when (-)-2-methyl butan-1-ol is heated with conc. HCI (+)-1-chloro-2- methyl butane is obtained .The reaction is an example of :

Options

A. inversion
B. resolution
C. racemisation
D. retention

Answer

To solve the question regarding the reaction of (-)-2-methyl butan-1-ol with concentrated HCl to yield (+)-1-chloro-2-methyl butane, we need to analyze the process step by step.

### Step-by-Step Solution:

**Step 1: Identify the Reactants**
- The reactant is (-)-2-methyl butan-1-ol, which has the following structure:
- CH3-CH(CH3)-CH2-CH2OH
- This compound has a chiral center at the second carbon atom.

**Hint 1:** Look at the structure of the reactant to identify the chiral center and the functional groups present.

**Step 2: Understand the Reaction Conditions**
- The reaction is carried out in the presence of concentrated HCl, which is a strong acid. This indicates that the alcohol (-)-2-methyl butan-1-ol will undergo protonation.

**Hint 2:** Recall that concentrated acids can protonate alcohols, making them better leaving groups.

**Step 3: Protonation of the Alcohol**
- The hydroxyl group (-OH) of the alcohol gets protonated to form water, which is a better leaving group:
- CH3-CH(CH3)-CH2-CH2OH + HCl → CH3-CH(CH3)-CH2-CH2OH2^+ + Cl^-

**Hint 3:** Remember that protonation converts the -OH group into a better leaving group (water).

**Step 4: Formation of the Carbocation**
- Upon losing water, a carbocation is formed at the second carbon:
- CH3-CH(CH3)-CH2^+-CH3

**Hint 4:** Carbocation formation is a key step in many substitution reactions.

**Step 5: Nucleophilic Attack**
- The chloride ion (Cl^-) from HCl acts as a nucleophile and attacks the carbocation. Since the nucleophile can attack from either side, but due to the nature of the reaction, it will lead to retention of configuration:
- CH3-CH(CH3)-CH2-CH2Cl

**Hint 5:** Consider how nucleophiles attack carbocations and the implications for stereochemistry.

**Step 6: Product Formation**
- The product formed is (+)-1-chloro-2-methyl butane. The stereochemistry is retained, meaning that the configuration of the chiral center does not change, but the optical activity does.

**Hint 6:** Retention of configuration means that the stereochemistry remains the same, but the optical activity may change.

**Conclusion: Identify the Type of Reaction**
- The reaction is an example of a **retention reaction** where the configuration at the chiral center is retained, leading to a change in optical rotation but not in stereochemistry.

**Final Answer:** The reaction is an example of **retention of configuration**.