Biomolecules

For the given reactions: -

(i) The oxidation of the hydroboration product with hydrogen peroxide (H₂O₂) and hydroxide (OH⁻) leads to the formation of a diol. The main product will be CH₃–CH₂–CH₂–CH₂OH, 1-propanol.

(ii) Phenol (C₆H₅OH) reacts with aqueous NaOH to form phenoxide ion (C₆H₅O⁻). When phenol reacts with CO₂ in the presence of H⁺, it undergoes carboxylation to form salicylic acid (2-hydroxybenzoic acid).

Main product: Salicylic acid (C₆H₄(OH)COOH)

Step 1: Reaction of glucose with HI Glucose undergoes complete reduction when treated with hydroiodic acid (HI) in the presence of heat. This reaction removes all oxygen functionality from glucose, converting it into -hexane. The reaction is as follows: Step 2: Reaction of glucose with Acetic Anhydride When glucose is treated with acetic anhydride , all the hydroxyl groups (-OH) get acetylated, leading to the formation of glucose pentaacetate. This reaction is shown below: 

Let's analyze each statement:

A) It is an aldohexose:
This is true. Glucose is a six-carbon sugar (hexose) with an aldehyde functional group (aldo-).

B) On heating with HI it forms n-hexane:
This is also true. HI is a strong reducing agent that removes all oxygen atoms from glucose, reducing it to n-hexane.

C) It exists in furanose form:
This is partially true but misleading. While glucose can exist in both pyranose (six-membered ring) and furanose (five-membered ring) forms, the pyranose form is overwhelmingly predominant in solution. The furanose form exists only in minor amounts.

D) It does not give Schiff's test:
This is false. Glucose does give Schiff's test because its open-chain form contains an aldehyde group (in equilibrium with cyclic forms). Schiff's reagent reacts with aldehydes to produce a pink-magenta color.

Conclusion:
While statements A, B, and C have elements of truth, statement D is completely incorrect. The statement that is not true is (D).

Final Answer:
The incorrect statement is (D).

To solve this question, let's analyze the assertion and reason provided.

Assertion (A):
All naturally occurring -amino acids except glycine are optically active.

Reason (R):
Most naturally occurring amino acids have L-configuration.

1. Optically Active Amino Acids:
Optical activity in amino acids arises from the presence of a chiral center at the -carbon (the carbon attached to the amino group, carboxyl group, and side chain). For an amino acid to be optically active, the -carbon must not be superimposable on its mirror image (i.e., it must be a chiral center).

2. Glycine and Optical Activity:
Glycine (NH₂CH₂COOH) is the only naturally occurring -amino acid that is not optically active. This is because its -carbon is not chiral (it is attached to two hydrogen atoms, making it achiral). Therefore, the assertion that all naturally occurring -amino acids except glycine are optically active is correct.

3. The L-Configuration of Amino Acids:
Most naturally occurring amino acids have the L-configuration at the chiral -carbon. The L-configuration is the stereochemical form found in proteins. However, the L-configuration does not directly determine optical activity; it is the presence of the chiral center that does.

4. Evaluation of Reason (R):
The reason is true, as most naturally occurring amino acids indeed have the L-configuration. However, the reason does not explain why amino acids are optically active. The optical activity arises from the chiral nature of the amino acid, not just its configuration.

5. Conclusion:
The assertion is correct, but the reason, while true, does not explain the optical activity. Therefore, the assertion is true, but the reason is not fully explanatory.

Final Answer:
The correct option is that the assertion is true, but the reason is false.

1. Reactions of D-Glucose with the following reagents:

(a) Reaction with HCN:
When D-glucose reacts with hydrogen cyanide (HCN), a condensation reaction takes place, leading to the formation of a cyanohydrin. The reaction proceeds as follows:

D-glucose + HCN → D-glucose cyanohydrin (a sugar with a hydroxyl and a cyano group attached to the same carbon atom).

Since glucose is an aldose, the reaction occurs at the aldehyde group, forming the corresponding cyanohydrin.

Reaction mechanism:
In this case, the nucleophilic cyanide ion (CN⁻) attacks the electrophilic carbonyl carbon of the aldehyde group, leading to the formation of the cyanohydrin.

(b) Reaction with Br₂ water:
When D-glucose is treated with bromine water (Br₂ in water), an oxidation reaction occurs. The aldehyde group in D-glucose is oxidized to a carboxyl group, converting D-glucose into D-gluconic acid. The reaction proceeds as follows:

D-glucose + Br₂/H₂O → D-gluconic acid (C₆H₁₂O₆) + HBr

Reaction mechanism:
The bromine (Br₂) oxidizes the aldehyde group to a carboxylic acid group (-COOH), while the bromine itself is reduced to bromide ions (Br⁻).

(a) When D-glucose reacts with HCN, it forms cyanohydrin, as the nucleophilic cyanide attacks the electrophilic carbonyl carbon in glucose.
(b) When D-glucose reacts with
(c) When D-glucose reacts with acetic anhydride ( ), it forms glucose pentaacetate.}

To solve the problem, we need to identify the vitamin responsible for the coagulation of blood.

1. Understanding Coagulation:
Coagulation refers to the process of blood clotting, where blood changes from a liquid to a gel, preventing excessive bleeding after injury. This process involves a series of proteins in the blood, known as clotting factors, which are activated in a cascade to form a blood clot.

2. Role of Vitamin K:
Vitamin K plays a crucial role in blood coagulation. It is required for the synthesis of certain clotting factors in the liver, particularly prothrombin (factor II) and factors VII, IX, and X. These clotting factors are essential for the formation of a blood clot. Without sufficient vitamin K, blood clotting is impaired, leading to excessive bleeding.

3. Final Answer:
The vitamin responsible for blood coagulation is Vitamin K.

To solve the problem, we need to identify the disaccharide from the given options: Glucose, Lactose, Amylose, and Fructose.

1. Understanding Disaccharides:
Disaccharides are carbohydrates composed of two monosaccharide units joined by a glycosidic bond. Common disaccharides include sucrose, lactose, and maltose. These sugars can be broken down into their monosaccharide components through hydrolysis.

2. Examining the Options:
- Glucose: Glucose is a monosaccharide, not a disaccharide. It is a simple sugar that serves as a primary energy source in living organisms. - Lactose: Lactose is a disaccharide, formed by the combination of one molecule of glucose and one molecule of galactose. It is commonly found in milk. - Amylose: Amylose is a polysaccharide, a component of starch, and consists of long chains of glucose units. It is not a disaccharide. - Fructose: Fructose is a monosaccharide, not a disaccharide. It is commonly found in fruits and is used as a sweetener.

3. Final Answer:
The disaccharide from the given options is Lactose.

To solve the problem, we need to explain the difference between native and denatured proteins.

1. Understanding Proteins:
Proteins are large, complex molecules made up of amino acid chains, which fold into specific three-dimensional structures that determine their function. The structure of a protein plays a critical role in its biological activity, and changes to this structure can significantly affect its function.

2. Native Protein:
Native proteins are proteins that have their natural, functional three-dimensional structure. In this state, the protein's secondary, tertiary, and sometimes quaternary structures are intact, allowing the protein to perform its specific biological function. These structures are stabilized by various forces, including hydrogen bonds, ionic interactions, hydrophobic interactions, and disulfide bridges.

3. Denatured Protein:
Denatured proteins are proteins whose three-dimensional structure has been altered, usually due to external factors such as heat, extreme pH, or the presence of certain chemicals. Denaturation disrupts the non-covalent bonds and sometimes even covalent bonds (like disulfide bridges) that maintain the protein's structure. This results in the unfolding or misfolding of the protein, which leads to the loss of its biological activity.

4. Key Differences:
- Structure: Native proteins have a well-defined, functional three-dimensional structure, while denatured proteins have an altered or unfolded structure.
- Function: Native proteins are functional, as their structure allows them to perform specific biological roles, while denatured proteins are typically non-functional due to their altered shape.
- Stability: Native proteins are stable under normal physiological conditions, while denatured proteins are often unstable and can lose their ability to bind to substrates or catalyze reactions.

5. Final Answer:
The main difference between native and denatured proteins is that native proteins retain their natural three-dimensional structure, allowing them to function properly, while denatured proteins have lost their native structure and thus their biological activity, typically due to environmental stress factors like heat or pH changes.