(Solved): Draw the organic molecule(s) which is(are) formed in the following reaction. ...

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Draw the organic molecule(s) which is(are) formed in the following reaction. Do not include molecules like $${\mathrm{H}}_2\mathrm{O}$$ or $$\mathrm{HCl}$$. (You have 2 chances until the answer will be given; you have already tried 0 times) Draw only one molecule in each drawing box. Draw the organic molecule(s) which is(are) formed in the following reaction. Do not include molecules like $${\mathrm{H}}_2\mathrm{O}$$ or $$\mathrm{HCl}$$. (You have 2 chances until the answer will be given; you have already tried 0 times) Draw only one molecule in each drawing box. Draw the organic molecule(s) which is(are) formed in the following reaction. Do not include molecules like $${\mathrm{H}}_2\mathrm{O}$$ or $$\mathrm{HCl}$$. (You have 2 chances until the answer will be given; you have already tried 0 times) Draw only one molecule in each drawing box. A. (Worth: 2 marks) The following is a nucleophilic substitution reaction of R-3-bromo-2-methylhexane with $$\mathrm{SH}$$. The obtical rotation of the solution after the reaction was complete was $$0.0^{?}$$ Select which of the following statements is correct: The mechanism for this reaction is $${\mathrm{S}}_{\mathrm{N}}1$$ because the product is a racemic mixture of enantiomers because the solution optical rotation was $$0^{?}$$ $${\mathrm{S}}_{\mathrm{N}}1$$ because the carbon attached to the leaving group, the bromine atom, is a primary carbon (a carbon with one carbons attached to it $${\mathrm{S}}_{\mathrm{N}}2$$ because the nucleophile, $$\mathrm{SH}$$ reacts via this mechanism due to its size $${\mathrm{S}}_{\mathrm{N}}2$$ because the steric crowding around the carbon attached to the leaving group, the bromine atom, is a minimum. The following is a nucleophilic substitution reaction of $$\mathrm{R}?3$$-bromohexane with $${\mathrm{SH}}^{?}$$. The experimental rate law for this reaction is $$\mathrm{Rate}=k$$ [R-3-bromohexane] Select which of the following statements is correct: The mechanism for this reaction is $${\mathrm{S}}_{\mathrm{N}}1$$ because the carbon attached to the leaving group, the bromine atom, is a secondary carbon (a carbon with two carbons attached to it $${\mathrm{S}}_{\mathrm{N}}1$$ because the experiment rate law is first order $${\mathrm{S}}_{\mathrm{N}}2$$ because the steric crowding around the carbon attached to the leaving group, the bromine atom, is a maximum. $${\mathrm{S}}_{\mathrm{N}}2$$ because the nucleophile, $$?\mathrm{SH}$$ reacts via this mechanism due to its size (c) University of Western Sydney 2022 Contact us Priv ABN 53014069881 CRICOS Provider No: $$00917\mathrm{k}$$ This question has several parts. Each part you will only have one chance to give the correct answer. A. (Worth: 3 marks) The following molecule has a stereoisomer which is referred to as its enantiomer Draw a representation of this molecule's stereoisomer in the box below Draw only one molecule in each drawing box. B. (Worth: 3 marks) Draw the stereoisomer of this molecule. (You have 2 chances until the answer will be given; you have already tried 0 times) Draw only one molecule in each drawing box. B. (Worth: 3 marks) Draw the stereoisomer of this molecule. (You have 2 chances until the answer will be given; you have already tried 0 times) Draw only one molecule in each drawing box. Draw the molecule in the box below that has the IUPAC name, R-4-ethyl-2-methyloctane (You have three chances until the answer will be given; you have already tried 0 times) Draw only one molecule in each drawing box.