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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/chemistry/edexcel/nmr-spectroscopy/flashcards","label":"NMR spectroscopy","subjectCode":"a_level_chemistry","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"542f45ea-8d1b-4629-a9f7-fdb502d98ec4","title":"NMR spectroscopy Flashcards","topicId":"114ace27-7a5c-5843-a52e-17a953a70064","topicSlug":"nmr-spectroscopy","cards":[{"id":"334f0852-292b-4eac-8142-886bff5227f3","cardType":"basic","points":1,"front":"What physical property must a nucleus possess to be detected by NMR spectroscopy?","back":"Nuclear spin","mcqOptions":[{"text":"Nuclear spin","correct":true,"explanation":"NMR detects nuclei with non-zero nuclear spin because they possess magnetic properties that interact with an applied magnetic field and radiofrequency radiation."},{"text":"Nuclear charge","correct":false,"explanation":"Nuclear charge alone does not determine NMR detection; the nucleus must have non-zero nuclear spin. Isotopes such as $^{12}C$ are not detected because their nuclear spin is zero."},{"text":"Electron spin","correct":false,"explanation":"NMR concerns the spin of atomic nuclei, not the spin of electrons. A nucleus must possess non-zero nuclear spin to interact with the magnetic field and produce a signal."},{"text":"Atomic mass","correct":false,"explanation":"Atomic mass does not make a nucleus NMR-active. Detection depends on non-zero nuclear spin, which is why $^{13}C$ is detected but $^{12}C$ is not."}]},{"id":"22edc052-8a7c-4779-931e-8a6e9c86db52","cardType":"basic","points":1,"front":"Identify two nuclei commonly used in A-Level NMR spectroscopy.","back":"

$^1H$ (protons)
$^{13}C$ (carbon-13)

","mcqOptions":[{"text":"

$^1H$ (protons)
$^{13}C$ (carbon-13)

","correct":true,"explanation":"Proton NMR uses $^1H$, while carbon-13 NMR uses $^{13}C$. These nuclei are commonly studied because they have suitable nuclear spin and are important in organic structures."},{"text":"

$^{12}C$ (carbon-12)
$^{16}O$ (oxygen-16)

","correct":false,"explanation":"Both $^{12}C$ and $^{16}O$ have nuclear spin zero, so they are not normally detected by NMR. The commonly used nuclei are $^1H$ and $^{13}C$."},{"text":"

$^{14}N$ (nitrogen-14)
$^{35}Cl$ (chlorine-35)

","correct":false,"explanation":"Although these nuclei can have nuclear magnetic properties, they are not the standard A-Level NMR nuclei. A-Level spectroscopy commonly focuses on $^1H$ and $^{13}C$."},{"text":"

$^{19}F$ (fluorine-19)
$^{31}P$ (phosphorus-31)

","correct":false,"explanation":"$$^{19}F$ and $^{31}P$ are NMR-active nuclei, but they are not the two nuclei usually required at A-Level. The expected pair is $^1H$ and $^{13}C$."}]},{"id":"f6893dae-ec8d-4134-bcb3-9fc07a9836a9","cardType":"basic","points":2,"front":"Why does propanone ($CH_3COCH_3$) produce only two signals in $^{13}C$ NMR?","back":"The molecule is symmetrical, making the two methyl carbons equivalent.","mcqOptions":[{"text":"The molecule is symmetrical, making the two methyl carbons equivalent.","correct":true,"explanation":"The two $CH_3$ groups are related by symmetry, so their carbon atoms are in the same environment and give one signal. The carbonyl carbon gives the second signal."},{"text":"The molecule is unsymmetrical, making the two methyl carbons different.","correct":false,"explanation":"Propanone is symmetrical, so its two methyl carbons are equivalent and produce one signal. Together with the carbonyl carbon, this gives two signals rather than three."},{"text":"The molecule is symmetrical, making the carbonyl carbon and methyl carbons equivalent.","correct":false,"explanation":"Symmetry makes the two methyl carbons equivalent, but it does not make them equivalent to the carbonyl carbon. The carbonyl carbon remains in a distinct environment, giving two signals overall."},{"text":"The molecule is symmetrical, making all three carbon atoms equivalent.","correct":false,"explanation":"Symmetry relates the two methyl carbons only; the carbonyl carbon has a different bonding environment. Therefore propanone produces one methyl signal and one carbonyl signal."}]},{"id":"cd58affa-790f-4177-aaee-5584ad1e87de","cardType":"cloze","points":null,"front":"Chemical shift, symbol $\\delta$, is measured in [ ] relative to the reference compound Tetramethylsilane (TMS).","back":"Chemical shift, symbol $\\delta$, is measured in ppm relative to the reference compound Tetramethylsilane (TMS).","mcqOptions":[{"text":"ppm","correct":true,"explanation":"Chemical shift is reported in parts per million, written ppm, relative to TMS. This provides a standardised, field-independent scale for comparing NMR signals."},{"text":"Hz","correct":false,"explanation":"Hz measures the frequency difference between signals and depends on the spectrometer’s magnetic field strength. Chemical shift is instead reported as ppm relative to TMS."},{"text":"%,","correct":false,"explanation":"A percentage expresses a proportion out of 100, not an NMR chemical shift. The chemical shift scale uses ppm, meaning parts per million, relative to TMS."},{"text":"mol dm$^{-3}$","correct":false,"explanation":"mol dm$^{-3}$ is a concentration unit and has no role in reporting chemical shift. NMR chemical shifts are measured in ppm relative to the TMS reference."}]},{"id":"b88819bf-c38a-4dbc-94c7-3a05e8b381f8","cardType":"cloze","points":null,"front":"Chemical shift, symbol $\\delta$, is measured in ppm relative to the reference compound [ ].","back":"Chemical shift, symbol $\\delta$, is measured in ppm relative to the reference compound Tetramethylsilane (TMS).","mcqOptions":[{"text":"Tetramethylsilane (TMS)","correct":true,"explanation":"Tetramethylsilane, abbreviated TMS, is assigned a chemical shift of zero and provides the reference against which other NMR signals are measured in ppm."},{"text":"Deuterated chloroform (CDCl$_3$)","correct":false,"explanation":"Deuterated chloroform is commonly used as an NMR solvent, not the standard chemical-shift reference. Chemical shifts are measured relative to Tetramethylsilane (TMS)."},{"text":"Propanone (CH$_3$COCH$_3$)","correct":false,"explanation":"Propanone is an analyte that produces its own carbon environments and signals, so it cannot provide a universal zero reference. The reference compound is Tetramethylsilane (TMS)."},{"text":"Ethanol (CH$_3$CH$_2$OH)","correct":false,"explanation":"Ethanol gives several NMR signals because it contains different environments, making it unsuitable as the standard reference. Chemical shift is measured relative to Tetramethylsilane (TMS)."}]},{"id":"947f1e9b-d296-4145-9df9-1e3daa1f7a75","cardType":"basic","points":2,"front":"How do electronegative atoms (e.g. O, Cl) affect the chemical shift of nearby nuclei?","back":"They withdraw electron density (deshielding), which increases the $\\delta$ value.","mcqOptions":[{"text":"They withdraw electron density (deshielding), which increases the $\\delta$ value.","correct":true,"explanation":"Electronegative atoms pull electron density away from nearby nuclei, reducing shielding. The nuclei then experience a greater effective magnetic field and resonate at a higher chemical shift."},{"text":"They donate electron density (shielding), which decreases the $\\delta$ value.","correct":false,"explanation":"This reverses the effect of electronegative atoms. Atoms such as O and Cl withdraw electron density, causing deshielding and an increased $\\delta$ value."},{"text":"They withdraw electron density (deshielding), which decreases the $\\delta$ value.","correct":false,"explanation":"The electron-withdrawing and deshielding part is correct, but deshielding shifts the signal downfield to a higher chemical shift. Therefore the $\\delta$ value increases."},{"text":"They donate electron density (deshielding), which increases the $\\delta$ value.","correct":false,"explanation":"Electron donation causes shielding rather than deshielding. Electronegativity withdraws electron density, and this deshielding effect increases the nearby nucleus’s $\\delta$ value."}]},{"id":"802cbe43-b90a-41b1-9213-618c3076bb59","cardType":"basic","points":2,"front":"State two reasons why Tetramethylsilane (TMS) is used as an NMR reference.","back":"

Chemically inert
Volatile (easy to remove)
All nuclei in one environment (one sharp signal)

","mcqOptions":[{"text":"

Chemically inert
Volatile (easy to remove)
All nuclei in one environment (one sharp signal)

","correct":true,"explanation":"TMS does not react with samples, can be removed easily because it is volatile, and all twelve hydrogen atoms are equivalent, producing one sharp reference signal."},{"text":"

Highly reactive
Non-volatile (difficult to remove)
All nuclei in several environments (many signals)

","correct":false,"explanation":"Each property is opposite to the reason TMS is useful. TMS is chemically inert, volatile, and has one equivalent nuclear environment that gives a single sharp signal."},{"text":"

Soluble in water
High boiling point (difficult to remove)
All nuclei in one environment (one sharp signal)

","correct":false,"explanation":"A single signal is useful, but water solubility and a high boiling point are not desired reference properties. TMS is chemically inert and volatile, so it is easy to remove."},{"text":"

Paramagnetic
Volatile (easy to remove)
All nuclei in several environments (many signals)

","correct":false,"explanation":"TMS is diamagnetic and gives one sharp signal because all its nuclei are equivalent. Paramagnetism and several environments would make it unsuitable as a clean reference."}]},{"id":"81d5966b-d45b-4481-91da-63a251ff6e91","cardType":"cloze","points":null,"front":"In $^{13}C$ NMR, the [ ] indicates the number of carbon environments, but peak height (or area) is usually not used for counting.","back":"In $^{13}C$ NMR, the number of peaks indicates the number of carbon environments, but peak height (or area) is usually not used for counting.","mcqOptions":[{"text":"number of peaks","correct":true,"explanation":"Each distinct carbon environment normally produces one $^{13}C$ NMR signal, so counting the number of peaks reveals the number of different carbon environments."},{"text":"peak height","correct":false,"explanation":"Peak height is not normally used to count carbon environments because $^{13}C$ signal intensities are affected by relaxation and other experimental factors. Count distinct peaks instead."},{"text":"number of protons","correct":false,"explanation":"The number of protons is not shown directly by the number of $^{13}C$ peaks. Each peak represents a distinct carbon environment, regardless of how many protons are attached."},{"text":"peak area","correct":false,"explanation":"Unlike idealised proton spectra, $^{13}C$ peak areas are generally unreliable for counting atoms or environments. The number of distinct peaks indicates the carbon environments."}]},{"id":"ac666362-7280-4a50-ad11-f664f620d249","cardType":"cloze","points":null,"front":"In $^{13}C$ NMR, the number of peaks indicates the number of carbon environments, but peak [ ] is usually not used for counting.","back":"In $^{13}C$ NMR, the number of peaks indicates the number of carbon environments, but peak height (or area) is usually not used for counting.","mcqOptions":[{"text":"height (or area)","correct":true,"explanation":"Peak intensity depends on factors such as relaxation and instrument conditions, so it is unreliable for counting environments. The number of distinct chemical shifts is the useful measurement."},{"text":"position","correct":false,"explanation":"Peak position identifies a carbon environment through its chemical shift, so excluding it would remove the main evidence used to count distinct environments."},{"text":"width","correct":false,"explanation":"Peak width is affected by relaxation and experimental conditions, but it is not normally the feature meant here. Counting relies on distinct chemical shifts instead."},{"text":"number","correct":false,"explanation":"The number of peaks is precisely the usual basis for counting carbon environments. Treating it as unusable reverses the key interpretation of a $^{13}C$ NMR spectrum."}]},{"id":"242ef507-315b-4ff8-8843-a17f5a9d2ef5","cardType":"basic","points":1,"front":"In $^{13}C$ NMR, which functional groups typically appear between 160-220 ppm?","back":"Carbonyl groups ($C=O$), such as aldehydes, ketones, acids, or esters.","mcqOptions":[{"text":"Carbonyl groups ($C=O$), such as aldehydes, ketones, acids, or esters.","correct":true,"explanation":"A carbonyl carbon is strongly deshielded because the $C=O$ bond withdraws electron density, placing aldehyde, ketone, acid, and ester carbons in this high-ppm region."},{"text":"Aromatic rings ($C=C$), such as benzene or substituted benzenes.","correct":false,"explanation":"Aromatic carbons are deshielded and can occur at relatively high shifts, but they generally lie below this range. Carbonyl carbons occupy approximately 160-220 ppm."},{"text":"Alcohol and ether groups ($C-O$), such as alcohols or ethers.","correct":false,"explanation":"Alcohol and ether carbons bonded to oxygen are shifted downfield compared with alkyl carbons, but usually appear around 50-90 ppm rather than 160-220 ppm."},{"text":"Alkane groups ($C-C$), such as alkanes or cycloalkanes.","correct":false,"explanation":"Alkane carbons are relatively shielded and normally appear at low chemical shifts, commonly below 50 ppm. The 160-220 ppm region instead indicates carbonyl carbons."}]},{"id":"b7e20ad8-f356-4890-8e7e-91cc785062f5","cardType":"basic","points":1,"front":"What does the integration (area under a signal) in $^1H$ NMR represent?","back":"The relative number of protons in that environment.","mcqOptions":[{"text":"The relative number of protons in that environment.","correct":true,"explanation":"Signal integration compares the areas of proton resonances, and those areas are proportional to the relative numbers of hydrogen nuclei occupying the corresponding environments."},{"text":"The chemical shift of protons in that environment.","correct":false,"explanation":"Chemical shift indicates shielding and the proton’s chemical environment, but it does not measure how many protons produce a signal. Integration provides the relative proton count."},{"text":"The number of neighbouring protons around that environment.","correct":false,"explanation":"Signal splitting reveals neighbouring, magnetically coupled protons and gives multiplicity information. It does not determine the number of protons represented by the signal."},{"text":"The frequency of absorption for that environment.","correct":false,"explanation":"The frequency of a signal reflects its chemical shift and magnetic environment, not the population of protons. Relative proton numbers are obtained from integrated area."}]},{"id":"7b258eff-d0d8-4d08-ad84-b7105ef7ea26","cardType":"cloze","points":null,"front":"The $n+1$ rule states that a signal is split into $n+1$ peaks, where $n$ is the number of [ ] on adjacent carbon atoms.","back":"The $n+1$ rule states that a signal is split into $n+1$ peaks, where $n$ is the number of equivalent neighbouring protons on adjacent carbon atoms.","mcqOptions":[{"text":"equivalent neighbouring protons","correct":true,"explanation":"Spin-spin coupling occurs with equivalent neighbouring protons on adjacent carbon atoms. If there are $n$ such protons, the signal is split into $n+1$ lines."},{"text":"adjacent carbon atoms","correct":false,"explanation":"The atom itself does not determine splitting; neighbouring proton spins do. Counting adjacent carbon atoms would give the wrong $n$ value and therefore the wrong multiplicity."},{"text":"all nearby hydrogen atoms","correct":false,"explanation":"All nearby hydrogen atoms are not necessarily coupled equivalently, and protons on the same carbon may be equivalent to each other. The rule counts equivalent neighbours on adjacent carbons."},{"text":"carbon atoms in the molecule","correct":false,"explanation":"The number of carbon atoms in the molecule does not control ordinary proton splitting. Only magnetically equivalent neighbouring protons participating in coupling contribute to $n$."}]},{"id":"3b82e94c-7b68-4902-8eed-4ee54e070b99","cardType":"cloze","points":null,"front":"The $n+1$ rule states that a signal is split into $n+1$ peaks, where $n$ is the number of equivalent neighbouring protons on [ ].","back":"The $n+1$ rule states that a signal is split into $n+1$ peaks, where $n$ is the number of equivalent neighbouring protons on adjacent carbon atoms.","mcqOptions":[{"text":"adjacent carbon atoms","correct":true,"explanation":"For the usual first-order splitting rule, coupling is counted through neighbouring carbon atoms. The relevant $n$ is the number of equivalent protons attached across those adjacent carbons."},{"text":"equivalent neighbouring protons","correct":false,"explanation":"Protons are the particles whose spin causes splitting, so the rule does not count adjacent carbon atoms as the numerical neighbours. It counts equivalent neighbouring protons instead."},{"text":"the whole carbon skeleton","correct":false,"explanation":"The whole molecule may contain many carbon atoms that cannot couple significantly to the signal. Only protons on directly adjacent carbon atoms are normally considered."},{"text":"the same carbon atom","correct":false,"explanation":"A proton’s own carbon atom does not provide the neighbouring spins responsible for its splitting. Coupling is normally considered with equivalent protons attached to adjacent carbon atoms."}]},{"id":"9eb1865d-f13a-4159-821d-72c2dbb8ae81","cardType":"basic","points":1,"front":"Predict the splitting pattern for a $CH_3$ group adjacent to a $CH_2$ group.","back":"Triplet ($2+1=3$)","mcqOptions":[{"text":"Triplet ($2+1=3$)","correct":true,"explanation":"The three methyl protons couple with the two equivalent protons of the neighbouring $CH_2$ group. Applying $n+1$ gives $2+1=3$, which is a triplet."},{"text":"Quartet ($3+1=4$)","correct":false,"explanation":"A quartet would result when the observed group couples to three equivalent neighbouring protons, as a $CH_2$ signal does beside a $CH_3$ group. Here, $n=2$."},{"text":"Doublet ($1+1=2$)","correct":false,"explanation":"A doublet results from coupling to one equivalent neighbouring proton. The adjacent $CH_2$ group supplies two protons, so the methyl signal has three lines."},{"text":"Quintet ($4+1=5$)","correct":false,"explanation":"A quintet requires four equivalent neighbouring protons. The adjacent $CH_2$ group has only two, giving $2+1=3$ lines for the methyl signal."}]},{"id":"58dbeb61-e78f-4da3-a908-4337faf7b3c9","cardType":"basic","points":1,"front":"Predict the splitting pattern for a $CH_2$ group adjacent to a $CH_3$ group.","back":"Quartet ($3+1=4$)","mcqOptions":[{"text":"Quartet ($3+1=4$)","correct":true,"explanation":"The two methylene protons couple with the three equivalent protons of the neighbouring $CH_3$ group. Applying $n+1$ gives $3+1=4$, producing a quartet."},{"text":"Triplet ($2+1=3$)","correct":false,"explanation":"A triplet results when the observed group couples to two equivalent neighbouring protons, as a $CH_3$ signal does beside a $CH_2$ group. Here, $n=3$."},{"text":"Doublet ($1+1=2$)","correct":false,"explanation":"A doublet would require only one equivalent neighbouring proton. The adjacent methyl group contributes three equivalent protons, so the methylene signal has four lines."},{"text":"Quintet ($4+1=5$)","correct":false,"explanation":"A quintet requires four equivalent neighbouring protons, but the adjacent $CH_3$ group supplies three. Therefore the correct calculation is $3+1=4$."}]},{"id":"209715fa-948e-48a5-a5f9-26b436afa7d5","cardType":"basic","points":2,"front":"How can $D_2O$ (deuterium oxide) be used to identify $OH$ or $NH$ protons?","back":"Adding $D_2O$ causes exchangeable protons to be replaced, making their signal disappear.","mcqOptions":[{"text":"Adding $D_2O$ causes exchangeable protons to be replaced, making their signal disappear.","correct":true,"explanation":"Deuterium exchange replaces acidic or exchangeable $OH$ and $NH$ hydrogens with deuterium, which is not observed in ordinary $^1H$ NMR, so their signals disappear."},{"text":"Adding $D_2O$ causes all protons to be replaced, making their signals disappear.","correct":false,"explanation":"The exchange affects $OH$ and $NH$ protons rather than every hydrogen atom. Non-exchangeable C-H signals normally remain, so they do not all disappear after adding $D_2O$."},{"text":"Adding $D_2O$ causes exchangeable protons to be replaced, making a new signal appear.","correct":false,"explanation":"Deuterium does not normally produce an additional observable $^1H$ signal in place of the exchanged hydrogen. The diagnostic observation is loss of the exchangeable proton signal."},{"text":"Adding $D_2O$ causes exchangeable protons to remain, making their signal shift downfield.","correct":false,"explanation":"Adding $D_2O$ is used to test exchangeable protons, not to move their resonances predictably downfield. Replacement by deuterium removes their ordinary proton signal."}]},{"id":"935d1284-f166-4fa7-896a-d022a89bdf1d","cardType":"basic","points":2,"front":"Why are deuterated solvents (e.g. $CDCl_3$) used in $^1H$ NMR?","back":"They do not contain $^1H$ nuclei, so they do not produce interfering signals.","mcqOptions":[{"text":"They do not contain $^1H$ nuclei, so they do not produce interfering signals.","correct":true,"explanation":"Deuterium is not detected at the operating frequency used for proton NMR, so a deuterated solvent avoids a large solvent peak obscuring the sample’s proton signals."},{"text":"They increase proton coupling, so they produce more clearly split signals.","correct":false,"explanation":"Deuterated solvents are not used to increase splitting. They are used because deuterium is not detected as $^1H$, preventing the solvent from producing interfering proton signals."},{"text":"They make all protons equivalent, so they produce only one sample signal.","correct":false,"explanation":"The solvent does not make every proton chemically equivalent. Its deuterium is not detected at the proton frequency, so solvent signals do not interfere with the sample spectrum."},{"text":"They prevent spin-spin coupling, so they do not produce split sample signals.","correct":false,"explanation":"A deuterated solvent does not remove spin-spin coupling within the sample. It avoids interference because its deuterium nuclei are not detected as $^1H$ nuclei."}]},{"id":"2802a38d-6538-418d-ba3c-c360edc64605","cardType":"cloze","points":null,"front":"A signal split into a doublet (intensity ratio $\\text{[...]}$) indicates one neighbouring proton.","back":"A signal split into a doublet (intensity ratio $\\boxed{1:1}$) indicates one neighbouring proton.","mcqOptions":[{"text":"$$1:1$","correct":true,"explanation":"One neighbouring proton can have either spin orientation, producing two equally probable local magnetic environments. The signal therefore splits into a doublet with a $1:1$ intensity ratio."},{"text":"$$1:2:1$","correct":false,"explanation":"The $1:2:1$ ratio comes from Pascal’s triangle for two neighbouring protons and corresponds to a triplet. One neighbouring proton instead produces a $1:1$ doublet."},{"text":"$$1:3:3:1$","correct":false,"explanation":"A $1:3:3:1$ ratio is the intensity pattern of a quartet caused by three equivalent neighbouring protons. A doublet caused by one neighbour has a $1:1$ ratio."},{"text":"$$1:1:1$","correct":false,"explanation":"A $1:1:1$ pattern incorrectly assumes three equally intense lines. A doublet contains only two lines, which have equal probabilities and therefore a $1:1$ intensity ratio."}]},{"id":"6b75d689-84fa-4276-b892-9048fb523aa1","cardType":"cloze","points":null,"front":"A signal split into a doublet (intensity ratio $1:1$) indicates [ ] neighbouring proton.","back":"A signal split into a doublet (intensity ratio $1:1$) indicates one neighbouring proton.","mcqOptions":[{"text":"one","correct":true,"explanation":"The $n+1$ rule states that one neighbouring proton splits a signal into two peaks. These two equally probable peaks form a doublet with a $1:1$ ratio."},{"text":"two","correct":false,"explanation":"Two neighbouring equivalent protons produce three lines by the $n+1$ rule, giving a $1:2:1$ triplet. A $1:1$ doublet indicates one neighbouring proton."},{"text":"three","correct":false,"explanation":"Three neighbouring equivalent protons would produce a quartet with a $1:3:3:1$ ratio. A doublet containing two equal lines is caused by one neighbouring proton."},{"text":"no","correct":false,"explanation":"With no neighbouring protons, no spin-spin splitting occurs and the signal is a singlet. A $1:1$ doublet requires coupling to one neighbouring proton."}]},{"id":"86719dd5-7e6b-49b1-85ac-94665fe8a6b9","cardType":"basic","points":2,"front":"A molecule has formula $C_2H_6O$ and shows only one singlet in $^1H$ NMR. Identify the molecule.","back":"Methoxymethane ($CH_3OCH_3$)","mcqOptions":[{"text":"Methoxymethane ($CH_3OCH_3$)","correct":true,"explanation":"Methoxymethane has two identical methyl groups related by symmetry, so all six hydrogen atoms occupy the same proton environment and produce a single singlet."},{"text":"Ethanol ($CH_3CH_2OH$)","correct":false,"explanation":"Ethanol has the required molecular formula, but its methyl, methylene and hydroxyl hydrogens occupy three different environments. It therefore does not produce only one singlet."},{"text":"Ethanal ($CH_3CHO$)","correct":false,"explanation":"Ethanal contains distinct methyl and aldehyde proton environments and has molecular formula $C_2H_4O$, not $C_2H_6O$. The molecule described is methoxymethane."},{"text":"Methanol ($CH_3OH$)","correct":false,"explanation":"Methanol has only one carbon and molecular formula $CH_4O$, so it cannot match the stated formula. The symmetrical $C_2H_6O$ isomer giving one singlet is methoxymethane."}]},{"id":"d55e1b85-a231-4926-8050-5e26ba67cc25","cardType":"basic","points":1,"front":"How many proton environments are present in ethanol ($CH_3CH_2OH$)?","back":"Three","mcqOptions":[{"text":"Three","correct":true,"explanation":"Ethanol has three chemically distinct proton groups: the $CH_3$ hydrogens, the $CH_2$ hydrogens and the $OH$ hydrogen. It therefore has three proton environments."},{"text":"Two","correct":false,"explanation":"Counting only the two carbon-containing groups overlooks the hydroxyl proton. The $CH_3$, $CH_2$ and $OH$ hydrogens occupy three distinct proton environments."},{"text":"One","correct":false,"explanation":"The six hydrogen atoms are not all chemically equivalent because they occur in methyl, methylene and hydroxyl groups. These groups give three proton environments, not one."},{"text":"Four","correct":false,"explanation":"Treating the two hydrogens in $CH_2$ as separate environments causes overcounting; in ordinary ethanol they are equivalent. Including $CH_3$ and $OH$ gives three environments."}]},{"id":"4178bf24-0771-40fd-818a-a40f93822fc1","cardType":"basic","points":1,"front":"What effect does \"shielding\" by electrons have on the $\\delta$ value of a nucleus?","back":"It decreases the $\\delta$ value (shifts the signal to the right).","mcqOptions":[{"text":"It decreases the $\\delta$ value (shifts the signal to the right).","correct":true,"explanation":"Greater electron density shields the nucleus from the external magnetic field. A shielded nucleus resonates upfield, which is towards the right and at a lower chemical shift."},{"text":"It increases the $\\delta$ value (shifts the signal to the left).","correct":false,"explanation":"Increasing chemical shift is the effect of deshielding, not shielding. Electron shielding moves the signal upfield towards the right, decreasing its $\\delta$ value."},{"text":"It does not change the $\\delta$ value (leaves the signal in the same position).","correct":false,"explanation":"Shielding does affect resonance position because electron density reduces the effective magnetic field experienced by the nucleus. This decreases $\\delta$ and shifts the signal to the right."},{"text":"It decreases the $\\delta$ value (shifts the signal to the left).","correct":false,"explanation":"The direction has been reversed: a lower chemical shift is plotted towards the right of an NMR spectrum. Shielding decreases $\\delta$, so the signal moves right rather than left."}]},{"id":"5e0aa670-2e11-4921-8c37-59241fdcc129","cardType":"basic","points":1,"front":"What is the peak intensity ratio for a triplet in $^1H$ NMR?","back":"$1:2:1$","mcqOptions":[{"text":"$1:2:1$","correct":true,"explanation":"A triplet is produced by two equivalent neighbouring protons. The three possible combined spin outcomes have relative probabilities given by Pascal’s triangle as $1:2:1$."},{"text":"$1:1:1$","correct":false,"explanation":"The three lines of a triplet are not equally probable because two different spin combinations contribute to the central line. Its correct intensity ratio is $1:2:1$."},{"text":"$2:1:2$","correct":false,"explanation":"This overstates the relative intensity of both outer lines. Pascal’s triangle gives a triplet one central line twice as intense as each outer line: $1:2:1$."},{"text":"$1:3:1$","correct":false,"explanation":"This ratio incorrectly makes the central line three times as intense as either outer line. For two neighbouring protons, the statistical spin combinations produce $1:2:1$."}]},{"id":"95449c0c-b072-4a1e-8301-d8fe958c6f40","cardType":"basic","points":2,"front":"Compare the information provided by integration vs splitting in $^1H$ NMR.","back":"

Integration: number of protons in the environment itself.
Splitting: number of protons on neighbouring carbons.

","mcqOptions":[{"text":"

Integration: number of protons in the environment itself.
Splitting: number of protons on neighbouring carbons.

","correct":true,"explanation":"The integrated area of a signal is proportional to the number of protons producing it. Splitting instead reveals the number of adjacent protons through the $n+1$ rule."},{"text":"

Integration: number of protons on neighbouring carbons.
Splitting: number of protons in the environment itself.

","correct":false,"explanation":"These roles have been reversed. Integration measures the relative number of protons producing a signal, while splitting provides information about protons in neighbouring environments."},{"text":"

Integration: number of proton environments in the molecule.
Splitting: total number of protons in the molecule.

","correct":false,"explanation":"The number of proton environments is found by counting distinct signals, not by integration. Integration measures protons within an environment, while splitting indicates neighbouring protons."},{"text":"

Integration: position of the signal in the spectrum.
Splitting: type of functional group in the molecule.

","correct":false,"explanation":"Chemical shift, rather than integration, indicates the electronic environment and possible functional groups. Integration counts protons in an environment, while splitting shows neighbouring proton numbers."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"NMR spectroscopy","title":"NMR spectroscopy Flashcards","topbar":{"backHref":"/a-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"114ace27-7a5c-5843-a52e-17a953a70064","walled":"$L39"}],"$L3a","$L3b"]