Organic Chemistry: Structure and Function
8th Edition
ISBN: 9781319079451
Author: K. Peter C. Vollhardt, Neil E. Schore
Publisher: W. H. Freeman
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Chapter 7, Problem 41P
Interpretation Introduction
Interpretation:The formation of slight excess inversion alcohol product from hydrolysis of
Concept introduction:
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5A
Explain which of the following statements are true and which are false.
i. N,N'-dicyclohexylurea (DCU) is the dehydrating reagent that converts a
carboxylic acid to the corresponding anhydride.
ii. The fact that carbonyl compounds have a lower boiling point than
alcohols with the same C atoms is because the molecular dipole moment of
carbonyl compounds is less than that of alcohols.
iii. 3-Methylbut-2-en-2-ol is the only tautomeric structure of 3-
methylbutan-2-one.
Elimination of HBr from 2-bromobutane affords a mixture of 1-butene and 2-butene. With sodium ethoxide as base, 2-butene
constitutes 81% of the alkene products, but with potassium tert-butoxide, 2-butene constitutes only 67% of the alkene
products. Offer an explanation for this difference.
iii) 2-Bromo-2-cyclopropylpropane will undergo an SN1 reaction called solvolysis in methanol to give several products, two of which are shown below. Use curly arrows to show how the formation of these two products occurs mechanistically.
Chapter 7 Solutions
Organic Chemistry: Structure and Function
Ch. 7.1 - Prob. 7.1ECh. 7.2 - Prob. 7.2ECh. 7.3 - Prob. 7.3ECh. 7.3 - Prob. 7.5TIYCh. 7.4 - Prob. 7.7TIYCh. 7.5 - Prob. 7.9TIYCh. 7.6 - Prob. 7.10ECh. 7.7 - Prob. 7.11ECh. 7.7 - Prob. 7.12ECh. 7.7 - Prob. 7.13E
Ch. 7.7 - Prob. 7.15TIYCh. 7.8 - Prob. 7.16ECh. 7.8 - Prob. 7.17ECh. 7.9 - Prob. 7.18ECh. 7.9 - Prob. 7.19ECh. 7.9 - Prob. 7.20ECh. 7.9 - Prob. 7.21ECh. 7.9 - Prob. 7.22ECh. 7 - Prob. 25PCh. 7 - Prob. 26PCh. 7 - Prob. 27PCh. 7 - Prob. 28PCh. 7 - Prob. 29PCh. 7 - Prob. 30PCh. 7 - Prob. 31PCh. 7 - Prob. 32PCh. 7 - Prob. 33PCh. 7 - Prob. 34PCh. 7 - Prob. 35PCh. 7 - Prob. 36PCh. 7 - Prob. 37PCh. 7 - Prob. 38PCh. 7 - Prob. 39PCh. 7 - Prob. 40PCh. 7 - Prob. 41PCh. 7 - Prob. 42PCh. 7 - Prob. 43PCh. 7 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - Prob. 46PCh. 7 - Prob. 47PCh. 7 - Prob. 48PCh. 7 - Prob. 49PCh. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - Prob. 53PCh. 7 - Prob. 54PCh. 7 - Prob. 55PCh. 7 - Prob. 56PCh. 7 - Prob. 57PCh. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - Prob. 60PCh. 7 - Prob. 61PCh. 7 - Prob. 62PCh. 7 - Prob. 63PCh. 7 - Prob. 64PCh. 7 - Prob. 65PCh. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Prob. 68PCh. 7 - Prob. 69PCh. 7 - Prob. 70P
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- OH H₂SO4 HỌC–CH2–CH2CH3 H,C=CH—CH,—CH3 + H₂O Acid-catalyzed dehydration of 1-butanol yields 1-butene as the major product. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism. Arrow-pushing Instructions H2C=CH=CH2CH3 NOC XT + OH₂ HỌC–CH—CH, CH3 I H H-O-Harrow_forwardGive all the monobromination products of 2-methylpropane (or isobutane) in presence of heat or energy. Identify the major product and propose a mechanism leading to the formation of the major product. Provide a reaction in the termination step.arrow_forwardCompound X has molecular formula C,H14. Hydrogenation of compound X produces 2,4-dimethylpentane. Hydroboration- oxidation of compound X produces a racemic mixture of 2,4-dimethyl-1-pentanol (shown below). Predict the major product(s) obtained when compound X is treated with aqueous acid (H3O*). HO 2,4-Dimethyl-1-pentanolarrow_forward
- Consider the monosubstituted benzene reagent X. Nitration of X yields product Y. Product Y is analysed by 1H and 13C NMR spectroscopy (see Figures 1 to 4). Determine the structure of X and Y on the basis of the spectral data provided.arrow_forwardGive the structures of the products you would expect when each alcohol reacts with(1) HCl, ZnCl2; (2) HBr; (3) PBr3; (4) P>I2; and (5) SOCl2.(a) butan-1-ol (b) 2-methylbutan-2-ol(c) 2,2-dimethylbutan-1-ol (d) cis-3-methylcyclopentanolarrow_forwardX Upon ozonolysis, Compound X produces two compounds: Compound Y and Compound Z. Compound Y can also be prepared from the following synthetic route: PCC 1. R₂BH, THF 1. Mg. Et₂O PCC Compound Y 2. CH₂Cl₂ 2. NaOH, HO CH₂Cl₂ 3. H₂O* From this information, draw the structures of Compounds X, Y, and Z. For Compounds X and Z, different substituents are possible. For grading purposes, just use hydrogens as the substituents. Br مرد →] ►arrow_forward
- 10. Compound X (C4H9Br) reacts by heating with NaOH in H2O to form Y. The compound Y then undergoes acid catalysed hydration by H2SO, in 180°C to form 2-methyl prop-1-ene. (e) Determine the structure of X and Y. (f) Predict a MAJOR product when compound Y reacts with H2SO4 in 140 C. (g) Draw a structural isomer of X. Name the isomer using IUPAC nomenclature. (h) Describe a chemical test to distinguish between compound Y and 1-butanol.arrow_forward2. There are two isomeric alkenes that can produce the first alcohol shown below as the major product with one set of reagents, but with other reagents, the same two alkenes will produce the second alcohol as the major product. H3C OH H3C, ОН H3C. CH3 (a) Draw the structure of the two alkenes described above. (b) Provide the reagents required to obtain each of the two alcohols. (c) Write out the arrow-pushing mechanism that produces the second alcohol.arrow_forwardPlease be clear in your writing, solve step by step Compound A (C9H12), when hydrogenated by catalysis on Pd / C, absorbs 3 equivalents of H2 to give compound B (C9H18). Ozonolysis of compound A gives cyclohexanone (C6H10O). Compound A reacting with NaNH2 / NH3 followed by addition of CH3Br gave compound C (C10H14). What are the structures of compounds A, B, and C?arrow_forward
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