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Chapter 23: Q23P (page 1226)

Show the products that result from hydrolysis of amygdalin in dilute acid. Can you suggest why amygdalin might be toxic to tumor (and possibly other) cells?

Short Answer

Expert verified

is released from amygdalin. It is potent and cytotoxic and neurotoxic in nature.

Step by step solution

01

Step-1. Hydrolysis of amygdalin in dilute acid:

Amygdalin is a naturally occurring chemical compound and is a cyanogenic glycoside as toxic cyanide ion can get released from it on its hydrolysis. Amygdalin on hydrolysis with dilute acid gives two molecules of glucose and benzaldehyde cyanohydrin as the products and further benzaldehyde cyanohydrin on hydrolysis gives hydrogen cyanide as the product.

Amygdalin Hydrogen cyanide

02

Step-2. Cyanide poisoning from amygdalin:

Amygdalin can cause cyanide poisoning as after its ingestion it is hydrolysed to give a molecule of hydrogen cyanide and two molecules of glucose. Enzymatic degradation of amygdalin occurs in three stages. Hydrogen cyanide is toxic to neurological system as well as it is cytotoxic meaning, cell killing. Due to cyanide poisoning, liver may get damaged and blood pressure could get drop very low.

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Most popular questions from this chapter

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.

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.

In 1891, Emil Fischer determined the structures of glucose and seven other D-aldohexoses using only simple chemical reactions and clever reasoning about stereochemistry and symmetry. He received the Nobel Prize for this work in 1902. Fischer has determined that D-glucose is an aldohexose, and he used Ruff degradation to degrade it to (+)-glyceraldehyde. Therefore, the eight D-aldohexose structures shown in Figure 23-3 are the possible structures for glucose.

Pretend that no names are shown in Figure 23-3 except for glyceraldehyde, and sue the following results to prove which of these structures represent glucose, mannose, arabinose, and erythrose.

(a)Upon Ruff degradation, glucose and mannose gives the same aldopentose: arabinose.Nitric acid oxidation of arabinose gives an optically active aldaric acid. What are the two possible structures of arabinose?

(b) Upon Ruff degradation, arabinose gives the aldotetrose erythrose. Nitric acid oxidation of erythrose gives an optically inactive aldaric acid, meso-tartaric acid. What is the structure of erythrose?

(c) Which of the two possible structures of arabinose is correct? What are the possible structures of glucose and mannose?

(d) Fischer鈥檚 genius was needed to distinguish between glucose and mannose. He developed a series of reactions to convert the aldehyde group of an aldose to an alcohol while converting the terminal alcohol to an aldehyde. In effect, he swapped the functional groups on the ends. When he interchanged the functional groups on D-mannose, he was astonished to find that the product was still D-mannose. Show how this information completes the proof of the mannose structure, and show how it implies the correct glucose structure.

(e) When Fischer interchanged the functional groups on D-glucose, the product was an unnatural L sugar. Show which unnatural sugar he must have formed, and show how it completes the proof of the glucose structure.

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.

Ruff degradation of D-arabinose gives D-erythrose. The Kiliani-Fischer synthesis converts D-erythrose to a mixture of D-arabinose and D-ribose. Draw out these reactions and give the structure of D-ribose.
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