Why there is a difference in yield when alanine and valine is prepared via amidomalonate synthesis need to be explained. Concept introduction: Amidomalonate synthesis is a process in which a halide is converted to a aminoacid with two new carbon atoms (comes from amidomalonate). By use of this method a new alkyl groups can be introduced. A general scheme of this synthesis is shown below, Steps involved in amidomalonater synthesis are, Deprotonation of α carbon in amidomalonate Alkylation of enolate Hydrolysis of ester using aqueous acid Decarboxylation at high temperature Please remember that if the primary alkyl halide reactivity is more than secondary alkyl halide due to less steric hindrance.

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Chapter 25, Problem 57PP

Interpretation Introduction

Interpretation:

Why there is a difference in yield when alanine and valine is prepared via amidomalonate synthesis need to be explained.

Concept introduction:

Amidomalonate synthesis is a process in which a halide is converted to a aminoacid with two new carbon atoms (comes from amidomalonate). By use of this method a new alkyl groups can be introduced. A general scheme of this synthesis is shown below,

Steps involved in amidomalonater synthesis are,

Deprotonation of α carbon in amidomalonate

Alkylation of enolate

Hydrolysis of ester using aqueous acid

Decarboxylation at high temperature

Please remember that if the primary alkyl halide reactivity is more than secondary alkyl halide due to less steric hindrance.

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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