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Consider the following eight reactions. How many of them produce an alkyl halide as one of the final products? (a) C2H5OH + SOCl2 -> (b) R-I + AgF -> (c) EtOH + NaI + H2SO4 -> (d) EtOH + NaI + H3PO4 -> (e) CH3-CH=CH2 + HBr -> (f) CH3-CH2-Cl + NaI (in acetone) -> (g) cyclopentyl-CH(Cl)-CH3 + moist Ag2O -> (h) CH3-O-C2H5 + HI ->

1. 5
2. 6
3. 4
4. 7

Correct answer: 5

Solution

(a) EtOH + SOCl2 -> C2H5Cl (ethyl chloride) + SO2 + HCl. YES, alkyl halide. (b) R-I + AgF -> R-F + AgI. YES, alkyl halide (fluoride). (c) EtOH + NaI + H2SO4 -> C2H5I + NaHSO4 + H2O. YES, alkyl halide. (d) EtOH + NaI + H3PO4: H3PO4 is a weaker acid than H2SO4 and does not protonate ethanol well enough for NaI substitution. This reaction does NOT give alkyl iodide efficiently. NO. (e) CH3-CH=CH2 + HBr -> CH3-CHBr-CH3 (Markovnikov). YES, alkyl halide. (f) C2H5Cl + NaI in acetone -> C2H5I + NaCl (Finkelstein reaction). YES, alkyl halide. (g) Cyclopentyl-CH(Cl)-CH3 + moist Ag2O -> cyclopentyl-CH(OH)-CH3 + AgCl. Product is alcohol, NOT alkyl halide. NO. (h) CH3-O-C2H5 + HI -> CH3I + C2H5OH or CH3OH + C2H5I. YES, alkyl halide (iodide). Count of YES: (a),(b),(c),(e),(f),(h) = 6. Count of NO: (d),(g) = 2. Answer = 6.

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