22. Analytical techniques
A section of Chemistry, 9701
Listing 10 of 112 questions
An aldehyde, an alkane and a carboxylic acid, all of similar volatility, are mixed together. The mixture is then analysed in a gas chromatograph. The gas chromatogram produced is shown. absorption time / mins Z X Y The separation of the compounds depends on their relative solubilities in the stationary phase. The stationary phase is a liquid alcohol. Complete the table to suggest which compound in the mixture is responsible for each peak X, Y and Z. Explain your answer by reference to the intermolecular forces of the compounds. peak organic compound explanation X Y Z  A student calculates the areas underneath the three peaks in the chromatogram. peak X Y Z area / mm2 The area underneath each peak is proportional to the mass of the respective compound. Calculate the percentage by mass in the original mixture of the compound responsible for peakZ.  % of mixture responsible for peak Z = The mass spectrum of a halogenoalkane containing one chlorine atom or bromine atom will show an additional peak at M+2. State the isotopes of chlorine and bromine responsible for M+2 peaks. chlorine bromine  The mass spectrum of bromochloromethane, CH2BrCl, has a molecular ion peak, M, at an m / e value of 128. It also has M+2 and M+4 peaks. Suggest the identity of the molecular ions that give rise to these peaks. M peak M+2 peak M+4 peak  Halogenoalkanes can be formed from the reaction of an alkene with a hydrogen halide. Methylpropene reacts with hydrogen bromide to form 2-bromo-2-methylpropane. H2C CH3 CH3 C + HBr methylpropene 2-bromo-2-methylpropane H3C CH3 Br CH3 C Draw the mechanism of this reaction. Include all relevant curly arrows, dipoles and charges.  1-bromo-2-methylpropane is also formed in this reaction. Explain why 2-bromo-2-methylpropane will be the major product in this reaction. Explain what is meant by the term partition coefficient, Kpartition. The partition coefficient of organic compound H between dichloromethane and water is 4.75. ● 2.50 g of compoundH was dissolved in water and made up to 100 cm3 in a volumetric flask. ● 50 cm3 of this aqueous solution were shaken with 10 cm3 of dichloromethane. Calculate the mass of compoundH that was extracted into the dichloromethane.  mass of compound H extracted = g 
9701_w18_qp_41
THEORY
2018
Paper 4, Variant 1
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An aldehyde, an alkane and a carboxylic acid, all of similar volatility, are mixed together. The mixture is then analysed in a gas chromatograph. The gas chromatogram produced is shown. absorption time / mins Z X Y The separation of the compounds depends on their relative solubilities in the stationary phase. The stationary phase is a liquid alcohol. Complete the table to suggest which compound in the mixture is responsible for each peak X, Y and Z. Explain your answer by reference to the intermolecular forces of the compounds. peak organic compound explanation X Y Z  A student calculates the areas underneath the three peaks in the chromatogram. peak X Y Z area / mm2 The area underneath each peak is proportional to the mass of the respective compound. Calculate the percentage by mass in the original mixture of the compound responsible for peakZ.  % of mixture responsible for peak Z = The mass spectrum of a halogenoalkane containing one chlorine atom or bromine atom will show an additional peak at M+2. State the isotopes of chlorine and bromine responsible for M+2 peaks. chlorine bromine  The mass spectrum of bromochloromethane, CH2BrCl, has a molecular ion peak, M, at an m / e value of 128. It also has M+2 and M+4 peaks. Suggest the identity of the molecular ions that give rise to these peaks. M peak M+2 peak M+4 peak  Halogenoalkanes can be formed from the reaction of an alkene with a hydrogen halide. Methylpropene reacts with hydrogen bromide to form 2-bromo-2-methylpropane. H2C CH3 CH3 C + HBr methylpropene 2-bromo-2-methylpropane H3C CH3 Br CH3 C Draw the mechanism of this reaction. Include all relevant curly arrows, dipoles and charges.  1-bromo-2-methylpropane is also formed in this reaction. Explain why 2-bromo-2-methylpropane will be the major product in this reaction. Explain what is meant by the term partition coefficient, Kpartition. The partition coefficient of organic compound H between dichloromethane and water is 4.75. ● 2.50 g of compoundH was dissolved in water and made up to 100 cm3 in a volumetric flask. ● 50 cm3 of this aqueous solution were shaken with 10 cm3 of dichloromethane. Calculate the mass of compoundH that was extracted into the dichloromethane.  mass of compound H extracted = g 
9701_w18_qp_43
THEORY
2018
Paper 4, Variant 3
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The structure of butamben is shown. H2N O O butamben Butamben can act as a base. Complete the equation for a reaction in which butamben acts as a base. H2N O O +  Predict whether butamben is a stronger or weaker base than ammonia. Give a reason for your answer. Complete the reaction scheme below to show the structural formulae of the products formed when butamben is treated separately with the stated reagent. H2N O O Br2HNO2+ HCl 5 °C NaOHunder reflux  The proton NMR spectrum of butamben in CDCl 3 contains one or more peaks that show a triplet splitting pattern. State the number of peaks in the spectrum that show a triplet splitting pattern. On the diagram of butamben below, circle the protons responsible for the peak or peaks you identified in . H N H C C O O H H H H H H H H H C C H H H C H  Describe and explain how the proton NMR spectrum of butamben in D2O would differ from the proton NMR spectrum of butamben in CDCl 3. The mass spectrum of butamben includes peaks at m/e 92 and 57. Identify the fragments responsible for these peaks. m/e 92 = m/e 57 =  
9701_w19_qp_41
THEORY
2019
Paper 4, Variant 1
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The structure of butamben is shown. H2N O O butamben Butamben can act as a base. Complete the equation for a reaction in which butamben acts as a base. H2N O O +  Predict whether butamben is a stronger or weaker base than ammonia. Give a reason for your answer. Complete the reaction scheme below to show the structural formulae of the products formed when butamben is treated separately with the stated reagent. H2N O O Br2HNO2+ HCl 5 °C NaOHunder reflux  The proton NMR spectrum of butamben in CDCl 3 contains one or more peaks that show a triplet splitting pattern. State the number of peaks in the spectrum that show a triplet splitting pattern. On the diagram of butamben below, circle the protons responsible for the peak or peaks you identified in . H N H C C O O H H H H H H H H H C C H H H C H  Describe and explain how the proton NMR spectrum of butamben in D2O would differ from the proton NMR spectrum of butamben in CDCl 3. The mass spectrum of butamben includes peaks at m/e 92 and 57. Identify the fragments responsible for these peaks. m/e 92 = m/e 57 =  
9701_w19_qp_43
THEORY
2019
Paper 4, Variant 3
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The proton NMR spectrum of compoundE in the solvent CDCl 3 is shown. The molecular formula of compoundE is C9H10O2. δ / ppm Explain why CDCl 3 is used as a solvent instead of CHCl 3. Explain why TMS is added to give the small peak at chemical shift δ = 0. CompoundE is hydrolysed by hot NaOH, giving two organic products only. One of these products is ethanol. Name the functional group in compoundE that is hydrolysed by hot NaOH. Describe and explain the splitting patterns of the peaks at δ = 1.4 and δ = 4.3. splitting pattern at δ = 1.4 reason for splitting pattern at δ = 1.4 splitting pattern at δ = 4.3 reason for splitting pattern at δ = 4.3  Each molecule of compoundE contains five protons which give rise to the peaks between δ=7.0 and δ=8.5. Identify the functional group in compoundE which contains these protons. Give the structural formula of compoundE.  The mass spectrum of compoundE includes fragment ions with m/e values of 29 and 77. Give the formulae of these fragment ions. fragment ion with m/e = 29 fragment ion with m/e = 77  
9701_w20_qp_41
THEORY
2020
Paper 4, Variant 1
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The proton NMR spectrum of compoundE in the solvent CDCl 3 is shown. The molecular formula of compoundE is C9H10O2. δ / ppm Explain why CDCl 3 is used as a solvent instead of CHCl 3. Explain why TMS is added to give the small peak at chemical shift δ = 0. CompoundE is hydrolysed by hot NaOH, giving two organic products only. One of these products is ethanol. Name the functional group in compoundE that is hydrolysed by hot NaOH. Describe and explain the splitting patterns of the peaks at δ = 1.4 and δ = 4.3. splitting pattern at δ = 1.4 reason for splitting pattern at δ = 1.4 splitting pattern at δ = 4.3 reason for splitting pattern at δ = 4.3  Each molecule of compoundE contains five protons which give rise to the peaks between δ=7.0 and δ=8.5. Identify the functional group in compoundE which contains these protons. Give the structural formula of compoundE.  The mass spectrum of compoundE includes fragment ions with m/e values of 29 and 77. Give the formulae of these fragment ions. fragment ion with m/e = 29 fragment ion with m/e = 77  
9701_w20_qp_43
THEORY
2020
Paper 4, Variant 3
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CompoundT is made by a three-stage synthesis. In stage1, phenylethanoicacid reacts with a suitable reagent to form compoundR. O OH O Cl phenylethanoic acid R stage 1 Suggest a suitable reagent for stage1. In stage 2, compoundR reacts with ethylamine to form compoundS. + C2H5NH2 O Cl R O S stage 2 N H Name the functional group formed in stage2. Identify the other product formed in stage2. In stage3, compoundS reacts with a suitable reagent to form compoundT. O S stage 3 N H T N H State the formula of a suitable reagent for stage3. Name the type of reaction that occurs in stage3. The relative abundance of the molecular ion peak in the mass spectrum of ethylamine is 62. Calculate the relative abundance of the M+1 peak in the mass spectrum of ethylamine.  relative abundance = The mass spectrum of compoundT contains several fragments. The m/e values of two of these fragments are 29 and 91. Draw the structures of the ions responsible for these peaks. m/e structure of ion  The proton (1H) NMR spectrum of compoundT shows hydrogen atoms in different environments. Six of these environments are shown on the structure using letters a, b, c, d, e and f. N H a b c e d f Use the letters a, b, c, d, e and f to answer the questions that follow. The questions relate to the proton (1H) NMR spectrum of T. Proton d does not cause splitting of the peaks for protons c or e under the conditions used. Each answer may be one, or more than one, of the letters a, b, c, d, e and f. Identify the proton or protons with a chemical shift (δ) in the range 6.0 to 9.0.  Identify the proton or protons whose peak will disappear if D2O is added.  Identify the proton or protons whose peak is a triplet.  Identify the proton or protons with the lowest chemical shift (δ).  
9701_w21_qp_41
THEORY
2021
Paper 4, Variant 1
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Alanine, H2NCH(CH3)CO2H, and glutamic acid, H2NCH(CH2CH2CO2H)CO2H, are two naturally occurring amino acids. H2NCH(CH3)CO2H exists as two optical isomers. Draw three-dimensional structures of these two optical isomers.  The proton (1H) NMR spectrum of either alanine in D2O or glutamic acid in D2O is shown. / ppm State whether this is the spectrum of alanine in D2O or the spectrum of glutamicacid in D2O. Explain your answer by reference to the number of peaks and splitting patterns. The mass spectrum of glutamicacid, H2NCH(CH2CH2CO2H)CO2H, is obtained. State the m/e value of the molecular ion peak in this spectrum. The spectrum has peaks with m/e values of 88 and 131. Draw the structures of the ions responsible for these peaks. m/e structure of ion  At pH 11 alanine exists as H2NCH(CH3)CO2 – ions and glutamic acid exists as H2NCH(CH2CH2CO2 –)CO2 – ions. A mixture of alanine and glutamicacid at pH 11 is subjected to electrophoresis. State how the mixture can be maintained at pH11 during electrophoresis. Draw a fully labelled diagram for the apparatus that would be used to carry out this electrophoresis. Your diagram should include the position of the mixture of alanine and glutamic acid at the start of the electrophoresis experiment. Identify the electrode that each amino acid travels towards during electrophoresis at pH11. alanine glutamic acid  In a particular electrophoresis experiment at pH 11, the glutamic acid travels 3.4 cm. Alanine travels a shorter distance. Explain the factors that account for the difference in the distances travelled. 
9701_w21_qp_42
THEORY
2021
Paper 4, Variant 2
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CompoundT is made by a three-stage synthesis. In stage1, phenylethanoicacid reacts with a suitable reagent to form compoundR. O OH O Cl phenylethanoic acid R stage 1 Suggest a suitable reagent for stage1. In stage 2, compoundR reacts with ethylamine to form compoundS. + C2H5NH2 O Cl R O S stage 2 N H Name the functional group formed in stage2. Identify the other product formed in stage2. In stage3, compoundS reacts with a suitable reagent to form compoundT. O S stage 3 N H T N H State the formula of a suitable reagent for stage3. Name the type of reaction that occurs in stage3. The relative abundance of the molecular ion peak in the mass spectrum of ethylamine is 62. Calculate the relative abundance of the M+1 peak in the mass spectrum of ethylamine.  relative abundance = The mass spectrum of compoundT contains several fragments. The m/e values of two of these fragments are 29 and 91. Draw the structures of the ions responsible for these peaks. m/e structure of ion  The proton (1H) NMR spectrum of compoundT shows hydrogen atoms in different environments. Six of these environments are shown on the structure using letters a, b, c, d, e and f. N H a b c e d f Use the letters a, b, c, d, e and f to answer the questions that follow. The questions relate to the proton (1H) NMR spectrum of T. Proton d does not cause splitting of the peaks for protons c or e under the conditions used. Each answer may be one, or more than one, of the letters a, b, c, d, e and f. Identify the proton or protons with a chemical shift (δ) in the range 6.0 to 9.0.  Identify the proton or protons whose peak will disappear if D2O is added.  Identify the proton or protons whose peak is a triplet.  Identify the proton or protons with the lowest chemical shift (δ).  
9701_w21_qp_43
THEORY
2021
Paper 4, Variant 3
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Gas-liquid chromatography involves a stationary phase and a mobile phase. Name, or describe in detail, a suitable substance that could be used for each phase. stationary mobile  A mixture of three organic compounds is separated by gas-liquid chromatography. The chromatogram obtained is shown in Fig.9.1. The amount of each substance is proportional to the area under its peak. retention time recorder response A C B Explain the meaning of retention time. Calculate the percentage of B in the mixture. Show your working.  percentage of B = % Complete Table9.1 to give the number of peaks in the carbon-13 NMR spectrum of each of the five isomers of C5H10O2 that has an ester group. Table 9.1 structural formula number of peaks CH3CH2CH2CO2CH3 CH3CH2CO2CH2CH3 CH3CO2CH2CH2CH3 (CH3)2CHCO2CH3 CH3CO2CH(CH3)2  State the number of peaks that would be seen in the proton (1H) NMR spectrum of methylbutanoate, CH3CH2CH2CO2CH3. Name all the splitting patterns seen in this spectrum. number of peaks splitting patterns  D and E are both esters with the molecular formula C5H10O2. Their proton (1H) NMR spectra are shown in Fig.9.2 and Fig.9.3. chemical shift, D chemical shift, E Table 9.2 environment of proton example typical chemical shift range, δ / 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 3.2–4.0 attached to alkene =CHR 4.5–6.0 Deduce the structures of the two esters D and E and draw their displayed formulae in the boxes below. D C5H10O2 E C5H10O2  The spectrum of D includes a quartet at δ4.1. Identify the protons responsible for this quartet on your structure in by labelling these protons with the letter F. Explain why this peak is split into a quartet.  The spectrum of E has a doublet at δ1.1. Identify the protons responsible for this doublet on your structure in by labelling these protons with the letter G. Explain why this peak has a chemical shift of 1.1.  
9701_w22_qp_41
THEORY
2022
Paper 4, Variant 1
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Questions Discovered
112