Maharashtra Board Class 12 Chemistry Chapter 14 Biomolecules Solutions

12th Chemistry Chapter 14 Exercise Biomolecules Solutions Maharashtra Board

Biomolecules Class 12 Exercise Question Answers Solutions Maharashtra Board

Chemistry Class 12 Chapter 14 Exercise Solutions

1. Select The Most Correct Choice.

 

Question i. CH2OH-CO-(CHOH)4-CH2OH is an example of
a. Aldohexose
b. Aldoheptose
c. Ketotetrose
d. Ketoheptose
Answer:
(d) Ketoheptose
In simple words: The given chemical formula represents a molecule with a ketone group and seven carbon atoms, fitting the definition of a ketoheptose.

🎯 Exam Tip: Identify the functional group (aldehyde or ketone) and the number of carbon atoms to correctly classify saccharides.

 

Question ii. Open chain formula of glucose does not contain
a. Formyl group
b. Anomeric hydroxyl group
c. Primary hydroxyl group
d. Secondary hydroxyl group
Answer:
(b) Anomeric hydroxyl group
In simple words: The open chain structure of glucose lacks an anomeric carbon, which is only present in its cyclic hemiacetal forms.

🎯 Exam Tip: Remember the key functional groups present in the open-chain and cyclic forms of monosaccharides to avoid common misconceptions.

 

Question iii. Which of the following does not apply to CH2NH2 - COOH
a. Neutral amino acid
b. L - amino acid
c. Exists as zwitterion
d. Natural amino acid
Answer:
(d) Natural amino acid
In simple words: The given structure CH2NH2-COOH represents glycine, which is a neutral amino acid and exists as a zwitterion, but it is not a natural amino acid (it can be synthesized).

🎯 Exam Tip: Understand the classification of amino acids (neutral, acidic, basic, essential, non-essential) and their properties like zwitterion formation.

 

Question iv. Tryptophan is called essential amino acid because
a. It contains an aromatic nucleus.
b. It is present in all the human proteins.
c. It cannot be synthesized by the human body.
d. It is an essential constituent of enzymes.
Answer:
(c) It cannot be synthesised by human body.
In simple words: Tryptophan is an essential amino acid because the human body cannot produce it, and it must be obtained through diet.

🎯 Exam Tip: Learn the definition of essential amino acids: those that cannot be synthesized by the body and must be acquired from food.

 

Question v. A disulfide link gives rise to the following structure of protein.
a. Primary
b. Secondary
c. Tertiary
d. Quaternary
Answer:
(c) Tertiary
In simple words: Disulfide links are covalent bonds that contribute to the specific three-dimensional folding of a single polypeptide chain, characteristic of tertiary protein structure.

🎯 Exam Tip: Differentiate between the various levels of protein structure and the types of bonds/interactions that stabilize each level.

 

Question vi. RNA has
a. A - U base pairing
b. P-S-P - S backbone
c. double helix
d. G - C base pairing
Answer:
(a) A - U base pairing
In simple words: RNA typically uses adenine (A) and uracil (U) for base pairing, unlike DNA which uses adenine (A) and thymine (T).

🎯 Exam Tip: Recall the key structural differences between DNA and RNA, especially regarding their characteristic bases and typical structural forms.

 

2. Give Scientific Reasons:

 

Question i. The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test.
Answer:
(1) In disaccharide sucrose, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a nonreducing sugar. As there is no free aldehyde group, it does not reduce Tollen's reagent to metallic silver. Hence, sucrose gives negative Tollen's test.
(2) While the disaccharide maltose is a reducing sugar because a free aldehyde group can be produced at C₁ of second sugar molecule. It is a reducing sugar. It reduces Tollen's reagent to shining silver mirror. Hence, Maltose gives positive Tollen's test.
In simple words: Sucrose is a non-reducing sugar because its glucose and fructose units are linked at their anomeric carbons, leaving no free aldehyde or ketone groups to react. Maltose, however, has a free anomeric carbon that can open to form an aldehyde, making it a reducing sugar that reacts positively with Tollen's reagent.

🎯 Exam Tip: The presence of a free anomeric carbon (or the ability to form one in equilibrium) determines whether a sugar is reducing or non-reducing, which is crucial for Tollen's test.

 

Question ii. On complete hydrolysis DNA gives equimolar quantities of adenine and thymine.
Answer:
On complete hydrolysis DNA yields 2-deoxy-D-ribose, adenine, thymine, guanine, cystosine and phosphoric acid. Since adenine always forms two hydrogen bonds with thymine, the hydrolysis of DNA gives equimolar quantities of adenine and thymine.
In simple words: DNA's double helix structure relies on specific base pairing where adenine (A) always pairs with thymine (T), meaning they are always present in equal amounts.

🎯 Exam Tip: Recall Chargaff's rules: in DNA, the amount of adenine (A) equals thymine (T), and the amount of guanine (G) equals cytosine (C).

 

Question iii. α-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.
Answer:

Formula	Molecular mass	Melting point
CH3-CH-COOH | NH2 Amino acid	89	293.5 °C
CH3-NH2	87	-55 °C
CH3 - COOH	88	-7.9 °C

α-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass due to the presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In aqueous solution, proton transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called zwitter ion.
\[\text{H}_2\text{N}-\text{CH}-\text{COOH} \implies \text{H}_3\text{N}^+ – \text{CH} – \text{COO}^-\]
\[\quad\quad|\quad\quad\quad\quad\quad\quad\quad|\text{R}\]
\[\quad\quad\text{R}\]
\[\quad\quad\text{Amino acid}\quad\quad\quad\quad\text{Zwitter ion}\]
\[\quad\quad\quad\quad\quad\quad\quad\quad\quad(\text{dipolar ion})\]
In simple words: Amino acids exist as zwitterions (dipolar ions) where a proton transfers from the carboxyl group to the amino group, forming strong electrostatic forces that require more energy to break, leading to higher melting points.

🎯 Exam Tip: The zwitterionic nature of amino acids is key to understanding their physical properties like high melting points and solubility.

 

Question iv. Hydrolysis of sucrose is called inversion.
Answer:
\[\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+, \text{invertase}} \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6\]
\[\text{sucrose}\quad\quad\quad\quad\quad\quad\quad\text{D}(+)\text{ glucose}\quad\text{D}(-)\text{ fructose}\]
Sucrose is dextro rotatory. On hydrolysis it gives equimolar mixture of D – (+) glucose and D – (-) fructose. Since the laevorotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.7°), the hydrolysis product has net laevorotation. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro ( + ) to laevo (-) and the product is called as invert sugar and so the hydrolysis of sucrose is called inversion.
In simple words: Sucrose, which is dextrorotatory, undergoes hydrolysis to yield an equimolar mixture of glucose (dextrorotatory) and fructose (levorotatory); because fructose's levorotation is stronger, the resulting solution becomes levorotatory, hence the change in optical rotation is called inversion.

🎯 Exam Tip: Understand that "inversion" refers to the change in the direction of optical rotation (from positive to negative) when sucrose is hydrolyzed into glucose and fructose.

 

Question v. On boiling, egg albumin becomes opaque white.
Answer:
Upon boiling the egg, denaturation occurs. During denaturation, secondary and tertiary structures are destroyed, but primary structure remains intact. Egg contains soluble globular proteins, which forms insoluble fibrous proteins (opaque) on boiling egg.
In simple words: Boiling egg albumin causes its soluble globular proteins to denature, unfolding their secondary and tertiary structures, which leads to their aggregation into insoluble fibrous proteins that appear opaque.

🎯 Exam Tip: Denaturation is the process where proteins lose their native three-dimensional structure due to external factors like heat, leading to loss of function and often insolubility.

 

3. Answer The Following

 

Question i. Some of the following statements apply to DNA only, some to RNA only and some to both. Lable them accordingly.
a. The polynucleotide is double stranded. ( )
b. The polynucleotide contains uracil. ( )
c. The polynucleotide contains D-ribose ( ).
d. The polynucleotide contains Guanine ( ).
Answer:
(1) The polynucleotide is double stranded. (DNA)
(2) The polynucleotide contains uracil. (RNA)
(3) The polynucleotide contain D-ribose (RNA)
(4) The polynucleotide contains Guanine (DNA, RNA)
In simple words: DNA is double-stranded and contains guanine, while RNA contains uracil and D-ribose, and guanine is found in both nucleic acids.

🎯 Exam Tip: Differentiate between DNA and RNA based on their strandedness, constituent sugars, and specific nitrogenous bases (thymine in DNA, uracil in RNA).

 

Question ii. Write the sequence of the complementary strand for the following segments of a DNA molecule.
a. 5' - CGTTTAAG – 3'
b. 5' - CCGGTTAATACGGC – 3'
Answer:
(1) DNA molecule : 5′ – CGTTTAAG – 3'
The complementary strand runs in opposite direction from the 3' end to the 5' end. It has the base sequence decided by complementary base pairs A – T and C – G.
Original strand : 5'-CGTTTAAG-3'
                  ------|||
Complementary strand: 3'-GCAAATTC-5'
(2) DNA molecule : 5′ – CCGGTTAATACGGC – 3'
The complementary strand runs in opposite direction from the 3' end to the 5' end. It has the base sequence decided by complementary base pairs A – T and C – G.
Original strand : 5'-CCGGTTAATACGGC-3'
                  --------
Complementary strand: 3'--GGCCAATTATGCCG-5'
In simple words: DNA complementary strands are formed by pairing adenine (A) with thymine (T) and cytosine (C) with guanine (G), running in an antiparallel direction (5' to 3' and 3' to 5').

🎯 Exam Tip: Remember the base pairing rules (A-T, C-G) and the antiparallel nature of DNA strands when determining complementary sequences.

 

Question iii. Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.
Answer:
(1) Dipeptide formed from alanine and glycine

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एलेनिन और ग्लाइसिन से बनने वाले एक डायपेप्टाइड के संश्लेषण को दर्शाता है। ग्लाइसिन का कार्बोक्सिल समूह एलेनिन के अमीनो समूह के साथ प्रतिक्रिया करके एक पेप्टाइड बंधन बनाता है, जिससे पानी का अणु निकलता है। परिणामी उत्पाद ग्लाइसिलालेनिन (Gly-Ala) है, जो एक डायपेप्टाइड है।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक और संभावित डायपेप्टाइड के संश्लेषण को दर्शाता है, जिसमें एलेनिन का कार्बोक्सिल समूह ग्लाइसिन के अमीनो समूह के साथ प्रतिक्रिया करता है। इस प्रक्रिया में एक पेप्टाइड बंधन बनता है और पानी का अणु बाहर निकलता है। यह आरेख यह भी दिखाता है कि एलेनिन से ग्लाइसिन में डायपेप्टाइड का निर्माण कैसे होता है।

(2) Dipeptide formed from alanine and tyrosine

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एलेनिन और टायरोसिन से बनने वाले एक डायपेप्टाइड के संश्लेषण को दर्शाता है। एलेनिन का कार्बोक्सिल समूह टायरोसिन के अमीनो समूह के साथ प्रतिक्रिया करता है, जिससे एक पेप्टाइड बंधन बनता है और पानी का अणु निकलता है। परिणामी उत्पाद एलानिलटायरोसिन (Ala-Tyr) है, जो एक डायपेप्टाइड है।

(3) Dipeptide formed from glycine and tyrosine.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख ग्लाइसिन और टायरोसिन से बनने वाले एक डायपेप्टाइड के संश्लेषण को दर्शाता है। टायरोसिन का कार्बोक्सिल समूह ग्लाइसिन के अमीनो समूह के साथ प्रतिक्रिया करता है, जिससे एक पेप्टाइड बंधन बनता है और पानी का अणु निकलता है। परिणामी उत्पाद टायरोसिलग्लाइसिन (Tyr-Gly) है, जो एक डायपेप्टाइड है।
In simple words: Dipeptides are formed by linking two amino acids via a peptide bond, where the carboxyl group of one amino acid reacts with the amino group of another, releasing a water molecule.

🎯 Exam Tip: To draw dipeptides, identify the amino and carboxyl groups involved in peptide bond formation, remembering that the bond is an amide linkage with the elimination of water.

 

Question iv. Give two pieces of evidence for the presence of the formyl group in glucose.
Answer:
(1) Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of – CHO (formyl group) in glucose.
\[\text{CHO}\quad\quad\quad\quad\text{CH} = \text{NOH}\]
\[|\quad\quad\quad\quad\quad\quad\quad\quad|\]
\[(\text{CHOH})_4\quad\xrightarrow{\text{NH}_2-\text{OH}}\quad(\text{CHOH})_4\]
\[|\quad\text{hydroxyl amine}\quad|\]
\[\text{CH}_2\text{OH}\quad\quad\quad\quad\text{CH}_2\text{OH}\]
\[\text{glucose}\quad\quad\quad\quad\quad\quad\quad\text{glucose oxime}\]
(2) Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid which shows carbonyl group in glucose is aldehyde (formyl group) group.
\[\text{CHO}\quad\quad\quad\quad\text{COOH}\]
\[|\quad\quad\quad\quad\quad\quad\quad|\\]
\[(\text{CHOH})_4\quad\xrightarrow{\text{(O)}}\quad(\text{CHOH})_4\]
\[|\quad\text{Bromine}\quad|\]
\[\text{CH}_2\text{OH}\quad\text{water}\quad\text{CH}_2\text{OH}\]
\[\text{Glucose}\quad\quad\quad\quad\text{Gluconic acid}\]
In simple words: The formyl group (aldehyde) in glucose is evidenced by its reaction with hydroxylamine to form an oxime, and its mild oxidation with bromine water to form gluconic acid, indicating the presence of an aldehyde functional group.

🎯 Exam Tip: Chemical reactions that are specific to aldehydes (like oxime formation or mild oxidation to carboxylic acids) are definitive proofs for the presence of a formyl group.

 

4. Draw A Neat Diagram For The Following:

 

Question i. Haworth formula of glucopyranose
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पायरान (Pyran) के मूल संरचना को दर्शाता है, जो छह-सदस्यीय विषमचक्रीय वलय है जिसमें एक ऑक्सीजन परमाणु होता है। यह पायरान रिंग को β-साइड को नीचे की ओर इंगित करते हुए दिखाता है।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र α-D-(+) ग्लूकोपाइरानोस और β-D-(+) ग्लूकोपाइरानोस के हॉवर्थ (Haworth) सूत्रों को दर्शाता है। दोनों संरचनाएँ एक छह-सदस्यीय वलय के रूप में ग्लूकोज के चक्रीय रूप हैं, जिसमें ऑक्सीजन परमाणु शीर्ष पर होता है। α-एनोमर में, C1 पर -OH समूह नीचे की ओर होता है, जबकि β-एनोमर में, C1 पर -OH समूह ऊपर की ओर होता है। यह आरेख ग्लूकोज के एनोमर के हॉवर्थ सूत्र की व्याख्या करता है।
In simple words: The Haworth formula for glucopyranose depicts glucose as a six-membered ring (pyranose), showing the relative positions of hydroxyl groups and distinguishing between alpha and beta anomers based on the orientation of the -OH group at the anomeric carbon (C1).

🎯 Exam Tip: When drawing Haworth formulas, pay close attention to the orientation of -OH groups (up or down) and the position of the anomeric carbon (C1) to correctly distinguish between alpha and beta anomers.

 

Question ii. Zwitter ion
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ज़्विटर आयन के निर्माण को दर्शाता है। एक अमीनो एसिड में, कार्बोक्सिल समूह एक प्रोटॉन दान कर सकता है, जबकि अमीनो समूह एक प्रोटॉन को स्वीकार कर सकता है, जिससे एक द्विध्रुवीय आयन बनता है जिसमें दोनों धनात्मक और ऋणात्मक आवेश एक ही अणु के भीतर मौजूद होते हैं। यह प्रोटॉन स्थानांतरण एक ज़्विटर आयन का निर्माण करता है।
In simple words: A zwitterion is a dipolar ion, specifically formed by amino acids, where the carboxyl group loses a proton (becoming negative) and the amino group gains a proton (becoming positive) within the same molecule, resulting in an overall neutral charge.

🎯 Exam Tip: The formation of a zwitterion is a crucial characteristic of amino acids, explaining their high melting points and solubility due to strong electrostatic interactions.

 

Question iii. Haworth formula of maltose
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र माल्टोस का हॉवर्थ सूत्र दर्शाता है, जो दो α-D-ग्लूकोस इकाइयों से बना एक डिसैकेराइड है जो α-1,4 ग्लाइकोसिडिक बंधन द्वारा जुड़ा होता है। पहली ग्लूकोस इकाई का C1 कार्बन दूसरे ग्लूकोस इकाई के C4 कार्बन से जुड़ा होता है।
In simple words: The Haworth formula of maltose illustrates it as a disaccharide composed of two α-D-glucose units linked by an α-1,4-glycosidic bond, showing their cyclic pyranose structures and relative stereochemistry.

🎯 Exam Tip: For disaccharides, accurately depicting the glycosidic bond (e.g., α-1,4) between the monosaccharide units is essential in their Haworth structural representations.

 

Question iv. Secondary structure of the protein
Answer:
The structure of proteins can be studied at four different levels i.e. primary, secondary, tertiary and quaternary levels. Each level is more complex than the previous one.
(1) Primary structure of proteins :
(a) Representation by structural formula
\[\text{....... - NH-CH-C-NH-CH-C-NH-CH-C-NH-CH-C-.......}\]
\[\quad\quad|\quad||\quad\quad\quad|\quad||\quad\quad\quad|\quad||\quad\quad\quad|\quad||\text{O}\]
\[\text{N-terminal R'}\quad\text{O R''}\quad\text{O R'''}\quad\text{O R'}\quad\text{C-terminal}\rightarrow\]
(b) Representation with amino acid symbols
Ala-Gly - Ser - Tyr - Gly - Gly - Lys .......
$\leftarrow$ N-terminal C - terminal
Primary structure of proteins is the sequence of constituent a-amino acid residues linked by peptide bonds. Any change in the sequence of amino acid residue creates different protein molecule. Primary structure of proteins is represented by writing the three letter symbols of amino acid residues as per their sequence in the concerned protein. The symbols are separated by dashes. According to the convention, the N-terminal amino acid residue as written at the left end and the C-terminal amino acid residue at the right end.
(2) Secondary structure of proteins : The three-dimensional arrangement of localized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.
α-Helix : In α-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clockwise spiral known as α-helix. The characteristic features of α-helical structure of protein are :

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्रोटीन के α-हेलिक्स बैकबोन की संरचना को दर्शाता है, जो एक दाएँ हाथ की कुंडलित संरचना है। हाइड्रोजन बंधन (N-H और C=O समूहों के बीच) हेलिक्स को स्थिर करते हैं। R समूह बाहर की ओर निकलते हैं, और प्रत्येक मोड़ में लगभग 3.6 अमीनो एसिड होते हैं।

(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and α-keratin in hair are proteins with almost entire α-helical secondary structure.
β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-pleated sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्रोटीन की β-प्लीटेड शीट संरचना को दर्शाता है, जहाँ कई पॉलीपेप्टाइड श्रृंखलाएँ एक-दूसरे के बगल में समानांतर या प्रतिसमानांतर फैशन में संरेखित होती हैं। हाइड्रोजन बंधन आसन्न श्रृंखलाओं के N-H प्रोटॉन और C=O ऑक्सीजन समूहों के बीच बनते हैं, जिससे एक स्थिर, प्लीटेड संरचना बनती है। R समूह शीट के तल के ऊपर और नीचे की ओर उन्मुख होते हैं।

• The C = O and N – H bonds lie in the planes of the sheet.
• Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
• The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र प्रोटीन की तृतीयक संरचना को दर्शाता है, जो एक पॉलीपेप्टाइड श्रृंखला की समग्र तीन-आयामी तह को संदर्भित करता है। इस जटिल संरचना को विभिन्न अंतःक्रियाओं द्वारा स्थिर किया जाता है, जैसे हाइड्रोजन बंधन, द्विध्रुवीय-द्विध्रुवीय आकर्षण, इलेक्ट्रोस्टैटिक आकर्षण और लंदन परिक्षेपण बल। इसमें डाइसल्फाइड बंधन भी शामिल हैं, जो इस जटिल तह में महत्वपूर्ण भूमिका निभाते हैं।
The three-dimensional shape acquired by the entire polypeptide chain of a protein is called its tertiary structure. The structure is stabilized and has attractive interaction with the aqueous environment of the cell due to the folding of the chain in a particular manner. Tertiary structure gives rise to two major molecular shapes i.e. globular and fibrous proteins. The main forces which stabilize a particular tertiary structure include hydrogen bonding, dipole-dipole attraction (due to polar bonds in the side chains), electrostatic attraction (due to the ionic groups like -COO¯, NH+ in the side chain) and also London dispersion forces. Finally, disulfide bonds formed by oxidation of nearby – SH groups (in cysteine residues) are the covalent bonds which stabilize the tertiary structure.
(4) Quaternary structure of proteins The two or more polypeptide chains with folded tertiary structures forms complex protein. The spatial arrangements of these polypeptide chains with respect to each other is known as quaternary structure. Each individual polypeptide chain is called a subunit of the overall protein. For example: Haemoglobin consists of four subunits called haeme held together by intermolecular forces in a compact three dimensional shape. Haemoglobin can do its function of oxygen transport only when all the four subunits are together.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख प्रोटीन संरचना के चार स्तरों का एक सारांश प्रस्तुत करता है: प्राथमिक (अमीनो एसिड अनुक्रम), माध्यमिक (α-हेलिक्स और β-प्लीटेड शीट), तृतीयक (पॉलीपेप्टाइड श्रृंखला की समग्र 3D तह), और चतुर्धातुक (कई तृतीयक इकाइयों का जटिल)। यह प्रोटीन की जटिल पदानुक्रमित संरचना को समझने में सहायता करता है।
In simple words: Secondary protein structure refers to localized folding patterns like alpha-helices and beta-pleated sheets, stabilized by hydrogen bonds between the backbone atoms of a polypeptide chain, while tertiary structure is the overall 3D shape of a single polypeptide, stabilized by various interactions including disulfide bonds. Quaternary structure involves the arrangement of multiple polypeptide subunits.

🎯 Exam Tip: Be able to describe and sketch the alpha-helix and beta-pleated sheet, identifying the hydrogen bonding patterns that stabilize these secondary structures.

 

Question v. AMP
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एडिनोसिन मोनोफॉस्फेट (AMP) की रासायनिक संरचना को दर्शाता है। AMP एक न्यूक्लियोटाइड है जिसमें एक एडेनिन बेस, एक राइबोज शुगर और एक फॉस्फेट समूह होता है। फॉस्फेट समूह राइबोज शुगर के 5' कार्बन से जुड़ा होता है।
In simple words: AMP (Adenosine Monophosphate) is a nucleotide composed of adenine (a nitrogenous base), ribose (a five-carbon sugar), and a single phosphate group, forming a fundamental building block of RNA.

🎯 Exam Tip: Remember the components of a nucleotide (base, sugar, phosphate) and how they differ between DNA and RNA (deoxyribose vs. ribose, thymine vs. uracil).

 

Question vi. dAMP
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र डीऑक्सीराइबोस एडेनिन मोनोफॉस्फेट (dAMP) की रासायनिक संरचना को दर्शाता है। dAMP एक डीऑक्सीराइबोन्यूक्लियोटाइड है जिसमें एक एडेनिन बेस, एक 2-डीऑक्सीराइबोस शुगर और एक फॉस्फेट समूह होता है। फॉस्फेट समूह डीऑक्सीराइबोस शुगर के 5' कार्बन से जुड़ा होता है।
In simple words: dAMP (deoxyadenosine monophosphate) is a deoxyribonucleotide, a building block of DNA, consisting of adenine, 2-deoxyribose sugar, and a single phosphate group.

🎯 Exam Tip: Distinguish between deoxyribonucleotides (for DNA) and ribonucleotides (for RNA) by the sugar component (deoxyribose vs. ribose) and associated bases.

 

Question vii. One purine base from nucleic acid
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एडेनिन (Adenine) की रासायनिक संरचना को दर्शाता है। एडेनिन एक प्यूरीन बेस है जिसमें एक दोहरी रिंग संरचना होती है, जिसमें नाइट्रोजन और कार्बन परमाणु होते हैं। यह डीएनए और आरएनए दोनों में पाया जाने वाला एक महत्वपूर्ण न्यूक्लियोबेस है।
In simple words: Adenine is a purine base found in both DNA and RNA, characterized by its double-ring structure.

🎯 Exam Tip: Identify the purine bases (adenine and guanine) and pyrimidine bases (cytosine, thymine, uracil) and their characteristic ring structures.

 

Question viii. Enzyme catalysis
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एंजाइम उत्प्रेरण की प्रक्रिया को दर्शाता है। सबसे पहले, एक सब्सट्रेट (Substrate) एक एंजाइम (Enzyme) की सक्रिय साइट (Active site) से जुड़ता है। यह एंजाइम-सब्सट्रेट कॉम्प्लेक्स (Enzyme-substrate complex) बनाता है। इसके बाद, एंजाइम रासायनिक प्रतिक्रिया को उत्प्रेरित करता है, सब्सट्रेट को उत्पाद (Product) में परिवर्तित करता है। अंत में, उत्पाद एंजाइम से अलग हो जाता है, जिससे एंजाइम अपरिवर्तित रहता है और नए सब्सट्रेट अणुओं को उत्प्रेरित करने के लिए तैयार रहता है।
In simple words: Enzyme catalysis is a process where an enzyme binds to a substrate at its active site, forming an enzyme-substrate complex, which then facilitates a chemical reaction to convert the substrate into products, after which the enzyme is released unchanged and ready to catalyze more reactions.

🎯 Exam Tip: Key points of enzyme catalysis include substrate specificity, the formation of an enzyme-substrate complex, and the enzyme's ability to lower activation energy without being consumed in the reaction.

12th Chemistry Digest Chapter 14 Biomolecules Intext Questions And Answers

Try - This (Textbook Page No 298)

Question 1. Observe the following structural formulae carefully and answer the questions. CHO | (CHOH)4 | CH2OH (glucose) CH2OH | CO | (CHOH)3 | CH2OH (fructose) CHO | (CHOH)3 | CH2OH (ribose)
(1) How many OH groups are present in glucose, fructose and ribose respectively?
(2) Which other functional groups are present in these three compounds?
Answer:
(1) Glucose contains five hydroxyl (- OH) groups.
Fructose contains five hydroxyl (- OH) groups.
Ribose contains four hydroxyl (- OH) groups.
(2) Glucose contains aldehyde (- CHO) as other functional group.
Fructose contains ketonic group (-\(C=O\)) as other functional group.
Ribose contains aldehyde (- CHO) as other functional group.
In simple words: Glucose, fructose, and ribose all contain multiple hydroxyl groups. Glucose and ribose also have an aldehyde group, while fructose has a ketone group.

🎯 Exam Tip: Understanding the functional groups present in different monosaccharides is crucial for identifying their chemical properties and reactions.

Use Your Brain Power! (Textbook Page No 299)

Question 1. Give IUPAC names to the following monosaccharides. (1) CHO | CHOH | CH2OH (Threose) (2) CHO | (CHOH)3 | CH2OH (Ribose) (3) CH2OH | CO | (CHOH)4 | CH2OH (Ketoheptose)
Answer:
(1) Aldotriose
(2) Aldopentose
(3) Ketoheptose
In simple words: The IUPAC names for these monosaccharides are determined by the number of carbon atoms and the type of carbonyl group present (aldehyde for aldo-, ketone for keto-).

🎯 Exam Tip: Mastering IUPAC nomenclature for common monosaccharides helps in accurate chemical communication and understanding carbohydrate structure.

Problem 14.1: (Textbook Page No 300)

Question 1. An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. Molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (- OH) groups must be present in the original compound?
Answer:Solution: In acetylation reaction H atom of an (- OH) group is replaced by an acetyl group (-\(COCH_3\)). This results in an increase in molecular mass by \([(12 + 16 + 12 + 3 \times 1) - 1]\) that has, 42 u. In the given alcohol, increase in molecular mass \(= 174 u - 90 u = 84 u\)
Number of - OH groups \(= \frac{84u}{42u} = 2\)
In simple words: Each hydroxyl group replaced by an acetyl group increases the molecular mass by 42 units; since the total mass increase was 84 units, the compound must have had two hydroxyl groups.

🎯 Exam Tip: Acetylation is a common method to determine the number of hydroxyl groups in organic compounds. Remember the mass change per -OH group acetylated.

Use Your Brain Power! (Textbook Page No 301)

Question 1.
(1) Write structural formula of glucose showing all the bonds in the molecule.
(2) Number all the carbons in the molecules giving number 1 to the (- CHO) carbon.
(3) Mark the chiral carbons in the molecule with asterisk (*).
(4) How many chiral carbons are present in glucose?
Answer:Refer structural formula of glucose for (1) (2) and (3).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्लूकोज की खुली श्रृंखला संरचना को दर्शाता है। इसमें कार्बन परमाणु 1 से 6 तक क्रमांकित हैं, जिसमें कार्बन-1 पर एल्डिहाइड समूह और कार्बन-2 से 6 पर हाइड्रॉक्सिल समूह हैं। यह अणु के चिरल केंद्रों को भी उजागर करता है।
(4) There are 4 chiral carbon atoms present in glucose.
In simple words: Glucose has a straight-chain structure with an aldehyde group at the top and multiple hydroxyl groups. Carbons 2, 3, 4, and 5 are chiral, meaning there are four chiral centers in total.

🎯 Exam Tip: Identifying chiral carbons in a glucose molecule is fundamental to understanding its stereochemistry and optical activity. Practice numbering carbons correctly.

Use Your Brain Power! (Textbook Page No 306)

Question 1.
(1) Is galactose an aldohexose or a ketohexose?
(2) Which carbon in galactose has different configuration compared to glucose?
(3) Draw Haworth formulae of α-D-galactose and β-D-galactose.
(4) Which disaccharides among sucrose, maltose and lactose is/are expected to give positive Fehling test?
(5) What are the expected products of hydrolysis of lactose?
Answer:
1. Galactose is an aldohexose.
2. Fourth carbon in galactose has different configuration compared to glucose.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह α-D-गैलेक्टोपाइरानोज और β-D-गैलेक्टोपाइरानोज के हावर्थ सूत्र को दर्शाता है। ये चक्रीय संरचनाएं ग्लूकोज के साथ गैलेक्टोज के विन्यास अंतर को, विशेष रूप से कार्बन-4 पर, चित्रित करती हैं।
3. (Drawing of α-D-Galactose and β-D-Galactose Haworth formulae would be here as described above).
4. Maltose and lactose are expected to give positive Fehling solution test.
5. The expected products of hydrolysis of lactose are D-(+) glucose and D-(+) galactose.
In simple words: Galactose is an aldohexose that differs from glucose in configuration at carbon-4. Maltose and lactose are reducing sugars that test positive with Fehling's reagent, and lactose hydrolyzes into glucose and galactose.

🎯 Exam Tip: Remember the structural differences between common monosaccharides like glucose and galactose, and their impact on properties like Fehling's test. Haworth projections are key for cyclic sugar structures.

Can You Think? (Textbook Page No 307)

Question 1. When you chew plain bread, chapati or bhaakari for long time, it tastes sweet. What could be the reason?
Answer:When chapati, bread or bhakari are chewed for long time the pulp mixes with saliva and carbohydrate component in them diseminates and gives the sweet taste.
In simple words: Prolonged chewing allows enzymes in saliva to break down complex carbohydrates in bread into simpler, sweet-tasting sugars.

🎯 Exam Tip: This question highlights the role of salivary amylase in carbohydrate digestion, a basic concept in biochemistry. The sweetness observed is due to the formation of monosaccharides or disaccharides.

Use Your Brain Power! (Textbook Page No 309)

Question 1. Tryptophan and histidine have the structures (I) and (II) respectively. Classify them into neutral? acidic/basic &amino acids and justify your answer. (Hint: Consider involvement of lone pair in resonance).
ℹ️ चित्र व्याख्या (Diagram Explanation): (I) यह ट्रिप्टोफैन की संरचना को दर्शाता है, जिसमें एक इंडोल रिंग (एक बेंजीन रिंग और एक पांच सदस्यीय नाइट्रोजन युक्त रिंग का संलयन) और एक अमीनो एसिड साइड-चेन होती है। (II) यह हिस्टिडाइन की संरचना को दर्शाता है, जिसमें एक इमिडाज़ोल रिंग (पांच सदस्यीय नाइट्रोजन युक्त हेटरोसाइक्लिक रिंग) और एक अमीनो एसिड साइड-चेन होती है।
Answer:
(I) N CH2-CH-COOH NH2 H In tryptophan, nitrogen atom present in cyclic structure cannot donate pair of electrons as it is stabilized by resonance. The other amino group and carboxylic group present in the side chain neutralize each other. Tryptophan has equal number of amino and carboxylic groups. Hence, tryptophan is a neutral amino acid.
(II) N N H CH2-CH-COOH NH2 In histidine, amino groups are more in number than carboxyl groups therefore histidine is basic in nature.
In simple words: Tryptophan is a neutral amino acid because its indole nitrogen's lone pair is involved in resonance, making it non-basic, and its amino and carboxyl groups balance each other. Histidine is basic due to the presence of an extra basic nitrogen in its imidazole ring.

🎯 Exam Tip: The acidity/basicity of amino acids is determined by the nature of their side chains. Resonance stabilization of lone pairs on nitrogen atoms can influence their basicity, as seen in tryptophan's indole ring.

Can You Think? (Textbook Page No 309)

Question 1. Compare the molecular masses of the following compounds and explain the observed melting points.

Formula	Molecular mass	Melting point
\(CH_3-CH-COOH\) \(|\) \(NH_2\)	89	293.5 °C
\(C_5H_{11}-NH_2\)	87	-55 °C
\(C_3H_7-COOH\)	88	-7.9 °C

Answer:Above compounds have same molecular masses but they have different melting points, α-amino acids have higher melting points compared to the corresponding amines or carboxylic acids of comparable masses. This property is due to the presence of both carboxylic group (acidic) and amino group (basic) in the molecule. In aqueous solution, protons transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called - Zwitter ion. \(CH_3-CH-COOH\) \(|\) \(NH_2\)
\( \iff \)
\(H_3N^+-CH-COO^-\) \(|\) \(CH_3\) Zwitter ion (dipolar ion)
In simple words: Despite similar molecular masses, alpha-amino acids have much higher melting points than corresponding amines or carboxylic acids because they exist as zwitterions (dipolar ions) with strong electrostatic forces between molecules.

🎯 Exam Tip: The zwitterionic nature of amino acids is crucial for understanding their physical properties, especially their high melting points and solubility in polar solvents. This concept is fundamental to protein structure.

Use Your Brain Power! (Textbook Page No 310)

Question 1.
(1) Write the structural formula of dipeptide formed by combination of carboxyl group of alanine and amino group of glycine.
(2) Name the resulting dipeptide.
(3) Is this dipeptide same as glycylalanine or its structural isomer?
Answer:
(1) \(H_2N-CH-COOH + H_2N - CH_2 - COOH\) \(|\) \(CH_3\) alanine glycine
\(- H_2O\)
\(H_2N-CH-CO-NH-CH_2 - COOH\) \(|\) \(CH_3\) peptide bond
(2) ala-glycine. OR ala-gly
(3) It is a structural isomer.
In simple words: When alanine's carboxyl group reacts with glycine's amino group, a dipeptide called alanyl-glycine forms. This dipeptide is a structural isomer of glycyl-alanine, as the sequence of amino acids is reversed.

🎯 Exam Tip: Understanding peptide bond formation and isomerism is vital. Remember that the N-terminal and C-terminal ends dictate the naming and distinctness of dipeptides composed of the same amino acids.

Question 54. Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.
Answer:(See Question iii from "3. Answer the following" section on Page 6 for a detailed answer with schematic representations.)
In simple words: The possible dipeptides from alanine, glycine, and tyrosine can be formed by combining any two in either order (e.g., Gly-Ala, Ala-Gly, Gly-Tyr, Tyr-Gly, Ala-Tyr, Tyr-Ala), each linked by a peptide bond.

🎯 Exam Tip: For problems involving combinations of amino acids, systematically list all possible permutations (e.g., AB and BA for two amino acids) to ensure all dipeptides are identified.

Problem 14.3 : (Textbook Page No 311)

Question 1. Chymotrypsin is a digestive enzyme that hydrolyzes those amide bonds for which the carbonyl group comes from phenylalanine, tyrosine or tryptophan. Write the symbols of the amino acids and peptides smaller than pentapeptide formed by hydrolysis of the following hexapeptide with chymotrypsin. Gly-Tyr-Gly-Ala-Phe-Val
Answer:Solution: In the given hexapeptide hydrolysis by chymotrypsin can take place at two points, namely, Phe and Tyr. The carbonyl group of these residues is towards the right side, that is, toward the C-terminal. Therefore the hydrolysis products in required range will be : Gly-Tyr, Gly-Ala-Phe and Val (a dipeptide) (a tripeptide) (α-amino acid)
In simple words: Chymotrypsin cleaves peptide bonds after phenylalanine or tyrosine residues. In the hexapeptide Gly-Tyr-Gly-Ala-Phe-Val, this results in the fragments Gly-Tyr (dipeptide), Gly-Ala-Phe (tripeptide), and Val (alpha-amino acid).

🎯 Exam Tip: Understanding enzyme specificity is key. For chymotrypsin, remember its preference for aromatic amino acid residues (Phe, Tyr, Trp) on the carbonyl side of the peptide bond during hydrolysis.

Problem 14.4: (Textbook Page No 311)

Question 1. Write down the structures of amino acids constituting the following peptide. \(CH_3-CH-CO - NH - CH - CO - NH - CH - COOH\) \(|\) \(NH_2\) \(|\) \(CH_2OH\) \(|\) \(CH_2SH\)
Answer:Solution: The given peptide has two amide bonds linking three amino acids. The structures of these amino acids are obtained by adding one \(H_2O\) molecule across the amide bond as follows : HOH \(CH_3 - CH - CO + NH-CH - CO+NH - CH - COOH\) \(|\) \(NH_2\) \(|\) \(CH_2SH\) \(|\) \(CH_2OH\)
\( \implies \)
\(CH_3-CH-COOH + H_2N-CH-COOH + H_2N - CH - COOH\) \(|\) \(NH_2\) \(|\) \(CH_2SH\) \(|\) \(CH_2OH\)
In simple words: To find the constituent amino acids of a peptide, we perform hydrolysis by adding water molecules across each peptide bond. This yields the individual amino acids: Alanine, Cysteine, and Serine, identifiable by their side chains.

🎯 Exam Tip: Hydrolysis of a peptide chain breaks the amide bonds, releasing the individual amino acid monomers. Identifying the R-groups after hydrolysis reveals the specific amino acids. Remember to write the amino acids in their un-bonded form.

Use Your Brain Power! (Textbook Page No 313)

Question 1. A protein chain has the following amino acid residues. Show and label the interactions that can be present in various pairs from these giving rise to tertiary level structure of protein.
(1) - HN-CH - CO - (Alanine)
(2) - HN - CH - CO - (Phenylalanine)
(3) - HN - CH - CO - (Asparagine)
(4) - HN - CH - CO - (Valine)
Answer:Tertiary level structure from amino residues.
(1) - HN-CH - CO - NH - CH - CO - NH - CH - CO - NH-CH-CO- ......
(Alanine)
(Phenylalanine)
(Asparagine)
(Valine)
(2) HN-CH-CO - NH - CH-CO-NH - CH-CO-NH-CH-CO-
(Valine)
(Asparagine)
(Phenylalanine)
(Alanine)
(3) -HN-CH-CO-NH - CH - CO - NH-CH-CO - NH - CH - CO -
(Alanine)
(Asparagine)
(Valine)
(Phenylalanine)
(4) HN-CH-CO-NH - CH - CO - NH - CH - CO - NH - CH - CO -
(Asparagine)
(Alanine)
(Phenylalanine)
(Valine)
In simple words: Tertiary protein structure arises from interactions between amino acid side chains, including hydrogen bonds (between polar groups), hydrophobic interactions (between nonpolar groups), ionic bonds (between charged groups), and disulfide bonds (between cysteine residues).

🎯 Exam Tip: Memorize the types of interactions (hydrogen bonding, hydrophobic, ionic, disulfide) that stabilize the tertiary structure of proteins. Relate specific amino acid side chains to the types of interactions they can form.

Can You Tell? (Textbook Page No 313)

Question 1. What is the physical change observed when (a) egg is boiled, (b) milk gets curdled on adding lemon juice?
Answer:
(a) When egg is boiled, coagulation of eggwhite (insoluble fibrous proteins) takes place. This is a common example of denaturation.
(b) When lemon juice is added to milk, it gets curdled due to the formation of lactic acid. This is another example of denaturation.
In simple words: Boiling an egg causes its proteins to coagulate and become opaque (denaturation), while adding lemon juice to milk causes its proteins to curdle due to pH change and lactic acid formation (also denaturation).

🎯 Exam Tip: Both examples illustrate protein denaturation, a process where the secondary, tertiary, and quaternary structures are disrupted, leading to loss of biological function and often visible changes like coagulation or curdling.

Can You Tell? (Textbook Page No 315)

Question 1. What is the single term that answers all the following questions?
(1) What decides whether you are blue eyed or brown eyed?
(2) Why does wheat grain germinate to produce wheat plant and not rice plant?
(3) Which acid molecules are present in nuclei of living cells?
Answer:
(1) Nucleic acid (DNA)
(2) Nucleic acid (DNA)
(3) Nucleic acid (DNA + RNA)
In simple words: Nucleic acids, specifically DNA and RNA, are the master molecules responsible for heredity, determining traits like eye color, directing species-specific development, and storing genetic information in cell nuclei.

🎯 Exam Tip: This question emphasizes the central role of nucleic acids (DNA and RNA) in genetics, heredity, and the fundamental processes of life. Their functions are essential for all living organisms.

Use Your Brain Power! (Textbook Page No 317)

Question 1. Draw structural formulae of nucleosides formed from the following sugars and bases.
(1) D - ribose and guanine
(2) D - 2 - deoxyribose and thymine
Answer:
(1) D-ribose and guanine
ℹ️ चित्र व्याख्या (Diagram Explanation): यह D-राइबोज और गुआनिन के बीच प्रतिक्रिया को दर्शाता है, जिसके परिणामस्वरूप एक राइबोन्यूक्लियोसाइड का निर्माण होता है। राइबोज की चक्रीय संरचना गुआनिन के N-9 परमाणु के साथ जुड़ती है, जिससे एक ग्लाइकोसिडिक बंधन बनता है और एक न्यूक्लियोसाइड बनता है।
(2) D - 2 - deoxyribose and thymine
ℹ️ चित्र व्याख्या (Diagram Explanation): यह D-2-डीऑक्सीराइबोज और थाइमिन के बीच प्रतिक्रिया को दर्शाता है, जिसके परिणामस्वरूप एक डीऑक्सीराइबोन्यूक्लियोसाइड का निर्माण होता है। डीऑक्सीराइबोज की चक्रीय संरचना थाइमिन के N-1 परमाणु के साथ जुड़ती है, जिससे एक ग्लाइकोसिडिक बंधन बनता है और एक डीऑक्सीन्यूक्लियोसाइड बनता है।
In simple words: Nucleosides are formed by linking a sugar (ribose or deoxyribose) to a nitrogenous base (like guanine or thymine) through a glycosidic bond, with water being eliminated in the process.

🎯 Exam Tip: Be able to draw the structures of common nucleosides. Focus on the C1' of the sugar linking to N9 of purines (like guanine) or N1 of pyrimidines (like thymine).

Problem 14.5 (Textbook Page No 318)

Question 1. Draw a schematic representation of trinucleotide segment ACT of a DNA molecule.
Answer:Solution: In DNA molecule sugar is deoxyribose. The base 'A' in the given segment is at 5' end while the base T at the 3' end. Hence the schematic representation of the given segment of DNA is
ℹ️ चित्र व्याख्या (Diagram Explanation): यह DNA अणु के एक त्रिन्यूक्लियोटाइड खंड ACT के योजनाबद्ध प्रतिनिधित्व को दर्शाता है। इसमें तीन न्यूक्लियोटाइड (एडेनिन, साइटोसिन, थाइमिन) हैं, जो फॉस्फेट-डीऑक्सीराइबोज बैकबोन से जुड़े हुए हैं, जिसमें 5' सिरे पर एडेनिन और 3' सिरे पर थाइमिन है।
In simple words: A DNA trinucleotide segment "ACT" consists of Adenine, Cytosine, and Thymine bases, each linked to a deoxyribose sugar, which are then connected by phosphate groups to form a chain from the 5' to 3' end.

🎯 Exam Tip: When drawing DNA segments, remember the 5' to 3' orientation, the deoxyribose sugar, and the phosphodiester bonds that link nucleotides. Also, correctly place the nitrogenous bases for each position.

Problem 14.6 : (Textbook Page No 320)

Question 1. Write the sequence of the complementary strand for the following portion of a DNA molecule : 5 -ACGTAC-3
Answer:Solution: The complementary strand runs in opposite direction from the 3' end to the 5' end. It has the base sequence decided by complementary base pairs A - T and C - G. Original strand 5'-ACGTAC-3'
↓↓↓↓↓↓
Complementary strand 3'-TGCATG-5'
In simple words: The complementary DNA strand pairs Adenine (A) with Thymine (T) and Cytosine (C) with Guanine (G), running in the opposite 3' to 5' direction. So, for 5'-ACGTAC-3', the complementary strand is 3'-TGCATG-5'.

🎯 Exam Tip: Always remember Chargaff's rules for base pairing (A-T, C-G) and the antiparallel nature of DNA strands (5' to 3' vs. 3' to 5') when determining complementary sequences.

Problem 14.2: (Text Page No 303)

Question 1. Assign D/L configuration to the following monosaccharides.
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह थ्रिओज के फिशर प्रक्षेपण सूत्र को दर्शाता है, जिसमें एक एल्डिहाइड समूह शीर्ष पर और हाइड्रॉक्सिल समूह कार्बन-2 और कार्बन-3 पर होते हैं। सबसे निचले चिरल कार्बन पर हाइड्रॉक्सिल समूह दाहिनी ओर है।
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह राइबोज के फिशर प्रक्षेपण सूत्र को दर्शाता है, जिसमें एक एल्डिहाइड समूह शीर्ष पर और कई हाइड्रॉक्सिल समूह होते हैं। सबसे निचले चिरल कार्बन (कार्बन-4) पर हाइड्रॉक्सिल समूह बाईं ओर है।
Answer:Solution: D/L configuration is assigned to Fischer projection formula of monosaccharide on the basis of the lowest chiral carbon.
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह थ्रिओज के फिशर प्रक्षेपण सूत्र को दर्शाता है, जिसमें एक एल्डिहाइड समूह शीर्ष पर और हाइड्रॉक्सिल समूह कार्बन-2 और कार्बन-3 पर होते हैं। सबसे निचले चिरल कार्बन पर हाइड्रॉक्सिल समूह दाहिनी ओर है। Threose has two chiral carbons C-2 and C-3. The given Fischer projection formula of threose has OH groups at the lowest C -3 chiral carbon on the right side.
\( \implies \) It is D-threose.
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह राइबोज के फिशर प्रक्षेपण सूत्र को दर्शाता है, जिसमें एक एल्डिहाइड समूह शीर्ष पर और कई हाइड्रॉक्सिल समूह होते हैं। सबसे निचले चिरल कार्बन (कार्बन-4) पर हाइड्रॉक्सिल समूह बाईं ओर है। Ribose has three chiral carbons C - 2, C - 3 and C -4. The given Fischer projection formula of ribose has - OH group at the lowest C -4 chiral carbon on the left side.
\( \implies \) It is L-ribose
In simple words: The D/L configuration of monosaccharides is determined by the position of the hydroxyl group on the lowest chiral carbon in the Fischer projection. If it's on the right, it's D; if on the left, it's L. Therefore, threose is D-threose, and ribose is L-ribose.

🎯 Exam Tip: Accurately identifying the lowest chiral carbon in a Fischer projection and the position of its hydroxyl group is crucial for correctly assigning D or L configuration. This is a fundamental concept in carbohydrate chemistry.

12th Std Chemistry Questions And Answers: