An organic acid (A), `C_(5)H_(10)O_(2)` reacts with `Br_(2)` in the presence of phosphorous to give (B). Compound (B) contains an asymmetric carbon atom and yield (C) on dehydrabromination. Compound (C) does not show geometric isomerism and on decarboxylation give an alkene (D) which on ozonolysis gives (E) and (F), compound (E) gives a positive schiff's test but (F) does not. Give structures of (A) to (F) with reasons.

Answer

To solve the problem step-by-step, we will identify the compounds (A) to (F) based on the given information and reactions.

### Step 1: Identify Compound (A)
Given that compound (A) is an organic acid with the formula `C5H10O2`, we can deduce its structure. A possible structure for an organic acid with this formula could be:
- **Structure of (A)**: CH3-CH2-CH2-COOH (pentanoic acid)

### Step 2: Reaction of (A) with `Br2` in the presence of phosphorus
When compound (A) reacts with bromine (`Br2`) in the presence of phosphorus, a bromination reaction occurs. The bromine adds to the carbon chain adjacent to the carboxylic acid group.
- **Structure of (B)**: CH3-CH2-CH2-CHBr-COOH

### Step 3: Identify the asymmetric carbon in (B)
In compound (B), the carbon atom bonded to the bromine, the carboxylic acid group, a hydrogen atom, and the rest of the carbon chain is asymmetric. This confirms that (B) contains an asymmetric carbon atom.

### Step 4: Dehydrobromination to form (C)
Dehydrobromination involves the elimination of HBr from compound (B). This results in the formation of a double bond.
- **Structure of (C)**: CH3-CH2-CH=CH-COOH

### Step 5: Analyze compound (C) for geometric isomerism
Compound (C) does not show geometric isomerism because the two methyl groups are on the same side of the double bond, thus it is a cis isomer.

### Step 6: Decarboxylation of (C) to form (D)
Decarboxylation involves the removal of the carboxylic acid group (CO2) from compound (C), leading to the formation of an alkene.
- **Structure of (D)**: CH3-CH2-CH=CH2 (butene)

### Step 7: Ozonolysis of (D) to form (E) and (F)
Ozonolysis of the alkene (D) will cleave the double bond and form two products:
- **Structure of (E)**: CH3-CO-CH3 (acetone, a ketone)
- **Structure of (F)**: HCHO (formaldehyde, an aldehyde)

### Step 8: Analyze the Schiff's test results
Compound (E) (acetone) does not give a positive Schiff's test, while compound (F) (formaldehyde) does give a positive Schiff's test, confirming the presence of the aldehyde functional group.

### Summary of Structures
- **(A)**: CH3-CH2-CH2-COOH (pentanoic acid)
- **(B)**: CH3-CH2-CH2-CHBr-COOH
- **(C)**: CH3-CH2-CH=CH-COOH
- **(D)**: CH3-CH2-CH=CH2 (butene)
- **(E)**: CH3-CO-CH3 (acetone)
- **(F)**: HCHO (formaldehyde)