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An IIT-JEE advanced (2012) chemistry question

I faced this question. But I am having some problems understanding the solution to this:

  1. How can the negative charge on the peptide vanish on $\pu{pH = 7}$? Suppose, we have some $\ce{H+}$ donor on either $\ce{R1}$ or $\ce{R2}$. Then the $\ce{H+}$ donor can make the carboxyl radical neutral. But by donating a $\ce{H+}$ ion, the donor becomes negatively charged. Maybe, the donor can take the $\ce{H+}$ from the $\ce{NH3+}$ radical. But the result, a completely neutral tetrapeptide, doesn't satisfy the condition of the question.

Need a hint.

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    $\begingroup$ Basically, positive charge at pH=7 means there are more basic groups than acidic groups. This would make the peptide accept more protons than it ejects, giving a net positive charge $\endgroup$
    – Pritt says Reinstate Monica
    May 22, 2017 at 2:21
  • $\begingroup$ @Pritt_Balagopal That means I misunderstood the question. $\endgroup$
    – Mockingbird
    May 22, 2017 at 2:25
  • $\begingroup$ @Pritt Balagopal Then I assume option 4,5,6,9 are correct. $\endgroup$
    – Mockingbird
    May 22, 2017 at 2:26
  • $\begingroup$ In theory, yes. Although option 6 is not quite sure. Amides are soo weakly basic, I doubt if that counts too. $\endgroup$
    – Pritt says Reinstate Monica
    May 22, 2017 at 2:29
  • $\begingroup$ @pritt option 6 has no amide. are u talking about option 5? I am confused between option 5 and option 8. -OH radical isn't that acidic. So the amine group alone can make the tetrapeptide basic? $\endgroup$
    – Mockingbird
    May 22, 2017 at 2:54

1 Answer 1

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To sum up: of the side chains listed, only the aminoalkyl ones will be positively charged at pH 7, so four peptides -- 4, 6, 8 and 9 -- will have a net positive charge at pH 7.

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