In the following sequence of reactions,
`CH_(3)CH_(2)CH_(2)Br overset("KOH"(Alc))(rarr) " (A) " overset("HBr")(rarr) " (B) " overset("KOH"(aq))(rarr)` (C ) The product (C) is

Options

A. Propene
B. Propyne
C. Propan-1-ol
D. Propan-2-ol

Answer

To solve the given sequence of reactions, we will break down each step to identify the products formed at each stage.

### Step 1: Reaction of CH₃CH₂CH₂Br with KOH (Alcoholic)
The starting compound is 1-bromopropane (CH₃CH₂CH₂Br). When it reacts with alcoholic KOH, which acts as a strong base, it undergoes elimination (dehydrohalogenation) to form an alkene.

- **Mechanism**: The base (KOH) abstracts a proton (H) from the β-carbon (the carbon adjacent to the one bonded to bromine), leading to the formation of a double bond. The bromine atom leaves as a bromide ion (Br⁻).

- **Product (A)**: The product formed is propene (CH₃CH=CH₂).

### Step 2: Reaction of Product A (Propene) with HBr
Next, propene reacts with HBr. This is an electrophilic addition reaction where HBr adds across the double bond.

- **Mechanism**: The double bond in propene acts as a nucleophile and attacks the H⁺ from HBr, forming a carbocation. The more stable carbocation will form due to Markovnikov's rule, which states that the hydrogen will add to the carbon with more hydrogens already attached.

- **Carbocation Formation**: The more stable secondary carbocation (CH₃-CH⁺-CH₃) is formed.

- **Product (B)**: The bromide ion (Br⁻) then attacks the carbocation, resulting in the formation of 2-bromopropane (CH₃CHBrCH₃).

### Step 3: Reaction of Product B (2-Bromopropane) with KOH (Aqueous)
Finally, 2-bromopropane reacts with aqueous KOH. In this case, KOH acts as a nucleophile and promotes a substitution reaction.

- **Mechanism**: The hydroxide ion (OH⁻) from KOH attacks the carbon atom that is bonded to the bromine, leading to the displacement of the bromide ion (Br⁻).

- **Product (C)**: The final product is 2-propanol (CH₃CHOHCH₃).

### Final Answer
The product (C) is **2-propanol (CH₃CHOHCH₃)**.

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