An alkene 'X' (molecular formula `C_(5)H_(10)`) on ozonolysis gives a mixture of two compounds 'Y' and 'Z'. Compound 'Y' gives positive Fehling's test and also forms iodoform on treatement with `I_(2)" and "NaOH`. Compound 'Z' does not give Fehling's test but forms iodoform. Identify the compounds X, Y and Z. Write the reaction for ozonolysis and formation of iodoform from Y and Z.

Options

A. `{:(" X"," Y"," Z"),(C_(6)H_(5)COCH_(3),CH_(3)CHO,CH_(3)COCH_(3)):}`
B. `{:(" X"," Y"," Z"),(CH_(3)-CH=underset(CH_(3))underset(|)C-CH_(3),CH_(3)CHO,CH_(3)COCH_(3)):}`
C. `{:(" X"," Y"," Z"),(CH_(3)CH_(2)CH=CH_(2),CH_(3)CH_(2)CHO,HCHO):}`
D. `{:(" X"," Y"," Z"),(CH_(3)-CH=CH-CH_(3),CH_(3)CHO,CH_(3)CHO):}`

Explanation

To solve the problem, we need to identify the alkene 'X' with the molecular formula C5H10, and the compounds 'Y' and 'Z' produced from its ozonolysis. We will also write the reactions for ozonolysis and the formation of iodoform from 'Y' and 'Z'.

### Step 1: Identify the Alkene 'X'
The molecular formula C5H10 corresponds to an alkene. There are several possible alkenes with this formula, but we need to find one that can yield two products upon ozonolysis.

One possible structure for 'X' is 2-pentene (CH3-CH=CH-CH2-CH3).

### Step 2: Ozonolysis of Alkene 'X'
Ozonolysis involves the cleavage of the double bond in the alkene and the formation of carbonyl compounds.

For 2-pentene:
\[ \text{CH}_3\text{CH}=\text{CH}\text{CH}_2\text{CH}_3 + \text{O}_3 \rightarrow \text{CH}_3\text{CHO} + \text{CH}_3\text{COCH}_3 \]

Here, we get two products:
- **Compound Y**: Acetaldehyde (CH3CHO)
- **Compound Z**: Acetone (CH3COCH3)

### Step 3: Analyze Compound 'Y'
Compound 'Y' (Acetaldehyde) gives a positive Fehling's test, indicating it is an aldehyde. It also forms iodoform when treated with iodine and sodium hydroxide.

**Iodoform Reaction for Y**:
\[ \text{CH}_3\text{CHO} + \text{I}_2 + \text{NaOH} \rightarrow \text{CHI}_3 + \text{NaCHO} \]

### Step 4: Analyze Compound 'Z'
Compound 'Z' (Acetone) does not give a positive Fehling's test, indicating it is a ketone. However, it does form iodoform.

**Iodoform Reaction for Z**:
\[ \text{CH}_3\text{COCH}_3 + \text{I}_2 + \text{NaOH} \rightarrow \text{CHI}_3 + \text{NaCOCH}_3 \]

### Summary of Compounds
- **X**: 2-Pentene (CH3-CH=CH-CH2-CH3)
- **Y**: Acetaldehyde (CH3CHO)
- **Z**: Acetone (CH3COCH3)

### Final Reactions
1. Ozonolysis:
\[ \text{CH}_3\text{CH}=\text{CH}\text{CH}_2\text{CH}_3 + \text{O}_3 \rightarrow \text{CH}_3\text{CHO} + \text{CH}_3\text{COCH}_3 \]

2. Iodoform Reaction for Y:
\[ \text{CH}_3\text{CHO} + \text{I}_2 + \text{NaOH} \rightarrow \text{CHI}_3 + \text{NaCHO} \]

3. Iodoform Reaction for Z:
\[ \text{CH}_3\text{COCH}_3 + \text{I}_2 + \text{NaOH} \rightarrow \text{CHI}_3 + \text{NaCOCH}_3 \]