ORG.CHEM W/TEXT+SOLU.MANUAL
15th Edition
ISBN: 9780393252125
Author: KARTY
Publisher: W.W.NORTON+CO.
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 6, Problem 6.13YT
Interpretation Introduction
Interpretation:
The stronger acid is to be determined using pKa values.
Concept introduction:
A higher (i.e., more positive) value of Ka corresponds to a lower (i.e., more negative) value of pKa. Therefore, when comparing pKa values of two different acids, the acid with the lower pKa value is the stronger acid.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Calculate pk, values for the following acids.
(Enter your answer to three significant figures.)
a. Hydrocyanic acid, HCN (K₂ = 4.90x10-10):
b. Ethanol (K = 1.00×10-16):
Consider acid base reaction shown below.
Assuming that pyridine and acetic acid are mixed in equal concentrations and using pKa values of acetic acid and pKa value of pyridinium cation calculate what percent of pyridine is transformed to the pyridinium cation in this reaction.
Answer is numerical. Round the answer to the nearest whole number.
Partition Coefficient Lab
1. Draw the deprotonation reaction using sodium bicarbonate as the base. Be sure to draw in the chemical structures for benzoic acid and sodium benzoate.
2. Draw the protonation reaction of sodium benzoate using HCl as the acid. Be sure to draw in the chemical structures for benzoic acid and sodium benzoate.
3. Which phase (water or methylene chloride) do you expect to be the top layer? Why?
4. Which phase do we expect the sodium benzoate to be found in (water or methylene chloride) what about benzoic acid? Please explain your answer.
Chapter 6 Solutions
ORG.CHEM W/TEXT+SOLU.MANUAL
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
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- 2. The pKa values of certain acids are shown below. Rank the conjugate bases corresponding to each of these acids in decreasing order of leaving group ability (1 = best leaving group, 4 = worst leaving group) ACID pka LG ABILITY HF 3.1 F3CSO3H -6.0 PhSO3H -2.8 HBr -5.8arrow_forwardWhich bases can deprotonate acetylene? The pKa values of the conjugate acids are given in parentheses.a. CH3NH- (pKa= 40)b. CO32- (pKa = 10.2)c. CH2 = CH- (pKa = 44)d. (CH3)3CO- (pKa = 18)arrow_forwardWhich reaction mechanism of a Lewis acid reacting with a Lewis base is correct?arrow_forward
- Classify the following reactions as acidic anhydride or basic anhydride. Drag the appropriate items to their respective bins. • View Available Hint(s) Reset Help Li20 + H2O → 2LİOH MgO + H2O –→ Mg(OH)2 P203 + 3H20 → 2H3PO3 SO3 + H2O –→ H2SO4 N2O5 + H2O → 2HNO3 K20 + H2O → 2KOH Acidic anhydride Basic anhydridearrow_forwardThe synthesis of an azo dye is shown below. Please answer the following questions. NH2 1) Na2CO3 2) NaNO2, HCI B N-Ñ A Š3H ŠO3 Step 1 Step 2 ŠO3 2 1 1. Why you need to add Na2CO3 in step 1? (multiple answers) Oa)To decrease the solubility of compound 1 - "salting out" ob) To deprotonate compound 1, increase solubility in water Oc) To increase the reaction yield Od)To slow down the reaction rate Oe)To neutralize the HCIarrow_forwardUse the information in the pK table to rank the molecules in order of decreasing basicity. For exam and "2" for the next strongest base and so on. CH3NH₂ H HSO4 HS Molecules (Choose one) (Choose one) (Choose one) (Choose one) X CH1 NH₂ H₁₂ CH,NH, H₂O CH₂OH 48 38 36 33 15.7 15.5 pk₁ table a CH,NH, CH, SH 10.4 HCN NH H₂S CH,CO,H 10.6 9.4 9.2 7.00 4.76arrow_forward
- 20 Arrange these acids according to their expected pKą values. CICH₂CH₂COOH СІСН,СООН Highest pKa Lowest pKa Answer Bank Cl₂CHCOOH CH₂CH₂COOHarrow_forward1. Acid-base equilibrium a.. Draw the major resonance contributors of the product structures for each acid-base reaction below. b. Use the pKa data from lecture to estimate the pKa of each acid / conjugate acid. Calculate Keq for each reaction and draw the corresponding equilibrium arrow. c. i H acid 1 pka = ΝΗ acid pka = O=S= HO-S-OH O acid ipka = NLi base ONa base NH 人 base (equilibrium arrow) Kea 1 1 (equilibrium arrow) B Keg Keg =! (equilibrium arrow) 1 T 1 1 1 1 1 1 1 1 I 1 NLI e conjugate base conjugate base conjugate base 1 H conjugate acid ¦pka = conjugate acid pka = conjugate acid pka = 1arrow_forwardH Benzoic Acid pka = O 3-Hydrox pKa =_ HO ноarrow_forward
- C341 Chapters (Acid-base chemistry) 9.) Acid-base equilibrium. For each reaction, perform the following tasks: a.) Complete the following reactions using the appropriate curved arrows to demonstrate the flow of electrons pairs in each potential reaction. b.) Indicate if the reaction lies to the right or the left (circle the correct arrow direction). c.) Calculate an approximate Ken for each reaction. d.) Write the pKa values under each acid, so we can evaluate your progress better and award partial credit. d.) Finally, label the acid (A), base (B), conjugate acid (CA) and base (CB) correctly based on the direction you decided. Keq = Direction: OH eg HO Keq = Direction: Keq = Direction: LOH OH ONa لها LIⒸ CHgLi Page 5 of 7arrow_forwardStep 1. Protonation of carboxylic acid. Draw curved arrows. Select Draw Rings More Erase | 7|/ C H |- H HöH H H : 0 : H H Step 2. Nucleophilic attack by alcohol. Draw curved arrows. Select Draw Rings More Erase H Step 3. Proton Transfer H. CH :0arrow_forward1) Conjugate Acid/Base Pairs - Draw the conjugate Acid-Base pair for each reaction, and indicate if the reaction will happen or not by circling YES or NO. pka = 5 Acid-Base Pair pka 19 CH3 CH3 CH3 protonate/deprotonate + +H / -H +H/-H + +H/-H +H/-H + +H/-H Conjugate Acid-Base Pair Will they React? YES NO YES NO YES NO YES NOarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
Organic Chemistry: A Guided Inquiry
Chemistry
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Cengage Learning