In the following sequence of the reactions :
`CH_(3)CH_(2)CH_(2)I overset(KOH(Alc.))rarr A overset((i) Br_(2))underset((ii) alc. KOH)rarr B overset(Na//NH_(3))rarr C`
The end product C is :

Options

A. Alkene
B. Alkanol
C. Alkyne
D. Alkyl amine

Explanation

To determine the end product C from the given sequence of reactions starting with iodopropane (CH₃CH₂CH₂I), we will go through each step methodically.

### Step 1: Reaction with Alcoholic KOH
- **Starting Compound:** CH₃CH₂CH₂I (1-iodopropane)
- **Reagent:** KOH (alcoholic)
- **Type of Reaction:** Dehydrohalogenation
- **Process:** The alcoholic KOH will facilitate the elimination of HI (hydrogen iodide) from the iodopropane. This is a beta-elimination reaction where the iodine (I) is removed from the alpha carbon and a hydrogen (H) is removed from the beta carbon.
- **Product A:** The product after this step will be propene (CH₃CH=CH₂).

### Step 2: Reaction with Br₂
- **Starting Compound:** Product A (CH₃CH=CH₂)
- **Reagent:** Br₂
- **Type of Reaction:** Electrophilic addition
- **Process:** The double bond in propene will react with bromine (Br₂), resulting in the addition of bromine across the double bond.
- **Product B:** The product after this step will be 1,2-dibromopropane (CH₃CHBrCH₂Br).

### Step 3: Reaction with Alcoholic KOH
- **Starting Compound:** Product B (CH₃CHBrCH₂Br)
- **Reagent:** KOH (alcoholic)
- **Type of Reaction:** Dehydrohalogenation
- **Process:** Again, alcoholic KOH will facilitate the elimination of HBr. In this case, two HBr molecules will be eliminated (one from each of the brominated carbons), resulting in the formation of a triple bond.
- **Product C:** The product after this step will be 1-butyne (CH₃C≡CCH₃).

### Step 4: Reaction with Sodium in NH₃
- **Starting Compound:** Product C (CH₃C≡CCH₃)
- **Reagent:** Sodium (Na) in liquid ammonia (NH₃)
- **Type of Reaction:** Reduction
- **Process:** Sodium in liquid ammonia will reduce the alkyne to a cis or trans alkene. The predominant product will be trans-alkene due to the mechanism of the reduction.
- **Final Product:** The final product will be trans-2-butene (CH₃CH=CHCH₃).

### Summary of the Reaction Sequence:
1. CH₃CH₂CH₂I + KOH (alc.) → CH₃CH=CH₂ (Propene)
2. CH₃CH=CH₂ + Br₂ → CH₃CHBrCH₂Br (1,2-Dibromopropane)
3. CH₃CHBrCH₂Br + KOH (alc.) → CH₃C≡CCH₃ (1-Butyne)
4. CH₃C≡CCH₃ + Na/NH₃ → trans-2-Butene

### Final Answer:
The end product C is trans-2-butene (CH₃CH=CHCH₃).

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