Case study (4 Marks)

Question 14 Marks

Read the passage given below and answer the following questions: Glucose is known as dextrose because it occurs in nature as the optically active dextrorotatory isomer. lt is essential constituent of human blood. The blood normally contains $65$ to $110mg$ of glucose per $100mL \ ($hence named Blood sugar$)$. The level may be much higher in diabetic persons. The urine of diabetic persons also contain considerable amount of glucose. ln combined form, it occurs in cane sugar and polysaccharides such as starch and cellulose. Glucose has an aldehyde group $(-\ce{CHO}),$ one primary alcoholic group $(-\ce{CH_2OH})$ and four secondary alcoholic groups $(-\ce{CHOH})$ in their structure. Due to the presence five hydroxyl groups $(-OH)$. glucose undergoes acetylation. Glucose also undergoes oxidation with mild oxidising agents like bromine water as well as with strong oxidising agents like nitric acid. Since glucose is readily oxidised, it acts as a strong reducing agent and reduces Tollen's reagent and Fehling solution. Glucose exists in two crystalline forms: $\alpha - D -$ glucose and $\beta - D - $ glucose. If either of the two forms is dissolved in water and allowed to stand, the specific rotation of the solution changes gradually, until a constant value is obtained. This change is called mutarotation. A statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion : A diabetic person carries a packet of glucose with him always.

Reason : Glucose increases the blood sugar level almost instantaneously.

Assertion : On oxidation with nitric acid, glucose as well as gluconic acid both yield saccharic acid.

Reason : The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free $-\text{CHO}$ group.

Assertion : Glucose reacts with acetyl chloride to form pentaacetyl glucose.

Reason : The formation of pentaacetyl derivative confirms the presence of five $-OH$ groups in glucose.

Assertion : A certain compound gives negative test with ninhydrin and positive test with Benedict's solution, the compound is an amino acid.

Reason : Glucose is a monosaccharide.

Assertion : The rapid interconversion of $\alpha - D - $ glucose and $\beta - D -$ glucose in solution is known as racemisation.

Reason : Hydrolysis reaction will take place when a mineral acid is treated with sugar.

Answer

$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.

$\ \text{CHO}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{COOH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{COOH}\\\ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\$\text{CHOH})_4\xrightarrow[\text{HNO}_3]{[\text{O}]}(\text{COOH})_4\xleftarrow[\text{[O]}]{\text{HNO}_3}(\text{CHOH})_4\\\ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \text{CH}_2\text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{COOH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_2\text{OH}\\\ \text{Glucose}\ \ \ \ \ \ \ \ \ \text{Saccharic acid}\ \ \ \ \ \ \ \text{Gluconic acid}$
Strong oxidising agents like nitric acid oxidises both the terminal $-\ce{CHO}$ and $-\ce{CH_2OH}$ groups of glucose to give the dibasic acid, saccharic acid.

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.

$(d)$ Assertion is wrong statement but reason is correct statement.

If a certain compound gives negative test with ninhydrin and positive test with Benedict's solution then the compound should be a monosaccharide.

$(d)$ Assertion is wrong statement but reason is correct statement.

The rapid interconversion of $\alpha -$ Dvgfucose and $\beta - D - $ glucose in solution is known as mutarotation.
Sugar gets hydrolysed with mineral acids.
$\text{C}_{12}\text{H}-{22}+\text{H}_2\text{O}\xrightarrow{\text{H}^+}\text{C}_6\text{H}_{12}\text{O}_6+\text{C}_6\text{H}_{12}\text{O}_6\\\ \ \ \ \text{Sugar}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Glucose}\ \ \ \ \ \text{Fructose}$

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Question 24 Marks

Read the passage given below and answer the following questions:
When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
The denaturation causes change in secondary and tertiary structures but primary structures remains intact. Examples of denaturation of protein are coagulation of egg white on boiling, curdling of milk, formation of cheese when an acid is added to milk.
The following questions are multiple choice questions. Choose the most appropriate answer:

Mark the wrong statement about denaturation of proteins.

The primary structure of the protein does not change.
Globular proteins are converted into fibrous proteins.
Fibrous proteins are converted into globular proteins.
The biological activity of the protein is destroyed.

Which structure(s) of proteins remains(s) intact during denaturation process?

Both secondary and tertiary structures.
Primary structure only.
Secondary structure only.
Tertiary structure only.

$\alpha$-helix and $\beta$-pleated structures of proteins are classified as:

Primary structure.
Secondary structure.
Tertiary structure.
Quaternary structure.

Cheese is a:

Globular protein.
Conjugated protein.
Denatured protein.
Derived protein.

Secondary structure of protein refers to:

Mainly denatured proteins and structure of prosthetic groups.
Three-dimensional structure, especially the bond between amino acid residues that are distant from each other in the polypeptide chain.
Linear sequence of amino acid residues in the polypeptide chain.
Regular folding patterns of continuous portions of the polypeptide chain.

Answer

(b) Primary structure only.

(b) Secondary structure.

Explanation:
Cheese is a denatured protein.

(d) Regular folding patterns of continuous portions of the polypeptide chain.

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Question 34 Marks

Read the passage given below and answer the following questions: The sequence of bases along the DNA and RNA chain establishes its primary structure which controls the specific properties of the nucleic acid. An RNA molecule is usually a single chain ofribose-containing nucleotide. On the basis of X-ray analysis of DNA, J.D., Watson and EH.C. crick (shared noble prize in 1962) proposed a three dimensional secondary structure for DNA. DNA molecule is a long and highly complex, spirally twisted, double helix, ladder like structure. The two polynucleotide chains or strands are linked up by hydrogen bonding between the nitrogeneous base molecules of their nucleotide monomers. Adenine (purine) always links with thymine (pyrimidine) with the help of two hydrogen bonds and guanine (purine) with cytosine (pyrimidine) with the help of three hydrogen bonds. Hence, the two strands extend in opposite directions, i.e., are antiparallel and complimentary.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion: DNA molecules and RNA molecules are found in the nucleus of a cell.

Reason: There are two types of nitrogenous bases, purines and pyrimidines. Adenine (A) and guanine (G) are substituted purines; cytosine (C), thymine (T) and uracil (U) are substituted pyrimidines.

Assertion: In both DNA and RNA, heterocyclic base and phosphate ester linkages are at C-1' and C-5' respectively of the sugar molecule.

Reason: Nucleotides and nucleosides mainly differ from each other in presence of phosphate units.

Assertion: The backbone of RNA molecule is a linear chain consisting of an alternating units of a heterocylic base, D-ribose and a phosphate.

Reason: The segment of DNA which acts as the instruction manual for the synthesis of protein is ribose.

Assertion: The double helical structure of DNA was proposed by Emil Fischer.

Reason: A nucleoside is an N-glycoside of heterocyclic base.

Assertion: In DNA, the complementary bases are, adenine and guanine; thymine and cytosine.

Reason: The phenomenon of mutation is chemical change in DNA molecule.

Answer

(d) Assertion is wrong statement but reason is correct statement.

Explanation:
DNA occurs in nucleus of the cell while RNA is found mainly in cytoplasm of the cell.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

Nucleosides contain only sugar and a base whereas nucleotides contain sugar, base and a phosphate group as well.

Explanation:
The segment of DNA which acts as the instruction mannual for the synthesis of protein is gene.

(d) Assertion is wrong statement but reason is correct statement.

Explanation:
The double helical structure of DNA was proposed by Watson and Crick.

(d) Assertion is wrong statement but reason is correct statement.

Explanation:
In DNA, the complementary bases are, adenine and thymine; guanine and cytosine.

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Question 44 Marks

Read the passage given below and answer the following questions: Carbohydrates are polyhydroxy aldehydes and ketones and those compounds which on hydrolysis give such compounds are also carbohydrates. The carbohydrates which are not hydrolysed are called monosaccharides. Monosaccharides with aldehydic group are called aldose and those which free ketonic groups are called ketose. Carbohydrates are optically active. Number of optical isomers $= 2^n$ Where $n =$ numberofasymmetric carbons. Carbohydrates are mainlysynthesised by plants during photosynthesis. The monosaccharides give the characteristic reactions of alcohols and carbonyl group $($aldehydes and ketones$).$ It has been found that these monosaccharides exist in the form of cyclic structures. In cyctization, the $-OH$ groups $($generally $C_5$ or $C_4$ in aldohexoses and $C_5$ or $C_6$ in ketohexoses$)$ combine with the aldehyde or keto group. As a result, cyclic structures of five or six membered rings containing one oxygen atom are formed, e.g., glucose forms a ring structure. Glucose contains one aldehyde group, one $IO$ alcoholic group and four $2^\circ$ alcoholic groups in its open chain structure. The following questions are multiple choice questions. Choose the most appropriate answer:

First member of ketos sugar is:

Ketotriose.
Ketotetrose.
Ketopentose.
Ketohexose.

In $\ce{CH_2OHCHOHCHOHCHOHCHOHCHO,}$ the number of optical isomers will be:

$16$
$8$
$32$
$4$

Some statements are given below:

Glucose is aldohexose.
Naturally occurring glucose is dextrorotatory.
Glucose contains three chiral centres.
Glucose contains one $1^\circ $ alcoholic group and four $2^\circ$ alcoholic groups.

Among the above, correct statements are:

$1$ and $2$ only
$3$ and $4$ only
$1, 2$ and $4$ only
$1, 2, 3$ and $4$

Two hexoses fonn the same osazone, find the correct statement about these hexoses.

Both of them must be aldoses.
They are epimers at $C-3.$
The carbon atoms $I$ and $2$ in both have the same configuration.
The carbon atoms $3, 4$ and $5$ in both have the same configuration.

Which of the following reactions of glucose can be explained only by its cyclic structure?

Glucose forms cyanohydrin with $\text{HCN}.$
Glucose reacts with hydroxylamine to form an oxime.
Pentaacetate of glucose does not react with hydroxylamine.
Glucose is oxidised by nitric acid to gluconic acid.

Answer

$(a)$ Ketotriose.

$(a)\ 16$

$(c)\ 1, 2$ and $4$ only

Glucose contains four chiral centres.

$(d)$ The carbon atoms $3, 4$ and $5$ in both have the same configuration.

In the formation of osazone, $C-1$ and $C-2$ react with phenylhydrazine to form phenylhydrazone. If $C-3, C-4, C-5$ have same configuration they will form same osazone even if they differ in configuration at $C-1$ or $C-2.$

$(c)$ Pentaacetate of glucose does not react with hydroxylamine.

Pentacetate of glucose does not react with hydroxylamine showing absence of free $-\text{CHO}$ group. This cannot be explained by open structure of glucose.

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Question 54 Marks

Read the passage given below and answer the following questions: Proteins are high molecular mass complex biomolecules of amino acids. The important proteins required for our body are enzymes, hormones, antibodies, transport proteins, structural proteins, contractile proteins etc. Except for glycine, all $\alpha$-amino acids have chiral carbon atom and most of them have L-configuration. The amino acids exists as dipolar ion called zwitter ion, in which a proton goes from the carboxyl group to the amino group. A large number of $\alpha$-amino acids are joined by peptide bonds forming polypeptides. The peptides having very large molecular mass (more than 10,000) are called proteins. The structure of proteins is described as primary structure giving sequence of linking of amino acids; secondary structure giving manner in which polypeptide chains are arranged and folded; tertiary structure giving folding, coiling or bonding polypeptide chains producing three dimensional structures and quaternary structure giving arrangement of sub-units in an aggregate protein molecule. A statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion: Except glycine, all naturally occurring CL-amino acids are optically active.

Reason: All naturally occurring CL-amino acids, except glycine, has at least one asymmetric carbon.

Assertion: All amino acids are optically active.

Reason: Amino acids contain asymmetric carbon atoms.

Assertion: In $\alpha$-helix structure, intramolecular H-bonding takes place whereas in $\beta$-pleated structure, intermolecular H-bonding takes place.

Reason: An egg contains a soluble globular protein called albumin which is present in the white part.

Assertion: Secondary structure of protein refers to regular folding patterns of continuos portions of the polypeptide chain.

Reason: Out of 20 amino acids, only 12 amino acids can be synthesised by human body.

Assertion: The helical structure of protein is stabilised by intramolecular hydrogen bond between -NH and carbonyl oxygen.

Reason: Sanger's reagent is used for the identification of N-tenninal amino acid of peptide chain.

Answer

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(d) Assertion is wrong statement but reason is correct statement.

Explanation:
All amino acids except glycine are optically active because they contain, asymmetric carbon atom. They exist in both D and L-forms. Most naturally occurring amino acids have L-configuration.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:
In $\alpha$-helix structure, the formation of hydrogen bonds takes place between -CO- and -NH groups, whereas in $\beta$-pleated structure, hydrogen bonds are formed between amide groups of two different chains.

Explanation:
Out of 20 amino acids, only 10 amino acids can be synthesised by human body.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

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Question 64 Marks

Read the passage given below and answer the following questions:
When a solution of an et$-$amino acid is placed in an electric field depending on the $pH$ of the medium, following three cases may happen.

In alkaline solution, $CL-$amino acids exist as anion $II,$ and there is a net migration of amino acid towards the anode.
In acidic solution, a$-$amino acids exist as cation $III,$ and there is a net migration of amino acid towards the cathode.
If $II$ and $III$ are exactly balanced there is no net migration; under such conditions any one molecule exists as a positive ion and as a negative ion for exactly the same amount of time, and any small movement in the direction of one electrode is subsequently cancelled by an equal movement back toward the other electrode. The $pH$ of the solution in which a particular amino acid does not migrate under the influence of an electric field is called the is oelectric point of that amino acid.

The following questions are multiple choice questions. Choose the most appropriate answer:

Arrange in order of increasing acid strengths.

$\ce{X > Z > Y}$
$\ce{Z < X < Y}$
$\ce{X > Y > Z}$
$\ce{Z > X > Y}$

In aqueous solutions, amino acids mostly exist as:

$\ce{NH_2 - CHR - COOH}$
$\ce{NH_2 - CHR - COO^-}$
$\stackrel{+}{\hbox{ N}}\text{H}_3\text{CHRCOOH}$
$\text{H}_3\stackrel{{+}}{\hbox{N}}\text{CHRCOO}^-$

Amino acids are least soluble:

at $\ce{pH 1}$
at $\ce{pH 7}$
At their isoelectric points.
None of these.

The $\text{pK}_{\text{a}_1}$ and $\text{pK}_{\text{a}_2}$ of an amino acid are $2.3$ and $9.7$ respectively. The is oelectric point of the amino acid is:

$12.0$
$7.4$
$6.0$
$3.7$

A tripeptide $(X)$ on partial hydrolysis gave two dipeptides $\ce{Cys-Gly}$ and $\ce{Glu-Cys.}$ Identify the tripeptide.

$\ce{Glu-Cys-Gly}$
$\ce{Gly-Glu-Cys}$
$\ce{Cys-Gly-Glu}$
$\ce{Cys-Glu-Gly}$

Answer

$(a) \ce{X > Z > Y}$

Carboxytic acids are stronger acids than $-\stackrel{{+}}{\hbox{N}}\text{H}_3,$ therefore $X$ is the strongest acid. Since $–\ce{COOH}$ has $-I$ effect which decreases with distance therefore, effect is more pronounced on $Z$ than on $Y.$
As a result $Z$ is more acidic than $Y,$ therefore, overall order of increasing acid strength is $\ce{X > Z > Y.}$

$(d)\ \text{H}_3\stackrel{{+}}{\hbox{N}}\text{CHRCOO}^-$

In aqueous solutions, amino acids mostly exist as zwitter ion or dipolar ion.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{R}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\stackrel{{+}}{\hbox{N}}\text{H}_3-\text{CH}-\text{COO}^-\\\ \ \ \ \ \ \ \ \ \ \text{Zwitter ion}$

$(c)$ At their isoelectric points.

Amino acids are least soluble at their isoelectric points. At a specific $pH,$ called isoelectric point, the positive and negative charges balance each other and the net charge becomes zero.
If there is a charge, the amino acid prefers to interact with water, rather than other amino acid molecules, this charge makes it more soluble.

$(c) 6.0$

Isoelectric point $=\frac{2.3+9.7}{2}=6$

$(a) \ce{Glu-Cys-Gly}$

Since the tripeptide on hydrolysis gave two dipeptides $\ce{Glu-Cys}$ and $\ce{Cys-Gly,}$
hence, cysteine must be in between glutamic acid and glycine as given below:
$\ \ \ \ \ \ \ \ \ \text{CH}_2\text{CH}_2\text{COOH}\\\ \ \ \ \ \ \ \ \ \ |\\\stackrel{{+}}{\hbox{N}}\text{H}_3\text{CH}-\text{C}-\text{NHCH}-\text{CNHCH}_2\text{CO}^-\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\ \ \ \ \ \ \text{HSCH}_2\ \ \text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Glu-Cys-Gly}$

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Question 74 Marks

Read the passage given below and answer the following questions:
Carbohydrates can exist in either of two conformations, as determined by the orientation of the hydroxyl group about the asymmetric carbon farthest from the carbonyl.

By convention, a monosaccharide is said to have $D-$configuration if the hydroxyl group attached to the asymmetric carbon atom adjacent to the $-\ce{CH_2OH}$ group is on the right hand side irrespective of the positions of the other hydroxyl groups. On the other hand, the molecule is assigned $L-$configuration if the $-OH$ group attached to the carbon adjacent to the $- \ce{CH_2OH}$ group is on the left hand side.
The following questions are multiple choice questions. Choose the most appropriate answer:

$D-$Glyceraldehyde and $L-$Glyceraldehyde are:

Epimers.
Enantiomers.
Anomers.
Conformational diasteriomers.

Which of the following monosaccharides, is the majority found in the human body?

$D-$type.
$L-$type.
Both of these.
None of these.

The two functional groups present in a typical carbohydrate are:

$-\ce{OH}$ and $-\ce{COOH}$
$-\ce{CHO}$ and $-\ce{COOH}$
$ > \ce{C= O}$ and $-\ce{OH}$
$-\ce{OH}$ and $-\ce{CHO}$

Monosaccharides contain:

Always six carbon atoms.
Always five carbon atoms.
Always four carbon atoms.
May contain $3$ to $7$ carbon atoms.

The correct corresponding order of names of four aldoses with configuration given below respectively, is:

$L-$erythrose, $L-$threose, $L-$erythrose, $D-$threose.
$D-$threose, $D-$erythrose, $L-$threose, $L-$erythrose.
$L-$erythrose, $L-$threose, $D-$erythrose, $D-$threose.
$D-$erythrose, $D-$threose, $L-$erythrose, $L-$threose.

Answer

$(b)$ Enantiomers.

$(a) D-$type.

$\ce{(c) > C= O}$ and $-\ce{OH}$

Carbohydrates are essentially polyhydroxy aldehydes and polyhydroxy ketones.
Thus, the two functional groups present are $ >C =O ($aldehyde or ketone$)$ and $-OH.$

$(d)$ May contain $3$ to $7$ carbon atoms.

$(d) D-$erythrose, $D-$threose, $L-$erythrose, $L-$threose.

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Question 84 Marks

Read the passage given below and answer the following questions:
Pentose and hexose undergo intramolecular hemiacetal or hemiketal formation due to combination of the $–\ce{OH}$ group with the carbonyl group. The actual structure is either of five or six membered ring containing an oxygen atom. In the free state all pentoses and hexoses exist in pyranose form $($resembling pyran$).$ However,inthe combined state some of them exist as five membered cyclic structures, called furanose $($resembling furan$).$

The cyclic structure of glucose is represented by Haworth structure:

$\alpha$ and $\beta D-$glucose have different configuration at anomeric $(C-1)$ carbon atom, hence are called anomers and the $C-1$ carbon atom is called anomeric carbon $($glycosidic carbon$).$
The six membered cyclic structure of glucose is called pyranose structure.
The following questionsare multiple choice questions. Choose the most appropriate answer:

$\alpha D(+)-$glucose and $\beta D(+)$ glucose are:

Enantiomers.
Conformers.
Epimers.
Anomers.

The following carbohydrate is:

A ketohexose.
An aldohexose.
An $n-$furanose.
An $\alpha-$pyranose.

In the following structure, anomeric carbon is:

$C-1$
$C-2$
$C-3$
$C-4$

The term anomers of glucose refers to:

Isomers of glucose that differ in configurations at carbons one and four $(C-1$ and $C-4).$
A mixture of $(D)-$glucose and $(L)-$glucose.
Enantiomers of glucose.
Isomers of glucose that differ in configuration at carbon one $(C-1).$

What percentage of $\beta-D-(+)$ glucopyranose is found at equilibrium in the aqueous solution?

$50\%$
$\approx100%$
$36\%$
$64\%$

Answer

$(d)$ Anomers.

$\alpha-D-( + )-$glucose and $\beta-D-( + )-$glucose differ in configuration at $C_1 ($i.e., anomeric or glycosidic carbon$)$ and hence are called anomers.

$(b)$ An aldohexose.

This structure is an example of pyranose and aldohexose. Here, the carbohydrate's structure is of the $\beta-$pyranose fonn.

$(a) C-1$

$C-1$ is the anomeric carbon.

$(d)$ Isomers of glucose that differ in configuration at carbon one $(C-1).$

Anomers are cyclic monosaccharides or glycosides that are epimers, differing from each other in the configuration at $C-1,$ if they are aldoses or in the configuration at $C-2$ if they are ketoses.

$(d) 64\%$

Ordinary glucose is $CL-$glucose, with a fresh aqueous solution having specific rotation, $[\alpha]_\text{D}=+111^\circ.$
On keeping the solution for sometime, $\alpha-$glucose slowly changes into an equilibrium mixture of $\alpha-$glucose $(36\%)$ and $\beta-$glucose$(64\%)$ and the mixture has specific rotation $+ 52.5^\circ$

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Chemistry Case study (4 Marks)

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