Step 1: Understanding the Concept:
Anticonvulsant therapy in veterinary medicine relies on drugs that stabilize neuronal membranes and prevent seizures.
Many anticonvulsants are prodrugs or undergo hepatic metabolism to form active metabolites that have different pharmacokinetic profiles than the parent compound.
Step 2: Detailed Explanation:
Let us analyze both statements in detail:
1. Assertion A: Primidone is a deoxybarbiturate which is converted by liver to phenobarbitone and phenylethyl malonamide.
This statement is correct.
Primidone is a 2-deoxybarbiturate anticonvulsant.
In the liver, it undergoes extensive metabolism via microsomal enzymes to form two active metabolites: phenobarbital (phenobarbitone) and phenylethylmalonamide (PEMA).
Both metabolites possess potent anticonvulsant activity.
Therefore, Assertion A is true.
2. Reason R: Its antiepileptic activity is mainly due to these active metabolites because half life of primidone is more than that of its active metabolites.
This statement is incorrect.
While the antiepileptic activity of primidone is indeed largely mediated by its active metabolites (especially phenobarbital), the reason given is pharmacokinetically incorrect.
The elimination half-life of the parent drug, primidone, is relatively short, typically ranging from $5$ to $15\text{ hours}$ in dogs.
In contrast, the half-life of its active metabolite phenobarbital is much longer, ranging from $30$ to $90\text{ hours}$ in dogs, and PEMA has a half-life of $15$ to $25\text{ hours}$.
Because the parent drug is cleared rapidly while the active metabolites accumulate and persist in systemic circulation, the metabolites drive the long-term therapeutic effects.
The statement incorrectly claims that the half-life of primidone is *longer* than that of its metabolites.
Therefore, Reason R is false.
Step 3: Final Answer:
Assertion A is true, but Reason R is false because the half-life of primidone is much shorter than that of its active metabolites (especially phenobarbital).
Therefore, the correct option is (C).