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Fichas por Emily Sutton, actualizado hace más de 1 año
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Creado por Emily Sutton
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| Pregunta | Respuesta |
| Alkanes (Questions and answers) | On the front of the flashcard is a question about alkanes, try and answer it. Then check your answer using the model answer on the back of the card. Good luck!!! |
| 1) Some of the hydrocarbons in kerosene have the formula C10H22. What is the name of the straight chain hydrocarbon with the formula C10H22? | Decane |
| ii) draw the skeletal formula of one branched chain isomer with the formula C10H22. | |
| iii) Explain why the straight chain isomer of C10H22 has a higher boiling point than any of its branched chain structural isomers. | Decane has more surface contact meaning the straight chains can get closer together. Therefore, decane has more London forces (induced dipole-dipole interactions), meaning more energy is required to break the intermolecular forces. |
| iv) Explain why the straight chain isomer of C10H22 is converted by the petroleum industry into its branched chain isomers. | Branched chains have more efficient combustion, meaning branched chains are easier to burn. |
| 2) Reaction mechanism shows the individual steps that take place during a reaction. Methane reacts with bromine in the presence of ultraviolet radiation to form several products. Two of these products are bromomethane and hydrogen bromide. Write an equation for the reaction between methane and bromine to make bromomethane and hydrogen bromide. | CH4 + Br2 -> CH3Br + HBr |
| ii) Name one other bromine-containing organic product which is formed when methane reacts with bromine. | Dibromomethane OR tribromomethane OR tetrabromomethane |
| iii) The mechanism for this reaction is called radical substitution. Describe the mechanism for the radical substitution of methane by bromine to make bromomethane. Use the mechanism to suggest why a small amount of ethane is also formed. In your answer, you should organise your answer and use the correct technical terms. | Initiation: Br2 -> 2Br⋅ Propagation: Br⋅ + CH4 -> HBr + CH3⋅ CH3⋅ + Br2 -> CH3Br + Br⋅ Termination: Br⋅ + CH3⋅ -> CH3Br OR Br⋅ + Br⋅ -> Br2 Ethane is made when two methyl radicals react (CH3⋅ + CH3⋅ -> C2H6) |
| 3) Crude oil is a source of hydrocarbons which can be used as fuels or for processing into petrochemicals. Octane, C8H18, is one of the alkanes present in petrol. Carbon dioxide is formed during the complete combustion of octane. C8H18 + 12½O2 → 8CO2 + 9H2O What is the general formula for an alkane? | CnH2n+2 |
| 4) Oil companies process hydrocarbons, such as octane, into branched and cyclic hydrocarbons that promote efficient combustion in petrol. Draw the skeletal formulae of a branched hydrocarbon and a cyclic hydrocarbon, each containing eight carbon atoms. | |
| 5) Butane, C4H10, reacts with chlorine to produce a chloroalkane with molecular formula C4H9Cl. The reaction is initiated by the formation of chlorine radicals from chlorine. What is meant by the term radical? | A species with an unpaired electron. |
| ii) State the conditions necessary to bring about the formation of the chlorine free radicals from Cl2. | Ultra violet light, high temperature at a minimum of 400 degrees and sunlight. |
| iii) State the type of bond fission involved in the formation of the chlorine radicals. | Homolytic fission. |
| iv) The chlorine radicals react with butane in several steps to produce C4H9Cl. Write equations for the two propagation steps. | C4H10 + Cl• (1) -> C4H9• + HCl C4H9• + Cl2 (1) -> C4H9Cl + Cl• |
| 6) Butane, C4H10, under certain conditions, reacts with Cl2 to form a mixture of chlorinated products. One possible product is C4H9Cl. C4H10 + Cl2 -> C4H9Cl + HCl ai) State the conditions. | UV, sunlight and high temperature (range 400 - 700 degrees) |
| ii) Write equations to show the mechanism of this reaction. initiation: ................................................................ propagation: ................................................................. ................................................................. | Initiation: Cl2 -> 2Cl∙ Propagation: C4H10 + Cl∙ -> HCl + C4H9∙ C4H9∙+ Cl2 -> C4H9Cl + Cl∙ |
| iii) Write one equation for a reaction that would terminate this mechanism. | Cl⋅ + Cl⋅ -> Cl2 OR Cl⋅ + C4H9⋅ -> C4H9Cl |
| iv) State the type of bond fission involved in the initiation step | Homolytic fission |
| b) One other possible product of the reaction between butane and chlorine is compound J, C4H8Cll2, shown below. i) Name compound J. | 2, 3-dichlorobutane |
| ii) Draw the skeletal formula of compound J. | |
| iii) In addition to compound J, suggest one other possible structural isomer of C4H8Cl2 that could have been formed in this reaction. | Any dichlorobutane, except 2, 3-dichlorobutane. |
| 7) Hexane reacts with Br2 in the presence of ultraviolet light. C6H14 + Br2 -> C6H13Br + HBr i) State the type of reaction. | Free radical substitution |
| ii) Identify the three possible structural isomers of the product, C6H13Br, that could be formed from this reaction with hexane. | 1-bromohexane, 2-bromohexane and 3-bromohexane. |
| 8) Alkanes can be separated from crude oil because they have different boiling points. Explain the variation in boiling points of alkanes. | As the hydrocarbon length increases, the boiling point also increases due to the increased number of electrons. This means there is a larger surface area and so more surface points of contact. Therefore they have more induced dipole-dipole interactions, making them stronger. Furthermore, as branching increases, as does boiling point. The larger the hydrocarbon, the more energy needed to break the stronger London forces. |
| 9) Isomer L, C5H10, reacts with Cl2 in the presence of UV light to produce the organic product C5HgCl. The reaction takes place in three stages: initiation, propagation and termination. i) The reaction is initiated by the fission of Cl2. State the type of fission involved. | Homolytic fission |
| ii) Write an equation to illustrate the fission of Cl2 in (i). | Cl2-> 2Cl∙ |
| iii) The fission of Cl2 leads to a chain reaction involving two propagation steps. Complete the equations for the two propagation steps. C5H10 + ......... -> ∙C5H9 + ......... ∙C5H9 + ......... -> ........... + .......... | (C5H10) + Cl∙ -> (C5H9∙) + HCl (C5H9∙) + Cl2 -> C5H9Cl + Cl∙ |
| This is the end of the questions. How did you do? If you made any mistakes be sure to check the model answer and understand where you went wrong. Well done! | You should now know about free radical substitution and the three steps involved (initiation, propagation and termination). This is a key section in alkanes so be sure to keep practicing the equations! |
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