IB DP Chemistry Reactivity 3.4 Electron-pair sharing reactions HL Paper 2- Exam Style Questions - New Syllabus
Question
Phenylethanone is a fragrant compound that occurs naturally in fruits such as bananas and apples.
(a) Phenylethanone may be synthesised in a two-stage process from phenylethene:
(i) Draw the structural formula of the intermediate compound \([X]\). [\(\,1\,\)]
(ii) Outline why the intermediate compound, \([X]\), can exhibit stereoisomerism. [\(\,1\,\)]
(iii) State the reagent required for the second stage of the synthesis, \(B\). [\(\,1\,\)]
(iv) Determine the compound that will be formed as a minor product in this two-stage synthesis, and outline why this will occur. [\(\,2\,\)]
(b) When heated with a mixture of concentrated sulfuric and nitric acids, phenylethanone is nitrated, in a similar manner to benzene, to form \(\,3\)-nitrophenylethanone.
(i) Write the formula of the electrophile produced in this acid mixture. [\(\,1\,\)]
(ii) Explain the mechanism of the reaction between phenylethanone and the nitrating agent, using curly arrows to represent the movement of electron pairs. [\(\,4\,\)]
(c) Chemists have a “rule of thumb” that raising the temperature by \(\,10\,^{\circ}\mathrm{C}\) doubles the reaction rate. Deduce the activation energy, in \(\,\mathrm{kJ\,mol^{-1}}\), assuming that a rise in temperature from \(\,25\,^{\circ}\mathrm{C}\) to \(\,35\,^{\circ}\mathrm{C}\) doubled the rate of this reaction. Use of the data booklet. [\(\,3\,\)]
Data (inserted from IB Chemistry Data Booklet): Gas constant \(R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}\);\; Arrhenius equation \(k=Ae^{-E_a/RT}\);\; \(T_1=298\ \mathrm{K}\) (for \(\,25\,^{\circ}\mathrm{C}\)), \(T_2=308\ \mathrm{K}\) (for \(\,35\,^{\circ}\mathrm{C}\)).
▶️ Answer/Explanation
(a)(i) \([X]\) is the acid-catalysed hydration product of phenylethene (Markovnikov addition): \(\displaystyle [X]=\mathrm{C_6H_5{-}CH(OH){-}CH_3}\) \(\;(1\text{-phenylethanol, secondary alcohol})\).
(a)(ii) The carbon bearing \(\mathrm{-OH}\) in \([X]\) is attached to four different groups \(\big(\mathrm{H},\,\mathrm{CH_3},\,\mathrm{C_6H_5},\,\mathrm{OH}\big)\) and is therefore a chiral centre \(\Rightarrow\) pairs of non-superimposable mirror-image stereoisomers (enantiomers).
(a)(iii) Oxidising agent for stage \(B\) (secondary alcohol \(\rightarrow\) ketone): \(\displaystyle \text{acidified }\mathrm{K_2Cr_2O_7}\ \text{(}\mathrm{H^+}/\mathrm{Cr_2O_7^{2-}}\text{)}\) or \(\displaystyle \text{acidified }\mathrm{KMnO_4}\ \text{(}\mathrm{H^+}/\mathrm{MnO_4^-}\text{)}\).
(a)(iv) Minor product (two-stage): if water adds in the anti-Markovnikov sense in stage \([X]\), the primary alcohol \(\mathrm{C_6H_5{-}CH_2{-}CH_2OH}\) can form and on oxidation gives \(\displaystyle \text{phenylethanoic acid (}\mathrm{C_6H_5{-}CH_2COOH}\text{)}\) or under milder oxidation \(\displaystyle \text{phenylethanal (}\mathrm{C_6H_5{-}CH_2CHO}\text{)}\). Reason: regioselectivity in the hydration step allows the hydroxyl group to add to the terminal carbon (anti-Markovnikov), leading to a primary alcohol that over-oxidises to the acid in standard dichromate conditions.
(b)(i) \(\displaystyle \text{Electrophile: }\mathrm{NO_2^{+}}\ \text{(nitronium ion)}\).
(b)(ii) Electrophilic aromatic substitution (nitration): • Generate \(\mathrm{NO_2^{+}}\) in the \(\mathrm{H_2SO_4/HNO_3}\) mixture. • Curly arrow from the delocalized \(\pi\)-electron cloud of the ring in phenylethanone to \(\mathrm{NO_2^{+}}\) to form a \(\sigma\)-complex (arenium ion). • Show the arenium ion with positive charge on the ring (resonance-stabilized). • Curly arrow from the \(\mathrm{C{-}H}\) bond on the substituted carbon to the ring to re-form aromaticity; \(\mathrm{HSO_4^-}\) (or \(\mathrm{H^+}\)/\(\mathrm{HSO_4^-}\)) removes \(\mathrm{H^+}\). • Product: \(\,3\)-nitrophenylethanone (meta-directing \(\mathrm{C{=}O}\) group in phenylethanone directs nitration to the meta position).
(c) Arrhenius approach with the “doubling per \(\,10\,^{\circ}\mathrm{C}\)” rule: Given \(\displaystyle \frac{k_2}{k_1}=2\) for \(T_1=298\ \mathrm{K}\) and \(T_2=308\ \mathrm{K}\), \[ \ln\!\left(\frac{k_2}{k_1}\right) \;=\; -\frac{E_a}{R}\!\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \;\Rightarrow\; E_a \;=\; R\,\ln 2\;\Big/\;\!\left(\frac{1}{T_1}-\frac{1}{T_2}\right). \] Numerical substitution: \[ R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}},\quad \ln 2=0.693,\quad \left(\frac{1}{298}-\frac{1}{308}\right)=1.0895\times 10^{-4}\ \mathrm{K^{-1}}. \] \[ E_a=\frac{(8.31)(0.693)}{1.0895\times 10^{-4}} =5.29\times 10^{4}\ \mathrm{J\,mol^{-1}} =\boxed{52.9\ \mathrm{kJ\,mol^{-1}}}. \] (Accept \(\,\approx 50\text{–}55\ \mathrm{kJ\,mol^{-1}}\).)