Spectroscopic Identification of Organic Compounds

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Spectroscopic Identification of Organic Compounds...

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Spectroscopic iden identificati tification on of o f organi or ganic c compounds c ompounds [28 marks]

 The structure of an unknown unknown compound A with empirical formula variety of analytical techniques. spectrum of A is shown below. 1a.  The mass spectrum

CH2

can be determined using information from a

[1 mark]

Deduce the formula of the molecular ion from the mass spectrum.

Markscheme C4 H+ 8 ; Penalize missing charge only once in (i) and (ii) .

Examiners report  This question focused on some of the fundamental spectroscopic spectroscopic techniques (MS, IR and 1 H NMR) used in analytical chemistry. The better candidates did well o n this question though few scored full ma rks. In (a), the most common mistake was omission of the positive charge in (i). One G2 comment stated that isotopic isotopic effects in ma ss spectra with regard to the determina tion of the molecular i on peak would confuse students. This generally was not the the case and although most got the C4 H8 formula a large majority of candidates simply did not read the question correctly which specifically asked for the formula of the molecular ion. In (ii) the the corre ct formulas of the fragments were usually given. In (b), C=C was usually cited as the correct structural formul a in (ii). The weaker ca ndidates struggled with explaining the doublet in (iii). Cyclobutane was obtained by a la rge number of candidates in (iv). In (c) (i), (i), the better candidates scored all three marks. In (ii), (ii), an understanding of the fingerprint region was poor ly conveyed. There were two parts to this question question – an outline of what happens on a m olecular l evel when radiation in the fingerpr int region is absorbed and how this region is used in chemical analysis. One G2 comment referred to the fact that the fingerprint region is not explicitly mentioned in the guide. Although this is true per se AS 3.2 does require a description of how the information from an IR spectrum ca n be used to identify bonds, and it would would be assumed that the fingerprint region would be discussed in the context of teaching teaching IR spectroscopy as part of the IB chem istry programme.

 The infrared (IR) spectrum of A is shown below.

1b.

Explain the presence of a do ublet in the high-resolution proton nuclear magnetic resonance (1H NMR) spectrum [3 marks] of A.

Markscheme produced by (Hs in) =

CH2

group;

adjacent C has 1 H atom; n

+ 1;

due to rela tive/(two) different orientations/alignment of spin of nuclei/protons/hydrogens (with applied/external magnetic field);

Examiners report  This question focused on some of the fundamental spectroscopic techniques (MS, IR and 1 H NMR) used in analytical chemistry. The better candidates did well o n this question though few scored full ma rks. In (a), the most common mistake was omission of the positive charge in (i). One G2 comment stated that isotopic effects in ma ss spectra with regard to the determina tion of the molecular i on peak would confuse students. This generally was not the case and although most got the C4 H8 formula a large majority of candidates simply did not read the question correctly which specifically asked for the formula of the molecular ion. In (ii) the corre ct formulas of the fragments were usually given. In (b), C=C was usually cited as the correct structural formul a in (ii). The weaker ca ndidates struggled with explaining the doublet in (iii). Cyclobutane was obtained by a la rge number of candidates in (iv). In (c) (i), the better candidates scored all three marks. In (ii), an understanding of the fingerprint region was poor ly conveyed. There were two parts to this question – an outline of what happens on a m olecular l evel when radiation in the fingerpr int region is absorbed and how this region is used in chemical analysis. One G2 comment referred to the fact that the fingerprint region is not explicitly mentioned in the guide. Although this is true per se AS 3.2 does require a description of how the information from an IR spectrum ca n be used to identify bonds, and it would be assumed that the fingerprint region would be discussed in the context of teaching IR spectroscopy as part of the IB chem istry programme.

1 1c. One isomer of A has only one signal in its H NMR spectrum. Deduce the structural formula of this isomer.

[1 mark]

Markscheme  ;

 Acce pt full or condensed structural formula .

Examiners report  This question focused on some of the fundamental spectroscopic techniques (MS, IR and 1 H NMR) used in analytical chemistry. The better candidates did well o n this question though few scored full ma rks. In (a), the most common mistake was omission of the positive charge in (i). One G2 comment stated that isotopic effects in ma ss spectra with regard to the determina tion of the molecular i on peak would confuse students. This generally was not the case and although most got the C4 H8 formula a large majority of candidates simply did not read the question correctly which specifically asked for the formula of the molecular ion. In (ii) the corre ct formulas of the fragments were usually given. In (b), C=C was usually cited as the correct structural formul a in (ii). The weaker ca ndidates struggled with explaining the doublet in (iii). Cyclobutane was obtained by a la rge number of candidates in (iv). In (c) (i), the better candidates scored all three marks. In (ii), an understanding of the fingerprint region was poor ly conveyed. There were two parts to this question – an outline of what happens on a m olecular l evel when radiation in the fingerpr int region is absorbed and how this region is used in chemical analysis. One G2 comment referred to the fact that the fingerprint region is not explicitly mentioned in the guide. Although this is true per se AS 3.2 does require a description of how the information from an IR spectrum ca n be used to identify bonds, and it would be assumed that the fingerprint region would be discussed in the context of teaching IR spectroscopy as part of the IB chem istry programme.

 The structure of an unknown compound A with empirical formula CH2 can be determined using information from a variety of analytical techniques.  The mass spectrum of A is shown below.

2a. Determine the relative molecular mass of the compound from the mass spectrum and deduce the formula of the [2 marks] molecular ion.

Markscheme 56;

C4 H+ 8 ; Penalize missing charge only once in (i) and (ii) .

Examiners report More than half of the candida tes obtained the molecular ma ss from the spectrum. About a third of the candidates identified C4 H8 as the molecular formula but only a few candidates remember ed to include the positive charge on the molecular ion and scored the second mark.

2b. Deduce the formulas of the fragments which give rise to peaks at m/z

m/z

= 27:

m/z

= 29:

= 27 and 29.

[1 mark]

Markscheme m/z

+ + + + = 27 : C2 H+ 3 /CH2 CH /CH2 =CH and m/z = 29 : C2 H5 /CH3 CH2 ;

Penalize missing charge only once in (i) and (ii) .

Examiners report About a third of the candidates identified the co rrect fragme nts. It was disappointing to see candidates suggesting fragments that did not match the masses of the peaks.

 The infrared (IR) spectrum of A is shown below.

2c. Identify the bond responsible fo r the IR absorption at B.

Markscheme C=C/carbon−carbon double bond;  Acce pt “alkenyl/a lkene”.

Examiners report Very well answered.

[1 mark]

2d. Deduce a structural formula co nsistent with the data.

[1 mark]

Markscheme CH3CH2CH=CH2;  Acce pt either a full or a condensed structural for mula.

Examiners report Only a few candidates deduc ed the corr ect structural formula co nsistent with the data.

Organic compo und X is 68.11% carbon, 13.74% hydrogen and 18.15 % oxygen by mass. 3a. Show that the empirical formula of compound X is C5 H12 O.

Markscheme

 / OWTTE;

 Acce pt mass of C5H12O = 88.17 so % of C  = ( % of H  =  (

12.12 88.17

60.05 88.17

) × 100 = 68.11,

) × 100 = 13.75 and % of O =  ( 16.00 ) × 100 = 18.15. 88.17

 Allow integer values for a tomic masses.

[1 mark]

Examiners report  The question on empirical formula in (a) posed no difficulty and even the weaker candidates scored the mark. In the spectroscopy question in (b), some of the better candidates manage d to score a ll 11 mar ks assigned to this extended response type question. In the MS, + was often omitted. Many students could not distinguish between observed io ns and lost fragments. Only one candidate mentioned the "fingerprint" region of IR but did not refer to the need to compare with library spectral data. For the IR the majority of candidates scored all three marks, though the weaker candidates frequently suggested NH bonds and CF bonds even though neither nitrogen nor fluori ne are pa rt of the empirical formula given in the stem of the question. Discussion of the 1HNMR spectrum proved the most challenging and many ca ndidates did not relate the splitting pattern to the specific car bon fragme nts. It was disappointing at HL seeing a number of candidates not including hydrogens in their final answer for the structural formula of 2methylbutan-2-ol.

1 3b.  The mass spectrum, infrared spectrum and details of the HNMR spectrum of compound X are given below.

Mass spectrum:

Infrared spectrum:

1HNMR spectrum:

Analyse these three spectra and, using relevant information, deduce the identity of the compound.

Mass spectrum:

Infrared spectrum:

1HNMR spectrum:

[11 marks]

Identity of X:

Markscheme Mass spectrum:

molecular ion peak at 88/M+

= 88 shows molecular

formula is C5 H12 O;

absorption at 73 due to (M–CH3)+ / X contains a methyl group as peak at M–15 / OWTTE; absorption at 59 due to (M–C2H5)+ / X contains an ethyl group as peak at M–29; Penalise once only if + charge omitted.  Acce pt that X contains a CHO group due to M–29 but in fact it cannot as there are too many hydrogen atoms in the compound for it to be an aldehyde. Infrared spectrum:

peak in range at 3200–3600 cm−1 shows it contains an OH group / OWTTE; (sharp) peaks just below 3000 cm−1 /in range 2850–3100 cm−1 due to C–H absorptions; lack of peak at approximately

1700 cm−1 shows it does not contain C=O;

absorption between 1050 and

1410 cm−1 due to C–O;

 Allow “due to alcohol ” instead of due to C–O.  Acce pt “absorption between 1050 and 1410 cm –1 due to ether or ester” although it cannot be either as there is only  one O a tom and it has been identified as bonded to H.

fingerprint region specific to compound but needs to be compared with library / OWTTE; 1H

NMR spectrum:

(12 protons are in) four different chemic al environments (in the ratio 1:2:6:3); singlet (with integration trace o f 1) due to OH pr oton; singlet (with integration trace of 6) suggests (two CH3 ) groups attached to a carbon atom with no Hs attached to it; quartet (with integration trace of 2) due to CH2 next to CH3 ; triplet (with integration trace of 3) due to CH3 next to CH2 ; Reference must be made to the association of the splitting pattern (singlet, triplet etc.) to the specific carbon fragments.

(X is) 2-methylbutan-2-ol/CH3 C H2C(CH3 )2 OH; No ECF througho ut 2(b).

Examiners report  The question on empirical formula in (a) posed no difficulty and even the weaker candidates scored the mark. In the spectroscopy question in (b), some of the better candidates manage d to score a ll 11 mar ks assigned to this extended response type question. In the MS, + was often omitted. Many students could not distinguish between observed io ns and lost fragments. Only one candidate mentioned the "fingerprint" region of IR but did not refer to the need to compare with library spectral data. For the IR the majority of candidates scored all three marks, though the weaker candidates frequently suggested NH bonds and CF bonds even though neither nitrogen nor fluori ne are pa rt of the empirical formula given in the stem of the question. Discussion of the 1HNMR spectrum proved the most challenging and many ca ndidates did not relate the splitting pattern to the specific car bon fragme nts. It was disappointing at HL seeing a number of candidates not including hydrogens in their final answer for the structural formula of 2methylbutan-2-ol.

4.  Two students were provided with three different isomers of C3 H6 O2 .

[6 marks]

 They were asked to suggest how the isomers could be distinguished and positively identified from each other using spectroscopic techniques. Student A said that they could be positively identified just from their infrared spectra. Student B said that they could be positively identified just from the number of pea ks and the areas under eac h peak in their 1 HNMR spectra. Evaluate these two claims and suggest how any possible limitations could be overcom e using the same spectroscopic technique.

Student A / Infrared:

Student B / 1HNMR:

Markscheme Student A / Infrared

propanoic acid can be distinguished from the other two by the (broad) absorption at 2500–3300 cm−1 /due to –OH absorption in carboxylic acids; not easy to distinguish between methyl ethanoate and ethyl methanoate because all absorb at 1700–1750 cm−1 / C=O / 1050–1410 cm−1 / C–O / 2850–3100 cm−1 / C–H / because they have same functional groups; can be distinguished from the pattern in the fingerprint regio n / by comparing with spectra of known samples; Student B / 1H NMR

methyl ethanoate can be distinguished from the other two as it will have two peaks of equa l area (due to the two –CH3 groups); propanoic acid and ethyl methanoate canno t be distinguished as both have three peaks / peaks in the ratio of 1: 2:3; chemical shift is also require d to distinguish them;  Absorption value or na me of func tional group requir ed for M1 and M2.

Examiners report  This question proved to be very challenging for most candidates as majority were unable to evaluate the two claims. Candidates appeare d to have some general understanding but were often lacking depth in understanding of the two analytical techniques. The answers were often too general a nd most did not suggest ways to overcome limitations using the same spectroscopic technique s. Limited understanding of the direc tive term ‘evaluate’ along with missing the ‘same’ spectroscopic technique in the que stion stem, penalized the most students from scoring even part mar ks.

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