Tetravalence Of Carbon Shapes Of Organic Compounds

Step 1: Analyzing the Reaction
The organic compound Q reacts with PhSNa (phenyl sodium sulfide). PhSNa is a nucleophile, which means it will likely undergo a nucleophilic substitution reaction with Q. The reaction would result in the formation of a phenylthiol derivative if Q contains a suitable leaving group (such as a halide).
Step 2: Understanding the Carius Test
The Carius test involves reacting the compound with sodium peroxide (Na₂O₂) and then adding BaCl₂. A positive Carius test indicates the presence of a halogen in the organic compound, because the halogen will react with BaCl₂ to form a precipitate of BaX₂ (where X is a halogen such as Cl, Br, or I).
Step 3: Identifying the Structure of Q
Upon analyzing the structures of the compounds in the images: - The first image shows a compound with a nitro group (NO₂) and a fluoro group (F) attached to a benzene ring. This compound does not have a halogen that could give a positive Carius test, as Fluorine (F) is not typically detected by the Carius test, and the nitro group is not reactive in this context.
- The second image shows a compound with a nitro group (NO₂), a methylsulfonyl group (MeS), and a chlorine (Cl) atom attached to the benzene ring. The chlorine atom is a halogen, which will give a positive Carius test because it can react with BaCl₂ to form a precipitate of BaCl₂.
Step 4: Conclusion
The compound Q in the second image, which contains a chlorine atom, will give a positive Carius test. The first image does not contain a halogen that would give such a test. Thus, the correct options for Q are:
A and B.