Question

please help solve question 21 and question 22. thank you

THE FOLLOWING REACTION SCHEME APPLIES TO QUESTIONS 21 AND 22 Jon + intos xor (2) Har Xothon QUESTION 21 [2] In the products of reaction (1), the halogen is on the less hindered carbon labelled (b) in the ether, and the oxygen remains on the more hindered carbon labelled (a). The reason for this is: 11 The reaction proceeds via an 2 mechanism, oxygen in the ether gets protonated first, then I attacks the lesser hindered carbon b. yielding the depicted alcohol and lodoethane. The reaction proceeds by E2 mechanism: The oxygen of the ether is protonated first, the jodide anion removes a proton from the carbon labelled ca double bond is formed between carbons and giving ethene, and the alcohol is eliminated. The HI regenerated then adds to ethene to give lodoethane. The reaction proceeds by El mechanism. The oxygen of the ether gets protonated first and the alcohol (CH..CHOH departs, leaving an ethyl cation CH CH" which combines with to form iodoethane The reaction proceeds via an S, 1 mechanism, oxygen in the ether gets protonated first, the alcohol (CH.CHOH departs, leaving an ethylcation CH,CH" which loses a proton to I to form ethene and H. These then combine in an addition reaction to form lodoethane C D QUESTION 22 [2] In the products of reaction (2), the halogen is on the more hindered carbon labelled (d) in the ether, and the oxygen remains on the less hindered carbon labelled (e). The reason for this is A B The reaction proceeds via an Su mechanism, oxygen in the ether gets protonated first the akohol CH CH.CH OH departs, leaving a tertiary butyl cation (CH) C which combines with Br to form 2-bromo-2-methylpropane. The reaction proceeds via an 5.2 mechanism, oxygen in the ether gets protonated first, then Br attacks the more hindered carbon d yielding the depicted 2-bromo-2- methylpropane and propan-1-ol. The reaction proceeds by E2 mechanism: The oxygen of the ether is protonated first, the Branion removes proton from the carbon labelled f, a double bond is formed between carbons and d, giving 2-methylbut-2-ene, and the alcohol is eliminated. The HBr regenerated then adds to 2-methylbut-2-ene to give -bromo-2-methylpropane The reaction proceeds by El mechanism. The oxygen of the ether gets protonated first and Br removes a proton from the carbon labelled a double bond forms between carbons and d and the alcohol CH CH.CH OH departs, leaving 2-methylbut-2-ene, which combines with the regenerated HBr to form 2 bromo-2-methylpropane. D