Alkenes Revision Notes
Subject: Chemistry | Level: A-Level | Exam Board: AQA
Unlock the secrets of carbon-based life with this essential guide to Organic Chemistry. From mastering IUPAC nomenclature to predicting complex reaction pathways, this module builds the foundation for understanding everything from fuels to pharmaceuticals."
Revision Notes & Key Concepts
Worked Examples
Worked Example
Question: Propene reacts with bromine. Draw the mechanism for this reaction. Include all partial charges and curly arrows. (4 marks)", "marks": 4, "solution": "Step 1: Draw propene and the Br-Br molecule. Induce a dipole on the Br-Br molecule as it approaches the electron-rich C=C bond. Step 2: Draw a curly arrow from the C=C double bond to the slightly positive Br atom ($\delta+$). Step 3: Draw a curly arrow from the Br-Br bond to the slightly negative Br atom ($\delta-$), breaking the bond. Step 4: Draw the intermediate carbocation structure with the positive charge on the central carbon (more stable secondary carbocation). Step 5: Draw a curly arrow from the lone pair on the bromide ion ($:Br^-$) to the positive carbon atom. Final Answer: The product is 1,2-dibromopropane.", "examiner_commentary": "This earns full marks because the curly arrows start exactly from the source of electrons (bond or lone pair) and end at the destination. A common error is drawing the arrow from the atom rather than the bond. The intermediate must show the correct stability (secondary carbocation preferred over primary)."
Solution:
Worked Example
Question: Explain why the boiling point of ethanol ($78^\circ C$) is significantly higher than that of ethane ($-89^\circ C$), despite having similar molecular mass. (3 marks)", "marks": 3, "solution": "Step 1: Identify the intermolecular forces in ethane. Ethane is non-polar and only has weak Van der Waals forces (London dispersion forces) between molecules. Step 2: Identify the intermolecular forces in ethanol. Ethanol contains an -OH group, which allows for hydrogen bonding between molecules. Step 3: Compare the strength. Hydrogen bonds are much stronger than Van der Waals forces, requiring significantly more heat energy to overcome. Final Answer: Ethanol has hydrogen bonding, which is stronger than the weak Van der Waals forces in ethane, thus requiring more energy to break.", "examiner_commentary": "Candidates must name the specific forces involved. Simply saying 'stronger bonds' is insufficient; you must specify 'intermolecular forces' or 'hydrogen bonds' between molecules."
Solution:
Worked Example
Question: Compound X has the molecular formula $C_4H_{10}O$. It reacts with acidified potassium dichromate(VI) to turn the solution from orange to green. It does not react with Tollen's reagent. Deduce the structure of X. (4 marks)
Solution: Step 1: Analyze the formula. $C_4H_{10}O$ fits the general formula for alcohols ($C_nH_{2n+2}O$). Step 2: Analyze the oxidation. The colour change (orange to green) confirms it is a primary or secondary alcohol (tertiary alcohols do not oxidise). Step 3: Analyze the Tollen's test. No reaction means the oxidation product is a ketone, not an aldehyde. Therefore, X must be a secondary alcohol. Step 4: Draw the secondary alcohol with 4 carbons. The -OH group must be on the second carbon. Final Answer: Butan-2-ol ($CH_3-CH(OH)-CH_2-CH_3$).
Practice Questions
Question: Name the following compound: CH3-CH(CH3)-CH2-CH=CH2
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Question: Describe a chemical test to distinguish between butane and but-2-ene. State the reagent and the observation for each.
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Question: Outline the mechanism for the reaction of ethane with chlorine in the presence of UV light to form chloroethane.
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Question: Explain why tertiary alcohols cannot be oxidised by acidified potassium dichromate(VI).
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Question: Calculate the atom economy for the production of ethene from the dehydration of ethanol: C2H5OH -> C2H4 + H2O.
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