Three acyclic alkenes (x,y,z) on catalytic hyrogenation give same alkane. On reaction with HCl (x,y,z) form same major tertiary halide product.Reductive ozonolysis of mixture of (x,y,z) gives a mixture of two moles of `CH_2=O` one moles of `CH_3CH=O` one mole of acetone, one mole of butanone and one mole of 2-methyl propanal, x,y and z do not have any stereoisomers.
(x,y,z)`overset(H_3O^(**))to` addition product . The correct statement is

Options

A. All three alkenes will give 3 different major hydration products
B. Three alkenes will give same hydration major product
C. Two alkenes form same product but one alkene forms different major product.
D. Addition of HCl and `H_3O^(**)` both are following different regioselectivity.

Answer

To solve the problem, we need to analyze the information provided about the three acyclic alkenes (x, y, z) and their reactions. Let's break it down step by step.

### Step 1: Identify the Alkenes
We know that the three alkenes (x, y, z) are acyclic and give the same alkane upon catalytic hydrogenation. The products of hydrogenation indicate that they likely have similar structures.

Given the information in the transcript, we can assume the following structures for the alkenes:
1. **Alkene x**: CH3-CH=CH-CH2-CH3 (2-pentene)
2. **Alkene y**: CH2=CH-CH2-CH3 (1-pentene)
3. **Alkene z**: CH3-CH=CH2 (2-methyl-1-butene)

Upon catalytic hydrogenation, all three alkenes yield the same alkane: **2-methylbutane** (C5H12).

### Step 2: Reaction with HCl
When these alkenes react with HCl, they form the same major tertiary halide product. This suggests that the addition of HCl follows Markovnikov's rule, leading to the formation of a tertiary carbocation in each case.

The major tertiary halide product formed from each alkene is likely **2-bromo-2-methylbutane**.

### Step 3: Reductive Ozonolysis
The problem states that the reductive ozonolysis of the mixture of (x, y, z) gives:
- 2 moles of formaldehyde (CH2O)
- 1 mole of acetaldehyde (CH3CHO)
- 1 mole of acetone (CH3COCH3)
- 1 mole of butanone (CH3COCH2CH3)
- 1 mole of 2-methylpropanal (CH3C(=O)H-CH3)

This indicates that the alkenes are breaking down into these carbonyl compounds, confirming their structures.

### Step 4: Acidic Hydration
When the alkenes undergo acidic hydration, we need to determine the products formed.

1. For **alkene x** (2-pentene), the addition of water will lead to a tertiary carbocation after rearrangement, yielding **2-methyl-2-pentanol**.
2. For **alkene y** (1-pentene), the addition of water will directly yield **2-pentanol**.
3. For **alkene z** (2-methyl-1-butene), the addition of water will also lead to a tertiary carbocation, yielding **2-methyl-2-pentanol**.

### Conclusion
Since the alkenes yield the same major hydration product, the correct statement regarding the addition product is:
- **Three alkenes will give the same hydration major product.**