`Me_3 C CH_2 COOH `is more acidic than `Me_3 SiCH_2 COOH.`

Answer

To determine whether `Me_3CCH_2COOH` is more acidic than `Me_3SiCH_2COOH`, we need to analyze the acidity of both compounds by considering the stability of the anions formed when they lose a proton (H⁺).

### Step-by-Step Solution:

1. **Identify the Structures**:
- `Me_3CCH_2COOH` has a tert-butyl group (`Me_3C` or `C(CH₃)₃`) attached to a carboxylic acid (`COOH`).
- `Me_3SiCH_2COOH` has a tert-butyl group attached to a silicon atom instead of carbon, followed by a carboxylic acid.

2. **Understanding Acidity**:
- Acidity is determined by the ability of a compound to donate a proton (H⁺). When the proton is lost, an anion is formed.
- The stability of this anion is crucial; the more stable the anion, the stronger the acid.

3. **Formation of Anions**:
- For `Me_3CCH_2COOH`, when it loses H⁺, it forms `Me_3CCH_2COO⁻`.
- For `Me_3SiCH_2COOH`, when it loses H⁺, it forms `Me_3SiCH_2COO⁻`.

4. **Analyzing Anion Stability**:
- The stability of the anion is influenced by the electronegativity of the atom bonded to the carboxylate group (the COO⁻ part).
- Carbon (C) is more electronegative than silicon (Si). This means that in `Me_3CCH_2COO⁻`, the carbon can better stabilize the negative charge on the oxygen due to its higher electronegativity.

5. **Comparing Electronegativity**:
- As we move down the group in the periodic table, electronegativity decreases. Therefore, silicon has a lower electronegativity than carbon.
- This lower electronegativity means that the anion formed from `Me_3SiCH_2COOH` is less stable than the anion formed from `Me_3CCH_2COOH`.

6. **Conclusion**:
- Since the anion from `Me_3CCH_2COOH` is more stable than that from `Me_3SiCH_2COOH`, it follows that `Me_3CCH_2COOH` is indeed more acidic than `Me_3SiCH_2COOH`.

### Final Answer:
The statement is **true**: `Me_3CCH_2COOH` is more acidic than `Me_3SiCH_2COOH`.