Question

X Incorrect. The synthesis above used bromobutane, but acetylene is the only allowed source of carbon atoms. Using the reagents given, identify a synthetic route for the production of bromobutane from acetylene. ? H Br The transformation above can be performed with some reagent or combination of the reagents listed below. Give the necessary reagent(s) in the correct order, as a string of letters (without spaces or punctuation, such as "EBF"). If there is more than one correct solution, provide just one answer. A B с NaNH2 HB, ROOR 1] R2BH; 2) H2O + NaOH F HBO H2, Lindlar's cat. H2SO4, H2O, HgSO I G NaOH H CH3CH2Br Hz. Pd ГАНВІ SHOW HENT SHOW ANSWER By accessing this Question Assistance, you will leam while you earn points based on the Point Potential Policy set by your instructor. Attempts: 3 of 3 used Part 3 x Incorrect. The synthesis above used bromoethane, but acetylene is the only allowed source of carbon atoms. Using the reagents given, identify a synthetic route for the production of bromoethane from acetylene. This transformation can be performed with some reagent or combination of the reagents listed below. Give the necessary reagent(s) in the correct order, as a string of letters (without spaces ar punctuation, such as "EBF"). If there is more than one correct solution, provide just one answer. с MeoNa NaNH2 -Виок D F E H2, Lindlar's cat. EtBr Br, hv G H I HBr, ROOR H2, Pt HBT AHEB

Answer 1

-H Br Product has 2 more carbons that reactant. So, a reagent which increases carbon number by 2 must be used. From the supplied options, H - CH3CH2Br does this. So, it must be reacted with the reactant somehow. H is an alkyl halide. The reactant is a terminal alkyne, which can easily be converted to a nucleophile by a base which can take up one of the H+ from the end of the triple bond. From the options, a base is in option A - NaNH2 NaNH This is a nucleophile. It can perform Sn2 H- substitutution on the alkyl halide, Н. Н4С H2 Br Н. H3C ECH H2, Lindlar's catalyst Triple bond can then be reduced to alkene by Lindlar's catalyst. This is done because it is easy to add Br to alkenes ( a Br is needed at the end in the final product.) H2 H3C -CH2 HBI, ROOR H2 H3C ECH2 Br HBr can then be added to the alkene. The Br needs to go to the carbon with most H atoms. So, anti Markovnikov addition is needed So, the order is : AHEB

Bromoethane from acetylene can be done following the last two steps of the previous question. H2, Lindlar's catalyst HC CH Acetylene H HBT bromoethane H2C=CH2 Н4С Br Here, no problem of Markovnikoff product arises since both the carbons of the double bond are equivalent. So, just HBr is enough to add H+ and Br- to the double bond. Order: EI Br OH To do: Convert Br to COOH. Note that this would need increase of 1 C atom. Again, only option C gives 2 carbon atoms, through an Sn2 reaction ( as done in first question). However,option A is the reagent for ozonolysis, which cleaves double and triple bonds. This can be used to convert 2 carbons added to 1 carbon added HC CNa Br CH 1) 03 2) H2O (Ozonolysis of terminal alkyne gives carboxylic acid +CO2) Order - CA + CO2 OH

OH -CH₃ OH Here, the chain length has to be extended, but not from the -OH already present in the reactant. If it extended from that -OH, then a branching would have occured from that carbon. So, the-OH must be removed, and the chain must extend from the-CH3 group written in red. However, just alkanes have very few reactions, and are hard to convert to other compounds. So, the-OH will be used to generate a functional group at the terminal -CH3, which can then be used for further reactions. One of the ways to do this is by eliminating water ( H+ from the terminal CH3, and -OH). This will give a double bond. For elimination, the likely reagent is C (conc H2SO4). Elimination also needs that there be an H on the beside the C to which H is attached. In the starting compound, there is H only on the CH3 side. So, no problem of selectivity arises. Also, after making the double bond, two C needs to be increased in the chain. Hence, the double bond must be further modified so that the reagent H can react with it ( implying, that the double bond must be converted to terminal alkyl halide). OH conc H2SO4 HBT, ROOR Br HC=CNa -CH2 H2, Lindlar's cat CH 1. BH3, THF 2. H2O2. NaOH OH -CH2

( in the last question, reagent G, adds H and OH to a double bond in anti-Markovnikov manner. So, OH goes to the carbon which has most H atoms. This reaction is known as Hydroboration oxidation)