a) The first step in the preparation of Hagemann’s ester, an aldol-like condensation between ethylacetoacetate and formaldehyde to produce a α, β-unsaturated product and the structure of the product is to be shown. b) The structure of the product obtained in the second step which is a Michael reaction between the product of the first step and ethyl acetoactate is to be given. Concept introduction: In the first step in the preparation of Hagemann’s ester involving a mixed aldol like condensation reaction, ethyl acetoacetate yields an enolate anion which attacks formaldehyde to yield an α, β-unsaturated product. In the second step α, β-unsaturated product obtained in the first step acts as a donor and ethyl acetoacetate acts as the acceptor (Michael reaction) to yield the product. To give: a) The first step in the preparation of Hagemann’s ester, an aldol-like condensation between ethylacetoacetate and formaldehyde to produce α, β-unsaturated product and the structure of the product. b) The structure of the product obtained in the second step which is a Michael reaction between the product of the first step and ethyl acetoactate.

Question

LTS Solid: AT=Te-Ti Trial 1 Trial 2 Trial 3 Average ΔΗ Mass water, g 24.096 23.976 23.975 Moles of solid, mol 0.01763 001767 0101781 Temp. change, °C 2.9°C 11700 2.0°C Heat of reaction, J -292.37J -170.473 -193.26J AH, kJ/mole 16.58K 9.647 kJ 10.85 kr 16.58K59.64701 KJ mol 12.35k Minimum AS, J/mol K 41.582 mol-k Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn & Show your calculations for: AH in J and then in kJ/mole for Trial 1: qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J qsin = qrxn = 292.35 292.37J AH in J = 292.375 0.2923kJ 0.01763m01 =1.65×107 AH in kJ/mol = = 16.58K 0.01763mol mol qrx Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con Kelvin.) AS=AHIT (1.65×10(9.64×103) + (1.0 Jimai

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Question

Chapter 23.SE, Problem 71AP

Interpretation Introduction

Interpretation:

a) The first step in the preparation of Hagemann’s ester, an aldol-like condensation between ethylacetoacetate and formaldehyde to produce a α, β-unsaturated product and the structure of the product is to be shown.

b) The structure of the product obtained in the second step which is a Michael reaction between the product of the first step and ethyl acetoactate is to be given.

Concept introduction:

In the first step in the preparation of Hagemann’s ester involving a mixed aldol like condensation reaction, ethyl acetoacetate yields an enolate anion which attacks formaldehyde to yield an α, β-unsaturated product.

In the second step α, β-unsaturated product obtained in the first step acts as a donor and ethyl acetoacetate acts as the acceptor (Michael reaction) to yield the product.

To give:

a) The first step in the preparation of Hagemann’s ester, an aldol-like condensation between ethylacetoacetate and formaldehyde to produce α, β-unsaturated product and the structure of the product.

b) The structure of the product obtained in the second step which is a Michael reaction between the product of the first step and ethyl acetoactate.

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