2 \xrightarrow{CCl_4}\)[/tex]
This reaction is an electrophilic addition of bromine (Br_2) to propene ([tex]\(CH_2=CH-CH_3\)[/tex]) in the presence of carbon tetrachloride (CCl_4) as a solvent. The bromine molecules add across the double bond, resulting in the formation of 1,2-dibromopropane ([tex]\(CH_2Br-CHBr-CH_3\)[/tex]).
Product:
[tex]\[ CH_2Br-CHBr-CH_3 \][/tex]
These are the products of the given reactions:
i. [tex]\(CH_3Cl + HCl\)[/tex]
ii. [tex]\(CH_3-CHCl-CH_3\)[/tex]
iii. [tex]\(CH_3-CH_2-CH_2Br\)[/tex]
iv. [tex]\(CH_2Br-CHBr-CH_3\)[/tex]
