NJC J1 Alkene tutorial soutions

1 4 Alkenes Tutorial Structure and naming 1 Give the IUPAC names of the following compounds. (a) (CH3)2CHCH=C(CH3)2 C CC CH3 C H C CH3 H H H H H HH 5 4 123 2,4-dimethylpent-2-ene (c) 1 2 5 43 Br 4-bromocyclopentene 2 Draw the structural formula for each of the following compounds. (a) 1-bromo-3-ethylcyclopentene 1 2 5 4 3Br CH2CH3 (b) 3,6-dimethyl-4-propylhepta-1,4-diene C CC C H CH H C H H C H CH3H CH3 CH2CH2CH3 H H H 71 2 543 6 (c) trans-hex-2-ene 54 2 3C C H3C CH2CH2CH3H H 6 1 Electrophilic addition 3 (2021/P1/16) Which statement about propene explains how it reacts with bromine? A Electrons in the carbon−carbon 𝜋 bond are donated to an electrophile. B Electrons in the carbon−carbon 𝜎 are donated to an electrophile. C The sp2 hybridised carbon is an electrophile and accepts a pair of electrons. D The 𝜎 bond between the sp2 hybridised carbon atoms is weak and readily broken. 4. For each of the following reactions, (i) give the structures of the product(s) formed in each of the following reactions. (ii) identify with explanation the major product where applicable. (iii) state the observation during reaction where applicable. Br

2 (a) + I is less electronegative than Cl, hence δ+I is the electrophile and the more stable carbocation with greater number of electron-donating alkyl groups attached to the electron deficient carbocation to reduce its electron deficiency. This more stable carbocation is formed more quickly in the rate-determining step, leading to a higher yield of the product. C CH H H I CH3 + C CH H H CH3 I + (more stable carbocation) 4 (b) + + The halohydrin(-Br & -OH) products are formed as major products due to the presence of huge amount of H2O (since it is a solvent). It is more likely for H2O to carry out the nucleophilic attack in step 2, rather than Br ‒ due to greater amount of H 2O in the reaction system as compared to Br−. is more stable than as the former has more electron -donating alkyl group bonded to the carbocation to reduce its electron deficiency. This more stable carbocation is formed more abundantly in the rate -determining step, leading to a higher yield of the product. Observations: Orange aqueous bromine is decolourised. + ICl CCl4 dark C C H H CH3 H C C H H I Cl CH3 H CC CH3 H ICl H H + Br2 (aq)C C H H CH3 H C C H H Br OH CH3 H C C H H OH Br CH3 H C C H H Br Br CH3 H C C H H Br CH3 H C C H H Br CH3 H CCl4 ICl major product minor product major product Common mistake for question 4b: When asked to draw mechanism for such question, many students drew the OH − attacking the carbocation in the 2nd step, instead of H2O. This is incorrect. Pls check the notes that H2O is the nucleophile instead. There is a further attack by Br − on the H atom to obtain the final product and HBr. C C H H Br OH CH3 H

3 5 Explain each of the following observations, as fully as you can, by making reference to the reaction mechanism of alkenes: a) Addition of pure hydrogen iodide to but-1-ene gives a mixture of two optical isomers in equal molar proportions. Electrophilic addition In the second step, I– has equal likelihood to attack the trigonal planar carbocation from above or below the plane, and there are four different groups attached to C in the product. Hence there are two optical isomers of equal proportion. Note: C C HH CH2CH3 H H is formed as well but the resulting product C C HH CH2CH3 H H I does not have a chiral carbon. C C HH H CH2CH3 slow H I C C HH H CH2CH3 H + + I− + − carbocation + I− fast C C H H H I H CH2CH3C C HH H CH2CH3 H + * Learning points/ question 5a. Laboratory synthesis with alkene as the reactant (in electrophilic addition) always end up with producing an optically inactive mixture, even with a chiral carbon present, since a racemic mixture will always be formed.

4 5 b) Addition of liquid bromine to a solution of ethene in methanol gives a mixture of 1,2-dibromoethane and CH 2BrCH2OCH3. However, when ethene is shaken with an aqueous solution containing both bromine and sodium chloride, 1-bromo-2-chloroethane is also formed, but not 1,2-dichloroethane. Electrophilic addition In aqueous solution of Br2 and NaCl, the same carbocation is formed. + Cl− fast C C H H Br H H ClC C HH Br H H + 1,2-dichloroethane cannot be formed as C l− is a nucleophile and it does not react with eth ene to formed carbocation in the first step. Thinking process to solve such question Step 1: write out all reactants present (in this question, NaCl, Br2, H2O) Step 2: Have a table as such: E+(attacked in step 1) Nu− (attack carbocation in step 2) Product + Br in Br2 Br− CH2BrCH2Br Cl− CH2BrCH2Cl H2O CH2BrCH2OH - Na+ cannot be E+ since it only forms ionic bond. - Cl− cannot be E+ since it is electron rich, not electron deficient. - H2O cannot be E+ since O−H bond in H2O is strong and does not break readily.

5 6 (2018 P3 Q4d) Abietic acid is a major component of many resins and has the following structure . Draw the structure of the major product when abietic acid is reacted with an excess of HBr. State how many extra chiral centres are formed. H HHO O Br Br * * 2 additional chiral centres are formed. 7. When compound R is reacted with iodine in the presence of aqueous sodium hydrogencarbonate, iodine does not add across the carbon–carbon double bond of compound R as might be expected. Instead, a compound S, which has a molecular formula, C8H11O2I, is formed as shown: NaHCO3 function as a base during this reaction. Suggest an explanation for the observation. R undergoes electrophilic addition with iodine followed by acid-base reaction in alkaline medium (HCO3−) to form

6 I O O-+ The -COO− attack the carbocation to form compound S. Oxidation 8 For each of the following, identify the alkene which forms the products upon vigorous oxidation by hot acidified KMnO4, giving its structure and IUPAC name. Draw the cis-trans isomers of the alkene (if any). (i) CH3CH2CO2H only CH3CH2CO2H is formed from CH3CH2CH= fragment 5 4 2 3 C C H CH2CH3 H 61 CH3CH2 5 4 2 3C C CH2CH3H H 6 1 CH3CH2 cis-hex-3-ene trans-hex-3-ene Learning points/ Thinking process for question 5 (and similar application question you see in future) - Product is given, you are supposed to deduce a logical sequence of steps to get this product. - There is a double bond in R, I2 is present and no more double bond in S. Electrophilic addition (otherwise what else?) must have occur. - If you are wondering which carbon does I add across in step 1 of electrophilic addition, look at the product. I is added to the top carbon, hence the carbocation should be on the bottom carbon. - Looking at the product, the bottom carbocation is likely to be attacked by O− in step 2 of electrophilic addition. - NaHCO3 can react with CH 2CO2H to give CH 2CO2−, which happens to be the nucleophile to react with the carbocation in step 2. More advanced way of looking at it: When a product has 1 additional ring than a reactant, an intramolecular reaction must have taken place. This will hint towards a nucleophile being in the same molecule as R. Thinking process/comments for question 8: Working backwards from the product of vigorous oxidation to derive the 4 groups attached to the C=C of an alkene. - Refer to pg 15 of Alkene notes if unsure

7 (ii) (CH3)2CO and CH3CH2COOH (CH3)2CO is formed from (CH3)2C= fragment CH3CH2CO2H is formed from CH3CH2CH= fragment 54 2 3C C H3C CH2CH3H3C H1 2-methylpent-2-ene (iii) CO2 and CH3COCH2CH2CH3 CO2 is formed from CH2= fragment CH3COCH2CH2CH3 is formed from (CH3)(CH3CH2CH2)C= frag