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Published on: 28/05/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
What are anomers ? Give two points of difference between two anomer of glucose.
2.
B-complex is an often prescribed Vitamin. What is complex about it ? What is its usefulness?
3.
Enumerate two reactions of glucose which cannot be explained by its open chain structure.
4.
How will you prove that all the carbon atoms of glucose are in straight chain?
5.
Enumerate the reactions of D-glucose which cannot be explained by the open chain structure.
1.
The pair of optical Isomers which differ in the orientation of H and OH gp only at C1 Carbon atom are called anomers. Difference between two anomers of glucose :
| α-D (+) glucose | β-D (+) glucose |
| (1) The specific rotation is + 111°. | (1) The specific rotation is + 19.2°. |
| (2) The – OH gp at C1 is below the plane. | (2) The – OH gp at C-1 is above the plane |
2.
It is a type of Vitamin which contains B1, B2, B6 and B12. It required to release energy from food and to promote healthy skin and muscles. Its deficiency causes beri-beri (Vitamin B1 ) and Anaemia (Vitamin B12).
3.
(i) Glucose does not give Schiff‘s Test although it contains aldehyde group.
(ii) Glucose does not form crystaline product with NaHSO3.
4.
The reaction of glucose with HI gives n-hexane and it proves that all sin-carbon-atom are in straight chain.
\(Ho{ CH }_{ 2 }-(CHOH)_{ 4 }-CHO\overset { HI }{ \underset { Heat }{ \longrightarrow } } { CH }_{ 3 }-\underset { n-hezane }{ ({ CH }_{ 2 })_{ 4 } } -{ CH }_{ 3 }\).
5.
Open structure of D-glucose could not explain the following reactions:
(i) Despite having the aldehyde group, glucose does not give Schiffs test and 2, 4-DNP test.
(ii) Glucose does not react with sodium hydrogen sulphite to form addition product.
(iii) The pentaacetate of glucose does not react with hydroxyl amine showing the absence of free -CHO group.
(iv) When glucose is heated with methanol in the presence of dry HCI gas, it forms two isomeric monomethyl derivatives known as \(\alpha \)-D-glucoside(m.p. = 165°C) and \(\beta \)-D-glucoside (m.p = 107°C). Since only one molecule of methanol is used for the formation of methyl glucoside, these must be hemiacetals.
These results show that glucose does not have open chain form structure.
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