The correct answer is option D, which states that all of the following statements are true for the compound (R)-2-butanol. Let's break down each statement and explain why it is true.

a) This compound is chiral:
- A molecule is considered chiral if it cannot be superimposed on its mirror image. In other words, it lacks a plane of symmetry.
- (R)-2-butanol has four different substituents attached to the carbon atom bearing the hydroxyl group. These substituents are a hydrogen atom, a methyl group (CH3), an ethyl group (C2H5), and a propyl group (C3H7).
- Due to this asymmetry, (R)-2-butanol does not possess a plane of symmetry, making it a chiral compound.

b) This compound is optically active:
- Optically active compounds are chiral compounds that rotate the plane of polarized light.
- (R)-2-butanol is optically active because it is chiral.
- The direction and amount of rotation depend on factors such as the concentration of the compound, the path length of the light, and the wavelength of the light.

c) This compound has an enantiomer:
- Enantiomers are pairs of molecules that are mirror images of each other but cannot be superimposed.
- Since (R)-2-butanol is chiral, it has a specific configuration at the chiral carbon atom.
- The mirror image of (R)-2-butanol, which would have the opposite configuration at the chiral carbon, is its enantiomer. This enantiomer is called (S)-2-butanol.
- (R)-2-butanol and (S)-2-butanol are enantiomers of each other.

In summary, (R)-2-butanol meets all the criteria for options a, b, and c. It is chiral because it lacks a plane of symmetry, it is optically active because it can rotate the plane of polarized light, and it has an enantiomer, which is (S)-2-butanol. Therefore, the correct answer is option D, "All of the above."