Sugar which does not give reddish brown precipitate with Fehling’s reagent, is: [2024]
Sucrose
Lactose
Glucose
Maltose
(1)
Sucrose does not have an aldehydic or ketonic group. Hence sucrose is a non-reducing sugar and does not give reddish brown precipitate with Fehling’s reagent. All monosaccharides (such as glucose) are reducing sugars.
Match List I with List II
List I | List II | ||
A. | -Glucose and -Galactose | I. | Functional isomers |
B. | -Glucose and -Glucose | II. | Homologous |
C. | -Glucose and -Fructose | III. | Anomers |
D. | -Glucose and -Ribose | IV. | Epimers |
Choose the correct answer from the options given below: [2024]
A-III, B-IV, C-II, D-I
A-III, B-IV, C-I, D-II
A-IV, B-III, C-II, D-I
A-IV, B-III, C-I, D-II
(4)
A. Epimers are diastereomers that differ in configuration at only one chiral centre. -Glucose and -Galactose are C-4 epimers.
B. Anomers are epimers which differ in configuration at acetal or hemiacetal carbon. -D-glucose and -D-glucose are anomers.
C. Glucose is an aldohexose and fructose is a ketohexose. These are thus functional isomers.
D. Formula of glucose () differs by — unit with the formula of ribose (). So, Homologns
The incorrect statement about Glucose is: [2024]
Glucose remains in multiple isomeric form in its aqueous solution
Glucose is one of the monomer unit in sucrose
Glucose is an aldohexose
Glucose is soluble in water because of having aldehyde functional group.
(4)
1. Glucose exists as equilibrium mixture of -D-glucose, open chain structure and -D-glucose.
2. Sucrose is a disaccharide formed by glycosidic linkage between C1 of -D-glucose and C2 of -D-fructose.
3. Glucose is an aldohexose.
4. Water solubility of glucose is because it forms extensive hydrogen bonding with water.
Match List I with List II [2024]
LIST I | LIST II |
---|---|
A. Glucose// | I. Gluconic acid |
B. Glucose/ | II. No reaction |
C. Glucose// | III. -hexane |
D. Glucose/Bromine water | IV. Saccharic acid |
Choose the correct answer from the options given below:
A-IV, B-I, C-III, D-II
A-I, B-IV, C-III, D-II
A-II, B-IV, C-III, D-I
A-III, B-II, C-I, D-IV
(3)
(A) Despite having the aldehyde group, glucose does not give Schiff’s test and it does not form the hydrogensulphite addition product with . This shows glucose exists in cyclic structure.
Note: In question, should be replaced by .
Which of the following is the correct structure of L-Glucose? [2024]
(2)
The incorrect statement regarding the given structure is: [2024]
has 4-asymmetric carbon atom
will coexist in equilibrium with 2 other cyclic structures
can be oxidized to a dicarboxylic acid with water
despite the presence of –CHO does not give Schiff’s test
(3)
1.
2. Two cyclic forms of glucose exist in equilibrium with its open chain structure.
3. Bromine water is a mild oxidizing agent, it oxidizes easily oxidisable aldehyde group to carboxylic acid, but cannot oxidize alcoholic group.
4. Despite having the presence of –CHO group, glucose does not give Schiff’s test because under the reaction conditions, it exists as cyclic form and free –CHO group is thus not available.
Which of the following is the correct structure of L-fructose? [2025]
[IMAGE 443]
[IMAGE 444]
[IMAGE 445]
[IMAGE 446]
(3)
L-fructose is enantiomer of D-fructose.
[IMAGE 447]
Given below are two statements : [2025]
Statement I : Wet cotton clothes made of cellulose-based carbohydrate takes comparatively longer time to get dried than wet nylon polymer-based clothes.
Statement II : Intermolecular hydrogen bonding with water molecule is more in nylon-based clothes than in the case of cotton clothes.
In the light of above statements, choose the Correct answer from the options given below
Statement I is false but Statement II is true.
Statement I is true but Statement II is false.
Both Statement I and Statement II are true.
Both Statement I and Statement II are false.
(2)
Cellulose has more number of OH bonds, hence more hydrogen bonding with water and takes more time to dry.
Given below are two statements : [2025]
Statement I : D-(+)-glucose + D-(+)-fructose sucrose
sucrose D-(+)-glucose + D-(+)-fructose
Statement II : Invert sugar is formed during sucrose hydrolysis.
In the light of the above statements, choose the correct answer from the options given below –
Both Statement I and Statement II are true.
Statement I is false but Statement II are true.
Statement I is true but Statement II is false.
Both Statement I and Statement II are false.
(2)
Hydrolysis of sucrose gives D(+)-glucose and D(–)-fructose (not D(+)-fructose as given in statement I).
Identify the number of structure/s from the following which can be correlated to D-glyceraldehyde. [2025]
[IMAGE 448]
one
two
four
three
(4)
In Fischer structure keep the most oxidized carbon (i.e. CHO in given question) at top and all the carbons on vertical line. Now if OH of last chiral carbon is to the right, the compound is said to have ‘D’ configuration. Compounds A, B and D have ‘D’ configuration.
[IMAGE 449]