Q.Give the plausible explanation for the following:

The concept first

There are just three rules for turning the open-chain Fischer projection into a Haworth pyranose ring.

Rule 1 — the "right–down" rule. Any $-\mathrm{OH}$ written on the right in the Fischer projection is drawn below the ring in Haworth; anything on the left is drawn above.

Rule 2 — the D-configuration. In a D-sugar the terminal $\mathrm{CH_2OH}$ (C-6, attached at C-5) points up.

Rule 3 — $\alpha$ versus $\beta$. The new stereocentre created on ring closure is C-1, the anomeric carbon.

• $\boldsymbol\beta$: the C-1 $\mathrm{OH}$ is on the same side as the $\mathrm{CH_2OH}$ — i.e. up.
• $\boldsymbol\alpha$: the C-1 $\mathrm{OH}$ is down (opposite to $\mathrm{CH_2OH}$).

Step-by-step

Step 1 — recall the Fischer projection of D-(+)-glucose. Reading down from C-1 (the $\mathrm{CHO}$):

The memory aid many students use is "right–left–right–right", the lone left-hand OH being at C-3.

Step 2 — apply the right–down rule to C-2, C-3, C-4.

• C-2: OH on the right $\Rightarrow$ OH below the ring, H above.
• C-3: OH on the left $\Rightarrow$ OH above the ring, H below.
• C-4: OH on the right $\Rightarrow$ OH below the ring, H above.

Step 3 — place the $\mathrm{CH_2OH}$. D-sugar $\Rightarrow$ $\mathrm{CH_2OH}$ at C-5 points up. (Both the question's diagrams and every standard Haworth drawing agree on this.)

Step 4 — set the anomeric carbon. We want the $\beta$ anomer, so the C-1 $\mathrm{OH}$ must be cis to the $\mathrm{CH_2OH}$ — i.e. OH above, H below.

Step 5 — assemble the required pattern.

$\text{C-1: OH \textbf{up}},\quad \text{C-2: OH \textbf{down}},\quad \text{C-3: OH \textbf{up}},\quad \text{C-4: OH \textbf{down}},\quad \text{C-5: }\mathrm{CH_2OH}\ \textbf{up}$ …

This question tests your knowledge of the classification and applications of important organic compounds in everyday life, particularly in the pharmaceutical and food industries. Each substance has a specific role determined by its chemical structure and properties.

Let's identify each compound and its function:

A. Antioxidant → III. Butylated hydroxy toluene (BHT)

Antioxidants prevent oxidative degradation of fats and oils in food by inhibiting free-radical chain reactions. BHT is a phenolic compound widely used in the food industry to prevent rancidity. The hydroxyl group on the aromatic ring donates hydrogen atoms to free radicals, breaking the oxidation chain and preserving freshness.

B. Food preservative → IV. Sodium benzoate

Food preservatives extend shelf life by inhibiting microbial growth (bacteria, fungi, yeasts). Sodium benzoate ($\ce{C6H5COONa}$) is one of the most common preservatives, especially effective in acidic foods like pickles, jams, and carbonated beverages. In acidic conditions it converts to benzoic acid, which penetrates microbial cell walls and disrupts their metabolism.

C. Sucralose → II. Artificial sweetener …

The key to this question is understanding two classifications at once: which amino acids are essential (must come from diet) and which are basic (have a net positive charge at physiological pH). You need the overlap.

Amino acids are basic when their side chain contains an extra amino group or a guanidino group — something that can grab a proton. The three standard basic amino acids are arginine (Arg), lysine (Lys), and histidine (His). But not all of these are essential for adults.

1. Identify the basic amino acids.

   At pH ~7.4, the side chains of Arg (guanidino group, pKa ~12.5), Lys (ε-amino group, pKa ~10.5), and His (imidazole group, pKa ~6.0) can carry a positive charge. So the set of basic amino acids is {Arg, Lys, His}.

2. Identify the essential amino acids.

   The nine essential amino acids for humans are: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. (Histidine is essential for infants and often considered conditionally essential for adults, but standard lists include it.)

3. Find the intersection.

   From the basic set {Arg, Lys, His}, which are essential?

   • Lysine is essential.
   • Histidine is essential.
   • Arginine is not essential for adults (the body can synthesize it), though it is essential for children. In most exam contexts, arginine is classified as non-essential for adults.

   So the essential basic amino acids are histidine and lysine. But look at the options — none say "His, Lys". Option (A) says "Arg, Lys". Why would that be correct?

4. Check each option against the standard NCERT/NEET classification. …

Concept & Intuition

Glucose is a six-carbon sugar with an aldehyde group at C1 and a primary alcohol group at C6 (the terminal $-\mathrm{CH_2OH}$). The other four carbons carry secondary alcohol groups ($-\mathrm{CHOH}$).

• Mild oxidation (like bromine water) selectively attacks the most easily oxidised group — the aldehyde — turning it into a carboxylic acid. This gives gluconic acid.
• Stronger oxidation (like hot nitric acid) is powerful enough to also oxidise the primary alcohol at C6 to a carboxylic acid. The result is a dicarboxylic acid called saccharic acid. The secondary alcohol groups ($-\mathrm{CHOH}$) are not oxidised under these conditions (they would require even harsher reagents). So the two steps target different functional groups: first the aldehyde, then the primary alcohol.

Step-by-step reasoning

1. Identify the functional groups in glucose

   Glucose has the formula $\mathrm{CHO(CHOH)_4CH_2OH}$. The key groups are:

   • An aldehyde: $-\mathrm{CHO}$ at C1.
   • A primary alcohol: $-\mathrm{CH_2OH}$ at C6.
   • Four secondary alcohols: $-\mathrm{CHOH}$ at C2–C5.

2. First conversion: glucose → gluconic acid

   Reagent: mild oxidising agent (e.g., $\mathrm{Br_2/H_2O}$).

   Reaction: The aldehyde group is oxidised to a carboxylic acid:

The product is gluconic acid: $\mathrm{HOOC(CHOH)_4CH_2OH}$.

Functional group involved: $-\mathrm{CHO}$.

3. Second conversion: gluconic acid → saccharic acid Reagent: strong oxidising agent (e.g., hot $\mathrm{HNO_3}$). Reaction: The primary alcohol at C6 is now oxidised to a carboxylic acid: …

The concept first: how to read a Haworth projection

A Haworth projection is the cyclic hemiacetal drawn as a flat ring seen edge-on. Three questions unlock it, in this order:

1. How many atoms in the ring? Six (5 C + O) = pyranose; five (4 C + O) = furanose. Glucose, an aldohexose, cyclises through C-1 and C-5 giving a six-membered pyranose. Fructose, a ketohexose, cyclises through C-2 and C-5 giving a five-membered furanose. So ring size alone separates glucose from fructose.
2. Which way does $\ce{CH2OH}$ (on C-5) point? Up ⇒ D-sugar; down ⇒ L-sugar.
3. Which way does the $\ce{OH}$ on the anomeric carbon (C-1 for glucose) point?

• Down (i.e. trans / on the opposite side from $\ce{CH2OH}$) ⇒ $\alpha$.
• Up (cis to $\ce{CH2OH}$) ⇒ $\beta$.

Step-by-step

Step 1 — Count the ring. The drawing is a hexagon with the oxygen at the top-right vertex: five carbons + one oxygen ⇒ pyranose. Immediately, both fructofuranose options fall away — fructose does not give a six-membered ring in these options.

Step 2 — Fix the D/L label. C-5 (top-left carbon, next to the ring O) carries $\ce{CH2OH}$ upward. ⇒ D-sugar. …

Concept and Intuition

Amino acids are the fundamental building blocks of proteins. Each amino acid has a central carbon atom (the alpha-carbon) bonded to four different groups:

1. An amino group ($-\text{NH}_2$)
2. A carboxyl group ($-\text{COOH}$)
3. A hydrogen atom ($-\text{H}$)
4. A unique side chain, also known as the R-group

The R-group is what distinguishes one amino acid from another and determines its chemical properties. To identify sulphur-containing amino acids, we need to look for the presence of a sulphur atom within this R-group.

There are 20 standard amino acids commonly found in proteins. Among these, only two contain sulphur atoms in their side chains: Cysteine and Methionine.

Step-by-step Analysis

1. Identify Sulphur-Containing Amino Acids:

   Out of the 20 standard amino acids, Cysteine and Methionine are the only two that contain sulphur in their R-groups.

2. Cysteine:

   Cysteine has a thiol group ($-\text{SH}$) in its side chain. This thiol group is highly reactive and can form disulfide bonds ($-\text{S}-\text{S}-$) with another cysteine residue, which is crucial for stabilizing protein structures.

   The structure of Cysteine is:

Here, the sulphur atom is part of the $-\text{CH}_2\text{SH}$ (thiomethyl) group.

3. Methionine:

Methionine contains a thioether group ($-\text{S}-\text{CH}_3$) in its side chain. Unlike cysteine, the sulphur in methionine is not directly bonded to a hydrogen atom, making it less reactive and unable to form disulfide bonds. Methionine is also unique as it is typically the first amino acid incorporated into a polypeptide chain during protein synthesis.

The structure of Methionine is:

Here, the sulphur atom is part of the $-\text{CH}_2\text{CH}_2\text{SCH}_3$ (ethyl methyl thioether) group.

> [!IMPORTANT]
> Cysteine contains a thiol ($-\text{SH}$) group, while Methionine contains a thioether ($-\text{S}-\text{CH}_3$) group. Both contain sulphur.

4. Addressing the Options (A, B, C, D): …

The field of chemistry, particularly medicinal chemistry and food chemistry, involves a vast array of substances designed for specific purposes. Understanding the function of these chemicals is crucial. This problem tests your knowledge of common examples of drugs and food additives and their classifications. Each substance has a distinct chemical structure that dictates its biological activity or physical property, allowing it to serve a particular role.

Here's how we can match each substance to its correct category:

1. Analyze Cimetidine:

   Cimetidine is a drug primarily used to treat heartburn and stomach ulcers. It works by blocking histamine H2 receptors in the stomach, which reduces the production of stomach acid. Substances that neutralize or reduce stomach acid are classified as antacids.

   • Therefore, Cimetidine matches with III. Antacid.

2. Analyze Alitame:

   Alitame is a high-potency artificial sweetener. It is a dipeptide derivative, significantly sweeter than sucrose (table sugar), and is used as a sugar substitute in various food products.

   • Therefore, Alitame matches with I. Artificial sweetener.

3. Analyze Morphine:

   Morphine is a powerful opioid drug used to relieve severe pain. It acts on the central nervous system to produce its pain-relieving effects. Drugs that alleviate pain are known as analgesics.

   • Therefore, Morphine matches with II. Analgesic.

4. Analyze Iodoform: …

The question asks you to match a given sub-structure to its parent amino acid. The key is to recognize the unique chemical group that defines each amino acid’s side chain. Among the options, proline has a pyrrolidine ring, arginine has a guanidinium group, tyrosine has a phenol ring, and histidine has an imidazole ring. The sub-structure in question is a five-membered ring containing two nitrogen atoms — that is the imidazole ring, exclusive to histidine.

Let’s walk through the reasoning step by step:

1. Identify the sub-structure’s features.

   The given sub-structure is a five-membered ring with two nitrogen atoms (one at position 1 and one at position 3) and three carbon atoms. This is the imidazole ring. It is aromatic and can exist in two tautomeric forms (with the hydrogen on either nitrogen).

2. Recall the side chains of the listed amino acids.

   • (A) Proline: Its side chain is a pyrrolidine ring — a five-membered ring with one nitrogen (a secondary amine). Not a match.
   • (B) Arginine: Its side chain is a straight chain ending in a guanidinium group (three nitrogens, no ring). Not a match.
   • (C) Tyrosine: Its side chain contains a phenol ring (a six-membered aromatic ring with one oxygen). Not a match.
   • (D) Histidine: Its side chain is exactly the imidazole ring. This is the only amino acid among the 20 standard ones that contains an imidazole group.

3. Confirm by elimination. …

The key is to recognise what cold, dilute, aqueous KMnO$_4$ does. This is Baeyer’s reagent — it performs syn dihydroxylation of alkenes, converting each C=C into a vicinal diol (two adjacent –OH groups). It does not cleave the double bond (that requires hot, concentrated KMnO$_4$ or ozonolysis). So every double bond in the molecule gets two –OH groups added across it.

Let’s look at the structure of 5-allylcyclohex-3-ene-1-ol.

1. Draw the molecule. The parent is cyclohex-3-ene-1-ol: a six-membered ring with a double bond between C3 and C4, and an –OH group at C1. At C5, there is an allyl substituent: –CH$_2$–CH=CH$_2$. So the molecule contains two separate double bonds — one in the ring (C3=C4) and one in the allyl side chain (the terminal CH$_2$=CH–).

2. Apply cold KMnO$_4$ to each double bond.

   • The ring double bond (C3=C4) gets two –OH groups added, one on C3 and one on C4.
   • The allyl double bond (CH$_2$=CH–) gets two –OH groups added, one on the terminal carbon and one on the middle carbon of the side chain.

3. Count the –OH groups in the product.

   • The original molecule already has one –OH at C1.
   • The ring dihydroxylation adds two new –OH groups.
   • The side-chain dihydroxylation adds two more –OH groups. …

Why fructose cyclizes differently from glucose

Fructose is a ketohexose — the carbonyl sits at C-2, not C-1. When it cyclizes into a five-membered furanose ring, the C-5 hydroxyl attacks the C-2 ketone, making C-2 the new anomeric center. This is fundamentally different from aldoses like glucose, where C-1 becomes anomeric.

The furanose ring of D-fructose therefore contains carbons 2, 3, 4, 5, and the ring oxygen. Carbon-1 and carbon-6 both end up as $\mathrm{CH_2OH}$ groups outside the ring — C-1 attached to C-2 (the anomeric carbon) and C-6 attached to C-5.

Building the Haworth projection step by step

1. Identify the ring atoms and numbering.

   The furanose ring runs C-2 (anomeric) → C-3 → C-4 → C-5 → O → back to C-2. In a standard Haworth projection, we draw the ring with oxygen at the back (top), C-2 to the right of oxygen, C-3 at the bottom right, C-4 at the bottom left, and C-5 to the left of oxygen.

2. Place the exocyclic $\mathrm{CH_2OH}$ groups.

   • C-1 is bonded to C-2 (the anomeric carbon, upper right position). So $\mathrm{CH_2OH}$ from C-1 projects from the anomeric carbon.
   • C-6 is bonded to C-5 (upper left position). So $\mathrm{HOH_2C}$ (or $\mathrm{CH_2OH}$) projects from C-5.

   This immediately tells us the upper-left and upper-right substituents are both $\mathrm{CH_2OH}$ groups.

3. Determine the anomeric configuration at C-2.

   In D-sugars, the α-anomer has the anomeric OH on the opposite side of the ring from the C-6 $\mathrm{CH_2OH}$ group (which defines the D-configuration). Since C-6 projects upward from C-5 in D-fructose, the α-OH at C-2 must point downward (below the ring plane).

   The anomeric carbon (C-2) therefore carries the C-1 $\mathrm{CH_2OH}$ substituent and an OH group; in α-D-fructofuranose, that OH is down.

4. Assign the remaining hydroxyls using Fischer → Haworth rules.

   In the D-fructose Fischer projection:

   • C-3: OH on the right → projects down in Haworth (below the ring). …

The key here is understanding what HI does to a sugar like glucose. HI is a powerful reducing agent, especially under prolonged heating. It doesn’t just replace one or two –OH groups — it goes all the way. Every hydroxyl group in glucose gets replaced by hydrogen, and the aldehyde group at C1 gets reduced to a methyl group as well. So the entire carbon skeleton becomes a straight-chain alkane.

Let’s walk through it step by step.

1. Glucose’s structure Glucose is an aldohexose: a six-carbon straight chain with an aldehyde group at C1 and a hydroxyl group on every other carbon. Its open-chain form is:

2. What HI does

   HI replaces each –OH group with –I in the first step, but under prolonged heating, the C–I bonds are further reduced to C–H bonds. The aldehyde group (–CHO) is also reduced: first to –CH₂OH, then to –CH₃. So every oxygen atom is removed and replaced by hydrogen.

3. The result on each carbon

   • C1 (originally –CHO) becomes –CH₃.
   • C2 through C5 (originally –CHOH–) each become –CH₂–.
   • C6 (originally –CH₂OH) becomes –CH₃.

   So the product is a straight chain of six carbons with no functional groups:

4. That’s n-hexane …