See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Define the categories: - Polar protic (PP): high dielectric constant/dipole moment AND has O-H or N-H bonds capable of hydrogen bond donation - Polar aprotic (PA): high dielectric constant/dipole moment but NO O-H or N-H bonds (cannot donate H-bonds) - Non-polar aprotic (NPA): low dielectric constant, no O-H or N-H bonds - Non-polar protic (NPP): low dielectric constant but has O-H or N-H bonds (rare category) Step 2 - Analyze compound A (DMSO, dimethyl sulfoxide): DMSO has a highly polar S=O bond giving it a large dipole moment (dielectric constant ~47). It has NO O-H or N-H bonds - the oxygen is only bonded to sulfur, and the two methyl groups have only C-H bonds. Therefore DMSO cannot donate hydrogen bonds. Classification: Polar Aprotic (PA). Step 3 - Analyze compound B (methanesulfinamide, CH3S(=O)NH2): This molecule has a polar S=O group making it polar overall, AND it has an NH2 group which contains N-H bonds capable of donating hydrogen bonds. Therefore it is both polar and protic. Classification: Polar Protic (PP). Step 4 - Analyze compound C (toluene, methylbenzene): Toluene consists of a benzene ring with a methyl substituent. The molecule is essentially nonpolar (dielectric constant ~2.4) due to the symmetric electron distribution of the aromatic ring and the nonpolar methyl group. It has only C-H bonds, so it cannot donate hydrogen bonds (aprotic). Classification: Non-polar Aprotic (NPA). Step 5 - Analyze compound D (DMF, N,N-dimethylformamide): DMF has a highly polar amide C=O group giving a large dipole moment (dielectric constant ~37). The nitrogen bears two methyl groups, so there are NO N-H bonds available for hydrogen bond donation. The carbonyl oxygen cannot donate H-bonds either. Therefore DMF is polar but cannot donate H-bonds. Classification: Polar Aprotic (PA). Step 6 - Why other classifications fail: - A cannot be PP because it lacks any O-H or N-H bonds - B cannot be PA because the NH2 group provides protic character - C cannot be PA or PP because its dielectric constant is too low to be considered polar - D cannot be PP because both N-H bonds are replaced by N-methyl groups, eliminating H-bond donor ability Therefore, the correct answer is {"A": "PA", "B": "PP", "C": "NPA", "D": "PA"}.