13. An introduction to AS Level organic chemistry
A section of Chemistry, 9701
Listing 10 of 507 questions
When 0.42 g of a gaseous hydrocarbon A is slowly passed over a large quantity of heated copper(oxide, CuO, A is completely oxidised. The products are collected and it is found that 1.32 g of CO2 and 0.54 g of H2O are formed. Copper is the only other product of the reaction. Calculate the mass of carbon present in 1.32 g of CO2. Use this value to calculate the amount, in moles, of carbon atoms present in 0.42 g of A. Calculate the mass of hydrogen present in 0.54 g of H2O. Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.42 g of A. It is thought that A is an alkene rather than an alkane. Use your answers to and to deduce whether this is correct. Explain your answer. Analysis of another organic compound, B, gave the following composition by mass: C, 64.86%; H, 13.50%, O, 21.64%. Use these values to calculate the empirical formula of B. The empirical and molecular formulae of B are the same. B is found to be chiral. Draw displayed formulae of the two optical isomers of this compound, indicating with an asterisk (*) the chiral carbon atom. There are three other structural isomers of B which are not chiral but which contain the same functional group as B. In the boxes below, draw the structural formulae of these isomers.
9701_w11_qp_23
THEORY
2011
Paper 2, Variant 3
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The structural formulae of six different compounds, P – U, are given below. CH3CH=CHCH2CH3 CH3CH2COCH2CH3 CH2=CHCH2CH2CH3 P Q R CH3CH2CH2CH2CH2OH HOCH2CH2CH(OH)CH3 CH3CH2CH2OCH2CH3 S T U What is the empirical formula of compound T? Draw the skeletal formula of compound S. Compounds S and U are isomers. What type of isomerism do they show? Two of the six formulae P – U can each be drawn in two forms which are known as stereoisomers. Which two compounds have formulae that can be drawn in two forms? What type of stereoisomerism does each show? Identify each compound by its letter. compound type of stereoisomerism Compound S can be converted into compound R. What type of reaction is this? What reagent would you use for this reaction? Write the structural formula of the compound formed when T undergoes the same reaction using an excess of the reagent you have used in . Compound P may be converted into compound Q in a two-step reaction. CH3CH=CHCH2CH3 step 1 intermediate step 2 CH3CH2COCH2CH3 P Q What is the structural formula of the intermediate compound formed in this sequence? Outline how step 1 may be carried out to give this intermediate compound. What reagent would be used for step 2?
9701_w11_qp_23
THEORY
2011
Paper 2, Variant 3
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Each of the three organic compounds, V, W, and X, has the empirical formula CH2O. The number of carbon atoms in each of their molecules is shown in the table. compound number of C atoms V W X V gives a brick red precipitate when warmed with Fehling’s reagent; W and X do not. W is a fruity smelling liquid. In X, the carbon atoms are bonded directly to one another. X gives an effervescence when shaken with Na2CO3; V and W do not. Give the structural formula of V. What functional group is present in W? Give the structural formula of W. When X is heated under reflux with acidified K2Cr2O7, the product, Y, gives no reaction with 2,4-dinitrophenylhydrazine reagent. Give the structural formula of X. Give the structural formula of Y, the compound formed from X. When X is warmed with a little concentrated sulfuric acid, a small amount of a cyclic compound, Z, is formed. Z has the molecular formula C6H8O4. Suggest a displayed formula for Z. What type of reaction occurs when Z is formed from X?
9701_w11_qp_23
THEORY
2011
Paper 2, Variant 3
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Compound C has the molecular formula C7H14O. Treating C with hot concentrated acidified KMnO4produces two compounds, D, C4H8O, and E, C3H4O3. The results of four tests carried out on these three compounds are shown in the following table. test reagent result of test with compound C compound D compound E Br2decolourises no reaction no reaction Nafizzes no reaction fizzes I2+ OH–no reaction yellow precipitate yellow precipitate 2,4-dinitrophenylhydrazine no reaction orange precipitate orange precipitate State the functional groups which the above four reagents test for. Br2NaI2+ OH–2,4-dinitrophenylhydrazine Based upon the results of the above tests, suggest structures for compounds D and E. D, C4H8O E, C3H4O3 Compound C exists as two stereoisomers. Draw the structural formula of each of the two isomers, and state the type of stereoisomerism involved. type of stereoisomerism
9701_w11_qp_41
THEORY
2011
Paper 4, Variant 1
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Proteins exist in an enormous variety of sizes and structures in living organisms. They have a wide range of functions which are dependent upon their structures. The structure and properties of an individual protein are a result of the primary structure – the sequence of amino acids that form the protein. Proteins are described as condensation polymers. Write a balanced equation for the condensation reaction between two glycine molecules, H2NCH2CO2H. Draw the skeletal formula for the organic product. X-ray analysis has shown that in many proteins there are regions with a regular arrangement within the polypeptide chain. This is called the secondary structure and exists in two main forms. State the two forms of secondary structure found in proteins. Draw a diagram to illustrate one form of secondary structure. There are around 20 different common amino acids found in humans most of which have the same general structure. C R H2N H CO2H The nature of the group R affects which bonds are formed as the secondary structure of the protein is further folded to give the tertiary structure. Complete the table indicating the type of tertiary bonding that each pair of the amino acid residues is likely to produce. residue 1 residue 2 type of tertiary bonding –HNCH(CH2CH2CH2CH2NH2)CO– –HNCH(CH2CH2CO2H)CO– –HNCH(CH3)CO– –HNCH(CH3)CO– –HNCH(CH2SH)CO– –HNCH(CH2SH)CO– –HNCH(CH2OH)CO– –HNCH(CH2CO2H)CO–
9701_w11_qp_41
THEORY
2011
Paper 4, Variant 1
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Compound C has the molecular formula C7H14O. Treating C with hot concentrated acidified KMnO4produces two compounds, D, C4H8O, and E, C3H4O3. The results of four tests carried out on these three compounds are shown in the following table. test reagent result of test with compound C compound D compound E Br2decolourises no reaction no reaction Nafizzes no reaction fizzes I2+ OH–no reaction yellow precipitate yellow precipitate 2,4-dinitrophenylhydrazine no reaction orange precipitate orange precipitate State the functional groups which the above four reagents test for. Br2NaI2+ OH–2,4-dinitrophenylhydrazine Based upon the results of the above tests, suggest structures for compounds D and E. D, C4H8O E, C3H4O3 Compound C exists as two stereoisomers. Draw the structural formula of each of the two isomers, and state the type of stereoisomerism involved. type of stereoisomerism
9701_w11_qp_42
THEORY
2011
Paper 4, Variant 2
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Proteins exist in an enormous variety of sizes and structures in living organisms. They have a wide range of functions which are dependent upon their structures. The structure and properties of an individual protein are a result of the primary structure – the sequence of amino acids that form the protein. Proteins are described as condensation polymers. Write a balanced equation for the condensation reaction between two glycine molecules, H2NCH2CO2H. Draw the skeletal formula for the organic product. X-ray analysis has shown that in many proteins there are regions with a regular arrangement within the polypeptide chain. This is called the secondary structure and exists in two main forms. State the two forms of secondary structure found in proteins. Draw a diagram to illustrate one form of secondary structure. There are around 20 different common amino acids found in humans most of which have the same general structure. C R H2N H CO2H The nature of the group R affects which bonds are formed as the secondary structure of the protein is further folded to give the tertiary structure. Complete the table indicating the type of tertiary bonding that each pair of the amino acid residues is likely to produce. residue 1 residue 2 type of tertiary bonding –HNCH(CH2CH2CH2CH2NH2)CO– –HNCH(CH2CH2CO2H)CO– –HNCH(CH3)CO– –HNCH(CH3)CO– –HNCH(CH2SH)CO– –HNCH(CH2SH)CO– –HNCH(CH2OH)CO– –HNCH(CH2CO2H)CO–
9701_w11_qp_42
THEORY
2011
Paper 4, Variant 2
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Different analytical techniques are used to build up a picture of complex molecules. Each technique on its own provides different information about complex molecules but together the techniques can give valuable structural information. Complete the table, identifying the technique which can provide the appropriate structural information. structural information analytical technique three-dimensional arrangement of atoms and bonds in a molecule chemical environment of protons in a molecule identity of amino acids present in a polypeptide One general method of separating organic molecules is chromatography. Briefly explain the chemical principles involved in each of the following techniques. paper chromatography thin-layer chromatography A combination of mass spectrometry and NMR spectroscopy is often enough to determine the structure of a simple organic compound. The organic compound N produced a mass spectrum in which the ratio of the M:M+1 peaks was 5.9:0.20, and which had an M+2 peak of similar height to the M peak. Calculate how many carbon atoms are present in one molecule of N. Deduce which element, other than carbon and hydrogen, is present in N. Explain how many atoms of this element are present in one molecule of N. The NMR spectrum of N is shown. 1H multiplet 6H doublet chemical shift, ppm absorption of energy State the empirical formula of N and, using the NMR data, suggest the structural formula of N, explaining your reasons.
9701_w11_qp_43
THEORY
2011
Paper 4, Variant 3
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Questions Discovered
507