Chapter 27: Amino Acids, Peptides and Proteins

Amino Acids Answers

Qu 1:	The non-polar side chains are alkyl groups such as methyl, isopropyl and isobutyl.
	As we learnt much earlier, alkanes are comparatively unreactive as they don't contain any polar covalent bonds.
Qu 2:	L-cysteine is unusual amongst the 20 common L-amino acids as it has the R-configuration.
	If we assign the priorities to the groups, (review ?) we see that +NH3 > CH2SH > CO2- > H  (remember, first point of difference : S > O) (note: For the other L-amino acids, the side chain R groups is of lower priority than the carboxylate ( CO2- ) group).
	assign priorities (based on atomic weight at the first point of difference) convert to wedge-hash in order to safely manipulate the image... with low priority group at the back, the sense of high to low is clockwise so it is R Use the 3D model to help if you can't do the manipulation of paper or to confirm the answer.
Qu 3:	α-Amino acids differ because of the proximity of the N substituent to the carboxyl group.
	The ammonium group,  +NH3 is able to stabilise the conjugate base since it is an electron withdrawing group.
Qu 4:	Both of the N atoms are part of an aromatic system, but N1 is like a pyridine N atom whereas N2 is like a pyrrole N.
	Remember basicity can be analysed by thinking about lone pair availability (more available = more basic). For N1 the lone pair is in an sp2 hybrid orbital and is not part of the aromatic system. Hence the protonated form is still aromatic. For N2, the lone pair is in a p orbital and most importantly is involved in the aromatic sextet.  Protonation of this lone pair will create a non-aromatic system so resulting in an unfavourable loss of aromatic stability. Hence N2 is not very basic and protonation at N1 will be favoured. (see also Implications of Aromaticity)
Qu 5:	Isoelectronic point, pI, is the pH at which the amino acid is predominantly of neutral charge.
	(a)  6.0   the pI = 1/2 (pKa1 +  pKa2) = 1/2 ( 2 + 10 ) (b)  3.0   with the acidic side chain present, pI = 1/2 (pKa1 + pKa3) = 1/2 ( 2 + 4 )  (c)  9.0   with the basic side chain present, pI = 1/2 (pKa2 + pKa3) = 1/2 ( 10 + 8 )