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Concept explainers
(a)
Interpretation: The product of Perkin condensation has to be written
Concept introduction: For the given condensation reactions,
In first case, acetic anhydride is treated with a base and a proton is abstracted. The enolate ion thus formed attacks carbonyl carbon of benzaldehyde. Loss of water molecule results in the formation of the product of Perkin condensation.
In second case, water is added to the product of Perkin condensation and it causes hydrolysis of the ester to form an acid.
In third case, diethyl malonate is treated with a base and enolate ion is formed. Enolate ion attacks a molecule of benzaldehyde and forms an intermediate species. Loss of a water molecule results in the formation of the product of Knoevenagel reaction.
In fourth case, the product of Knoevenagel reaction on acidic hydrolysis gives
(b)
Interpretation: The compound formed after addition of water to the product of Perkin condensation has to be written.
Concept introduction: For the given condensation reactions,
In first case, acetic anhydride is treated with a base and a proton is abstracted. The enolate ion thus formed attacks carbonyl carbon of benzaldehyde. Loss of water molecule results in the formation of the product of Perkin condensation.
In second case, water is added to the product of Perkin condensation and it causes hydrolysis of the ester to form an acid.
In third case, diethyl malonate is treated with a base and enolate ion is formed. Enolate ion attacks a molecule of benzaldehyde and forms an intermediate species. Loss of a water molecule results in the formation of the product of Knoevenagel reaction.
In fourth case, the product of Knoevenagel reaction on acidic hydrolysis gives carboxylic acid and on heating undergoes decarboxylation.
(c)
Interpretation: The product of Knoevenagal condensation has to be written
Concept introduction: For the given condensation reactions,
In first case, acetic anhydride is treated with a base and a proton is abstracted. The enolate ion thus formed attacks carbonyl carbon of benzaldehyde. Loss of water molecule results in the formation of the product of Perkin condensation.
In second case, water is added to the product of Perkin condensation and it causes hydrolysis of the ester to form an acid.
In third case, diethyl malonate is treated with a base and enolate ion is formed. Enolate ion attacks a molecule of benzaldehyde and forms an intermediate species. Loss of a water molecule results in the formation of the product of Knoevenagel reaction.
In fourth case, the product of Knoevenagel reaction on acidic hydrolysis gives carboxylic acid and on heating undergoes decarboxylation.
(d)
Interpretation: A compound formed after product of Perkin condensation is heated in an acidic medium has to be written
Concept introduction: For the given condensation reactions,
In first case, acetic anhydride is treated with a base and a proton is abstracted. The enolate ion thus formed attacks carbonyl carbon of benzaldehyde. Loss of water molecule results in the formation of the product of Perkin condensation.
In second case, water is added to the product of Perkin condensation and it causes hydrolysis of the ester to form an acid.
In third case, diethyl malonate is treated with a base and enolate ion is formed. Enolate ion attacks a molecule of benzaldehyde and forms an intermediate species. Loss of a water molecule results in the formation of the product of Knoevenagel reaction.
In fourth case, the product of Knoevenagel reaction on acidic hydrolysis gives carboxylic acid and on heating undergoes decarboxylation.
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Chapter 17 Solutions
ORGANIC CHEMISTRY-W/S.G+SOLN.MANUAL
- Please answer the questions in the photos and please revise any wrong answers. Thank youarrow_forward(Please be sure that 7 carbons are available in the structure )Based on the 1H NMR, 13C NMR, DEPT 135 NMR and DEPT 90 NMR, provide a reasoning step and arrive at the final structure of an unknown organic compound containing 7 carbons. Dept 135 shows peak to be positive at 128.62 and 13.63 Dept 135 shows peak to be negative at 130.28, 64.32, 30.62 and 19.10.arrow_forward-lease help me answer the questions in the photo.arrow_forward
- For the reaction below, the concentrations at equilibrium are [SO₂] = 0.50 M, [0] = 0.45 M, and [SO3] = 1.7 M. What is the value of the equilibrium constant, K? 2SO2(g) + O2(g) 2SO3(g) Report your answer using two significant figures. Provide your answer below:arrow_forwardI need help with this question. Step by step solution, please!arrow_forwardZn(OH)2(s) Zn(OH)+ Ksp = 3 X 10-16 B₁ = 1 x 104 Zn(OH)2(aq) B₂ = 2 x 1010 Zn(OH)3 ẞ3-8 x 1013 Zn(OH) B4-3 x 1015arrow_forward
- Help me understand this by showing step by step solution.arrow_forwardscratch paper, and the integrated rate table provided in class. our scratch work for this test. Content attribution 3/40 FEEDBACK QUESTION 3 - 4 POINTS Complete the equation that relates the rate of consumption of H+ and the rate of formation of Br2 for the given reaction. 5Br (aq) + BrO3 (aq) + 6H (aq) →3Br2(aq) + 3H2O(l) • Your answers should be whole numbers or fractions without any decimal places. Provide your answer below: Search 尚 5 fn 40 * 00 99+ 2 9 144 a [arrow_forward(a) Write down the structure of EDTA molecule and show the complex structure with Pb2+ . (b) When do you need to perform back titration? (c) Ni2+ can be analyzed by a back titration using standard Zn2+ at pH 5.5 with xylenol orange indicator. A solution containing 25.00 mL of Ni2+ in dilute HCl is treated with 25.00 mL of 0.05283 M Na2EDTA. The solution is neutralized with NaOH, and the pH is adjusted to 5.5 with acetate buffer. The solution turns yellow when a few drops of indicator are added. Titration with 0.02299 M Zn2+ requires 17.61 mL to reach the red end point. What is the molarity of Ni2+ in the unknown?arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
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