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← JEE Main 2016 · 9 Apr · Shift 1
ChemistrySingle correct+4 / -1Question 58 of 90

JEE Main 2016 · 9 Apr · Shift 1, Chemistry question 58

Consider the following sequence for aspartic acid :

The pI (isoelectric point) of aspartic acid is :

Ionisation of aspartic acid: the fully protonated form (pK₁ = 1.88, α-CO₂H) → zwitterion with the side-chain CH₂CO₂H (pK_R = 3.65) → dianion form (pK₂ = 9.60, α-NH₃⁺) → fully deprotonated form, shown as Fischer projections.
Choose one option
Answer and worked solution

Official answer: D

The neutral zwitterion lies between the first two dissociations, so pI=pK1+pKR2=1.88+3.652≈2.77\text{pI}=\frac{pK_1+pK_R}2=\frac{1.88+3.65}2\approx2.77.

Topics:Amino acids and proteins
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