Answer :
Explanation:
Let's solve the reaction step by step:
1. **Starting Compound**: CH3-CH-CH3 with a Cl attached to the second carbon (CH3-CH(Cl)-CH3).
2. **Reagent**: alc. KOH (alcoholic potassium hydroxide).
- This reagent is known for causing an elimination reaction (dehydrohalogenation), where the HCl (hydrogen chloride) is removed.
- The product of this reaction is an alkene.
**Reaction**:
\[
CH3-CH(Cl)-CH3 \xrightarrow{\text{alc. KOH}} CH3-CH=CH2 + HCl
\]
**Compound A**: CH3-CH=CH2 (propene).
3. **Next Reaction**: A (propene) is treated with HBr in the presence of peroxide.
- HBr and peroxide follow the anti-Markovnikov rule (addition of HBr across the double bond with the bromine attaching to the less substituted carbon).
**Reaction**:
\[
CH3-CH=CH2 \xrightarrow{\text{HBr, peroxide}} CH3-CH2-CH2Br
\]
**Compound B**: CH3-CH2-CH2Br (1-bromopropane).
4. **Final Reaction**: B (1-bromopropane) is treated with NaI in dry ether.
- This reaction is a nucleophilic substitution (SN2) where iodine replaces bromine.
**Reaction**:
\[
CH3-CH2-CH2Br \xrightarrow{\text{NaI, dry ether}} CH3-CH2-CH2I + NaBr
\]
**Compound C**: CH3-CH2-CH2I (1-iodopropane).
**Summary**:
- **A**: CH3-CH=CH2 (propene).
- **B**: CH3-CH2-CH2Br (1-bromopropane).
- **C**: CH3-CH2-CH2I (1-iodopropane).
Explanation:
let's solve the problem step by step in a shorter way:
The given compound is 2-chloropropane.
alc. KOH is an eliminating reagent i.e. used to do elimination reaction.
so A. is CH3-CH=CH2 (propene)
NOW,
in 2nd reaction ,
peroxide with HBr is used i.e. it will perform antimarkownikoff rule and Br will get attached to -Ch2 grp.
B. is CH3-CH2-CH2-Br (1-bromopropane)
NOW,
In 3rd reaction,
it is Finkelstein reaction, here Halogen gets replaced by Iodine .
so, C is CH3-CH2-CH2-I (1-iodopropane)
SUMMARY
A- CH3-CH=CH2(Propene)
B-CH3-CH2-CH2-Br(1-bromopropane)
C-CH3-CH2-CH2-I (1-iodopropane)