Question

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

Total number of $ sp^2 $ hybridised carbon atoms in product D is _____.

Hint:

Remember the reaction mechanisms and stereochemistry/regiochemistry of each step. \( PCl_5 \) replaces \( C=O \) with \( CCl_2 \). \( NaNH_2/NH_3 \) promotes elimination to form alkynes. Hydroboration-oxidation of terminal alkynes gives aldehydes or ketones depending on regioselectivity, which is anti-Markovnikov with \( B_2H_6 \). Finally, count the number of carbon atoms directly bonded to three other atoms or involved in a double bond to determine \( sp^2 \) hybridization.

Answer 1

Correct Answer: 7

To identify the structure of the final product (D) and count the number of sp² hybridized carbon atoms, we proceed as follows:

1. Step 1: Conversion to Compound A
   The starting compound, acetophenone (Ph-C(=O)-CH₃), reacts with PCl₅ to form Compound A, an α-chloro ketone (Ph-C(=O)-CH₂Cl).
2. Step 2: Elimination to Compound B
   Heating Compound A with excess NaNH₂ induces a double elimination, forming an alkyne, Compound B, phenylacetylene (Ph-C≡CH).
3. Step 3: Hydroboration-Oxidation to Compound C
   Hydroboration of phenylacetylene with BH₃ followed by oxidation with H₂O₂/NaOH yields an aldehyde, Compound C, phenylacetaldehyde (Ph-CH=CH₂).
4. Step 4: Determination of Compound D
   The hydroboration-oxidation of an alkyne typically yields an anti-Markovnikov aldehyde. Therefore, Compound D again is phenylacetaldehyde.

Now, we determine the number of sp² hybridized carbon atoms in Compound D. Phenylacetaldehyde (Ph-CH=CH₂) contains:
- 6 sp² hybridized carbons in the phenyl ring.
- 1 sp² carbon from the aldehyde group.
Total: 7 sp² hybridized carbon atoms.

The calculated number of sp² hybridized carbon atoms falls within the expected range of 7.

Answer 2

Step 1: Formation of A.
Acetophenone reacts with \( PCl_5 \) to give a geminal dichloride: \[ Ph-\overset{O}{\underset{||}{C}}-CH_3 \xrightarrow{PCl_5} Ph-CCl_2-CH_3 \quad (A) \] 
Step 2: Formation of B.
Reaction with 3 equivalents of \( NaNH_2/NH_3 \) leads to elimination of HCl and formation of an alkyne: \[ Ph-CCl_2-CH_3 \xrightarrow{3eq. \, NaNH_2/NH_3} Ph-C \equiv C-CH_3 \quad (B) \] 
Step 3: Formation of C.
Acidification does not change the structure of the alkyne. 
Thus, C is also \( Ph-C \equiv C-CH_3 \). 
Step 4: Formation of D (Hydroboration-oxidation).
Hydroboration-oxidation of a terminal alkyne with \( B_2H_6 \) followed by \( H_2O_2/OH^- \) proceeds with anti-Markovnikov regioselectivity, placing the hydroxyl group on the less substituted carbon after tautomerization. \[ Ph-C \equiv C-CH_3 \xrightarrow[1. \, B_2H_6]{2. \, H_2O_2/OH^-} Ph-\overset{OH}{C}=CH-CH_3 \xrightarrow{Tautomerization} Ph-CO-CH_2-CH_3 \quad (D) \] The final product D is 1-phenylpropan-2-one. 
Step 5: Determine the number of \( sp^2 \) hybridized carbon atoms in D.
The structure of 1-phenylpropan-2-one is: \[ \underbrace{\overset{sp^2}{C}_6H_5}_{\text{6 } sp^2 \text{ carbons}} - \overset{sp^3}{CH_2} - \overset{sp^2}{C} = O - \overset{sp^3}{CH_3} \] The phenyl ring has 6 \( sp^2 \) hybridized carbon atoms. The carbonyl carbon is also \( sp^2 \) hybridized. The \( CH_2 \) and \( CH_3 \) carbons are \( sp^3 \) hybridized. 
Total number of \( sp^2 \) hybridized carbon atoms in product D = 6 (from the phenyl ring) + 1 (carbonyl carbon) = 7.