Compounds \(\text{A}\) and \(\text{B}\) are isomers with formula \(\ce{C3H7X}\), where \(\ce{X}\) is a halogen. The mass spectrum for compound \(\text{A}\) is shown. Compounds \(\text{A}\) and \(\text{B}\) undergo substitution reactions in the presence of hydroxide ions, producing alcohols \(\text{C}\) and \(\text{D}\). Compound \(\text{D}\) can be oxidised to a ketone; compound \(\text{C}\) can also be oxidised, but does not produce a ketone. Compound \(\text{E}\) can be produced by refluxing 3-methylbutanoic acid, with one of the alcohols \(\text{C}\) or \(\text{D}\), in the presence of a catalyst. The \({ }^1 \text{H NMR}\) spectrum for compound \(\text{E}\) contains the following features. Draw the structure of compounds \(\text{A}\), \(\text{B}\) and \(\text{E}\). Explain your answer with reference to the information provided. (7 marks) --- 12 WORK AREA LINES (style=lined) --- → The mass spectrum has two peaks to the far right of the spectrum of similar height at 122 m/z and 124 m/z. This is due to the halogen having two isotopes with the same relative abundance. → Isotope \(\ce{X}\) must be bromine, whose isotopes are \(\ce{Br-79}\) and \(\ce{Br-81}\). The two molecular ion peaks both correspond correctly to the molar mass of the parent molecule, \(\ce{C3H7Br}\), depending on which isomer is present in the compound. \(MM\ce{(C3H7Br-79)} = 3 \times 12 + 1 \times 7 + 79 = 122\) \(MM\ce{(C3H7Br-81)} = 3 \times 12 + 1 \times 7 + 81 = 124\) → The two isomers of \(\ce{C3H7Br}\) are 1-bromopropane and 2-bromopropane and when they undergo a substitution reaction with \(\ce{OH^{-}}\), they will produce propan-1-ol and propan-2-ol respectively. → Only secondary alcohols will oxidise to produce a ketone. Hence compound \(\text{D}\) must be propan-2-ol and compound \(\text{C}\) must be propan-1-ol which can be oxidised to an aldehyde and the aldehyde can be oxidised to a carboxylic acid. → Therefore, compound \(\text{B}\) is 2-bromopropane and compound \(\text{A}\) is 1-bromopropane as during the substitution reaction the bromine atom is substituted with the hydroxide ion. → When 3-methylbutanoic acid is reacted with alcohol \(\text{C}\) (propan-1-ol) the following reaction takes place: → When 3-methylbutanoic acid is reacted with alcohol \(\text{D}\) (propan-2-ol) the following reaction takes place: → There are 6 unique hydrogen environments present in compound \(\text{E}\). Therefore compound \(\text{E}\) must be the ester produced in the reaction between 3-methylbutanoic acid and propan-1-ol (it has 6 hydrogen environments vs propan-2-ol ester which has 5). → Compound \(\text{E}\) is propyl 3-methylbutanoate. → This can be confirmed by the integration (ratio of hydrogens in each environment) and peak splitting columns (number of splits = number of adjacent hydrogens + 1) → The shift at 0.96 is due to environment 1 which has six hydrogen atoms and has one neighbouring hydrogen atom (produces a doublet). → The shift at 2.1 is due to environment 2 which has one hydrogen 1 atom and has 8 neighbouring hydrogen atoms a (produces a mutliplet of 9)(. → The shift at 4.0 is due to environment 4. This \(\ce{CH2}\) group is bonded to an oxygen atom corresponding to a large chemical shift between 3.2–5.0. It also has 2 neighbouring hydrogens and so produces a triplet. → Other answers could have included a further explanation regarding the integration and peak splitting of all the hydrogen environments and their relative chemical shifts. → The mass spectrum has two peaks to the far right of the spectrum of similar height at 122 m/z and 124 m/z. This is due to the halogen having two isotopes with the same relative abundance. → Isotope \(\ce{X}\) must be bromine, whose isotopes are \(\ce{Br-79}\) and \(\ce{Br-81}\). The two molecular ion peaks both correspond correctly to the molar mass of the parent molecule, \(\ce{C3H7Br}\), depending on which isomer is present in the compound. \(MM\ce{(C3H7Br-79)} = 3 \times 12 + 1 \times 7 + 79 = 122\) \(MM\ce{(C3H7Br-81)} = 3 \times 12 + 1 \times 7 + 81 = 124\) → The two isomers of \(\ce{C3H7Br}\) are 1-bromopropane and 2-bromopropane and when they undergo a substitution reaction with \(\ce{OH^{-}}\), they will produce propan-1-ol and propan-2-ol respectively. → Only secondary alcohols will oxidise to produce a ketone. Hence compound \(\text{D}\) must be propan-2-ol and compound \(\text{C}\) must be propan-1-ol which can be oxidised to an aldehyde and the aldehyde can be oxidised to a carboxylic acid. → Therefore, compound \(\text{B}\) is 2-bromopropane and compound \(\text{A}\) is 1-bromopropane as during the substitution reaction the bromine atom is substituted with the hydroxide ion. → When 3-methylbutanoic acid is reacted with alcohol \(\text{C}\) (propan-1-ol) the following reaction takes place: → When 3-methylbutanoic acid is reacted with alcohol \(\text{D}\) (propan-2-ol) the following reaction takes place: → There are 6 unique hydrogen environments present in compound \(\text{E}\). Therefore compound \(\text{E}\) must be the ester produced in the reaction between 3-methylbutanoic acid and propan-1-ol (it has 6 hydrogen environments vs propan-2-ol ester which has 5). → Compound \(\text{E}\) is propyl 3-methylbutanoate. → This can be confirmed by the integration (ratio of hydrogens in each environment) and peak splitting columns (number of splits = number of adjacent hydrogens + 1) → The shift at 0.96 is due to environment 1 which has six hydrogen atoms and has one neighbouring hydrogen atom (produces a doublet). → The shift at 2.1 is due to environment 2 which has one hydrogen 1 atom and has 8 neighbouring hydrogen atoms a (produces a mutliplet of 9)(. → The shift at 4.0 is due to environment 4. This \(\ce{CH2}\) group is bonded to an oxygen atom corresponding to a large chemical shift between 3.2–5.0. It also has 2 neighbouring hydrogens and so produces a triplet. → Other answers could have included a further explanation regarding the integration and peak splitting of all the hydrogen environments and their relative chemical shifts.
CHEMISTRY, M8 2024 HSC 19 MC
Which of the following compounds produces TWO doublets in the \({ }^1 \text{H NMR}\) spectrum?
Show Worked Solution
→ For two doublets to occur, the substance must have two chemically different hydrogen environments that both have one neighbouring hydrogen on an adjacent carbon atom.
→ This is displayed by option \(\text{A}\) where the hydrogen atoms in environments 1 and 3 both have only one neighbouring hydrogen atom each.
\(\Rightarrow A\)
♦ Mean mark 48%.
CHEMISTRY, M8 2023 HSC 36
An organic reaction pathway involving compounds \(\text{A, B,}\) and \(\text{C}\) is shown in the flow chart.
The molar mass of \(\text{A}\) is 84.156 g mol\(^{-1}\).
A chemist obtained some spectral data for the compounds as shown.
| \( \text{Data from} \ ^{1} \text{H NMR spectrum of compound C} \) | ||
| \( Chemical \ Shift \ \text{(ppm)} \) | \( Relative \ peak \ area \) | \( Splitting \ pattern \) |
| \(1.01\) | \(3\) | \(\text{Triplet}\) |
| \(1.05\) | \(3\) | \(\text{Triplet}\) |
| \(1.65\) | \(2\) | \(\text{Multiplet}\) |
| \(2.42\) | \(2\) | \(\text{Triplet}\) |
| \(2.46\) | \(2\) | \(\text{Quartet}\) |
| \( ^{1} \text{H NMR chemical shift data}\) | |
| \( Type \ of \ proton \) | \( \text{δ/ppm} \) |
| \( \ce{R - C\textbf{H}3,R - C\textbf{H}2 - R}\) | \(0.7-1.7\) |
| \( \left.\begin{array}{l}\ce{\textbf{H}3C - CO - \\-C\textbf{H}2 - CO -}\end{array}\right\} \begin{aligned} & \text { (aldehydes, ketones,} \\ &\text{carboxylic acids or esters) }\end{aligned}\) | \(2.0-2.6\) |
| \( \ce{R - C\textbf{H}O} \) | \(9.4-10.00\) |
| \( \ce{R - COO\textbf{H}} \) | \(9.0-13.0\) |
Identify the functional group present in each of compounds \(\text{A}\) to \(\text{C}\) and draw the structure of each compound. Justify your answer with reference to the information provided. (9 marks)
--- 28 WORK AREA LINES (style=lined) ---
Show Answers Only
Compound \(\text{A}\): Alkene
Compound \(\text{B}\): Secondary alcohol
Compound \(\text{C}\): Ketone
Reasoning as follows:
→ Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond \(\Rightarrow \) Alkene
→ Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.
→ The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.
→ The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.
→ Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.
→ Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.
→ Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.
→ The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.
→ Chemical shift and splitting patterns information indicate:
1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))
1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)
2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))
2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))
Show Worked Solution
Compound \(\text{A}\): Alkene
Compound \(\text{B}\): Secondary alcohol
Compound \(\text{C}\): Ketone
Reasoning as follows:
→ Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond \(\Rightarrow \) Alkene
→ Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.
→ The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.
→ The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.
→ Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.
→ Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.
→ Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.
→ The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.
→ Chemical shift and splitting patterns information indicate:
1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))
1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)
2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))
2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))
CHEMISTRY, M8 2019 HSC 14 MC
A molecule, `text{C}_(4) text{H}_(9) text{Cl}`, is analysed. The \(\ce{^1H NMR} \) spectrum of this molecule is shown.
What is the structural formula of this molecule?
CHEMISTRY, M8 2021 HSC 18 MC
The table lists the information from a proton NMR spectrum.
Which compound could have produced this spectrum?
- `text{1,2,2-trichlorobutane}`
- `text{1,3-dichloro-2-methylpropane}`
- `text{2-chloro-2-methylbutane}`
- `text{2,2-dichlorobutane}`