Proton decoupled 13C NMR spectrum of a bicyclooctane (C8H14), exhibits only two signals. The structure of the compound is A. B. C. D.

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The correct answer is $\boxed{\text{A}}$.

A bicyclooctane is a hydrocarbon with the molecular formula C8H14. It has a ring structure with eight carbon atoms and fourteen hydrogen atoms. The two signals in the proton decoupled 13C NMR spectrum of a bicyclooctane indicate that there are two types of carbon atoms in the molecule. The two types of carbon atoms are those that are in the ring and those that are not in the ring. The carbon atoms in the ring are all equivalent, so they give rise to one signal. The carbon atoms that are not in the ring are all equivalent, so they give rise to another signal.

The structure of $\text{A}$ is consistent with the two signals in the proton decoupled 13C NMR spectrum. The carbon atoms in the ring are all equivalent, so they give rise to one signal. The carbon atoms that are not in the ring are all equivalent, so they give rise to another signal.

The structures of $\text{B}$, $\text{C}$, and $\text{D}$ are not consistent with the two signals in the proton decoupled 13C NMR spectrum. In $\text{B}$, the carbon atoms in the ring are not all equivalent, so they would give rise to more than one signal. In $\text{C}$, the carbon atoms that are not in the ring are not all equivalent, so they would give rise to more than one signal. In $\text{D}$, there are three types of carbon atoms, so they would give rise to three signals.