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This question is about six isomers of C7H14O2 - AQA - A-Level Chemistry - Question 10 - 2017 - Paper 2
Question 10
This question is about six isomers of C7H14O2. 1. Give the full IUPAC name of isomer P. CH3CH2–C(COOH)(H)–CH3 2. A sample of P was mixed with an excess of oxygen ... show full transcript
Worked Solution & Example Answer:This question is about six isomers of C7H14O2 - AQA - A-Level Chemistry - Question 10 - 2017 - Paper 2
Step 1
Step 2
Write an equation for the combustion of P in an excess of oxygen and calculate the mass, in mg, of P used.
Answer
The general equation for the complete combustion of 2-methylpentanoic acid can be written as:
ightarrow CO_2 + H_2O $$ To calculate the mass of P used, we start with the change in volume of CO2: Initial volume of gas = 335 cm³ Final volume = 155 cm³ Therefore, volume of CO2 = 335 cm³ - 155 cm³ = 180 cm³. Using the ideal gas equation: $$ PV = nRT $$ We convert the volume of gas evolved to m³: $$ 180 ext{ cm}^3 = 0.00018 ext{ m}^3$$ Using values: P = 105 kPa, R = 8.31 J K⁻¹ mol⁻¹. We first convert 105 kPa to Pa, which is 105000 Pa: $$ n = rac{PV}{RT} = rac{(105000)(0.00018)}{(8.31)(298)} \ ightarrow n ext{ (mols of CO2 produced)} $$ After calculating, you would find the mol of 2-methylpentanoic acid used and convert to mass.Step 3
Use these spectra and Tables A and C in the Data Booklet to deduce the structure of Q.
Answer
Based on the provided IR spectrum (Figure 4) and HNMR (Figure 5), the structure of Q can be deduced as follows:
- The IR spectrum shows a characteristic peak for the C=O carbonyl group, indicating the presence of a carboxylic acid.
- The HNMR spectrum reveals splitting patterns suggestive of the number of neighboring protons, allowing the identification of specific functional groups.
The evidence points towards a structure that is consistent with a cyclic compound featuring a carboxylic acid.
Step 4
Justify this statement using Table C from the Data Booklet.
Answer
In the 13C NMR spectroscopy:
- R has two peaks due to the two unique carbon environments while S has three peaks due to three unique environments. The difference in peak count indicates that the carbon skeleton and connectivity differ between R and S despite similar molecular formulas.
Step 5
Justify this statement using the splitting patterns of the peaks.
Answer
In the 1H NMR spectroscopy:
- R displays peaks indicating splitting consistent with coupling between ethyl and methyl groups, revealing distinct hydrogen environments. For instance, R's peaks can be described as 'a triplet and a quartet' reflecting its structure.
- Conversely, S's distinct hydrogen environments yield different splitting patterns, illustrating how their structures can be inferred from these patterns.
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