See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: To convert propene (CH3CH=CH2) into 1-propanol (CH3CH2CH2OH), we need anti-Markovnikov addition of OH to the terminal carbon (C1) of the double bond, with retention of the carbon skeleton. Step 1 - Identify the target product: 1-propanol is CH3CH2CH2OH, where the OH group is on the terminal (less substituted) carbon. This is the anti-Markovnikov product. Step 2 - Analyze each reagent set: (a) KMnO4 (alkaline): Alkaline KMnO4 oxidizes alkenes to diols (syn-dihydroxylation) or, under vigorous conditions, cleaves the double bond giving carboxylic acids/ketones. It does NOT give a monoalcohol product; it gives a 1,2-diol or oxidative cleavage products. Does not give 1-propanol. (b) Osmium tetroxide (OsO4/CH2Cl2): OsO4 effects syn-dihydroxylation of the double bond, converting CH3CH=CH2 into a 1,2-diol (propane-1,2-diol), not 1-propanol. Does not give 1-propanol. (c) B2H6 (diborane) followed by alkaline H2O2: This is hydroboration-oxidation. Diborane adds to the alkene in an anti-Markovnikov, syn fashion, placing boron on the less hindered (terminal) carbon. Subsequent oxidation with alkaline H2O2 replaces boron with OH with retention of configuration. For propene, this gives 1-propanol (OH on C1). This is exactly the desired conversion. (d) O3/Zn: Ozonolysis with Zn (reductive workup) cleaves the C=C double bond to give two carbonyl compounds. For propene, this would give formaldehyde (HCHO) and acetaldehyde (CH3CHO), not 1-propanol. The carbon skeleton is broken. Step 3 - Conclusion: Only hydroboration-oxidation (B2H6 followed by alkaline H2O2) achieves anti-Markovnikov addition of water across the double bond, yielding 1-propanol from propene without breaking the carbon chain or over-oxidizing. Therefore, the correct answer is C.