RI 2024 Tutorial 14 RX Ans

-1- Raffles Institution Year 5 H2 Chemistry 2024 Tutorial 14: Halogen Derivatives (Suggested Answers) Discussion Questions – Suggested Answers 3a Preparation of CH3CH2Cl Reagent: PCl5 Condition: room temperature Equation: CH3CH2OH + PCl5  CH3CH2Cl + POCl3 + HCl Preparation of CH3CH2Br Reagent: PBr3 or HBr(g) Condition: heat under reflux if HBr is used Equations: 3CH3CH2–OH + PBr3  3CH3CH2–Br + H3PO3 CH3CH2OH + HBr  CH3CH2Br + H2O heat Preparation of CH3CH2I Reagent: red P, I2 Condition: room temperature Equations: 2P + 3I2  2PI3 3CH3CH2OH + PI3  3CH3CH2I + H3PO3 3b From the Data Booklet: BE(C–Cl) = 340 kJ mol−1; BE(C–Br) = 280 kJ mol−1; BE(C–I) = 240 kJ mol−1. The reactivity towards nucleophilic reagents increases from CH 3CH2Cl to CH3CH2Br to CH3CH2I. This is due to decreasing strength of the carbon–halogen bond in the following order: C–Cl > C–Br > C–I. 3c Q R Cl CH3HO NH2 CH3HO

-2- 4a Compare expt. 1 and 2. When [NaOH] is doubled, initial rate is doubled.  rate  [NaOH]  order of reaction with respect to NaOH = 1 Compare expt. 2 and 3. When [NaOH] and [(2-chloroethyl)benzene] is doubled, initial rate is quadrupled. Since order of reaction with respect to NaOH =1,  rate  [(2-chloroethyl)benzene]  order of reaction with respect to (2-chloroethyl)benzene = 1 OR Let the rate equation be: rate = k [NaOH] [(2-chloroethyl)benzene] n Compare expt. 2 and 3  n = 1  order of reaction with respect to (2-chloroethyl)benzene = 1 Hence the rate equation is: rate = k [NaOH] [(2-chloroethyl)benzene] Since the overall order of the reaction is 2, the mechanism involved in the reaction is S N2. 4b(i) (1-chloroethyl)benzene displays enantiomerism. Stereoisomers of (1-chloroethyl)benzene:

-3- 4b(ii) The mechanism is SN1. Step 1: Step 2: In step 1, the optically active (1-chloroethyl)benzene forms a carbocation via the heterolytic fission of the C–Cl bond. This carbocation is trigonal planar with respect to the positively charged carbon. In step 2, the OH  nucleophile can attack this positively charged carbon from either side of the trigonal plane with equal likelihood. This results in 50% of the product molecules formed with retention of configuration and the other 50% of the product molecules formed with inversion of configuration. Consequently, the product solution is a racemic mixture and hence is opticall y inactive. 4c In (a), (2-chloroethyl)benzene undergoes hydrolysis via the S N2 mechanism because it is a primary halogenoalkane with little steric hindrance for the OH nucleophile to attack the electron-deficient carbon from the side directly opposite the chlorine atom. In this case, (2-chloroethyl)benzene forms a less stable carbocation and hence the S N1 mechanism is not favoured