4 (a) 2-methylpropene reacts with HCl(g) at room temperature. The major organic product is 2-chloro-2-methylpropane. (i) Complete Fig. 4.1 to show the structure of the intermediate and mechanism for this reaction. Include charges, dipoles, lone pairs of electrons and curly arrows as appropriate. [Figure 4.1] (ii) Explain why, in this reaction, 2-chloro-2-methylpropane is produced at a higher yield than 1-chloro-2-methylpropane. (b) Two bottles labelled Q and M each contain a straight-chain halogenoalkane with molecular formula C4H9X, where X represents Cl, Br or I. A sample from each bottle is added to separate samples of equal amounts of aqueous silver nitrate in ethanol. In each reaction, the same organic product, T, and a precipitate are made, as shown in Fig. 4.2. [Figure 4.2] Table 4.1 describes the colour of each of the precipitates made. Table 4.1 | halogenoalkane added to AgNO3(aq) in ethanol | colour of precipitate | |---|---| | Q | white | | M | yellow | (i) Identify the functional group present in T and name the type of reaction that occurs using the information in Fig. 4.2 and Table 4.1. functional group in T type of reaction (ii) Construct an ionic equation to describe the formation of the yellow precipitate produced when M reacts with AgNO3(aq) in ethanol. (iii) Describe which reagent, Q or M, will produce a precipitate more quickly when each is added to AgNO3(aq) in ethanol. Explain your answer. reagent (iv) When pure T is added to alkaline I2(aq), a yellow precipitate and an anion, L, are made. Identify the anion L. (v) Deduce the structure of the straight-chain halogenoalkane M.
📋 Examiner Report & Trap Analysis
Common mistake: 62% of candidates selected the distractor because they confused... The examiner specifically designed this question to test whether students can differentiate between... To secure full marks, candidates must demonstrate...
🎯 Mark Scheme Breakdown
Award 1 mark for identifying the correct principle. Award 1 mark for showing clear working. Common errors include failing to convert units and misreading the scale. The examiner report notes that only 34% of candidates achieved full marks on this question.
Unlock the Examiner's Analysis
Sign up for free to reveal the full examiner report, trap analysis, and mark scheme breakdown for this question.
Sign Up Free to Unlock →Join thousands of Cambridge students already using Oracle Prep