In a nucleophilic substitution reaction: R $$-$$ Br + Cl$$^-$$ $$\xrightarrow{DMF}$$ R $$-$$ Cl + Br$$^-$$, which one of the following undergoes complete inversion of configuration?

The reaction given is a nucleophilic substitution: R-Br + Cl⁻ → R-Cl + Br⁻ in DMF solvent. DMF is a polar aprotic solvent, which favors the SN2 mechanism. In SN2 reactions, the nucleophile attacks from the back side, leading to complete inversion of configuration at the chiral carbon center. For inversion to occur, the carbon attached to the leaving group (Br) must be chiral and the reaction must proceed via SN2.

Now, let's analyze each option to determine which compound has a chiral carbon that undergoes SN2 and hence complete inversion.

Option A: C₆H₅CHC₆H₅Br, which is (C₆H₅)₂CHBr. The carbon bonded to Br is attached to two phenyl groups (C₆H₅), one hydrogen (H), and bromine (Br). Since the two phenyl groups are identical, this carbon is not chiral (it has two identical substituents). The compound is benzylic and secondary, so it undergoes SN1 in DMF due to resonance stabilization of the carbocation, leading to racemization, not inversion.

Option B: C₆H₅CCH₃C₆H₅Br, which is (C₆H₅)₂CBrCH₃. The carbon bonded to Br is attached to two phenyl groups (C₆H₅), one methyl group (CH₃), and bromine (Br). Again, the two phenyl groups are identical, so this carbon is not chiral. The compound is tertiary and undergoes SN1, leading to racemization.

Option C: C₆H₅CH₂Br, which is benzyl bromide. The carbon bonded to Br is CH₂Br, attached to two hydrogens (H), one benzyl group (C₆H₅CH₂-), and bromine (Br). With two identical H atoms, it is not chiral. Though primary and benzylic, it can undergo both SN1 (due to resonance) and SN2 in DMF. However, SN1 may occur, preventing complete inversion.

Option D: C₆H₅CH₂CH₂Br, which is 2-bromoethylbenzene. The carbon bonded to Br is CH₂Br, attached to two hydrogens (H), one ethyl group (C₆H₅CH₂CH₂-), and bromine (Br). Like option C, it has two identical H atoms, so it is not chiral. However, this carbon is primary and not benzylic (the phenyl group is attached to the adjacent carbon). In DMF, it undergoes pure SN2 without competing SN1, as the carbocation would not be stabilized. If this carbon were chiral (e.g., if one H were different), it would undergo complete inversion via SN2. Among the options, it is the only one that exclusively follows SN2, making it the correct choice for complete inversion if chirality were present.

Although none of the compounds have a chiral carbon, option D is selected because it undergoes a clean SN2 mechanism, which is necessary for complete inversion. The other options either lack chirality or undergo SN1, which does not guarantee inversion.

Hence, the correct answer is Option D.