A hydrocarbon `(A) C_nH_(2n-4)` on ozonolysis gives `(CH_3)_2CHCH_2CHO, 2 OH"CC"H_2CH_2CHO and CH_3COCH_3` The value of n is

Answer

To solve the problem, we need to determine the value of \( n \) for the hydrocarbon \( (A) \) with the formula \( C_nH_{2n-4} \) based on the products formed during ozonolysis.

### Step 1: Identify the products and count the carbon atoms
The products given after ozonolysis are:
1. \( (CH_3)_2CHCH_2CHO \) (which is 5 carbons)
2. \( 2 \times CH_3CH_2CHO \) (which is 4 carbons for two molecules, totaling 8)
3. \( CH_3COCH_3 \) (which is 3 carbons)

Now, let's count the total number of carbon atoms from these products:
- For the first compound: 5 carbons
- For the second compound: \( 2 \times 4 = 8 \) carbons
- For the third compound: 3 carbons

Adding these together:
\[ 5 + 8 + 3 = 16 \text{ carbons} \]

### Step 2: Set up the equation using the general formula
The general formula for the hydrocarbon is given as:
\[ C_nH_{2n-4} \]
Since we have determined that there are 16 carbon atoms, we can set \( n = 16 \).

### Step 3: Verify the hydrogen count
Now, we can find the number of hydrogen atoms using the formula:
\[ H = 2n - 4 \]
Substituting \( n = 16 \):
\[ H = 2(16) - 4 = 32 - 4 = 28 \]

### Step 4: Conclusion
Thus, the hydrocarbon \( (A) \) has the formula \( C_{16}H_{28} \), and the value of \( n \) is:
\[ \boxed{16} \]