The 'X' in the given reaction ?
`{:(CH_2OH),(|),(CH_2OH):} + "oxalic acid " overset(210^(@)C)to underset("(major product)")(X)`

Options

A. `{:(CH_2),(||),(CH_2):}`
B. `{:(CH-OH),(||),(CH_2):}`
C. `{:(CHO),(|),(CHO):}`
D. `{:(CH_2OH),(|),(CHO):}`

Answer

To solve the problem of determining the product 'X' when ethylene glycol reacts with oxalic acid at 210°C, we will follow these steps:

### Step 1: Identify the Reactants
The reactants are ethylene glycol (HOCH2CH2OH) and oxalic acid (HOOC-COOH). Ethylene glycol can be represented as CH2OH (for each hydroxymethyl group), and oxalic acid has the structure of two carboxylic acid groups connected by a carbon chain.

### Step 2: Understand the Reaction Type
The reaction between an alcohol (ethylene glycol) and a carboxylic acid (oxalic acid) typically leads to esterification. In this case, since we have a diol (ethylene glycol) and a dicarboxylic acid (oxalic acid), we can expect the formation of a cyclic ester (a lactone) through intramolecular esterification.

### Step 3: Draw the Structures
1. **Ethylene Glycol**:
- CH2OH - CH2OH
2. **Oxalic Acid**:
- HOOC - COOH

### Step 4: Mechanism of Reaction
1. The hydroxyl group of ethylene glycol will attack the carbonyl carbon of one of the carboxylic groups of oxalic acid.
2. This will lead to the formation of a tetrahedral intermediate.
3. An intramolecular proton transfer will occur, leading to the departure of water and the formation of a cyclic structure.
4. The remaining hydroxyl group will then attack the other carbonyl carbon, leading to a second intramolecular proton transfer and the formation of a cyclic ester.

### Step 5: Final Product
The final product 'X' will be a cyclic compound formed by the reaction of ethylene glycol and oxalic acid. The structure will resemble a five-membered ring with two carbonyl groups and an ether linkage.

### Conclusion
The major product 'X' formed from the reaction of ethylene glycol and oxalic acid at 210°C is a cyclic ester, specifically a lactone.

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