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Why would 1-iodo-2-nitrobenzene be difficult to synthesize using EAS? (EAS=
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- The nitro groups on the benzene ring in the reactant serve two purposes.One is to let you know what atoms in the reactant correspond to what atomsin the product. But what role do the nitro groups play electronically – whywould the reaction be much slower if these nitro groups weren’t attached tothose benzene carbons? Draw any relevant structures to support youranswer.Explain why trichloroacetic anhydride [(Cl3CCO)2O] is more reactive than acetic anhydride [(CH3CO)2O] in nucleophilic acyl substitution reactions.Possible alternative brominations include: Veratrole (1,2-dimethoxybenzene) to 1,2-dibromo-4,5-dimethoxybenzene; 4-Methylacetanilide to 2-bromo-4-methylacetanilide; 2-Methylacetanilide (made in experiment S.1) to 4-bromo-2-methylacetanilide; Vanillin to 5-bromovanillin; Acetanilide to 4-bromoacetanilide; a. b. C. d. e. EXPERIMENT S4: BROMINATION OF AROMATIC COMPOUNDS Certain other acetanilides made in experiment S.1 may also be used as precursors in this experiment. Estimated time: 1 afternoon Associated learning goals: Section 6, LG 6.6; Section 7, LG 7.2 and 7.4 Pre-lab report: complete the standard report form, and answer the following questions. In this experiment, molecular bromine (Br2) is generated from the redox reaction of potassium bromate with hydrobromic acid. Write a balanced equation for this process. Briefly outline the mechanism by which Br2 brominates your aromatic compound. Why do the bromine atoms end up at the positions indicated rather than anywhere else in the…
- How to synthesize 1-phenyl-1-bromopropane from benzene?Electrophilic aromatic substitution usually occurs at the 1-position of naphthalene, also called the a position. Predict the major products of the reactions of naphthalene with HNO3, H2SO4.b) Listed below are several hypothetical nucleophilic substitution reactions. None is synthetically useful because the product indicated is not formed at an appreciable rate. In each case provide an explanation for the failure of the reaction to take place as indicated. OMe HO + OMe + OH HO + CH; OH
- Consider the tetracyclic aromatic compound drawn below, with rings labeled as A, B, C, and (a) Which of the four rings is most reactive in electrophilic aromatic substitution? (b) Which of the four rings is least reactive in electrophilic aromatic substitution? (c) What are the major produces) formed when this compound is treated with one equivalent of Br2?● ● Aromatic Bromination and Green Chemistry Two step synthesis and identification of product Will use vacuum filtration and analysis with m.p., IR, and NMR Day 1: Protection of Aniline Nitrogen (Amide Formation) ● Modifying aniline through reaction with acetic anhydride Day 2: Aromatic Bromination Monobromination of an activated aromatic compound Evaluation of directing effects by the isomeric product obtained Modification of procedure from Luong et al, J. Chem Ed. 2012, 89, 1061-1063 NH₂ aniline NH NaBr, NaClO AcOH, EtOH acetanilide para NH ΝΗ Br ortho meta Which position is brominated? (ortho, meta, or para)Rank the compounds in order of increasing reactivity in electrophilic aromatic substitution. Briefly explain your answer. CóH5OH, C6H5SO3H, C6H5CI
- PHENOL: what effect and orientation does –OH group exhibit in electrophilic aromatic substitution reaction?CHEM 2252 Label the following compounds as aromatic, anti-aromatic, or non-aromatic (assume that all molecules are planar). In-class Assignment #4 Predict the product that would be formed when biphenyl reacts with a mixture of nitric and sulfuric acid. Draw a detailed, step-wise mechanism that shows formation of the electrophile and explains the observed regiochemistry. HNO3 H₂SO4The following compounds are given to you:2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane(i) Write the compound which is most reactive towards SN2 reaction.(ii) Write the compound which is optically active.(iii) Write the compound which is most reactive towards P-elimination reaction.