A compound 'D' `(C_(8)H_(16)O)` has following observation
(a) It reacts with Lucas reagent with in `5 min`.
(b) On heating in presence of acid given a compound X, which can decolorize `KMnO_(4)` as well as `Br_(2)//CCl_(4)`.
(c ) compound `Xoverset("ozonolysis")rarr Yoverset(I_(2)//NaOH)rarr2` moles of iodoform `+` sodium salt of dibasic acid `overset("Sodalime")rarr n -` butane
Then D, X, Y are

Options

A.
B.
C.
D.

Answer

To solve the problem step by step, we will analyze the given observations about the compound 'D' with the molecular formula \( C_8H_{16}O \).

### Step 1: Identify Compound D
Given that compound D reacts with Lucas reagent within 5 minutes, we can conclude that it is a secondary alcohol. This is because secondary alcohols react with Lucas reagent (a mixture of hydrochloric acid and zinc chloride) relatively quickly, while primary alcohols react much slower.

**Hint:** Lucas reagent is used to differentiate between primary, secondary, and tertiary alcohols based on their reactivity.

### Step 2: Determine Compound X
Upon heating compound D in the presence of acid, it produces compound X, which can decolorize both KMnO4 and Br2/CCl4. The ability to decolorize these reagents indicates that compound X is likely an alkene, as alkenes can undergo oxidation with KMnO4 (Baeyer test) and addition reactions with bromine.

**Hint:** Alkenes are characterized by the presence of a carbon-carbon double bond, which is responsible for their reactivity with bromine and permanganate.

### Step 3: Analyze Compound Y
Compound X undergoes ozonolysis to yield compound Y. Ozonolysis typically cleaves double bonds and can produce carbonyl compounds. The problem states that Y can yield 2 moles of iodoform and a sodium salt of a dibasic acid upon treatment with iodine and NaOH.

The iodoform test indicates that Y must contain a methyl ketone (a carbonyl group adjacent to a methyl group). The presence of a dibasic acid suggests that Y also has a structure that allows for the formation of two carboxylic acid groups upon decarboxylation.

**Hint:** The iodoform test is a qualitative test for methyl ketones and some alcohols that can be oxidized to methyl ketones.

### Step 4: Decarboxylation to Form n-butane
The sodium salt of the dibasic acid derived from Y can undergo decarboxylation (removal of CO2) using soda lime to yield n-butane. This indicates that Y must have a structure that, after losing two CO2 molecules, results in the straight-chain alkane n-butane.

**Hint:** Decarboxylation typically occurs at the terminal carbon of a carboxylic acid, leading to the formation of alkanes.

### Step 5: Construct the Structures
1. **Compound D**: Since D is a secondary alcohol, it could be something like 2-octanol.
2. **Compound X**: Upon dehydration of D, X could be an alkene like 2-octene.
3. **Compound Y**: Ozonolysis of 2-octene could yield a compound that contains a methyl ketone and a carboxylic acid, such as 3-oxooctanoic acid.

### Conclusion
Based on the analysis:
- **D** is likely 2-octanol (C8H16O).
- **X** is likely 2-octene.
- **Y** is likely 3-oxooctanoic acid.

Thus, the final answer is:
- **Compound D**: 2-octanol
- **Compound X**: 2-octene
- **Compound Y**: 3-oxooctanoic acid