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. In (b)(ii), (1-chloroethyl)benzene undergoes hydrolysis via the S N1 mechanism because it can form a carbocation which is resonance-stabilised. In this carbocation, the empty p orbital of the positively charged carbon overlaps with the  electron cloud of the benzene ring. This allows the  electrons of the benzene ring to delocalise over the positively charged carbon and disperse the positive charge, thus stabilising the carbocation. In this case, the SN2 mechanism is less favoured because it is a 2 halogenoalkane with one methyl and one very bulky phenyl group attached to the electron deficient C atom, thus hindering the backside attack of the OH nucleophile. 5a A nucleophile is a species which can donate an electron pair to an electron-deficient atom to form a covalent bond.

-4- 5b (i) Reagents and conditions: Excess conc NH3 in ethanol, sealed tube and heat (ii) Reagents and conditions Step 1: ethanolic KCN, heat (or heat under reflux) Step 2: H2, Ni catalyst, heat (iii) Reagents and conditions Step 1: ethanolic KOH, heat (or heat under reflux) Step 2: KMnO 4(aq), NaOH(aq), cold (iv) Br Step 1 Step 2 Step 3 Step 4 COOH HOOC CH2=CH2 Br Br CN NC Reagents and conditions Step 1: ethanolic KOH, heat (or heat under reflux) Step 2: Br2 in CCl4, room temperature, dark Step 3: ethanolic KCN, heat (or heat under reflux) Step 4: H2SO4(aq), heat (or heat under reflux)

-5- 5c The mechanism is SN2. (followed by deprotonation) Then, 5d S gives an immediate precipitate with AgNO3(aq).  The precipitate is AgBr and S is an ionic compound containing C8H20N+ and Br− ions. S is S is formed from simplified nucleophilic substitution reactions as shown below. NH3 + CH3CH2Br  CH3CH2NH2 + HBr CH3CH2NH2 + CH3CH2Br  (CH3CH2)2NH + HBr (CH3CH2)2NH + CH3CH2Br  (CH3CH2)3N + HBr (CH3CH2)3N + CH3CH2Br  (CH3CH2)4N+ Br− 6a The rate of formation of precipitate decreases in the following order: AgI (yellow) > AgBr (pale cream) > AgCl (white) The rate of the nucleophilic substitution reaction of the halogenobutanes decreases in the following order: 2-iodobutane > 2-bromobutane > 2-chlorobutane The nucleophilic substitution reaction involves the cleavage of the carbon-halogen bond and the rate of the reaction depends on the strength of the carbon-halogen bond. The weaker the carbon-halogen bond, the easier it can be cleaved and the faster is the rate of the nucleophilic substitution reaction. Hence the above order of rate of reaction arises because the strength of the carbon–halogen bond increases as follows: C–I < C–Br < C–Cl

-6- 6b In clorobenzene molecule, the p orbital containing the lone pair of electrons on chlorine overlaps with the  electron cloud of the benzene ring. This allows the lone pair of electrons in the p orbital of chlorine to delocalise into the  electron cloud of the benzene ring. This results in partial double bond character in the C– C l bond. This bond is strengthened and hence not easily broken Note: In addition, the electron-rich benzene ring repels the electron-rich nucleophile and also hinders the backside attack by the nucleophile on the electron-deficient carbon. Consequently, chlorobenzene does not under go any nucleophilic substitution reaction. 6c In 2-chlorobut-2-ene, the p orbital containing the lone pair of electrons on the Cl atom overlaps with the  electron cloud of the C=C bond. This allows the lone pair of electrons in the p orbital of chlorine to delocalise into the  electron cloud of the C=C bond. This results in partial double bond character in the C–Cl bond. This bond is strengthened and hence not easily broken. Consequently, 2-chlorobut-2-ene is resistant to nucleophilic substitution under the given conditions. There will not be any C l- ions present to react with Ag+ ions. Hence using 2-chlorobut-2-ene will lead to no obervable change in the experiment. Note: In addition, the electron-rich C=C bond repels the electron-rich nucleophile and hinders the backside attack by the nucleophile on the electron-deficient carbon. Consequently, 2-chlorobut-2-ene does not under go any nucleophilic substitution reaction. 7a Add NaOH(aq) to each sample in a test-tube and heat each mixture in a hot water bath. Cool each mixture and acidify each one with dilute HNO3. Then add AgNO3(aq) to each mixture. For (2-bromoethyl)benzene, a pale cream precipitate of AgBr will be observed. For 1-bromo-2-eth ylbenzene, a pale cream precipitate of AgBr will not be observed. 7b Add NaOH(aq) to each sample in a test-tube and heat each mixture in a hot water bath. Cool each mixture and acidify each one with dilute HNO3. Then add AgNO3(aq) to each mixture. For chloroethane, a white precipitate of AgCl will be observed. For iodoethane, a yellow precipitate of AgI will be obs