1

2

Q1. Spectral data for use in this question are provided below the Periodic Table (first item on the database).Compound Q has the molecular formula C4H8O2

(a) The infra-red spectrum of Q is shown below.

Identify the type of bond causing the absorption labelled R and that causing the absorption labelled S.

R ………S……… (2)

(b) Q does not react with Tollens’ reagent or Fehling’s solution. Identify a functional group which would react with these reagents and therefore cannot be present in Q. (1)

(c) Proton n.m.r. spectra are recorded using a solution of a substance to which tetramethylsilane (TMS) has been added.

(i) Give two reasons why TMS is a suitable standard.

(ii) Give an example of a solvent which is suitable for use in recording an n.m.r. spectrum. Give a reason for your choice. (4) (d) The proton n.m.r. spectrum of Q shows 4 peaks.

The table below gives δ values for each of these peaks together with their splitting patterns and integration values.

δ/ppm 2.20 2.69 3.40 3.84

Splitting pattern singlet triplet singlet triplet

Integration value 3 2 1 2

What can be deduced about the structure of Q from the presence of the following in its n.m.r. spectrum?

(i) The singlet peak at δ = 2.20; (ii) The singlet peak at δ = 3.40; (iii) Two triplet peaks (3) (e) Using your answers to parts (a), (b) and (d), deduce the structure of compound Q. (1)(Total 11 marks)

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Q2. Three sections of the proton n.m.r. spectrum of CH3CHClCOOH are shown below.

(a) Name the compound CH3CHClCOOH (1)

(b) Explain the splitting patterns in the peaks at δ 1.72 and δ 4.44 (2)

(c) Predict the splitting pattern that would be seen in the proton n.m.r. spectrum of the isomeric compound ClCH2CH2COOH (1)

(d) The amino acid alanine is formed by the reaction of CH3CHClCOOH with an excess of ammonia. The mechanism is nucleophilic substitution. Outline this mechanism, showing clearly the structure of alanine.(5)

(e) The amino acid lysine has the structure. Draw structures to show the product formed in each case when lysine reacts with

(i) an excess of aqueous HCl,

(ii) an excess of aqueous NaOH,

(iii) another molecule of lysine.(3)(Total 12 marks)

Q3. The conversion of compound A into compound B can be achieved in two steps as shown below.

The intermediate compound, X, has an absorption at 1650 cm–1

in its infra-red spectrum.

(a) Identify compound X. Explain your answer.(2)

(b) For each step in this conversion, give the reagents and essential conditions required and outline a mechanism.(11)

(c) Show how the number of peaks in their proton n.m.r. spectra would enable you to distinguish between compounds A and B.(2)(Total 15 marks)

4

Q4. The proton n.m.r. spectrum of compound X is shown below.

Compound X, C7H12O3, contains both a ketone and an ester functional group. The measured integration trace for the peaks in the n.m.r. spectrum of X gives the ratio shown in the table below.

Chemical shift, δ/ppm 4.13 2.76 2.57 2.20 1.26

Integration ratio 0.8 0.8 0.8 1.2 1.2

Refer to the spectrum, the information given above and the data below the Periodic Table provided to answer the following questions.

(a) How many different types of proton are present in compound X? (1)

(b) What is the whole-number ratio of each type of proton in compound X? (1)

(c) Draw the part of the structure of X which can be deduced from the presence of the peak at δ2.2 (1) (d) The peaks at δ4.13 and δ1.26 arise from the presence of an alkyl group. Identify the group and explain the splitting pattern. (3)

(e) Draw the part of the structure of X which can be deduced from the splitting of the peaks at δ2.76 and δ2.57. (1) (f) Deduce the structure of compound X. (2)(Total 9 marks)

5

Q5. Use the data given on the back of the Periodic Table (PT) to help you answer this question. Compounds A to G are all isomers with the molecular formula C6H12O2

(a) Isomer A, C6H12O2, is a neutral compound and is formed by the reaction between compounds X and Y in the presence of a small amount of concentrated sulphuric acid. X and Y can both be formed from propanal by different redox reactions. X has an absorption in its infra-red spectrum at 1750 cm

–1.

Deduce the structural formulae of A, X and Y. Give suitable reagents, in each case, for the formation of X and Y from propanal and state the role of concentrated sulphuric acid in the formation of A.(7) (b) Isomers B, C, D and E all react with aqueous sodium carbonate to produce carbon dioxide. Deduce the structural formulae of the three isomers that contain an asymmetric carbon atom. The fourth isomer has only three singlet peaks in its proton n.m.r. spectrum. Deduce the structural formula of this isomer and label it E.(4) (c) Isomer F, C6H12O2, has the structural formula shown below, on which some of the protons have been labelled.

A proton n.m.r. spectrum is obtained for F. Using Table 1 at the back of the Periodic Table (PT), predict a value of δ for the protons labelled a and also for those labelled b. State and account for the splitting patterns of the peaks assigned to the protons a and b.(6) (d) Isomer G, C6H12O2, contains six carbon atoms in a ring. It has an absorption in its infra-red spectrum at 3270 cm

–

1 and shows only three different proton environments in its proton n.m.r. spectrum. Deduce a structural formula for

G.(2)(Total 19 marks)

Q6. The three compounds CH3CH2CH2CH2OH, (CH3)3COH and CH3CH2CH2CHO can be distinguished by use of the following three reagents

1. potassium dichromate(VI) acidified with dilute sulphuric acid 2. Tollens’ reagent 3. ethanoic acid, together with a small amount of concentrated sulphuric acid.

(a) Identify which of these three organic compounds would reduce acidified potassium dichromate(VI). Give the structures of the organic products formed. Write a half-equation for the reduction of dichromate(VI) ions in acidic solution.(6) (b) Identify which one of these three organic compounds would reduce Tollens’ reagent. Give the structure of the organic product formed. Write a half-equation for the reduction of Tollens’ reagent.(3) (c) Identify which of these three organic compounds would react with ethanoic acid in the presence of concentrated sulphuric acid. In each case, give the structure of the organic product formed.(4) (d) State the number of peaks in the proton n.m.r. spectra of CH3CH2CH2CH2OH and of (CH3)3COH. (Analysis of peak splitting is not required.)(2)(Total 15 marks)

Q7. Compound Q has the molecular formula C4H7ClO and does not produce misty fumes when added to water.

(a) The infra-red spectrum of Q contains a major absorption at 1724 cm–1

. Identify the bond responsible for this absorption. (1)

(b) The mass spectrum of Q contains two molecular ion peaks at m/z = 106 and m/z = 108. It also has a major peak at m/z = 43.

(i) Suggest why there are two molecular ion peaks (ii) A fragment ion produced from Q has m/z = 43 and contains atoms of three different elements. Identify this fragment ion and write an equation showing its formation from the molecular ion of Q.

6

Fragment ion ......................................................Equation(3)

(c) The proton n.m.r. spectrum of Q was recorded. (i) Suggest a suitable solvent for use in recording this spectrum of Q. (ii) Give the formula of the standard reference compound used in recording proton n.m.r. spectra. (2) (d) The proton n.m.r. spectrum of Q shows three peaks. Complete the table below to show the number of adjacent, non-equivalent protons responsible for the splitting pattern.

(1) (e) Using the information in parts (a), (b) and (d), deduce the structure of compound Q.(1) (f) A structural isomer of Q reacts with cold water to produce misty fumes. Suggest a structure for this isomer.(1) (Total 9 marks)

Q8. Butenedioic acid, HOOCCH=CHCOOH, occurs as two stereoisomers. One of the isomers readily forms the acid anhydride C4H2O3 when warmed.

(a) Draw the structures of the two isomers of butenedioic acid and name the type of isomerism shown. Use the structures of the two isomeric acids to suggest why only one of them readily forms an acid anhydride when warmed. Draw the structure of the acid anhydride formed.(6) (b) Identify one electrophile which will react with butenedioic acid and outline a mechanism for this reaction.(4) (c) Write an equation for a reaction which occurs when butenedioic acid is treated with an excess of aqueous sodium hydroxide.(2) (d) Describe and explain the appearance of the proton n.m.r. spectrum of butenedioic acid.(3)(Total 15 marks)

Q9. (a) Ester X, CH3CH2COOCH3, can be produced by the reaction between propanoyl chloride and methanol. Name X and outline a mechanism for this reaction. Name the mechanism involved.(6)

(b) The proton n.m.r. spectrum of X is shown below together with that of an isomeric ester, Y. Deduce which of Spectrum 1 and Spectrum 2 is that obtained from X. Use Table 1 on the Data Sheet and the integration data on the spectra to help you to explain your deduction. Suggest a structure for Y.

(4) (Total 10 marks)

Peak 1 Peak 2 Peak 3

Integration value 3 3 1

Splitting pattern doublet singlet quartet

Number of adjacent, non-equivalent protons

1

7

Q10. Compounds C (first one) and D (second one), shown below, are isomers of C5H10O

(a) Name compound C.(1)

(b) Use Table 2 on the Data Sheet to help you to answer this question.

(i) Suggest the wavenumber of an absorption which is present in the infra-red spectrum of C but

not in that of D.

(ii) Suggest the wavenumber of an absorption which is present in the infra-red spectrum of D but not in that of C. (2)

(c) Deduce the number of peaks in the proton n.m.r. spectrum of C. (1) (d) Identify a reagent that you could use to distinguish between C and D. For each of C and D, state what you would observe when the compound is treated with this reagent. (3) (e) Compound E, CH3CH2CH2CH2CHO, is also an isomer of C5H10O . Identify a reagent which will react with E but not with C or D. State what you would observe when E is treated with this reagent. (2)(Total 9 marks)

Q11. Each of the parts (a) to (e) below concerns a different pair of isomers.

Draw one possible structure for each of the species A to J, using Table 2 on the Data Sheet where appropriate.

(a) Compounds A and B have the molecular formula C5H10

A decolourises bromine water but B does not.(2)

(b) Compounds C and D have the molecular formula C2H4O2 . Each has an absorption in its infra-red spectrum at about 1700 cm

–1 but only D has a broad absorption at 3350 cm

–1 (2)

(c) Compounds E and F are esters with the molecular formula C5 H10O2 . The proton n.m.r. spectrum of E consists of two singlets only whereas that of F consists of two quartets and two triplets.(2)

(d) Compounds G and H have the molecular formula C3H6Cl2 G shows optical activity but H does not (2)

(e) Compounds I and J have the molecular formula C6H12 . Each has an absorption in its infra-red spectrum at about 1650 cm

–1 and neither shows geometrical isomerism. The proton n.m.r. spectrum of I consists of a singlet

only whereas that of J consists of a singlet, a triplet and a quartet.(2)(Total 10 marks)

Q12. This question concerns four isomers, W, X, Y and Z, with the molecular formula C5H10O2

(a) The proton n.m.r. spectrum of W shows 4 peaks. The table below gives the chemical shifts, δ values, for each of these peaks, together with their splitting patterns and integration values.

8

δ/ppm 2.18 2.59 3.33 3.64

Splitting pattern singlet triplet singlet triplet

Integration value 3 2 3 2

State what can be deduced about the structure of W from the presence of the following in its n.m.r. spectrum.

(i) The singlet peak at δ = 2.18 ; (ii) The singlet peak at δ = 3.33 ; (iii) Two triplet peaks (iv) Hence, deduce the structure of W.(4) (b) The infra-red spectrum of X is shown below.

(i) What can be deduced from the broad absorption centred on 3000 cm

–1 in the infra-red spectrum

of X?

(ii) Given that the proton n.m.r. spectrum of X contains only two peaks with the integration ratio 9:1, deduce the structure of X.(2)

(c) Isomers Y and Z have the structures shown below.

Identify the two reagents you could use in a simple chemical test to distinguish between Y and Z. State what you would observe when each of Y and Z is tested with a mixture of these two reagents. (3)(Total 9 marks)

Q13. (a) The reaction of but-1-ene with chlorine produces 1,2-dichlorobutane, C4H8Cl2

(i) Given that chlorine exists as a mixture of two isotopes, 35

Cl and 37

Cl, predict the number of molecular ion peaks and their m/z values in the mass spectrum of C4H8Cl2

(ii) The mass spectrum of 1,2-dichlorobutane contains peaks at m/z = 77 and 79. Draw the structure of the fragment ion which produces the peak at m/z = 77 and write an equation showing its formation from the molecular ion.(6) (b) The reaction of but-2-ene with hydrogen chloride forms a racemic mixture of the stereoisomers of 2-chlorobutane. (i) Name the type of stereoisomerism shown by 2-chlorobutane and give the meaning of the term racemic mixture. State how separate samples of the stereoisomers could be distinguished. (ii) By considering the shape of the reactive intermediate involved in the mechanism of this reaction, explain how a racemic mixture of the two stereoisomers of 2-chlorobutane is formed.(7) (c) The reaction of but-2-ene with chlorine produces 2,3-dichlorobutane, C4H8Cl2 (i) State the number of peaks, their integration ratio and any splitting of peaks in the proton n.m.r. spectrum of 2,3-dichlorobutane. (ii) Compound S, an isomer of C4H8Cl2, produces a proton n.m.r. spectrum which consists only of a singlet, a triplet and a quartet with an integration ratio of 3:3:2 respectively.Compound T, also an isomer of C4H8Cl2, produces a proton n.m.r. spectrum which consists only of two singlets with an integration ratio of 3:1 . Draw the structures of S and of T. (6)(Total 19 marks)

9

ANSWERS TO NMR QUESTIONS A2

M1. (a) R: O‑H (alcohols) (1)

S: C=O or carbonyl (1) 2

(b) aldehyde (1) ‑CHO or RCHO (1)

1

(c) (i) Reason 1: TMS inert or non-toxic or volatile / easily removed Reason 2: single (intense) peak peak of 12 protons has 12 equivalent protons all protons in same environment OR peak / signal upfield of others highly shielded more shielded peak away from others or δ = 0 or low

not solvent, not cheap

any 2 reasons × (1)

(ii) Solvent: CDCl3 or CCl4 (NOT D2O) Reason: proton free (1)

allow no hydrogens (atoms)

NOT H+ / hydrogen ions

4

(d) (i) (1)

(ii) –OH (1)

(iii) –CH2–CH2– (1) 3

(e) (1) 1

[11]

10

M2. (a) 2-chloropropanoic acid (1)

1

(b) δ 1.72 Doublet next to CH (1)

δ 4.44 Quartet next to CH3 (1)

2

(c) Two triplets (1) 1

(d)

Allow SN1 5

(e) (i)

(ii)

(iii)

O

11

r anhydride 3

[12]

M3. (a) Identity of X; 2-methylpropene (1) Absorption at 1650 cm

–1 indicates an alkene present (1)

OR a chemical answer e.g. Br2 (aq) brown to colourless 2

(b) Reagents Step 1 KOH (allow NaOH) (1) alcoholic (1) warm (1)

Only allow solvent and warm if reagent correct

Step 2 HBr (1)

Mechanism: A → X

Or a carbocation mechanism

Mechanism X → B

11

(c) A gives three peaks (1) B gives one peak (1)

Allow one for “A has more peaks than B” when no number of peaks is given

2

[15]

M4. (a) 5 (1) 1

(b) 2:2:2:3:3 (1)

any order but not multiples 1

(c) (1) 1

12

(d) CH3CH2 or C2H5 or ethyl (1) δ4.13 (quartet) : CH2 peak split by CH3 / next to CH3 (1) δ1.26 (triplet) : CH3 peak split by CH2 / next to CH2 (1)

3

(e) CH2CH2 (1) 1

(f) (2)

allow (1) for CH3COCH2CH2OCOCH2CH3

or CH3COOCH2CH2COCH2CH3

Must be C7H12O3

2

[9]

M5. (a) X contains > C=O (1)

if X and Y reversed lose this mark but allow remaining max 6/7

X is CH3CH2COOH (1)

Y is CH3CH2CH2OH (1)

A is (1)

Conc H2SO4 : catalyst (1) 7

(b)

4

13

(c) 3.1 – 3.9 (1)

2.1 – 2.6 (1)

a: quartet (1) 3 adjacent H (1)

b: triplet (1) 2 adjacent H (1) 6

(d) 3269 cm–1

OH alcohol (1)

G is (1) 2

Notes

(a) first mark for C=O stated or shown in X

Ignore wrong names

Y CH3CH2CH2OH allow C3H7 in A if Y correct or vice versa Allow (1) for A if correct conseq to wrong X and Y

other oxidising agents: acidified KMnO4; Tollens; Fehlings

other reducing agents: LiAlH4; Na/ethanol; Ni/H2; Zn or Sn or Fe/HCl

(b) give (1) for carboxylic acid stated or COOH shown in each suggestion (1) for correct E any 2 out of 3 for B, C or D allow C3H7 for either the B or D shown on the mark scheme i.e. a correct structure labelled B, C or D or E will gain 2.

(c) protons a – quartet must be correct to score 3 adjacent H mark. Same for b

(d) allow (1) for any OH (alcohol) shown correctly in any structure – ignore extra functional groups. Structure must be completely correct to gain second mark

[19]

Organic points

(1) Curly arrows: must show movement of a

14

pair of electrons, i.e. from bond to atom or from lp to atom / space e.g.

(2) Structures

penalise sticks (i.e. ) once per paper

Penalise once per paper

allow CH3– or –CH3 or or CH3

or H3C–

M6. (a) K2Cr2O7/H2SO4 reuced by

CH3CH2CH2CH2OH (1)

oxidised to CH3(CH2)2CHO (1) and CH3(CH2)2COOH (1)

CH3CH2CH2CHO (1)

oxidised to CH3(CH2)2COOH (1)

Equation: Cr2O72–

+ 14H+ + 6e

– → 2Cr

3+ + 7H2O (1)

Note: Deduct one if all three compounds given as reducing agents. 6

(b) Tollens’ reduced by CH3CH2CH2CHO (1) oxidised to CH3(CH2)2COOH (1)

Equation [Ag(NH3)2]+ + e

– → Ag + 2NH3 (1)

3

(c) CH3CH2CH2CH2OH (1)

Product CH3CH2CH2CH2OOCCH3 (1)

(CH3)3COH

Product (CH3)3COOCCH3 (1) 4

15

(d) CH3CH2CH2OH has five peaks (1)

(CH3)3COH has two peaks (1) 2[15]

M7. (a) C=O (1)

or “carbonyl” 1

(b) (i) Cl has (2) isotopes (1)

Allow 35

Cl and 37

Cl without word isotope – but must be correct isotopes

must have 3 different elements, i.e. not C3H7+ but allow balanced equation

including C3H7+ for the equation mark

(ii) Fragmentation: (1)

must be an ion (*)

Equation: C4H7ClO+•

→ + C2H4Cl• (1)

(*) allow C2H3O+ or any form of it (i.e. CH2CHO+ or CH2COH

+) in equation,

be generous with position of + or •

if fragment ion completely wrong (not m/z = 43) no further marks 3

(c) (i) CDCl3 or CCl4 (1) or D2O, C6D6

(ii) Si(CH3)4 (1) or SiC4H12

2

(d) (1)

Peak 1 Peak 2 Peak

Integration value 3 3 1

Splitting pattern doublet singlet quartet

Number of adjacent, non-equivalent protons

1 0 3

1

(e) (1)

or CH3COCHClCH3

1

(f)

16

(1) 1

[9]

M8. (a)

1

1

NB The bonds shown in the structure must be correct

Isomerism: Geometric or cis-trans

If written answer is correct, ignore incorrect labelling of structures. If no written answer, allow correctly labelled structures.

1

Both COOH groups must be on the same side/ close together/ cis 1

No rotation about C=C axis 1

Structure

1

(b) Br2 / HBr / H2SO4 / H+ / Br

+ / NO2

+ (Mark M1)

17

NB If electrophile H+ / Br

+ / NO2

+ allow M1, M2 and M4

If the acid is incorrect, M2 and M3 can still be scored Allow M4 consequentially if repeat error from part (a)

4

(c) e.g. 2NaOH + HO2CCHCHCO2H → NaO2CCHCHCO2Na + 2H2O

Both H replaced 1

Balanced for atoms and charges 1

NB Allow ionic equations and 2NaOH + C4H4O4 → C4H2O4Na2+ 2H2O

Allow one if structure incorrect but molecular formula correct

Allow one for a correct equation showing one H replaced

(d) M1 Two peaks 1

M2 No splitting or singlets 1

M3 (Two) non-equivalent protons or two proton environments 1

M4 No adjacent protons 1

M5 Same area under the two peaks or same relative intensity 1

NB Doublet could score M1 and M3 or M5 (Max 2)

More than two peaks CE = 0

Apply the “list principle” to incorrect answers if more than 3 given

Max 3[15]

M9. X is methyl propanoate

1

M1 for arrow and lone pair,

18

4

M2 for arrow: addition-elimination 1

Spectrum 2: if thinks Spectrum 1 = X can only score for structure of Y

1

Y is CH3COOCH2CH3

1

The two marks for explanation are awarded for discussing one or more of the four peaks (not those for the CH3 of the ethyl groups) for stated δ values the integration or the splitting should be related to the structure: e.g. structure of X shows that at δ 3.7 – 4.1 (1) spectrum of X should have integration 3 / singlet (1)

or

at δ 2.1 – 2.6 (1) spectrum of X should have integration 2 / quartet (1)

Spectrum 2 has these [OR Spectrum 1 has at 3.7 – 4.1 (1) quartet / integration 2 (1) so not X at 2.1 – 2.6 (1) singlet / integration 3 (1) so not X]

2[10]

M10. (a) Pentan-2-one 1

(b) (i) 1680 – 1750 (cm–1

) 1

(ii) 3230 – 3550 or 1000 – 1300 (cm–1

) 1

(iii) 4 1

(c)

Reagent K2Cr2O7/H+

KMnO4 /H+

Na CH3COOH/ H2SO4

1

with C no reaction no reaction no reaction no reaction 1

with D goes green goes colourless effervescence smell 1

(penalise incomplete reagent e.g. K2Cr2O7 or Cr2O7 2–

/H+ then mark on)

(d)

Reagent Tollens Fehlings or Benedicts 1

with E silver (mirror)

red ppt or goes red (not red solution)

1

[9]

M11. (a) A any C5 alkene 1

B

19

1

(b) C

or CH3COOH or HCOOCH3

1

D

or HOCH2CHO 1

(c) E

1

F

1

(d) G

1

H

1

(e) I

20

1

J

NOT hex-3-ene 1[10]

M12. (a) (i)

or RCOCH3;

(or description in words) (ignore trailing bonds)

1

(ii) H3C—O or ROCH3;(allow 1 if both (i) and (ii) give CH3- or H3C– only) 1

(iii) CH2CH2 or two adjacent methylene groups; 1

(iv)

OR

CH3COCH2CH2OCH3; 1

(b) (i) OH in acids or (carboxylic) acid present

(ii)

(c)

reagent K2Cr2O7 /H+

KMnO4 /H+

Y no reaction no reaction

Z orange to green or turns green

purple to colourless or turns colourless

5

[9]

21

M13. (a) (i) 3 peaks or shown in a list 1

m/z = 126, 128 and 130 (56 +70/72/74)

(all 3 scores 2)

(if 56 wrong allow (x + 70/72/74) for 1(x cannot be zero)

(any two scores 1) 2

(ii)

(1) (1) (1)

allow wrong structure for structure dot can be anywhere

3

(b) (i) optical 1

equal mixture of enantiomers 1

(optically) inactive or effects cancel 1

plane polarised light use stereospecific reagent (QoL) 1

rotated in opposite/different directions (QoL) reacts with one isomer only

1

(ii) carbocation 1

planar

(must refer to carbocation or intermediate) 1

attack from either side equally likely

(must refer to carbocation /intermediate) 1

7 max

(c) (i) 2 peaks

(if 4 peaks allow splitting only) 1

ratio 6:2 or 3:1 1

doublet (6 or 3) 1

quartet (2 or 1) 1

(ii) S

22

1

T

1[19]