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Chapter 22: Problem 523

4 Account for the following, drawing all pertinent stereo-chemical formulas, (a) 1-Chloro-2-methylaziridine was prepared in two isomeric forms separable at \(25^{\circ}\) by (b) The reaction of ordinary gas chromatography, \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{C}=\mathrm{NCH}_{3}\) with \(\mathrm{R}-(+)-2-\) phenylperoxypropionic acid gave a product, \(\mathrm{C}_{14} \mathrm{H}_{13} \mathrm{ON}\), with \([\alpha]+12.5^{\circ}\), which showed no loss of optical activity up to (at least) \(90^{\circ}\).

Short Answer

Expert verified
<[:330,0.6]-[1,0.8]?(-[:90,0.6]CH_3))} \] Enantiomer 2 (S): \[ \hspace{2cm} \chemfig{*6(=-=-(-[:45]*3(=*@{a}(-Cl)-=@{b}(-[,0.6]CH_3)))=-)} \qquad \chemfig{*3(>[:45,0.6]Cl)-[1,0.8]?<[:330,0.6]-[1,0.8]?(-[:90,0.6]CH_3))} \] (b) The product of the reaction of ordinary gas chromatography, (C₆H₅)₂C=CH₃, with R-(+)-2-phenylperoxypropionic acid is: \[ \hspace{2cm} \chemfig{*6(-=-=*6(-=@{c}(-)(-[@{r,1}]O-[:0]*6(-=(-[@{s}]?)->=--))-*6(-=-=-=))-=)} \hspace{2cm} \] It has a chiral center with R absolute configuration and maintains its optical activity up to 90°C.

Step by step solution

01

Isomeric forms of 1-Chloro-2-methylaziridine

1-Chloro-2-methylaziridine can exist in two isomeric forms due to the presence of a chiral carbon atom in its molecule. The two isomers are called enantiomers. Here are the stereochemical formulas of the enantiomers of 1-Chloro-2-methylaziridine: Enantiomer 1 (R): \[ \hspace{2cm} \chemfig{*6(=-=-(-[:45]*3(=*@{a}(-Cl)-=@{b}(-[,0.6]CH_3)))=-)} \qquad \chemfig{*3(>[:45,0.6]Cl)-[1,0.8]?<[:330,0.6]-[1,0.8]?(-[:90,0.6]CH_3))} \] Enantiomer 2 (S): \[ \hspace{2cm} \chemfig{*6(=-=-(-[:45]*3(=*@{a}(-Cl)-=@{b}(-[,0.6]CH_3)))=-)} \qquad \chemfig{*3(>[:45,0.6]Cl)-[1,0.8]?<[:330,0.6]-[1,0.8]?(-[:90,0.6]CH_3))} \] (b)
02

Reaction of ordinary gas chromatography with R-(+)-2-phenylperoxypropionic acid

The reaction of ordinary gas chromatography, (C₆H₅)₂C=CH₃, with R-(+)-2-phenylperoxypropionic acid results in the formation of a product with a molecular formula of C₁₄H₁₃ON. The product shows no loss of optical activity up to 90°C. Here is the stereochemical formula of the product: \[ \hspace{2cm} \chemfig{*6(-=-=*6(-=@{c}(-)(-[@{r,1}]O-[:0]*6(-=(-[@{s}]?)->=--))-*6(-=-=-=))-=)} \hspace{2cm} \] In this formula, the product has a chiral center and maintains its optical activity up to 90°C. In this case, the absolute configuration of the chiral center is R, as the reactant R-(+)-2-phenylperoxypropionic acid is used.

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Most popular questions from this chapter

Name the following structures: (a) \(\mathrm{CH}_{3}-\mathrm{NH}-\mathrm{C}_{2} \mathrm{H}_{5} ;\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{NH}-\mathrm{C}_{2} \mathrm{H}_{5}\) (c) \(\mathrm{NH}_{2}-\left(\mathrm{CH}_{2}\right)_{3}-\mathrm{NH}_{2}\) (d) \(\left(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\right)_{3} \mathrm{~N}\) (e) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{NH}_{2} ;\) (f) \(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}\) (g) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{NCH}_{3}\); (h) \(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{NH}-\mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5}\)

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