Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
expand_more
expand_more
format_list_bulleted
Question
Chapter 6, Problem 6.82P
Interpretation Introduction
(a)
Interpretation:
Based on the understanding of charge stability, it is to be determined which of the following compounds is expected to have the most similar pKa value to that of formaldehyde.
Concept introduction:
Table 6-1 refers to various compounds and their Ka and pKa values. Based on the structural similarities in compounds listed in Table 6-1 and their
Interpretation Introduction
(b)
Interpretation:
The pKa value for formaldehyde is to be estimated based on the concept of charge stability and the reference to Table 6.1.
Concept introduction:
pKa value represents the strength of an acid. The lower the value, the stronger the acid. Structural similarity in conjugate bases of two compounds indicates that the pKa values for the two compounds must be in the similar range. The chemical behavior of a compound is governed by the functional group it possesses.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The pka of formaldehyde (H2C=0) is not listed in Table 6-1.
(a) Based on your understanding of charge stability, which of the following compounds (A-F) would you expect to have a
pKa most similar to that of formaldehyde?
C.
H
H3C-OH
H3C-CH3
H,C=CH2
HC=CH
H-C=N
HO.
A
В
E
F
(b) Using Table 6-1, as well as your knowledge of the factors that affect charge stability, estimate the pka of
formaldehyde.
6-46
Show how each compound might be synthesized by the SN2 displacement of an alkyl halide.
SCH,CH,
(a)
-CH₂OH
(c)
(d)
CHÍNH,
6-47 (a) Give twn synthecac fox (CI) qu
(b)
(e) H₂C=CH-CH₂CN
n
OLL BU
(1) H-C=C-CH₂CH₂CH₂
6-137. The mixed aldol reaction between propionaldehyde (propanal) and acetone gives an 85% yield of 4-hydroxy-2-
hexanone when run in THF at -78 °C with lithium diisopropylamide (LDA, a powerful base). The reaction is carried
out by first adding the ketone to the base, cooling the solution, and then adding the aldehyde.
(a) Why does this mixed system give essentially a single product?
(b) Why is there no self-condensation of the acetone?
(c) Why does the system not rapidly go on to dehydrate?
(d) Why is the ketone added to the base rather than vice versa?
ليك
debudes for ketones) has no α-hydrogen atoms. This is
Chapter 6 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
Ch. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - Prob. 6.10P
Ch. 6 - Prob. 6.11PCh. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.14PCh. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - Prob. 6.17PCh. 6 - Prob. 6.18PCh. 6 - Prob. 6.19PCh. 6 - Prob. 6.20PCh. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - Prob. 6.23PCh. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - Prob. 6.33PCh. 6 - Prob. 6.34PCh. 6 - Prob. 6.35PCh. 6 - Prob. 6.36PCh. 6 - Prob. 6.37PCh. 6 - Prob. 6.38PCh. 6 - Prob. 6.39PCh. 6 - Prob. 6.40PCh. 6 - Prob. 6.41PCh. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - Prob. 6.46PCh. 6 - Prob. 6.47PCh. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - Prob. 6.50PCh. 6 - Prob. 6.51PCh. 6 - Prob. 6.52PCh. 6 - Prob. 6.53PCh. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Prob. 6.57PCh. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - Prob. 6.60PCh. 6 - Prob. 6.61PCh. 6 - Prob. 6.62PCh. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Prob. 6.72PCh. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Prob. 6.83PCh. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - Prob. 6.86PCh. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.1YTCh. 6 - Prob. 6.2YTCh. 6 - Prob. 6.3YTCh. 6 - Prob. 6.4YTCh. 6 - Prob. 6.5YTCh. 6 - Prob. 6.6YTCh. 6 - Prob. 6.7YTCh. 6 - Prob. 6.8YTCh. 6 - Prob. 6.9YTCh. 6 - Prob. 6.10YTCh. 6 - Prob. 6.11YTCh. 6 - Prob. 6.12YTCh. 6 - Prob. 6.13YTCh. 6 - Prob. 6.14YTCh. 6 - Prob. 6.15YTCh. 6 - Prob. 6.16YTCh. 6 - Prob. 6.17YTCh. 6 - Prob. 6.18YTCh. 6 - Prob. 6.19YTCh. 6 - Prob. 6.20YT
Knowledge Booster
Similar questions
- 8-8 Identify the major product(s) of the following reactions. Include stereochemistry. (a) (b) (c) Br NaOEt NaOEt NaOEtarrow_forwardEt Et-N: Et Me ΘΗ Me: N Me Me pK₂ = 12.5 Me Me Me: `N° N Me Me Me H Et Et (a) Add electron-flow arrows to show how the electrons move in this reaction. Et pK₂ = 10 (b) Does the equilibrium for this reaction favor the starting materials, the products, or neither?arrow_forwardBr OH H₂O HO (14 pts) Provide a complete mechanism for the reaction in the box. Include all electron-pushing arrows, lone pairs, and formal charges. HO blo NO (S)arrow_forward
- N-Nitrosamines by themselves are not significant carcinogens. However, they are activated in the liver by a class of iron-containing enzymes (members of the cytochrome P-450 family). Activation involves the oxidation of a C-H bond next to the amine nitrogen to a C-OH group. OH ОН N=0 N=O Og 'N- H+ H N,+ cyt P-450 N-Nitroso- 2-Нydroxy-N- nitrosopiperidine An alkyl diazonium ion piperidine (a carcinogen) Show how this hydroxylation product can be transformed into an alkyl diazonium ion, an active alkylating agent and therefore a carcinogen, in the presence of an acid catalyst.arrow_forwardUse curved arrow formalism to draw the mechanism of the reaction shown below. Include intermediates and their relevant resonance forms. Be sure to account for all the products. Br CH3OH OCH3 + Joc 1w OCH 3arrow_forwardThe nitroso group, −N=O, is one of the few nonhalogens that is an ortho- and para-directing deactivator. Explain this behaviour by drawing resonance structures of the carbocation intermediates in ortho, meta, and para electrophilic reaction onnitrosobenzene, C6H5−N=O.arrow_forward
- Complete the synthetic sequence below by providing the structures of the reaction (questions 7-8). 1 mole of HOCH,CH,OH/H > B 1. LIAIH, 1.03 H* 2. Zn/H 2. Н.О. What is the structure of B? HO. (A) (D) OHarrow_forwardHO CH30 Na A + B O,N pK =7 (a) Provide structures for A and B (including nonbonding electron pairs and formal charges where necessary), where A is the conjugate acid and B is the conjugate base of the reaction. Answer: (b) Provide and arrow pushing mechanism to show how A and B are formed. Answer: (c) Will the reaction proceed from left to right (i.e. are A and B favoured at equilibrium)? Explain your answer. Answer: (d) Draw all important resonance contributors of the conjugate base B. Rank the contributors from most stable to least stable (if contributors are equally stable indicate this with an equals sign)arrow_forward20. Identify the product for each step in the following reaction sequence. Write complete equations showing the structure of all reactants and products for each step. (a) Isopropyl benzene + NBS/heat ---------->A (b) A+ t-BuOK ---------> B (c) B + 1)BH3 followed by 2)H2O2/NaOH -----> C (d) C + PCC ------> Darrow_forward
- H 5+++ CH3COH H H LOSO₂Ar -H A H Solvolysis of this bicyclic compound in acetic acid gives a mixture of products. The leaving group is the anion of a sulfonic acid, ArSO,H. A sulfonic acid is a strong acid, and its anion, ArSO3, is a weak base and a good leaving group. Draw curved arrows to show the movement of electrons in the following step of the reaction mechanism. Arrow-pushing Instructions AOC XT H H H H H OCCH3 A-Harrow_forward5. Use curved arrows to show the mechanism of the following acid-base reaction: "H ¡N-H H y N-H Harrow_forward12. Draw detailed mechanisms for the following reactions including all reaction intermediates, resonance forms, curved arrows, formal charges, and reaction arrows. Then, determine the enthalpy of each step and include a reaction coordinate diagram. Me Me Me HBr Br Me Me Me Me Me H2SO4 Me. OH Me Br2 Br α ∞ OH MeOH Mearrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
