Addition of mercuric acetate in the presence of water is called as oxymercuration.The adduct obtained gives alcohol on reduction with `NaBH_4` in alkaline medium.This is known as demercuration.Oxymercuration demercuration allows Markownikoff's addition of H, OH without rearrangement.The net result is the addition of `H_2O` Answer the following question :
`CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)(C)-CH=CH_2underset((ii)CH_3OH NaBH_4NaOH)overset((i)Hg(OAc)_2.THF)toAoverset(HI(Conc.))toB`
Product B is :

Explanation

To solve the problem, we will follow the steps of the reaction involving oxymercuration and demercuration, and then the subsequent reaction with concentrated HI.

### Step-by-Step Solution:

1. **Identify the Starting Material**:
The starting material is given as:
\[ \text{CH}_3-\text{C}(\text{CH}_3)(\text{C})-\text{CH}=\text{CH}_2 \]
This molecule has a double bond between the carbon atoms.

2. **Oxymercuration Reaction**:
The first step involves the addition of mercuric acetate (Hg(OAc)₂) in the presence of THF and water. This process is called oxymercuration. During this step, the mercuric acetate adds across the double bond in a Markownikoff manner, resulting in the formation of an intermediate organomercury compound. The addition of water leads to the formation of an alcohol.

The structure of the product after oxymercuration (let's call it A) will be:
\[ \text{CH}_3-\text{C}(\text{CH}_3)(\text{C})-\text{CH}(\text{OH})-\text{CH}_3 \]

3. **Demercuration Reaction**:
The next step involves the reduction of the organomercury compound using sodium borohydride (NaBH₄) in an alkaline medium (NaOH). This process is known as demercuration, which replaces the mercury with a hydrogen atom, yielding an alcohol.

The resulting product after demercuration (let's call it A) will be:
\[ \text{CH}_3-\text{C}(\text{CH}_3)(\text{C})-\text{CH}(\text{OCH}_3)-\text{CH}_3 \]

4. **Reaction with Concentrated HI**:
The product A is then treated with concentrated hydroiodic acid (HI). In this step, the hydroxyl group (OH) is replaced by an iodine atom (I) through a nucleophilic substitution reaction. This results in the formation of another product (let's call it B).

The final product B will be:
\[ \text{CH}_3-\text{C}(\text{CH}_3)(\text{C})-\text{CH}(\text{I})-\text{CH}_3 \]
Along with the formation of methyl iodide (CH₃I) as a byproduct.

### Final Answer:
Product B is:
\[ \text{CH}_3-\text{C}(\text{CH}_3)(\text{C})-\text{CH}(\text{I})-\text{CH}_3 \quad \text{and} \quad \text{CH}_3I \]

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