Among the carboxylic acid shown below, the ones that exhibit stereoisomerism and also form , cyclic anhydrides on heating are
`(I) "HOOCCH"(CH_3) CH_2CH_2COOH`
`(II) "HOOCCH"(C_3H_7)COOH`
`(III) "HOOCCH"(C_2H_5)CH_2COOH`
`(IV) "HOOCC"(CH_3)(C_2H_5)COOH`

Options

A. (I) and (II)
B. (I) and (III)
C. (II) and (III)
D. (II) and (IV)

Answer

To solve the problem, we need to analyze each of the given carboxylic acids to determine which ones exhibit stereoisomerism and can form cyclic anhydrides upon heating.

### Step 1: Analyze Compound I
**Structure:** HOOC-CH(CH₃)-CH₂-CH₂-COOH

1. Identify the chiral center: The carbon atom attached to four different groups (COOH, CH₃, CH₂, and H) is a chiral center.
2. Since it has a chiral center, it can exhibit stereoisomerism.
3. Heating this compound will lead to the formation of a cyclic anhydride by the elimination of water (H₂O).

**Conclusion:** Compound I exhibits stereoisomerism and can form a cyclic anhydride.

### Step 2: Analyze Compound II
**Structure:** HOOC-CH(C₃H₇)-COOH

1. Identify the chiral center: The carbon atom attached to COOH, C₃H₇, and H does not have four different substituents (the two COOH groups are identical).
2. Therefore, it does not have a chiral center and cannot exhibit stereoisomerism.
3. It can still form a cyclic anhydride upon heating, but it does not meet the stereoisomerism requirement.

**Conclusion:** Compound II does not exhibit stereoisomerism.

### Step 3: Analyze Compound III
**Structure:** HOOC-CH(C₂H₅)-CH₂-COOH

1. Identify the chiral center: The carbon atom attached to COOH, C₂H₅, CH₂, and H is a chiral center.
2. Thus, it can exhibit stereoisomerism.
3. Heating this compound will also lead to the formation of a cyclic anhydride by the elimination of water.

**Conclusion:** Compound III exhibits stereoisomerism and can form a cyclic anhydride.

### Step 4: Analyze Compound IV
**Structure:** HOOC-CH(CH₃)(C₂H₅)-COOH

1. Identify the chiral center: The carbon atom attached to COOH, CH₃, C₂H₅, and another COOH does not have four different substituents (the two COOH groups are identical).
2. Therefore, it does not have a chiral center and cannot exhibit stereoisomerism.
3. It can still form a cyclic anhydride upon heating, but it does not meet the stereoisomerism requirement.

**Conclusion:** Compound IV does not exhibit stereoisomerism.

### Final Answer
The compounds that exhibit stereoisomerism and can form cyclic anhydrides upon heating are **Compound I and Compound III**.