Q.Identify A, B, C in the following reaction sequence.
CH3-CH=CH2 --Br2/CCl4, room temperature--> A --Zn--> B --dil. alk. KMnO4--> C

Step 1. Form A. Propene (CH3-CH=CH2) with Br2 in CCl4 at room temperature is halogen addition (section 15.2.4), giving the vicinal dibromide A = 1,2-dibromopropane, CH3-CHBr-CH2Br.

Step 2. Form B. Treating this vicinal dihalide with Zn is exactly the dehalogenation reaction of section 15.2.2 (used there to make alkenes from vicinal dihalides, including this very example, 1,2-dibromopropane + Zn), regenerating the alkene: B = propene, CH3-CH=CH2 (plus ZnBr2).

Step 3. Form C. Treating propene (B) with dilute alkaline KMnO4 is the hydroxylation reaction of section 15.2.4 (again, this exact example -- propene + dilute alkaline KMnO4 -- is worked out in that section), giving the 1,2-diol: C = propane-1,2-diol, CH3-CH(OH)-CH2OH. …

Working through each step of the sequence using the specific halogenation, dehalogenation, and hydroxylation reactions of sections 15 …

• Assuming B must be something OTHER than propene since it 'looks like a new step' -- Zn-dehalogenation of a vicinal dihalide genuinely just regenerates the original alkene it came from. …