JEE Chemistry 2025 Question with Solution

Given: A reaction sequence starts from acetophenone and forms product $D$ after treatment with $PCl_5$, excess $NaNH_2/NH_3$, acidification, and then hydroboration-oxidation.

Find: The total number of $sp^2$ hybridised carbon atoms in product $D$.

From the extracted solution sequence:

1. Acetophenone reacts with $PCl_5$ to form a geminal dichloride.
2. Treatment with excess $NaNH_2/NH_3$ gives an alkyne.
3. Acidification does not change that alkyne.
4. Hydroboration-oxidation then gives the carbonyl compound $D$.

The relevant transformation shown is:

$$\text{Ph-CCl}_2-\text{CH}_3 \xrightarrow{3eq.\, NaNH_2/NH_3} \text{Ph-C} \equiv \text{C-CH}_3$$

So the alkyne intermediate is $\text{Ph-C} \equiv \text{C-CH}_3$.

Hydroboration-oxidation followed by tautomerization gives:

$$\text{Ph-C} \equiv \text{C-CH}_3 \xrightarrow[1.\, B_2H_6]{2.\, H_2O_2/OH^-} \text{Ph-CO-CH}_2-\text{CH}_3$$

Thus, product $D$ is 1-phenylpropan-1-one / propiophenone-type ketone structure as extracted in the solution working, and the count of $sp^2$ carbons is obtained directly from this structure.

Now count the $sp^2$ hybridised carbons:

• The phenyl ring contributes $6$ $sp^2$ carbons.
• The carbonyl carbon contributes $1$ $sp^2$ carbon.
• The alkyl side-chain carbons are $sp^3$.

Therefore,

$$6 + 1 = 7$$

So, the total number of $sp^2$ hybridised carbon atoms in product $D$ is $7$.

Given: The reaction sequence and the worked solution indicate successive conversion of acetophenone into intermediates $A$, $B$, $C$ and finally $D$.

Find: Number of $sp^2$ hybridised carbon atoms in $D$.

Step 1: Formation of $A$

$$\text{Ph-CO-CH}_3 \xrightarrow{PCl_5} \text{Ph-CCl}_2-\text{CH}_3 \; (A)$$

The carbonyl group is converted into a geminal dichloride.

Step 2: Formation of $B$

$$\text{Ph-CCl}_2-\text{CH}_3 \xrightarrow{3eq.\, NaNH_2/NH_3} \text{Ph-C} \equiv \text{C-CH}_3 \; (B)$$

Double dehydrohalogenation gives an alkyne.

Step 3: Formation of $C$ Acidification does not alter the carbon skeleton here, so $C$ remains the same alkyne.

Step 4: Formation of $D$ Hydroboration-oxidation of the alkyne, followed by tautomerization, gives the ketone:

$$\text{Ph-C} \equiv \text{C-CH}_3 \rightarrow \text{Ph-CO-CH}_2-\text{CH}_3 \; (D)$$

Step 5: Count $sp^2$ carbons in $D$ The structure contains:

• one phenyl ring = $6$ $sp^2$ carbons
• one carbonyl carbon = $1$ $sp^2$ carbon
• one $CH_2$ and one $CH_3$ carbon = $sp^3$

Hence,

$$\text{Total } sp^2 \text{ carbons} = 6 + 1 = 7$$

Therefore, the required numerical answer is $7$.

The first provided approach contains structural inconsistencies in the intermediate/product naming, but it also concludes the same final count. The second worked approach provides the defensible reaction sequence and is used for the answer determination.