Which of the following is not used in methyl alcohol poisoning :
**Core Concept**
Methyl alcohol (methanol) poisoning is a toxic condition caused by the ingestion of methanol, leading to metabolic acidosis, visual disturbances, and potential neurological damage. The treatment of methanol poisoning involves the administration of antidotes to inhibit the metabolism of methanol to its toxic metabolites.
**Why the Correct Answer is Right**
Methanol is metabolized by alcohol dehydrogenase to form formaldehyde, which is then converted to formic acid by aldehyde dehydrogenase. Formic acid is the primary toxic metabolite responsible for the clinical manifestations of methanol poisoning. The administration of fomepizole, a competitive inhibitor of alcohol dehydrogenase, is used to inhibit the metabolism of methanol to formic acid. Ethanol is also used to inhibit alcohol dehydrogenase, but it is less potent than fomepizole. Hemodialysis is used to remove methanol and its metabolites from the body.
**Why Each Wrong Option is Incorrect**
* **Option A:** Fomepizole is used in methanol poisoning, so this option is incorrect.
* **Option B:** Ethanol is used in methanol poisoning, so this option is incorrect.
* **Option D:** Hemodialysis is used in methanol poisoning, so this option is incorrect.
**Clinical Pearl / High-Yield Fact**
Fomepizole is a more potent inhibitor of alcohol dehydrogenase than ethanol and is the preferred treatment for methanol poisoning. It is essential to administer fomepizole promptly in cases of suspected methanol poisoning to prevent the formation of toxic metabolites.
**Correct Answer: C.