Which one of the following is correct structure for cytosine?

Step 1: Formation of Product 'A' (Acid-Catalyzed Dehydration)

The starting material is cyclohexylmethanol (a primary alcohol). It is treated with a strong acid ($H_3PO_4$) and heat ($120^\circ C$). This triggers a dehydration reaction, but with a crucial intermediate step.

1. Carbocation Formation: The acid protonates the $-OH$ group, turning it into water, which is a good leaving group. When the water molecule leaves, it forms a primary carbocation ($-CH_2^+$) attached to the cyclohexane ring.
2. Rearrangement (The Catch): Primary carbocations are highly unstable. To become more stable, the molecule undergoes a 1,2-hydride shift. A hydrogen atom from the adjacent carbon on the ring (which is a tertiary carbon) shifts over to the primary carbon. This moves the positive charge onto the ring itself, creating a highly stable tertiary carbocation. 3. Elimination (Zaitsev's Rule): Finally, a water molecule acts as a weak base and removes a neighboring proton to form a double bond. According to Zaitsev's rule, the molecule will form the most substituted (and therefore most stable) alkene possible. Removing a proton from inside the ring creates a highly stable trisubstituted double bond.

Result for A:

Product A is 1-methylcyclohexene. (It is a cyclohexane ring with a double bond, and a methyl group attached to one of the carbons sharing the double bond).

Step 2: Formation of Product 'P' (Hydroboration-Oxidation)

Now we take Product A (1-methylcyclohexene) and subject it to hydroboration-oxidation using $(BH_3)_2$ followed by $H_2O_2$ and $OH^-$.

As we established in the previous question, this reagent combo adds water (an $H$ and an $OH$) across the double bond following specific rules:

1. Anti-Markovnikov Regioselectivity: The $-OH$ group will attach to the least substituted carbon of the double bond. In 1-methylcyclohexene, one side of the double bond has a methyl group (tertiary), and the other side only has a hydrogen (secondary). Therefore, the $-OH$ attaches to the secondary carbon.
2. Syn-Addition Stereochemistry: The new $H$ and the new $-OH$ add to the exact same face of the ring. Because they crowd one side, the existing methyl group is pushed to the opposite face.

Result for P:

Product P is 2-methylcyclohexanol (specifically, trans-2-methylcyclohexanol, because the methyl group and the hydroxyl group will be pointing in opposite directions).