`3 -" Methylbutane" - 2 - ol +Hl overset(Delta)rarrX` Idenfity the position of carbon in the main chain to which nucleophile is added in the product 'X'?

Answer

To solve the problem, we need to identify the position of carbon in the main chain of 2-methylbutane-2-ol where the nucleophile (I-) attacks to form the product X. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Structure of 2-Methylbutane-2-ol
2-Methylbutane-2-ol is a tertiary alcohol. Its structure can be represented as follows:
- The main carbon chain consists of four carbon atoms (butane).
- The second carbon in the chain has a hydroxyl group (-OH) and a methyl group (-CH3) attached to it.

### Step 2: Identify the Type of Alcohol
Since the hydroxyl group is attached to a carbon that is connected to three other carbon atoms, we classify it as a tertiary alcohol. Tertiary alcohols are known to undergo reactions via the SN1 mechanism due to the stability of the resulting carbocation.

### Step 3: Reaction with HI
When 2-methylbutane-2-ol reacts with HI (hydroiodic acid), the following steps occur:
1. Protonation of the hydroxyl group (-OH) occurs, making it a better leaving group.
2. The -OH group leaves, forming a carbocation at the same carbon where the -OH was attached (the second carbon).

### Step 4: Formation of the Carbocation
The carbocation formed is a tertiary carbocation, which is stable. This stability allows the reaction to proceed via the SN1 mechanism.

### Step 5: Nucleophilic Attack
After the formation of the carbocation, the nucleophile (I-) attacks the carbocation. Since the carbocation was formed at the second carbon, the nucleophile will also attack this same carbon.

### Step 6: Identify the Product
The product formed will have the iodine atom attached to the second carbon, resulting in 2-iodo-2-methylbutane.

### Conclusion
The nucleophile (I-) attacks the second carbon in the main chain of 2-methylbutane-2-ol.

### Final Answer
The position of carbon in the main chain to which the nucleophile is added in the product X is the **2nd carbon**.

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