An alkene X on ozonolysis gives a mixture of simplest ketone (Y) and 3-Pentanone. The IUPAC name of the alkene X is
Show Hint
Reverse Ozonolysis Trick: Write the two carbonyl products facing each other: \( >\text{C=O} \quad \text{O=C}< \). Erase the O's and connect the carbons with a double bond: \( >\text{C=C}< \).
Step 1: Identify the Ozonolysis Products:
1. Simplest Ketone (Y): The simplest ketone is Propanone (Acetone).
Structure: \(\text{CH}_3\text{-C(=O)-CH}_3\)
2. 3-Pentanone: A 5-carbon ketone with the carbonyl group at the 3rd position.
Structure: \(\text{CH}_3\text{-CH}_2\text{-C(=O)-CH}_2\text{-CH}_3\)
Step 2: Determine Structure of Alkene X:
Ozonolysis cleaves a double bond \( \text{C=C} \) and places an Oxygen on each carbon atom to form two carbonyl compounds \( \text{C=O} + \text{O=C} \). To find the original alkene, we reverse this process: remove the oxygen atoms and join the carbonyl carbons with a double bond.
\[ [\text{Propanone}] = \text{(CH}_3)_2\text{C=O} \]
\[ [\text{3-Pentanone}] = \text{O=C(CH}_2\text{CH}_3)_2 \]
Joining them:
\[ \text{(CH}_3)_2\text{C} = \text{C(CH}_2\text{CH}_3)_2 \]
Expanded Structure:
\[ \text{CH}_3\text{-C(CH}_3)\text{=C(CH}_2\text{CH}_3)\text{-CH}_2\text{CH}_3 \]
Step 3: IUPAC Naming of X:
1. Select Principal Chain: The longest carbon chain containing the double bond.
Path: \(\text{CH}_3\text{-C-C-C-C}\) is a 5-carbon chain. (Parent: Pentene).
2. Numbering: Start from the end that gives the double bond the lowest number.
- From Left: \( \text{C}^1\text{H}_3\text{-C}^2(\text{CH}_3)\text{=C}^3(\text{CH}_2\text{CH}_3)\text{-C}^4\text{H}_2\text{-C}^5\text{H}_3 \) \(\rightarrow\) Double bond at C-2.
- From Right: Double bond starts at C-3.
Correct numbering is from the left.
3. Identify Substituents:
- At C-2: Methyl group (\(-\text{CH}_3\))
- At C-3: Ethyl group (\(-\text{CH}_2\text{CH}_3\))
4. Assemble Name: Substituents in alphabetical order (Ethyl before Methyl).
\(3-Ethyl-2-methylpent-2-ene\).
Final Answer: Option (C).
