Propyne reacts with `Br_(2)//HOH` to produce

Options

A. 1,1,2,2-Tetrabromopropane
B. Acetone
C. Acetaldehyde
D. `alpha` - Bromoacetone

Answer

To solve the problem of what product is formed when propyne reacts with `Br2/H2O`, we can follow these steps:

### Step 1: Understand the Reactants
Propyne is an alkyne with the formula C₃H₄, specifically it is 1-propyne (H₃C-C≡C-H). The reaction involves bromine (Br₂) and water (H₂O).

### Step 2: Identify the Reaction Type
The reaction of an alkyne with bromine in the presence of water typically leads to the formation of a bromo-alcohol. This is a halohydrin formation reaction where bromine adds across the triple bond.

### Step 3: Draw the Reaction Mechanism
1. The triple bond in propyne reacts with bromine, leading to the formation of a cyclic bromonium ion intermediate.
2. Water (H₂O) then attacks the more substituted carbon of the bromonium ion, leading to the formation of an enol.

### Step 4: Determine the Structure of the Enol
The enol formed from this reaction has the structure:
- CH₃-C(OH)=CBr-H (where the OH is on the more substituted carbon).

### Step 5: Tautomerization to Keto Form
The enol can undergo tautomerization, which is the rearrangement of the molecule to form a more stable keto form. The keto form of the product is:
- CH₃-C(=O)-CH₂Br (which is alpha-bromoacetone).

### Step 6: Final Product
The final product of the reaction is alpha-bromoacetone (CH₃C(=O)CH₂Br).

### Conclusion
The product of the reaction of propyne with `Br₂/H₂O` is alpha-bromoacetone.

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