13.1. Formulas, functional groups and the naming of organic compounds
A subsection of Chemistry, 9701, through 13. An introduction to AS Level organic chemistry
Listing 10 of 266 questions
The structure of compoundP is shown in Fig.7.1. H2N O O P Fig.7.1 P is optically active. Use an asterisk (*) to identify all chiral carbon atoms on the structure of P in Fig.7.1. Plane polarised light is passed through a pure sample of one enantiomer of P. This is then repeated with a pure sample of the other enantiomer of P. Describe the results of these two experiments, stating the similarities and differences of the results. P can be used to make compoundQ in a single step reaction. HN O O Q O Give the structural formula of the compound that is added to P to make Q and give the formula of the other product of this reaction. compound added to P other product  When an ester is treated with LiAl H4 in dry ether the ester linkage is cleaved by the addition of four hydrogen atoms and two alcohols are produced. Draw the structures of the compounds formed when Q is treated with an excess of LiAl H4 in dry ether.  Compare the relative basicities of compoundP, compoundQ and phenylamine. least basic most basic Explain your answer.  P can be used to make compoundR in a two-step reaction, shown in . two steps H2N O O H2N H2N OH O P R Identify the reagents and conditions used for the two steps of the reaction. step 1 step 2  Complete Table7.1 by drawing the structures of the organic products formed when R is treated separately with the reagents given. Table 7.1 reagent product HNO2at 4 °C an excess of Br2at room temperature  P can be used to produce compoundT. H2N OH O T In aqueous solution, T has a property called an isoelectric point. Explain what is meant by isoelectric point. T can polymerise under suitable conditions. No other monomer is involved in this reaction. Draw a section of the polymer chain formed that includes three T monomers. Identify the repeat unit on your diagram.  
9701_w22_qp_41
THEORY
2022
Paper 4, Variant 1
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The structure of compoundP is shown in Fig.7.1. H2N O O P Fig.7.1 P is optically active. Use an asterisk (*) to identify all chiral carbon atoms on the structure of P in Fig.7.1. Plane polarised light is passed through a pure sample of one enantiomer of P. This is then repeated with a pure sample of the other enantiomer of P. Describe the results of these two experiments, stating the similarities and differences of the results. P can be used to make compoundQ in a single step reaction. HN O O Q O Give the structural formula of the compound that is added to P to make Q and give the formula of the other product of this reaction. compound added to P other product  When an ester is treated with LiAl H4 in dry ether the ester linkage is cleaved by the addition of four hydrogen atoms and two alcohols are produced. Draw the structures of the compounds formed when Q is treated with an excess of LiAl H4 in dry ether.  Compare the relative basicities of compoundP, compoundQ and phenylamine. least basic most basic Explain your answer.  P can be used to make compoundR in a two-step reaction, shown in . two steps H2N O O H2N H2N OH O P R Identify the reagents and conditions used for the two steps of the reaction. step 1 step 2  Complete Table7.1 by drawing the structures of the organic products formed when R is treated separately with the reagents given. Table 7.1 reagent product HNO2at 4 °C an excess of Br2at room temperature  P can be used to produce compoundT. H2N OH O T In aqueous solution, T has a property called an isoelectric point. Explain what is meant by isoelectric point. T can polymerise under suitable conditions. No other monomer is involved in this reaction. Draw a section of the polymer chain formed that includes three T monomers. Identify the repeat unit on your diagram.  
9701_w22_qp_43
THEORY
2022
Paper 4, Variant 3
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Benzene can be used to make benzoic acid in the two-step process shown in . benzene methylbenzene CH3 step 1 step 2 benzoic acid COOH Give the reagents and conditions for step 1 and step 2. step 1 step 2 Methylbenzene and benzoic acid each have five different peaks in the carbon (13C) NMR spectrum. Table 7.1 hybridisation of the carbon atom environment of carbon atom example chemical shift range / ppm sp3 alkyl CH3–, –CH2–, –CHCC=Cpeakin the chemical shift range of and ● has peakin the chemical shift range of . The carbon (13C) NMR spectrum of benzoic acid: ● has peakin the chemical shift range of and ● has peakin the chemical shift range of . When treated with Cl 2 under suitable conditions, methylbenzene forms compound J. When treated with Cl 2 under different conditions with different reagents, methylbenzene forms compound K. Suggest and draw structures of compounds J and K in the boxes. The molecular formula of each compound is given. methylbenzene J C7H7Cl K C7H7Cl CH3 State the reagents and conditions required to form each product. to form compound J to form compound K When treated with a chlorine-containing reagent under suitable conditions, benzoic acid forms compound L. When treated with a different chlorine-containing reagent under different conditions, benzoic acid forms compound M. Suggest and draw structures of compounds L and M in the boxes. The molecular formula of each product is given. benzoic acid L C7H5ClO2 M C7H5ClO COOH State the reagents and conditions to form compound M from benzoic acid.
9701_w23_qp_42
THEORY
2023
Paper 4, Variant 2
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