An alcohol (A ) on dehydration gives (B) which adds bromine molecule to give (C ), (C ) on heating with sodamide gives (D) which on hydration in the presence of `Hg^(++)//H_(2)SO_(4)` gives (E ). E on reduction by lithium aluminium hydride gives (A). (E ) is also obtained on dry distillation of calcium salt of acetic acid. How many enolisable proton present in the product (E ) ?

Options

A. 2
B. 6
C. 3
D. 0

Explanation

To solve the problem step by step, we will analyze the transformations of the compounds mentioned in the question.

### Step 1: Identify Compound A
Given that compound A is an alcohol, we can assume it to be isopropyl alcohol (2-propanol), which can be represented as:
\[ \text{A} = \text{CH}_3\text{CHOHCH}_3 \]

### Step 2: Dehydration of A to form B
When alcohol A undergoes dehydration (removal of water), it forms an alkene. The dehydration of isopropyl alcohol leads to:
\[ \text{B} = \text{CH}_3\text{C}=\text{CH}_2 \]
This is propene.

### Step 3: Addition of Bromine to form C
The alkene B reacts with bromine (Br2) in an addition reaction, resulting in the formation of a dibromide:
\[ \text{C} = \text{CH}_3\text{C}(\text{Br})\text{CH}_2\text{Br} \]

### Step 4: Heating C with Sodium Amide to form D
When compound C is heated with sodamide (NaNH2), it undergoes dehydrohalogenation to form an alkyne:
\[ \text{D} = \text{CH}_3\text{C} \equiv \text{CH} \]
This is 1-butyne.

### Step 5: Hydration of D to form E
The alkyne D can be hydrated in the presence of mercuric sulfate (HgSO4) and sulfuric acid (H2SO4) to form a ketone:
\[ \text{E} = \text{CH}_3\text{C(=O)CH}_3 \]
This is acetone (propan-2-one).

### Step 6: Reduction of E to form A
When acetone (E) is reduced by lithium aluminum hydride (LiAlH4), it gives back the alcohol A:
\[ \text{A} = \text{CH}_3\text{CHOHCH}_3 \]

### Step 7: Dry Distillation of Calcium Salt of Acetic Acid
The problem states that E is also obtained by the dry distillation of the calcium salt of acetic acid. The calcium salt of acetic acid (Ca(CH3COO)2) upon dry distillation yields acetone (E).

### Step 8: Count Enolisable Protons in E
To find the number of enolisable protons in E (acetone), we look for protons that are adjacent to the carbonyl group (C=O). Acetone has the structure:
\[ \text{E} = \text{CH}_3\text{C(=O)CH}_3 \]
The two methyl groups (CH3) adjacent to the carbonyl each have three protons. Therefore, the total number of enolisable protons is:
- 3 (from one CH3) + 3 (from the other CH3) = 6 enolisable protons.

### Final Answer
The number of enolisable protons present in the product E is **6**.

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