Question

The problem involves a two-step reaction sequence starting from cyclohexanone.

Step 1: Reaction (1) The starting material is cyclohexanone (O=C1CCCCC1). Reagents: KOH, Δ (potassium hydroxide, heat). This is a typical aldol condensation reaction. Under basic conditions and heating, two molecules of cyclohexanone undergo condensation.

1. A base (KOH) deprotonates an alpha-hydrogen of one cyclohexanone molecule to form an enolate.
2. The enolate acts as a nucleophile and attacks the carbonyl carbon of another cyclohexanone molecule to form a β-hydroxy ketone (aldol adduct).
3. Under heating conditions (Δ), the β-hydroxy ketone undergoes dehydration to form an α,β-unsaturated ketone.

The product (A) formed from the aldol condensation of two cyclohexanone molecules followed by dehydration is 2-cyclohexylidenecyclohexanone. The reaction is: 2 C6H10O (Cyclohexanone) --(KOH, Δ)--> O=C1CCCCC1=C2CCCCC2 (Product A) + H2O

Structure of Product (A):

      O
     //
    C1
   /  \
  C6    C2
 /  \  //
C5    C3
 \   /
  C4
  ||
  C(of second ring)

SMILES: O=C1CCCCC1=C2CCCCC2

Step 2: Reaction (B) Reactant: Product (A) = 2-cyclohexylidenecyclohexanone Reagents: HCHO (formaldehyde), KOH, Δ. This is a crossed aldol condensation reaction with formaldehyde, followed by dehydration.

First, identify the acidic alpha-hydrogens in Product (A):

• The carbonyl carbon is C1.
• The alpha carbons are C2 and C6.
• Carbon C2 is part of the exocyclic double bond (C2=C(second ring)), so it has no alpha-hydrogens.
• Carbon C6 has two alpha-hydrogens. These are acidic and will be abstracted by the base (KOH).

Mechanism for reaction (B):

1. Enolate formation: KOH deprotonates an alpha-hydrogen from C6 of Product (A) to form an enolate.

         O
        //
       C1
      /  \
     C6-    C2
    /  \  //
   H    C3
    \  /
     C4
     ||
     C(of second ring)
   

   Product A + KOH <=> [Enolate at C6]- K+ + H2O The enolate can be represented as:

         O-
        / \
       C1=C6
      /    \
     C5     C2
      \    //
       C4  C3
       ||
       C(of second ring)
   

2. Nucleophilic attack: The enolate (carbanion at C6) attacks the electrophilic carbonyl carbon of formaldehyde (HCHO). This forms a β-hydroxy ketone (aldol adduct). The adduct is 6-(hydroxymethyl)-2-cyclohexylidenecyclohexanone.

         O
        //
       C1
      /  \
     C6-CH2OH C2
    /  \    //
   H    C3
    \  /
     C4
     ||
     C(of second ring)
   

3. Dehydration: Under heating conditions (Δ), the β-hydroxy ketone undergoes dehydration. The -OH group from the CH2OH (beta carbon) and a hydrogen from the alpha carbon (C6) are removed, forming a new double bond. The final product (B) is 6-methylene-2-cyclohexylidenecyclohexanone.

Structure of Product (B):

      O
     //
    C1
   /  \
  C6=CH2 C2
 /  \  //
C5    C3
 \   /
  C4
  ||
  C(of second ring)

SMILES: O=C1C(=C)CCCC1=C2CCCCC2

The final product (B) is 6-methylene-2-cyclohexylidenecyclohexanone.

The image provided only asks for "Product (2)", which corresponds to Product (B) in our analysis. The structure of Product (B) is shown above.

Answer 1

O=C1C(=C)CCCC1=C2CCCCC2

Answer 2

1. Step 1 (Reaction 1): Cyclohexanone undergoes base-catalyzed aldol condensation followed by dehydration to form 2-cyclohexylidenecyclohexanone (Product A).
2. Step 2 (Reaction B): Product A has acidic alpha-hydrogens only at C6. These hydrogens are abstracted by KOH to form an enolate. This enolate then attacks formaldehyde (HCHO) in a crossed aldol reaction. The resulting β-hydroxy ketone undergoes dehydration under heating (Δ) to form an α,β-unsaturated ketone, 6-methylene-2-cyclohexylidenecyclohexanone (Product B).