Consider the following reactions
`A underset((ii)H_(3)O^(+))overset((i)CH_(3)MgBr)to B underset(573K)overset(Cu)to 2`-methyl-2-butene
the mass percentage of carbon in A is ________.

Answer

To determine the mass percentage of carbon in compound A, we will follow these steps:

### Step 1: Identify the structure of compound B
From the information given, compound B is formed from the reaction of compound A with CH3MgBr and H3O+. The final product after the reaction with Cu at 573K is 2-methyl-2-butene.

### Step 2: Determine the structure of compound A
Since B is a tertiary alcohol (which can be dehydrated to form 2-methyl-2-butene), compound A must be a ketone. The structure of B suggests that it is derived from a ketone with four carbon atoms. Therefore, compound A is likely 3-pentanone (C5H10O).

### Step 3: Calculate the molar mass of compound A
The molecular formula for 3-pentanone is C5H10O. We will calculate its molar mass:
- Carbon (C): 5 atoms × 12 g/mol = 60 g/mol
- Hydrogen (H): 10 atoms × 1 g/mol = 10 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol

Total molar mass of C5H10O = 60 + 10 + 16 = 86 g/mol.

### Step 4: Calculate the mass of carbon in compound A
The mass of carbon in compound A is:
- Mass of carbon = Number of carbon atoms × Molar mass of carbon = 5 × 12 g/mol = 60 g/mol.

### Step 5: Calculate the mass percentage of carbon in compound A
The mass percentage of carbon in compound A is calculated using the formula:
\[ \text{Mass percentage of carbon} = \left( \frac{\text{Mass of carbon}}{\text{Total molar mass}} \right) \times 100 \]
Substituting the values:
\[ \text{Mass percentage of carbon} = \left( \frac{60 \text{ g/mol}}{86 \text{ g/mol}} \right) \times 100 \approx 69.77\% \]

### Final Answer
The mass percentage of carbon in compound A is approximately **69.77%**.

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