In the following sequence of reactions, identify the final product. Cyclopentance `overset(Cl_2//hv)rarr(I)overset(alc.KOH,Delta)rarr(II) overset((i) O_3)rarr(III)`

Options

A. pentanal
B. pentane -2- one
C. pentan -3-one
D. pentane -1,3,-dial

Answer

To solve the given problem, we will break down the reactions step by step.

### Step 1: Chlorination of Cyclopentane
- Cyclopentane (C5H10) is treated with Cl2 in the presence of light (hv). This leads to a free radical substitution reaction.
- The chlorine radical (Cl·) abstracts a hydrogen atom from cyclopentane, forming chlorocyclopentane (C5H9Cl) and another radical.

**Product after Step 1:** Chlorocyclopentane (C5H9Cl)

### Step 2: Elimination Reaction with Alcoholic KOH
- Chlorocyclopentane is then treated with alcoholic KOH and heated (Δ).
- In this step, KOH acts as a base and facilitates the elimination of HCl through a beta-elimination reaction. The hydrogen is removed from the beta carbon (the carbon adjacent to the one bearing the chlorine).
- This results in the formation of cyclopentene (C5H8).

**Product after Step 2:** Cyclopentene (C5H8)

### Step 3: Ozonolysis of Cyclopentene
- Cyclopentene undergoes ozonolysis (reaction with O3), which cleaves the double bond.
- In reductive ozonolysis, the double bond is broken, and each carbon that was part of the double bond is converted to a carbonyl group (aldehyde or ketone).
- Since cyclopentene is a cyclic compound, the cleavage will yield two aldehyde groups.

**Final Product after Step 3:** Pentane-1,5-dial (C5H10O2)

### Summary of the Steps:
1. **Chlorination of Cyclopentane** → Chlorocyclopentane (C5H9Cl)
2. **Elimination with Alcoholic KOH** → Cyclopentene (C5H8)
3. **Ozonolysis of Cyclopentene** → Pentane-1,5-dial (C5H10O2)

### Final Answer:
The final product of the sequence of reactions is **Pentane-1,5-dial**.

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