Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. They have been broadly classified into three groups - monosaccharides, oligosaccharides and polysaccharides. The carbohydrates may also be classified as either reducing or non-reducing sugars. An important monosaccharide glucose is an aldohexose and is correctly named as D(+)-glucose. It was assigned the open structure on the basis of its reactions with HI, \(\mathrm{NH_2OH}\), \(\mathrm{Br_2}\) water, \(\mathrm{(CH_3CO)_2O}\) and nitric acid.
Despite having the - CHO group, glucose does not give Schiff's test and hydrogen sulphite addition product with \(\mathrm{NaHSO_3}\). It was found that glucose forms a six-membered ring in which one of the - OH groups add to the - CHO group and form a cyclic hemiacetal structure.
Answer the following questions :
What type of carbohydrates are called reducing sugars ?
Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. They have been broadly classified into three groups - monosaccharides, oligosaccharides and polysaccharides. The carbohydrates may also be classified as either reducing or non-reducing sugars. An important monosaccharide glucose is an aldohexose and is correctly named as D(+)-glucose. It was assigned the open structure on the basis of its reactions with HI, \(\mathrm{NH_2OH}\), \(\mathrm{Br_2}\) water, \(\mathrm{(CH_3CO)_2O}\) and nitric acid.
Despite having the - CHO group, glucose does not give Schiff's test and hydrogen sulphite addition product with \(\mathrm{NaHSO_3}\). It was found that glucose forms a six-membered ring in which one of the - OH groups add to the - CHO group and form a cyclic hemiacetal structure.
Answer the following questions :
What type of carbohydrates are called reducing sugars ?
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Solution and Explanation
Sugars are grouped as reducing or non-reducing by their behaviour with mild oxidising agents. The two common reagents are Fehling's solution and Tollens' reagent.
Step 2: State the definition:
Carbohydrates that reduce Fehling's solution and Tollens' reagent are called reducing sugars.
They can do this because they have a free aldehydic or ketonic group, or a free hemiacetal -OH group that can open to give one.
They give a red precipitate of \(\mathrm{Cu_2O}\) with Fehling's solution and a silver mirror with Tollens' reagent.
Step 3: Examples:
Glucose, fructose, maltose and lactose are reducing sugars.
Sucrose is non-reducing, because the anomeric carbons of both its units are used in the glycosidic bond.
Final Answer:
Reducing sugars are carbohydrates that reduce Fehling's solution and Tollens' reagent because they have a free aldehyde, ketone or hemiacetal group.
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
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(ii) Secondary and tertiary amines
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(v) Aniline and N-methylanilineAccount for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
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Top CBSE CLASS XII Glucose Questions
- How do you explain the following ?
(a) Presence of an aldehyde group in glucose
(b) Presence of a primary alcoholic group in glucose - How do you explain the following ? (a) Presence of a carbonyl group in glucose (b) Presence of five --OH groups in glucose
- Assertion (A) : D (+) – Glucose is dextrorotatory in nature. Reason (R) : (+) represents dextrorotatory nature and D represents the configuration.
- How will you confirm the presence of five -- OH groups in a glucose molecule, which are attached to different carbon atoms ?
- How do you explain the presence of following in open structure of glucose ?
(a) all the six carbon atoms are in a straight chain.
(b) five \(-OH\) groups which are attached to different carbon atoms.
(c) an aldehyde group.
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