14. Hydrocarbons
A section of Chemistry, 9701
Listing 10 of 313 questions
The compound Advantame is a sweetener that tastes approximately 25 000 times sweeter than sucrose. RO HO OH CH3 H N O O O O Advantame N H Advantame is optically active. On the diagram of Advantame, circle all the chiral carbon atoms. The decomposition of Advantame produces three molecules, J, K and L. The RO–group in Advantame is unreactive. HO OH OH H N O O + O OH H2N J K + L RO Suggest possible reagents and conditions for this decomposition. Name the type of reaction occurring. Draw the structure of L in the box above. Aqueous bromine was added dropwise to a solution of J until the bromine was in excess. State what you would observe. J has the molecular formula C14H19O6N. Use this formula to write an equation for the reaction of excess aqueous sodiumhydroxide with one mole of J. State what you would observe when an excess of aqueous bromine is added to a solution of K. K can be polymerised. Draw the structure of the polymer showing two repeat units. The linkage between the monomer units should be fully displayed. Use the Data Booklet to help you answer this question. The carbon-13 NMR spectrum of K was recorded. O K OH H2N State how many different carbon environments are present in K. The chemical shifts, δ, due to two of the carbon atoms x and y present in K are given in the table. carbon atom δ / ppm x y On the structure of K, circle and label two carbon atoms which could correspond to x and y.
9701_m17_qp_42
THEORY
2017
Paper 4, Variant 2
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2-Chloropropanoic acid, CH3CHCl COOH, is used in many chemical syntheses. An equilibrium is set up when CH3CHCl COOH is added to water. Write the equation for this equilibrium. 0.150 mol of CH3CHCl COOH dissolves in 250 cm3 of distilled water to produce a solution of pH1.51. Calculate the pKa of CH3CHCl COOH.  pKa = An equal concentration of aqueous propanoic acid has pH2.55. Explain the difference in the pH of solutions of equal concentration of CH3CHCl COOH and propanoic acid. When CH3CHCl COOH reacts with aqueous NH3, alanine forms. alanine C H CH3 H2N COOH Alanine is an amino acid. Its isoelectric point is 6.1. State what is meant by isoelectric point. Give the structural formula of alanine at pH 2. Alanine exists as a pair of optical isomers. The structure of one optical isomer is shown in Fig.5.2. Draw the three-dimensional structure of the other optical isomer of alanine. optical isomer 1 C CH3 COOH H H2N optical isomer 2  Polymer C forms from the reaction between alanine and 4-aminobutanoic acid, H2N(CH2)3COOH. Draw a repeat unit of C. The functional group formed should be displayed.  State the type of polymerisation shown in . Scientists are investigating C as a replacement for polyin packaging. Suggest an advantage of using C instead of poly. A student studies the reaction of CH3CHCl COOH with aqueous NH3 to determine the reaction mechanism. The student finds that when CH3CHCl COOH and NH3 are added in a 1 : 1 stoichiometric ratio, the conjugate acid and base of the reactants are quickly formed. reaction 1 CH3CHCl COOH + NH3 → CH3CHCl COO– + NH4 + Identify the conjugate acid–base pairs in reaction1. conjugate acid–base pair I and conjugate acid–base pair II and  In an excess of NH3, CH3CHCl COO– undergoes a nucleophilic substitution reaction. reaction 2 CH3CHCl COO– + NH3 → CH3CH(NH2)COO– + H+ + Cl – A student investigates the rate of reaction2. The student mixes CH3CHCl COO– with a large excess of NH3. The graph in Fig.5.3 shows the results obtained. time / s 1000 1200 1400 1600 0.0250 0.0200 0.0150 0.0100 0.0050 0.0000 [CH3CHCl COO–] / mol dm–3 Use the graph in Fig.5.3 to show that reaction2 is first order with respect to [CH3CHCl COO–]. Explain why a large excess of NH3 needs to be used in order to obtain the results in Fig.5.3. The student measures the effect of changing the concentration of NH3 on the rate of reaction2. Table5.1 shows the results obtained. Table 5.1 experiment [CH3CHCl COO–] / mol dm–3 / mol dm–3 initial rate of reaction / mol dm–3 s–1 0.00120 0.00300 1.47 × 10–5 0.00120 0.00450 2.21 × 10–5 Use the information in Table 5.1 and in to determine whether the nucleophilic substitution reaction proceeds via an SN1 or an SN2 mechanism. Explain your answer. Describe the effect of an increase in temperature on the rate of reaction of CH3CHCl COO– and NH3. Explain your answer. When an excess of CH3CHCl COO– is used, further substitution reactions occur. One product has the formula C6H9NO4 2–. Suggest the structure of C6H9NO4 2–.  
9701_m22_qp_42
THEORY
2022
Paper 4, Variant 2
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Amino acids are molecules that contain —NH2 and —COOH functional groups. Glycine, H2NCH2COOH, is the simplest stable amino acid. The isoelectric point of glycine is 6.2. Define isoelectric point. Draw the structure of glycine at pH 4. shows two syntheses starting with glycine. hippuric acid O COOH N H N C2H5Br glycine U reaction 1 H2N COOH COOH reaction 2 reaction 3 an excess of LiAl H4 State the essential conditions for reaction 1. Identify the reagent used in reaction 2. Draw the structure of the organic product U that forms when hippuric acid reacts with an excess of LiAl H4 in reaction 3. A molecule of phenylalanine, R, can react with a molecule of glycine to form two dipeptides, S and T. S and T are structural isomers. glycine S and T R H2N + COOH COOH H2N Draw the structures of these dipeptides. The peptide bond formed should be shown fully displayed. S T A student proposes a synthesis of hippuric acid by the reaction of benzamide, C6H5CONH2, and chloroethanoic acid, Cl CH2COOH. The reaction does not work well because benzamide is a very weak base. Explain why amides are weaker bases than amines. The pKa of chloroethanoic acid is 2.86 whereas the pKa of ethanoic acid is 4.76. Explain the difference between these two pKa values. Compound V is another amino acid. The proton (1H) NMR spectrum of V shows hydrogen atoms in five different environments, a, b, c, d and e, as shown in . H H H H HOOC CH2CH2NH2 b V a c d e Table 6.1 environment of proton example chemical shift range, d / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl, –CH2–N 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0 Complete Table 6.2 for the proton (1H) NMR spectrum of V taken in CDCl 3. Table 6.1 gives some relevant data. Table 6.2 proton a b c d e chemical shift range, d / ppm name of splitting pattern multiplet Complete Table 6.3 by placing a tick (✓) to indicate any protons whose peaks are still present in the proton (1H) NMR spectrum of V taken in D2O. Table 6.3 proton a b c d e present in D2O
9701_m24_qp_42
THEORY
2024
Paper 4, Variant 2
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Questions Discovered
313