NJC Alkene Tutorial Answer 2023

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 abundantly 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