Alkenes

Among hydrogen halides the strongest acid is HI. As the size of halogen increase, bond length increase hence bond energy decreases. So, the HI releases ions easily and show high acidic nature.

The distance between the hydrogen and halogen atoms rises as the anion's size grows. For hydracids, it gets simpler to liberate H + ions as one move down in the halogen group.
So, the order will be HI>HBr>HCl>HF and HI is the strongest among them.

Discover More From This Chapter: Acids and Bases

To solve the problem, we need to determine which alkene (X) gives a mixture of Propan-2-one (acetone) and Methanal (formaldehyde) upon ozonolysis.

1. Understanding Ozonolysis:
Ozonolysis cleaves the double bond in alkenes and forms carbonyl compounds (aldehydes or ketones) at the positions where the double bond was present.

2. Analyze the products:
- Propan-2-one: (a ketone with 3 carbon atoms)
- Methanal: (an aldehyde with 1 carbon atom)

3. Work backwards to deduce the structure of X:
To get these two products, the double bond must be between a carbon with two methyl groups (to give acetone) and a terminal CH₂ group (to give methanal).

4. Structure of X:
The correct structure is that of 2-Methylpropene ( )
Upon ozonolysis:
(Propan-2-one) + (Methanal)

Final Answer:
The correct alkene X is 2-Methylpropene.

Step 1: Understanding the homologous series

The homologous series of alkanes consists of compounds differing by a -CH₂- group. The alkane next to methane is ethane (C₂H₆).

Step 2: Evaluating the given reactions

1. Reaction I: Wurtz Reaction

2CH₃Br + 2Na → _{dry ether} C₂H₆ + 2NaBr

This reaction successfully forms ethane, making it a correct method.

2. Reaction II: Decarboxylation of Carboxylic Acids

CH₃COOH + NaOH → _{CaO, Δ} CH₄ + Na₂CO₃

This reaction produces methane, not ethane, so it is incorrect.

3. Reaction III: Hydrogenation of Ethene

CH₂CH₂ + H₂ → _Pt C₂H₆

This reaction forms ethane, but it is not the best preparative method for homologous series alkane synthesis.

4. Reaction IV: Reduction of Alkyl Halides

CH₃CH₂Br + 2[H] → _Zn/H⁺ C₂H₆ + HBr

This reaction also produces ethane, making it a valid method.

Step 3: Evaluating the Given Options

• Option (1): Correct, as I (Wurtz reaction) and IV (alkyl halide reduction) are the valid methods.
• Option (2): Incorrect, as II produces methane instead of ethane.
• Option (3): Incorrect, as I is correct but III is not a conventional method for preparing alkanes in a homologous series.
• Option (4): Incorrect, as II does not yield ethane.

Thus, the correct answer is Option (1).

Step 1: Understanding the Reaction Sequence
1. Step 1: Acid-Catalyzed Hydration of Alkene
- The given compound is an alkene.
- In the presence of sulfuric acid ( ) and water, Markovnikov's rule applies, leading to the formation of a more substituted alcohol.
2. Step 2: Ozonolysis Reaction
- The alcohol undergoes ozonolysis ( , Zn/H_2O), which cleaves the double bond and forms aldehydes or ketones depending on the structure of the starting alkene.
Step 2: Identifying 'Y'
\flushleft - The given reactant is butene ( ).
- Hydration leads to the formation of butan-2-ol as the intermediate product.
- Ozonolysis of butan-2-ol results in acetone ( ) and formaldehyde ( ).
Step 3: Evaluating the Given Options
- Option (1): Incorrect, as it does not match the expected ozonolysis products.
- Option (2): Correct, as it represents acetone and formaldehyde.
- Option (3): Incorrect, as it does not correspond to the reaction mechanism.
- Option (4): Incorrect, as it represents a different structural outcome.
Thus, the correct answer is

Option (2).