Amino acids are molecules that contain $$\(-\mathrm{NH}_{2}\)$$ and -COOH functional groups. Glycine, $$\(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH}\)$$, is the simplest stable amino acid. Fig. 6.1 shows two syntheses starting with glycine. (i) State the essential conditions for reaction 1. ................................................................................................................................. (ii) Identify the reagent used in reaction 2 . ................................................................................................................................. (iii) Draw the structure of the organic product $$\(\mathbf{U}\)$$ that forms when hippuric acid reacts with an excess of $$\(\mathrm{LiAlH}_{4}\)$$ in reaction 3 . (iv) A molecule of phenylalanine, $$\(\mathbf{R}\)$$, can react with a molecule of glycine to form two dipeptides, $$\(\mathbf{S}\)$$ and $$\(\mathbf{T}\)$$. $$\(\mathbf{S}\)$$ and $$\(\mathbf{T}\)$$ are structural isomers. Draw the structures of these dipeptides. The peptide bond formed should be shown fully displayed.
Exam No:9701_m24_qp_42 Year:2024 Question No:6(b)
Answer:
Knowledge points:
13.2.1.1 homologous series
13.2.1.2 saturated and unsaturated
13.2.1.3 homolytic and heterolytic fission
13.2.1.4 free radical, initiation, propagation, termination (the use of arrows to show movement of single electrons is not required)
13.2.1.5 nucleophile, electrophile, nucleophilic, electrophilic
13.2.1.6 addition, substitution, elimination, hydrolysis, condensation
13.2.1.7 oxidation and reduction (in equations for organic redox reactions, the symbol [O] can be used to represent one atom of oxygen from an oxidising agent and the symbol [H] one atom of hydrogen from a reducing agent)
13.2.2.1 free-radical substitution
13.2.2.2 electrophilic addition
13.2.2.3 nucleophilic substitution
13.2.2.4 nucleophilic addition (in organic reaction mechanisms, the use of curly arrows to represent movement of electron pairs is expected; the arrow should begin at a bond or a lone pair of electrons)
Solution:
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