(a) In the following reaction scheme, draw the structures of the missing intermediates 5-8 and the product 9. Ph H₂N i COOH Me Me O Me DCC, DMAP CH₂Cl₂ Me Me 7 NH₂ NaBH₁ F₂CCOOH CH₂Cl₂ 20% piperidine CH₂Cl₂

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Chapter1: Chemical Foundations
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(a) In the following reaction scheme, draw the structures of the missing intermediates 5-8 and
the product 9.
Ph
H₂N
COOH
Me
Me
O Me
DCC, DMAP
CH₂Cl₂
Me
Me
7
H₂N
NH₂
NaBH₁
F₂CCOOH
CH₂Cl₂
(b) Using the tripeptide below, illustrate the structure as it appears in aqueous solution at the
isoelectric point and explain what this term means.
Under what conditions will the net charge of the tripeptide in solution become positive, and
under what conditions will it become negative?
OH
Me
HN-
Me
20%
piperidine
CH₂Cl₂
LOH
Transcribed Image Text:(a) In the following reaction scheme, draw the structures of the missing intermediates 5-8 and the product 9. Ph H₂N COOH Me Me O Me DCC, DMAP CH₂Cl₂ Me Me 7 H₂N NH₂ NaBH₁ F₂CCOOH CH₂Cl₂ (b) Using the tripeptide below, illustrate the structure as it appears in aqueous solution at the isoelectric point and explain what this term means. Under what conditions will the net charge of the tripeptide in solution become positive, and under what conditions will it become negative? OH Me HN- Me 20% piperidine CH₂Cl₂ LOH
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