Identify the structure of the final product (D) in the following sequence of the reactions : Ph-C(=O)-CH3 →A →B →C →D Total number of sp2 hybridised carbon atoms in product D is _______.

The correct answer is Option 7. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q75 – JEE-MAIN 2025

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Home Chemistry Organic Chemistry – Some Basic Principles and TechniquesQuestion #21

Identify the structure of the final product (D) in the following sequence of the reactions :

P h - C (= O) - C H_{3} \overset{P C l_{5}, Δ}{\to} A \overset{3 eq. N a N H_{2} / N H_{3}}{\to} B \overset{Acidify}{\to} C \overset{1. B_{2} H_{6} 2. H_{2} O_{2} / O H^{-}}{\to} D

Total number of

s p^{2}

hybridised carbon atoms in product D is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Reaction Sequence Analysis

Break down the overall transformation into individual steps, identifying the product of each reaction.

🧠 Full Solution Path

Step 1: Determine products A, B, and C✦ Active

The starting material is acetophenone ( $P h - C (= O) - C H_{3}$ ). 1. Reaction with $P C l_{5}$ replaces the carbonyl oxygen with two chlorine atoms, forming 1,1-dichloro-1-phenyl-ethane. $A = P h - C C l_{2} - C H_{3}$ 2. Reaction with 3 equivalents of $N a N H_{2} / N H_{3}$ : The first two equivalents perform double dehydrohalogenation to form phenylacetylene ( $P h - C \equiv C H$ ). The third equivalent deprotonates the terminal alkyne to form the acetylide anion. $B = P h - C \equiv C^{-} N a^{+}$ 3. Acidification of the acetylide anion protonates it to form phenylacetylene. $C = P h - C \equiv C H$

Step 2: Determine product D○ Expand

Product C, phenylacetylene ( $P h - C \equiv C H$ ), undergoes hydroboration-oxidation ( $1. B_{2} H_{6}; 2. H_{2} O_{2} / O H^{-}$ ). This is an anti-Markovnikov addition of water across the triple bond, leading to an enol that tautomerizes to an aldehyde.

P h - C \equiv C H \overset{1. B_{2} H_{6} 2. H_{2} O_{2} / O H^{-}}{\to} P h - C H_{2} - C H O

Thus, product D is phenylacetaldehyde ( $P h - C H_{2} - C H O$ ).

💡 Teacher's Secret Hint

Remember that hydroboration-oxidation of terminal alkynes yields aldehydes, while mercuric ion catalyzed hydration yields ketones.

Step 3: Count

s p^{2}

hybridized carbon atoms in D○ Expand

The structure of product D is phenylacetaldehyde ( $P h - C H_{2} - C H O$ ). - The phenyl group (benzene ring) contains 6 carbon atoms, all of which are $s p^{2}$ hybridized. - The carbonyl carbon in the aldehyde group ( $- C H O$ ) is double-bonded to oxygen and single-bonded to two other atoms, making it $s p^{2}$ hybridized. - The methylene carbon ( $- C H_{2} -$ ) is single-bonded to four other atoms, making it $s p^{3}$ hybridized. Therefore, the total number of $s p^{2}$ hybridized carbon atoms in product D is $6 + 1 = 7$ .

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Chemistry Question 75 – JEE-MAIN 2025

Clue 1 of 3: Reaction Sequence Analysis

🧠 Full Solution Path