In the following sequence of reaction , `CH_3CH_2CH_2CH_2Broverset(alc. KOH)rarrAoverset(HBr)rarrBoverset(aq.KOH)rarrC,` The product 'c' is ?

Options

A. Butan - 2 - ol
B. Butan - 1 - ol
C. Butyne
D. Butene

Answer

To solve the question regarding the sequence of reactions involving `CH3CH2CH2CH2Br`, we will analyze each step carefully.

### Step 1: Reaction with Alcoholic KOH
The first reaction involves `CH3CH2CH2CH2Br` (1-bromobutane) reacting with alcoholic KOH. This is a strong base that facilitates an elimination reaction (E2 mechanism).

- **Mechanism**: In the E2 mechanism, the base (alkoxide ion from KOH) abstracts a hydrogen atom from the β-carbon (the carbon adjacent to the one bonded to the bromine), while the bromine leaves as a bromide ion. This results in the formation of a double bond between the α-carbon and the β-carbon.

- **Product A**: The product of this reaction is But-1-ene (`CH3CH2CH=CH2`).

### Step 2: Reaction with HBr
Next, we take the product A (But-1-ene) and react it with HBr.

- **Mechanism**: According to Markovnikov's rule, when HBr adds to an alkene, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms (the less substituted carbon), while the bromine will attach to the more substituted carbon.

- **Product B**: The addition of HBr to But-1-ene will yield 2-bromobutane (`CH3CH2CHBrCH3`).

### Step 3: Reaction with Aqueous KOH
Finally, we take product B (2-bromobutane) and react it with aqueous KOH.

- **Mechanism**: In this case, aqueous KOH acts as a nucleophile and will perform a nucleophilic substitution reaction (either SN1 or SN2, but likely SN1 due to the stability of the secondary carbocation formed).

- **Product C**: The bromine atom will be replaced by a hydroxyl group (OH), leading to the formation of Butane-2-ol (`CH3CH(OH)CH2CH3`).

### Final Answer
The product C is **Butane-2-ol**.

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