14.2. Alkenes
A subsection of Chemistry, 9701, through 14. Hydrocarbons
Listing 10 of 191 questions
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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The antipyretic (fever-reducing) drug antifebrin can be made from benzene and ethanoic acid by the following route. What type of reaction is reaction I? Suggest the reagents and conditions for reaction I. Complete the following scheme showing the mechanism of reaction I, by drawing appropriate formulae in the three boxes. What type of reaction is reaction II? Suggest the reagents and conditions for reaction II. NO2 + + NO2 NH2 NHCOCH3 I CH3CO2H CH3COCl antifebrin III IV II Use Suggest the reagents and conditions for reaction III. Apart from the benzene ring, name the functional group in antifebrin. What reagents and conditions are needed to hydrolyse antifebrin? Examiner's Use
9701_s05_qp_4
THEORY
2005
Paper 4, Variant 0
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Use The following scheme shows some reactions of methylbenzene. CH3 Cl I II III CH3 CH2Cl E F IV CO2H COCl G Suggest reagents and conditions for reactions I to IV. I II III IV What type of reaction is each of the following? reaction I reaction III Examiner’s Use Compound F can be converted into 2-phenylethylamine in a two-stage process. Suggest a structure for the intermediate, H, in the box below, and suggest reagents and conditions for the steps V and VI. CH2Cl V VI CH2CH2NH2 H F reagents and conditions for step V reagents and conditions for step VI The compounds E, F and G react at different rates with nucleophilic reagents. Draw structures for the products of each compound with the following reagents. If no reaction occurs, write “no reaction” in the box. compound reagent cold water hot NaOHE F G
9701_s07_qp_4
THEORY
2007
Paper 4, Variant 0
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Use Electrophoresis can be used to separate amino acids which are produced by the hydrolysis of a polypeptide. Using glycine as an example, explain why the result of electrophoresis depends on pH. The diagram below shows the results of electrophoresis in neutral solution. At the start of the experiment a spot of a solution containing a mixture of amino acids P, Q, R and S was placed in the middle of the plate. Following electrophoresis the amino acids had moved to the positions shown in the lower diagram. before + – after + – P Q R S Which amino acid existed mainly as a zwitterion in the buffer solution? Explain your answer. Assuming amino acids R and S carry the same charge when in this buffer solution, which is likely to be the larger molecule? Explain your answer. Amino acids may also be separated by using two-dimensional paper chromatography. This involves putting a spot of the mixture on the corner of a piece of chromatography paper and allowing a solvent to soak up the paper. The paper is then dried, turned through 90° and placed in a second solvent. This method gives better separation than a one solvent method. Paper chromatography relies on partition between the solvent applied and another phase. What is this second phase? Examiner’s Use The table below shows the Rf values for some amino acids in two different solvents. amino acid Rf solvent 1 Rf solvent 2 A 0.1 0.2 B 0.0 0.4 C 0.3 0.0 D 0.8 0.9 E 0.6 0.5 Use the grid below to plot the positions of the amino acids after two-dimensional paper chromatography using solvent 1 followed by solvent 2. solvent 2 solvent 1 solvent front solvent front spot applied here Which amino acid travelled fastest in both solvents? Which amino acid did not move at all in solvent 2?
9701_s07_qp_4
THEORY
2007
Paper 4, Variant 0
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Use Both ethene and benzene react with bromine, but the mechanisms and the types of products of the two reactions are different. Br2 Br BrCH2CH2Br H2C CH2 + HBr + reaction I no heat, no light, no catalyst needed Br2 + reaction II heat and catalyst needed State the type of reaction undergone in each of reactions I and II. reaction I reaction II Examiner’s Use In each of reactions I and II, the intermediate is a bromine-containing cation. In each of the following boxes, draw the intermediate and use curly arrows to show how it is converted into the product. reaction I product intermediate reaction II product intermediate Why do ethene and benzene differ in their reaction with bromine?
9701_s08_qp_4
THEORY
2008
Paper 4, Variant 0
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Alkenes such as propene can be readily prepared from alcohols in a school or college laboratory by using the apparatus below. mineral wool soaked with the alcohol strong heat X propene water Give the name of an alcohol that can be used in this apparatus to prepare propene. Draw the skeletal formula of the alcohol you have named in . What type of reaction occurs in this case? During the reaction, the material X becomes black in colour. Suggest the identity of the black substance and suggest how it is produced during the reaction. At the end of the experiment, when no more propene is being produced, the delivery tube is removed from the water before the apparatus is allowed to cool. Suggest why this done. The material labelled X can be broken crockery, broken brick or pumice. Give the chemical formula of a compound that is present in one of these materials. State another reagent that could be used to produce propene from an alcohol. Give the structural formula of the organic product formed when propene reacts separately with each of the following substances. bromine cold, dilute manganate(ions hot, concentrated manganate(ions Propene may be polymerised. What is the essential condition for such a polymerisation? The disposal of waste polyis very difficult. Give one important reason for this.
9701_s10_qp_22
THEORY
2010
Paper 2, Variant 2
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Although standard electrode potentials are measured for solutions where the concentrations of ions are 1.0 mol dm–3, cells used as sources of battery power tend to operate with more concentrated solutions. This question concerns the electrode reactions involved in the hydrogen-oxygen fuel cell and the lead-acid car battery. In the hydrogen-oxygen fuel cell, H2and O2are fed onto two inert electrodes dipping into NaOH. V hydrogen oxygen NaOHThe following reactions take place. left hand electrode : H2+ 2OH–2H2O+ 2e– right hand electrode : O2+ 2H2O+ 4e– 4OH–Use the Data Booklet to calculate E o–– cell for this reaction. Construct an equation for the overall reaction. By using one of the phrases more positive, more negative or no change, deduce the effect of increasing [OH–] on the electrode potential of • the left hand electrode • the right hand electrode Hence deduce whether the overall Ecell is likely to increase, decrease or remain the same, when [OH–] increases. Explain your answer. Suggest one other reason why a high is used in the fuel cell. In the cells of a lead-acid car battery the following reactions take place. cathode: PbPb2++ 2e– anode: PbO2+ 4H++ 2e– Pb2++ 2H2OUse the Data Booklet to calculate E o–– cell for this reaction. Construct an equation for the overall reaction. The electrolyte in a lead-acid cell is H2SO4. Most of the Pb2+ions that are produced at the electrodes are precipitated as the highly insoluble PbSO4. Construct an equation for the overall cell reaction in the presence of H2SO4. By considering the effect of decreasing [Pb2+] on the electrode potentials of the cathode and the anode, deduce the effect of the presence of H2SO4in the electrolyte on the overall Ecell. State whether the Ecell will increase, decrease or remain the same. Overall Ecell will . Explain your answer.
9701_s10_qp_41
THEORY
2010
Paper 4, Variant 1
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Questions Discovered
191