Chemistry 9701/42 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) An aqueous solution of chromium(III) contains the green [Cr(H2O)6]3+ complex ion.
(i) Complete the electronic configuration of an isolated, gaseous Cr3+ ion.
(ii) Define the term complex ion.
. ……
. …… [1]
(b) [Cr(H2O)6]3+(aq) shows some similar chemical properties to [Co(H2O)6]2+(aq).
Samples of [Cr(H2O)6]3+ are reacted separately with either NaOH(aq), H2O2(aq), or excess
NH3(aq).
Use this information and the Data Booklet to suggest the formula of the chromium species formed. State the type of reaction taking place in each case.
reagent added to formula of chromium type of reaction
[Cr(H2O)6]3+(aq) species formed
NaOH(aq)
H2O2(aq)
an excess of NH3(aq)
[5]
(c) [Cr(H2O)6]2+ and [Cr2(O2CCH3)4(H2O)2] are both complexes of chromium(II) and have different colours.
Explain why the colours of these complexes are different.
. ……
. ……
. ……
. …… [2]
(d) The structure of [Cr2(O2CCH3)4(H2O)2] is shown. Ethanoate ions act as ligands in this complex.
The ethanoate ligand, CH3CO2–, is shown as O O.
O O
O O
H 2O Cr Cr OH2
O O
O O
(i) Water and ethanoate ions behave as different types of ligand in this complex.
Suggest an explanation for this statement.
. ……
. …… [1]
(ii) Deduce the coordination number of Cr and the geometry around each Cr atom in this structure.
coordination number . ……
[1]
(iii) State the type of bond between the two atoms in the Cr–Cr bond.
. …… [1]
(e) The [Cr2(O2CCH3)4(H2O)2] complex reacts with aqueous acid to form Cr2+(aq) ions.
Cr2+(aq) ions react with O2(aq) under acidic conditions. Cr3+(aq) ions are formed.
Use the Data Booklet to answer the following questions.
(i) Construct an ionic equation for the reaction of Cr2+(aq) with O2(aq) under acidic conditions.
. …… [2]
(ii) Calculate for the reaction in (e)(i).
= …… V [1]
[Total: 15]
Mark scheme
1(a)(i) [1s2] 2s22p63s23p63d3 1
1(a)(ii) (a molecule or ion formed by a central) metal atom/ion surrounded by / bonded to one or more ligands 1
1(b) 5
[Cr(H2O)6]3+(aq) formula of chromium species formed type of reaction
+ NaOH(aq) Cr(OH)3 or Cr(OH)3(H2O)3 precipitation
+ H2O2(aq) Cr2O72– / CrO42– redox / oxidation
+ excess NH3(aq) Cr(NH3)63+ ligand substitution
chromium species: one mark for each correct species
type of reaction: two correct for one mark and three correct for two marks
1(c) M1: ΔE is different OR energy gap between d-orbitals is different 2
M2: different frequency / wavelength is absorbed
OR different energy / light in visible region is absorbed
1(d)(i) ethanoate ions are bidentate whereas as H2O are monodentate 1
OR ethanoate ions form two dative bonds whereas H2O forms one (dative) bond
OR ethanoate ions donate two lone pairs whereas H2O donates one (lone) pair
1(d)(ii) (coordination number) six AND (geometry around Cr) octahedral 1
1(d)(iii) coordinate / (dative) covalent 1
1(e)(i) 4Cr2+ + O2 + 4H+ 4Cr3+ + 2H2O OR 2Cr2+ + O2 + 2H+ 2Cr3+ + H2O2 2
M1: correct species
M2: balancing
1(e)(ii) Eocell = 1.23 – (–0.41) = (+)1.64 V OR Eocell = 0.68 – (–0.41) = (+)1.09 V 1
value linked to (e)(i)
(a) State and explain the trend observed in the thermal stability of the Group 2 nitrates.
. ……
. ……
. ……
. ……
. …… [3]
(b) (i) Lead(II) nitrate, Pb(NO3)2, decomposes on heating in a similar manner to the Group 2 nitrates.
Write an equation for the decomposition of lead(II) nitrate.
. …… [1]
(ii) Suggest how the ease of decomposition of Pb(NO3)2 would compare to that of Ba(NO3)2.
Explain your answer. You may find it useful to refer to the Data Booklet.
. ……
. …… [1]
(c) (i) Barium ethanedioate, BaC2O4, decomposes on heating to produce barium oxide and a mixture of two different gases.
Construct an equation for the decomposition of barium ethanedioate.
. …… [1]
(ii) An impure sample of BaC2O4, of mass 0.500 g, is added to 50.0 cm3 of 0.0200 mol dm–3 acidified MnO4–(aq), an excess. A redox reaction takes place and all the BaC2O4 reacts.
The resulting solution, containing unreacted acidified MnO4–, is titrated with 0.0500 mol dm–3
Fe2+(aq).
The end-point is reached when 30.40 cm3 of 0.0500 mol dm–3 Fe2+(aq) has been added.
C2O42– 2CO2 + 2e–
MnO4– + 8H+ + 5e– Mn2+ + 4H2O
Fe2+ Fe3+ + e–
Calculate the percentage by mass of BaC2O4 in the 0.500 g impure sample. Show your working.
[Mr: BaC2O4, 225.3]
percentage by mass of BaC2O4 = …… [4]
(d) Barium hydroxide, Ba(OH)2, is completely dissociated in aqueous solution.
Calculate the pH of 0.120 mol dm–3 Ba(OH)2(aq) at 298 K.
pH = …… [2]
[Total: 12]
Mark scheme
2(a) M1: increases down the group 3
M2: radius / size of cation / M2+ increases OR charge density of cation / M2+ decreases
M3: less polarisation / less distortion of anion / NO3– ion OR less weakening of NO bond
2(b)(i) Pb(NO3)2 PbO + 2NO2 + ½O2 1
2(b)(ii) lead nitrate / Pb(NO3)2 would decompose more / easier 1
AND as Pb2+ is smaller / Pb2+ has larger charge density (so more polarising)
2(c)(i) BaC2O4 BaO + CO2 + CO OR BaC2O4 BaO + 2CO + ½O2 1
2(c)(ii) M1: [a] initial moles MnO4– = 0.0200 × 0.050 = 1.00 × 10–3 4
[b] moles Fe2+ = 0.050 × 0.0304 = 1.52 × 10–3
M2: [a] moles MnO4– unreacted = 1.52 × 10–3 / 5 = 3.04 × 10–4
[b] moles MnO4– reacted = 1.00 × 10–3 – 3.04 × 10–4 = 6.96 × 10–4
M3: moles C2O42- reacted = 6.96 × 10–4 × 5/2 = 1.74 × 10–3
M4: mass of BaC2O4 = 225.3 × 1.74 × 10–3 = 0.392 g
% Purity of BaC2O4 = 100 × 0.392/0.50 = 78.4
2(d) M1: [OH–] = 2 × 0.12 = 0.24 (mol dm–3) 2
[H+] = 1 × 10–14/0.24 = 4.17 × 10–14 / pOH = –log(0.24) OR 0.62
M2: pH = –log[H+] = 13.4 OR pH = 14 – 0.6 = 13.4
(a) (i) Define the term standard electrode potential.
. ……
. ……
. …… [2]
Three redox systems, A, B and C, are shown. The ligand 1,2-diaminoethane, H2NCH2CH2NH2, is represented by en.
A [Ru(H2O)6]3+ + e– [Ru(H2O)6]2+
B [Ru(NH3)6]3+ + e– [Ru(NH3)6]2+
C [Ru(en)3]3+ + e– [Ru(en)3]2+
Two electrochemical cells are set up to compare the standard electrode potentials, E o, of three half-cells. The diagrams show the relative potential of each electrode.
+ V – + V –
salt bridge salt bridge
Pt Pt Pt Pt
[Ru(en)3]3+ [Ru(NH3)6]3+ [Ru(H2O)6]3+ [Ru(en)3]3+
[Ru(en)3]2+ [Ru(NH3)6]2+ [Ru(H2O)6]2+ [Ru(en)3]2+
(ii) Use this information to complete the table by adding the labels A, B and C to deduce the order of E o for the three half-cells.
Eo redox system
most negative
least negative
[1]
(iii) The complex [Ru(en)3]3+ shows stereoisomerism. The ligand en is bidentate.
Draw three-dimensional diagrams to show the two isomers of [Ru(en)3]3+. Represent the ligand en by using .
N N
Name the type of stereoisomerism.
isomer 1 isomer 2
Ru Ru
type of stereoisomerism ……
[3]
(b) (i) An electrochemical cell consists of a Br2/Br – half-cell and a Ag+/Ag half-cell, under standard conditions.
Use the Data Booklet to calculate the . Deduce the direction of electron flow in the wire through the voltmeter between these two half-cells.
= …… V
direction of electron flow from …… to ����������������������������������������� [1]
(ii) Water is added to the Ag+/Ag half-cell in (b)(i).
Suggest the effect of this addition on the Ecell. Place a tick ( ) in the appropriate box.
less positive no change more positive
Explain your answer.
. ……
. ……
. …… [2]
(c) Silver bromide, AgBr, dissolves in an aqueous solution of S2O32– ions to form the complex ion
[Ag(S2O3)2]3–. The S2O32– ions act as monodentate ligands.
equilibrium 1 AgBr(s) + 2S2O32–(aq) [Ag(S2O3)2]3–(aq) + Br –(aq)
(i) Define the term ligand.
. ……
. …… [1]
(ii) Write an expression for the equilibrium constant, Kc, for equilibrium 1.
Kc =
[1]
(iii) Some additional data are given about the dissolution of AgBr in S2O32–(aq).
equilibrium constant numerical value solubility product, Ksp, of AgBr 5.4 × 10–13 stability constant, Kstab, of [Ag(S2O3)2]3– 2.9 × 1013
Use your answer to (c)(ii) and these data to calculate Kc for equilibrium 1. Include the units for Kc.
Kc = …… units …… [2]
(d) The numerical values for the stability constants, Kstab, of two other silver(I) complexes are given.
silver(I) complex numerical value of Kstab
[Ag(CN)2] –
5.3 × 1018
[Ag(NH3)2]+ 1.6 × 107
An aqueous solution containing Ag+ is added to a solution containing equal concentrations of
CN–(aq), NH3(aq) and S2O32–(aq). The mixture is left to reach equilibrium.
Deduce the relative concentrations of [Ag(CN)2]–, [Ag(NH3)2]+ and [Ag(S2O3)2]3– present in the resulting mixture. Explain your answer.
…… > …… > ……
highest concentration lowest concentration
. ……
. …… [2]
[Total: 15]
Mark scheme
3(a)(i) M1: voltage of an electrode / a half-cell compared to / connected to (standard) hydrogen electrode / half-cell 2
M2: (at concentration of) 1 mol / dm3 AND (pressure of) 1 atm / 101 kPa (or in Pa) AND 298 K / 25°C
3(a)(ii) 1
3(a)(iii) 3
OR
M1 / M2: two 3D isomers of [Ru(en)3]3+
M3: optical / enantiomerism
3(b)(i) Eocell = 1.07 – 0.80 = (+)0.27 V 1
AND direction of electron flow = Ag+ / Ag to Br2 / Br–
3(b)(ii) M1: Eocell 3rd box ticked 2
M2:
[Ag+] decreases AND so (Ag+ / Ag) equilibrium shifts to the left
OR
[Ag+] decreases AND E for (Ag+ / Ag) becomes less positive / more negative
3(c)(i) (a species) that uses / shares a lone pair of electrons to form a coordinate bond to a metal atom / ion 1
3(c)(ii) Kc = [Ag(S2O3)23–] [Br–] / [S2O32–]2 1
3(c)(iii) M1: Kc = Kstab × Ksp = 15.7 2
M2: 1 OR none / no units
3(d) M1: highest [Ag(CN)2]– [Ag(S2O3)2]3– [Ag(NH3)2]+ lowest 2
M2: Kstab [Ag(CN)2]– is highest / [Ag(CN)2]– is the most stable
OR higher Kstab forms the more stable complex
(a) (i) Define the term lattice energy.
. ……
. ……
. …… [2]
(ii)
Use the following data to calculate a value for the enthalpy change of solution of copper(II) chloride, CuCl 2(s). You might find it helpful to construct an energy cycle.
enthalpy change of hydration of Cl – = –378 kJ mol–1 enthalpy change of hydration of Cu2+ = –2099 kJ mol–1 lattice energy of CuCl 2(s) = –2824 kJ mol–1
enthalpy change of solution of CuCl 2(s) = …… kJ mol–1 [2]
(iii) The enthalpy change of hydration of Ca2+ is –1579 kJ mol–1.
Use the Data Booklet to suggest why there is a big difference in the values of ∆Hhyd for Ca2+ and Cu2+.
. ……
. ……
. …… [2]
(b) (i) Identify the substances formed at the anode and at the cathode during the electrolysis of saturated CaCl 2(aq).
at the anode . ……
at the cathode ……
[1]
(ii) Calcium can be produced by the electrolysis of molten calcium chloride, CaCl 2(l).
Calculate the mass, in g, of Ca formed when a current of 0.75 A passes through CaCl 2(l) for 60 minutes.
[Ar: Ca, 40.1]
mass of Ca = …… g [2]
(c) (i) Explain what is meant by the term entropy of a system.
. ……
. …… [1]
(ii) Place one tick ( ) in each row of the table to show the sign of each entropy change, ∆S.
process ∆S is negative ∆S is zero ∆S is positive
NaCl dissolving in water
water solidifying to ice
[1]
(iii) The evaporation of one mole of water has a standard Gibbs free energy change, ∆G o, of
+8.6 kJ at 25 °C.
Sketch a graph on the axes to show how ∆G o changes for this process between 25 °C and
150 °C at 101 kPa.
positive
G o
0 temperature / C
25 50 75 100 125 150
negative
[2]
(d) The reaction between A and B is feasible at low temperatures but is not feasible at high temperatures.
A + B C + D
Deduce the signs of ∆H and ∆S for this reaction and explain why the feasibility changes with temperature.
sign of ∆H = …… sign of ∆S = ……
. ……
. ……
. ……
. …… [2]
[Total: 15]
Mark scheme
4(a)(i) M1: energy change when 1 mole of a ionic compound is formed 2
M2: from its gaseous ions under standard conditions
4(a)(ii) ΔHsol = (–2099) + (2× – 378) – (–2824) 2
ΔHsol = –31 kJ mol–1
M1: use of ×2 as only multiplier
M2: correct signs and evaluation
4(a)(iii) M1: Cu2+ is smaller OR Cu2+ has a higher charge density 2
M2: Cu2+ attracts water molecules more / stronger
OR (Cu2+) forms stronger ion-dipole forces to water molecules
4(b)(i) anode: chlorine / Cl2 1
cathode: hydrogen / H2
4(b)(ii) M1: Q = 0.75 × 60 × 60 = 2700 C AND 96 500 or 193 000 used 2
M2: [a] moles of Ca = 2700 / 193 000 = 0.0140
[b] mass = 0.0140 × 40.1 = 0.56 g
4(c)(i) measure / degree of (dis)order / randomness (of a system) 1
OR the number of possible arrangements of the particles and their energy (in a given system)
4(c)(ii) 1
ΔS is negative ΔS is zero ΔS is positive
solid dissolving in water
water solidifying to ice
4(c)(iii) 2
two correct for 1 mark, three correct for two marks:
• starting at +8.6 kJ / in positive region close to the y-axis
• line passes through x-axis around 100°C
• negative gradient straight / curve line through the x-axis (no clear positive inflexions)
4(d) M1: ΔH negative / – , ΔS negative / – 2
M2: as temperature increase, ΔG becomes (more) positive / less negative ora
OR at low(er) T, (ΔH more negative than TΔS) so ΔG is negative
OR at high(er) T, (ΔH less negative than TΔS) so ΔG is positive
(a) Describe and explain the relative basicities of phenylamine, ethylamine and 4-nitrophenylamine.
…… > …… > ……
most basic least basic
. ……
. ……
. ……
. ……
. …… [4]
(b) The dye R can be synthesised from 4-nitrophenylamine in two steps.
4-nitrophenylamine Q
step 1
O2N NH2
step 2
R
OH
O2N N N
(i) Deduce and draw the structure of the organic salt Q in the box. [1]
(ii) Suggest reagents and conditions for step 1 and 2 in (b).
step 1 . ……
step 2 . ……
[2]
(c) Compound G can be synthesised from methylbenzene in three steps.
E F
CH3
step 1 step 2
step 3
G
CO2H
NO2
Br
(i) Give the systematic name of compound G.
. …… [1]
(ii) Deduce the identities of E and F and draw their structures in the boxes. [2]
(iii) Suggest reagents and conditions for each of steps 1 to 3 in (c).
step 1 . ……
step 2 . ……
step 3 . ……
[3]
[Total: 13]
Mark scheme
5(a) M1: 4
ethylamine phenylamine 4-nitrophenylamine
most basic least basic
M2 / 3 / 4: explanation two correct for one mark, three correct for two marks, four correct for three marks
• (basicity linked to) lone pair / p orbital on N AND being able accept / donate to / coordinate to a proton / H+
• ethyl / alkyl group is electron donating / has a positive inductive effect (so lone pair on N is more able to accept a
proton)
• (phenylamines are less basic than ethylamine as) p orbital / lone pair on N is delocalised (into the ring so less able to
accept a proton)
• (4-nitrophenylamine is less basic than phenylamine as) nitro / NO2 group is electron withdrawing (so lone pair on N is
less able to accept a proton)
5(b)(i) 1
5(b)(ii) M1: 2
step 1: HNO2, ⩽ 10°C OR NaNO2, HCl(aq), ⩽ 10°C
M2:
step 2: NaOH / alkaline AND 1-naphthol / α-naphthol / structure below
5(c)(i) 4-bromo-2-nitrobenzoic acid OR 4-bromo-2-nitro(-1-)benzenecarboxylic acid 1
5(c)(ii) 2
5(c)(iii) M1: step 1 conc. H2SO4 and conc. HNO3 3
M2: step 2 Br2 and AlBr3
M3: step 3 hot (alkaline / acidified) MnO4– / KMnO4
(a) There are four possible structural isomers of C8H10 that contain a benzene ring.
Draw the skeletal formulae of the four structural isomers in the appropriate boxes. The number of peaks observed in the carbon‑13 (13C) NMR spectrum of each compound is given.
isomer 1 isomer 2
three peaks in 13C NMR four peaks in 13C NMR
isomer 3 isomer 4
five peaks in 13C NMR six peaks in 13C NMR
[4]
(b) A three-step synthesis of X (C10H10O) from benzene is suggested as shown.
W
Cl (CH2)3CO2H
Al Cl 3 step 1
step 2
Cl
O
step 3 Al Cl 3
X (C10H10O)
(i) Step 1 is the alkylation of benzene by electrophilic substitution.
Use R–Cl to represent Cl (CH2)3CO2H.
Write an equation for the formation of an electrophile from R–Cl and Al Cl 3.
. …… [1]
(ii) Deduce and draw the structures of W and X in the boxes. [2]
(iii) Suggest the reagents and conditions for step 2.
. …… [1]
(iv)
Complete the mechanism for the reaction of benzene with the electrophile formed in (b)(i).
Include all relevant charges and curly arrows showing the movement of electron pairs.
Draw the structure of the intermediate.
intermediate
R
[3]
[Total: 11]
Mark scheme
6(a) 4
correct isomers and correct assignment to peaks: mark as • • • •
6(b)(i) RCl + AlCl3 R+ + AlCl4– 1
OR Cl(CH2)3COOH + AlCl3 +(CH2)3COOH + AlCl4–
6(b)(ii) 2
6(b)(iii) SOCl2 OR PCl5 1
ALLOW PCl3 AND heat
6(b)(iv) 3
M1: arrow to R+ OR arrow to positive carbon of +(CH2)3COOH
M2: correct structure of intermediate
M3: arrow from C-H bond into the ring AND H+
(a) In aqueous solution, chlorine dioxide, Cl O2, reacts with hydroxide ions as shown.
2Cl O2 + 2OH– → Cl O3– + Cl O2– + H2O
A series of experiments is carried out using different concentrations of Cl O2 and OH–. The table shows the results obtained.
[Cl O2] [OH–] initial rate experiment
/ mol dm–3 / mol dm–3 / mol dm–3 min–1
1 0.020 0.030 7.20 × 10–4
2 0.020 0.120 2.88 × 10–3
3 0.050 0.030 4.50 × 10–3
(i) Explain the term order of reaction.
. ……
. …… [1]
(ii) Use the data in the table to determine the order of reaction with respect to each reactant,
Cl O2 and OH–.
Explain your reasoning.
. ……
. ……
. ……
. ……
. ……
. …… [2]
(iii) Use your answer to (a)(ii) to construct the rate equation for this reaction.
rate = …… [1]
(iv)
Use your rate equation and the data from experiment 1 to calculate the rate constant, k, for this reaction.
Include the units of k.
k = …… units …… [2]
Question 7 continues on the next page.
(b) The decomposition of benzenediazonium ions, C6H5N2+, using a large excess of water, is a first‑order reaction.
The graph shows the results obtained.
0.025
0.020
0.015
[C6H5N2+]
/ mol dm–3
0.010
0.005
0
0 100 200 300 400 500 600 time / s
(i) Draw the structure of the organic product formed in this reaction.
[1]
(ii) Use the graph to determine the rate of reaction at 100 s. Show your working.
rate = …… mol dm–3 s–1 [1]
(c) Sketch a concentration–time graph for a zero-order reaction.
Use your graph to suggest how successive half-lives for a zero-order reaction vary as the concentration of a reactant decreases. Indicate this by placing a tick ( ) in the appropriate box in the table.
concentration
time
successive half-lives no change in successive half-lives decrease successive half-lives increase
[1]
[Total: 9]
Mark scheme
7(a)(i) the power to which a concentration of a reactant is raised in the rate equation / law 1
7(a)(ii) M1: (using expt 1 and 3) as [ClO2] × 2.5 rate × 6.25 so 2nd order 2
M2: (using expt 1 and 2) as [OH–] × 4 rate × 4 so 1st order
7(a)(iii) rate = k[ClO2]2[OH–] 1
7(a)(iv) M1: 2
k = rate / [ClO2]2[OH–]
k = 7.20 × 10-4 / (0.02)2(0.03)
k = 60
M2: mol–2 dm6 min–1
7(b)(i) 1
structure of phenol: C6H5OH OR
7(b)(ii) tangent drawn correctly AND rate = 0.015 / 260 = 5.8 × 10–5 1
ALLOW values consistent with tangent drawn at 100 sec
7(c) 1
AND half-life decreases (1st box)
(a) State and explain the relative rate of hydrolysis of acyl chlorides, alkyl chlorides and aryl chlorides.
…… > …… > ……
fastest slowest
. ……
. ……
. ……
. ……
. …… [3]
(b) The drug remifentanil is shown.
remifentanil
O O
O
O N
N
O
Remifentanil is completely hydrolysed under acidic conditions. Three different organic compounds are formed.
Draw the structures for these organic compounds in the boxes.
[3]
(c) Compound Y, C5H10O2, reacts with Na2CO3(aq) to evolve bubbles of gas. The proton (1H) NMR spectrum of compound Y in D2O is shown.
2 1 0
/ ppm
(i) Use this information to suggest a structure for Y.
[1]
(ii) Use the Data Booklet, the proton (1H) NMR spectrum and your answer to (c)(i) to complete the table.
chemical number of 1H atoms environment of proton splitting pattern shift (δ) responsible for the peak
0.95
1.90
2.20
[3]
[Total: 10]
Mark scheme
8(a) M1: 3
acyl chlorides > alkyl chlorides > aryl chlorides
fastest slowest
M2 / 3: Any two from:
• acyl chlorides carbon in C-Cl bond is more electron deficient since it is also attached to an oxygen atom OR C-Cl
bond is weakest / weakened in acyl chlorides since it is also attached to an oxygen atom / two electronegative atoms
• aryl chlorides (no hydrolysis) C-Cl bond is part of delocalised system / partially double bond character (so C-Cl bond is
stronger) OR lone pair / p-orbital on Cl delocalises with π ring (so C-Cl bond is stronger)
• alkyl chlorides C-Cl bond strengthened by electron donating effect / positive inductive effect of alkyl / R group (as
compared to acyl chlorides) OR carbon atom has a smaller δ+ and the C-Cl bond is stronger (than the C-Cl bond in
COCl) due to (the carbon) being only attached to one electronegative atom
8(b) 3
ALLOW amine salt for the third structure – mono or di ion
8(c)(i) 1
8(c)(ii) 3
chemical splitting pattern number of 1H atoms
environment of proton
shift (δ) (words required) responsible for the peak
0.95 alkane / CH3 doublet 6
1.90 alkane / CH multiplet 1
ALLOW alkyne
2.20 R / alkyl / CH2 next to doublet 2
C=O / COOH
mark as • • • • • •
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