See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The starting material is hex-1-en-5-ol: H2C=CH-CH2-CH2-CH(OH)-CH3 (a 6-carbon chain with a terminal alkene at C1-C2 and a hydroxyl group at C5). Step 2: Reaction with SOCl2/Pyridine to give (A). SOCl2 with pyridine converts the secondary alcohol (-OH at C5) to a chloride with retention of configuration (via SNi or SN2 with pyridine). The alkene remains intact. So (A) is: H2C=CH-CH2-CH2-CHCl-CH3 (5-chlorohex-1-ene). Step 3: Ozonolysis of (A) with O3/Zn/H2O to give (B), molecular formula C5H9ClO. Ozonolysis of a terminal alkene (H2C=CH-) with reductive workup (Zn/H2O) cleaves the double bond to give formaldehyde (HCHO) from the terminal =CH2 and an aldehyde from the internal carbon. The internal carbon (C2) becomes an aldehyde: OHC-CH2-CH2-CHCl-CH3. This fragment is: 4-chloropentanal = OHC-CH2-CH2-CHCl-CH3, which has formula C5H9ClO. This matches the given molecular formula for (B). Formaldehyde (HCHO) is the other fragment. Step 4: Reduction of (B) with NaBH4 to give (C). NaBH4 reduces the aldehyde (-CHO) to a primary alcohol (-CH2OH). So (B): OHC-CH2-CH2-CHCl-CH3 becomes (C): HOCH2-CH2-CH2-CHCl-CH3. This is: HO-CH2-CH2-CH2-CH(Cl)-CH3, i.e., 4-chloropentan-1-ol. Step 5: Match with options. (c) HO-CH2-CH2-CH2-CH-CH3 with Cl on the CH corresponds exactly to HOCH2CH2CH2CHClCH3 = 4-chloropentan-1-ol. Why other options fail: (a) has OH at C2 and Cl at C4 - this would require a different connectivity. (b) is a 7-carbon compound - too many carbons. (d) has Cl at C3 instead of C4 - wrong position. Therefore, the correct answer is C.