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Chapter 23: Question 23.45 (page 1248)

All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance forms of the amide groups are drawn, as is done for thymine at left. Redraw the hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in figure 23-24, using the polar resonance forms of the amides. Show how these forms help to explain why the hydrogen bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises between an electron-deficient hydrogen atom and electron-rich pair of nonbonding electrons.

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Answer:

Hydrogen bonding between guanine and cytosine

Hydrogen bonding between adenine and thymine

Step by step solution

01

Hydrogen bonding in adenine-thymine and guanine-cytosine pair:

Hydrogen bond arises between an electron-deficient hydrogen atom and electron-rich pair of non-bonding electrons. Adenine and guanine are purine bases whereas thymine and cytosine are pyrimidine bases. Electronegative atoms present in these bases have a negative charge or lone pair which is involved in hydrogen bonding with hydrogen and in each pair, one N-H is polarized more strongly because the nitrogen atom possesses a positive charge which further enhances the electronegativity of nitrogen.

Hydrogen bonding between guanine and cytosine

Hydrogen bonding between adenine and thymine

02

Reason for stronger hydrogen bonding between purine and pyrimidine bases:

Hydrogen bonds are created when hydrogen atom which is bonded to an electronegative atom approaches a nearby electronegative atom. These are characterised by strong intermolecular forces and more the electronegativity of hydrogen bond acceptor, more will be the hydrogen bond strength. In between the purine and pyrimidine base pairs, nitrogen atom possess positive charge and this will highly increase hydrogen bond acceptor strength and hydrogen bond strength. Negative charge on oxygen also increases hydrogen bond strength.

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Most popular questions from this chapter

Give an equation to show the reduction of Tollens reagent by maltose.

a) Draw D-allose, the C3 epimer of glucose.

b) Draw D-talose, the C2 epimer of D-galactose.

c) Draw D-idose, the C3 epimer of D-talose. Now compare your answers with Figure 23-3.

d) Draw the C4 鈥渆pimer鈥� of D-xylose. Notice that this 鈥渆pimer鈥� is actually an L-series sugar, and we have seen its enantiomer. Give the correct name for this L-series sugar.

Raffinose is a trisaccharide (C18H32O16) isolated from cottonseed meal. Raffinose does not reduce Tollens reagent, and it does not mutarotate. Complete hydrolysis of raffinose gives D-glucose, D-fructose, and D-galactose. When raffinose is treated with invertase, the products are D-fructose and a reducing disaccharide called melibiose. Raffinose is unaffected by treatment with a-galactosidase, but an 伪 -galactosidase hydrolyzes it to D-galactose and sucrose. When raffinose is treated with dimethyl sulfate and base followed by hydrolysis, the products are 2,3,4-tri-O-methylglucose, 1,3,4,6-tetra-O-methylfructose, and 2,3,4,6-tetra-O-methylgalactose. Determine the complete structures of raffinose and melibiose and give a systematic name for melibiose.

Fructose is found in many fruits. From memory, draw fructose in

  1. the Fischer projection of the open chain.
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  3. The Haworth projection of the most stable pyranose anomer

Question. (a) Figure 23-2 shows that the degradation of D-glucose gives D-arabinose, an aldopentose. Arabinose is most stable in its furanose form. Draw D-arabinofuranose.

(b) Ribose, the C2 epimer of arabinose, is most stable in its furanose form. Draw D-ribofuranose.

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