Organic Chemistry

The stability order of the given species is: I > III > II > IV

There is an increase in the stability of carbocations when an electron-donating group (+I effect) is present.

In 1^st the lone pair on oxygen stabilizes the carbocation as well as it is tertiary carbocation, hence more stable.

The order of the stability of the carbocation is tertiary alkyl > secondary alkyl > primary alkyl> methyl carbonium ion.

To solve the problem, we'll analyze the named reactions in List-I and identify the corresponding major products from List-II. Let's match each:

Based on these connections, we get:

• P (Aldol Condensation) → 3
• Q (Finkelstein Reaction) → 4
• R (Sandmeyer Reaction) → 5
• S (Williamson Ether Synthesis) → 2

Thus, the correct option is: P →3; Q→4 ; R →5; S →2

Given that plotting against results in a straight line, we need to find the ratio using the information provided:

• = molar conductivity.
• = limiting molar conductivity.
• = molar concentration.
• = dissociation constant of HX.

For a monobasic weak acid HX, the relation between molar conductivity and concentration is given by:

Rearranging the equation:

This equation is in the form , where:

So, the line has a y-intercept and a slope .

Thus, the ratio .

However, we seek , which simplifies our ratio to .

Therefore, the correct answer is .

Urea Solution and Osmotic Pressure Calculation

Mole of Urea = 0.2

Weight of Urea is:

Weight of solvent = 1000 g

The volume of the solution is:

50 ml solution contains:

The concentration of the solution is:

Using the formula for osmotic pressure:

Final Answer:

The osmotic pressure is 682.

Sucrose and Its Properties

Sucrose is a non-reducing disaccharide composed of glucose and fructose units joined together by a glycosidic bond. Unlike reducing sugars, sucrose cannot be oxidized by bromine water because it does not have a reducing group (such as an aldehyde or a ketone) that can undergo oxidation.

When sucrose undergoes acid hydrolysis, it breaks down into its constituent monosaccharides, glucose and fructose. This process occurs due to the cleavage of the glycosidic bond between the two monosaccharides. The resulting mixture of glucose and fructose is known as invert sugar.

Invert sugar is laevorotatory, meaning it rotates plane-polarized light to the left. This property arises from the combination of glucose and fructose, which in their free form have optical activities that are not completely canceled out.

Conclusion:

Based on the information provided, the disaccharide X is most likely sucrose.

Number of Heteroatoms in R is 9.

Number of Carbon atoms + Number of Heteroatoms = 51

Osmotic Pressure Calculation

The osmotic pressure (in Torr) of the resulting solution at 300 K is .

Mole of Urea = 0.2

Weight of Urea is:

Weight of solvent = 1000 g

Therefore, the volume of the solution is:

50 ml solution contains:

Therefore, the concentration of the solution is:

Finally, using the formula for osmotic pressure , we get:

Hybridization and Carbon Atoms in Major Product

All marked carbon atoms are hybridized.

There are 28 hybridized carbon atoms in the major product .

Therefore, the total number of hybridized carbon atoms is:

Chemical Reaction Sequence and Heteroatom Calculation

Chemical Reaction Sequence

This image shows a sequence of chemical reactions starting with a compound containing nitro groups. The process involves various reagents and results in several intermediate products:

Step 1: Reduction

The compound with two nitro groups ( ) undergoes a reduction using tin (Sn) and hydrochloric acid (HCl), converting the nitro groups into amino groups ( ). The resulting intermediate is shown as the product .

Step 2: Diazotization

The amino group is diazotized with sodium nitrite ( ) and hydrochloric acid at 0-5°C, forming a diazonium chloride ( ) group. The resulting compound is labeled as .

Step 3: Hydrolysis

In the third step, hydrolysis of the diazonium group leads to the hydroxyl group ( ) in the intermediate .

Step 4: Chlorination

Next, chlorination occurs, resulting in the substitution of the hydroxyl group ( ) with a chloro group ( ) to form .

Step 5: Bromination

In the final step, the compound undergoes bromination using to introduce three bromine atoms ( ) to the aromatic ring, resulting in the product labeled as .

Calculation of Heteroatoms in R

The final compound has three hydroxyl groups ( ), and three bromine atoms ( ), giving us:

• 3 hydroxyl groups contribute 3 oxygen atoms
• 3 bromine atoms contribute 3 bromine atoms

The total number of heteroatoms (atoms that are not carbon) in the structure of is 9, as given in the problem statement.

Compound S Analysis

Analysis of Compound S

(2) Compound S is shown below:

It contains a central benzene ring bonded to three branches. Each branch ends in a phenyl group with a hydroxyl (-OH) substituent.

The structure contains:

• 6 benzene rings (3 from the side arms, 1 central, and 2 in the upward direction)
• 3 azo groups (–N=N–) connecting phenyl rings
• 2 triazole rings (with 3 nitrogen atoms each)
• 3 hydroxyl groups (–OH)

Counting atoms:
Each benzene ring = 6 carbon atoms
Total benzene rings = 9 → carbon atoms

However, due to shared carbons at junctions, let’s assume actual unique carbon count = C
Let’s count the heteroatoms:

• 3 azo linkages → 6 nitrogen atoms
• 2 triazole rings → nitrogen atoms
• 3 hydroxyl groups → 3 oxygen atoms

Total heteroatoms =
Hence,

Reagent Sequence for Compound Transformation

1. The compound contains multiple functional groups, including an ester, a nitrile, and an alkyne.

Reagent Sequence for (C):

1. Step 1: Reduction of ester using forms an alcohol.
2. Step 2: Reduction of nitrile using gives aldehyde.
3. Step 3: Acidic hydrolysis ( ) converts alkyne to a ketone.
4. Step 4: Lindlar’s catalyst reduces alkyne to cis-alkene.

Reagent Sequence for (D):

Similar to (C), but the sequence starts with Lindlar’s catalyst to reduce the alkyne before performing reductions.

Final Conclusion:

The correct pathways result in Q with the desired functional groups.

(P) Stephen reaction:
Stephen reaction reduces nitriles (RCN) to aldehydes. The precursor is usually benzonitrile, which is derived from benzoic acid.
(Q) Sandmeyer reaction:
Used to substitute an amino group on an aromatic ring (from aniline) via diazotization. Requires a nitro compound as a precursor.
(R) Hoffmann bromamide degradation reaction:
Converts amides to amines with one fewer carbon. The amine product is Toluene.
(S) Cannizzaro reaction:
Occurs with aldehydes having no alpha-H (like benzaldehyde), which can be derived from oxidation of toluene.

• P → 2
• Q → 3
• R → 5
• S → 1

Final Answer:

(P): The compound contains an -amino acid backbone → reacts with ninhydrin to give a purple color and phenol group gives violet color with FeCl .
(Q): It is an aryl hydrazone derivative. Upon hydrolysis, it gives an amine and a carbonyl. These give positive ninhydrin test but no phthalein dye.
(R): Aniline salt reacts with diazonium salts to form azo dyes (yellow).
(S): Phenylhydrazine (NHNH ) reacts with glucose to give hydrazones.

• P → 2
• Q → 5
• R → 1
• S → 3

Final Answer:

Step 1: Identify the halides formed
- Compound : Cyclohexyl bromide (R–Br) - Compound : Cyclohexyl iodide (R–I), via Finkelstein reaction - Compound : Cyclohexyl fluoride (R–F), via Swarts reaction Step 2: C–X bond enthalpy trend
Bond enthalpy order for C–X bonds generally follows: So in this case: Why other options are incorrect: (A) C–X bond length increases with atomic size: (C) Reactivity towards depends on leaving group ability. Better leaving group → more reactive: (D) of conjugate acids: Final Answer Final Answer:

Step 1: Track yields through the reaction sequence

Initial moles of X = 16 mol

Step 2: Calculate molar mass of S

S is formed from R (via NaH in ether). Given molar mass of P is 210 and the reaction doesn't drastically change the skeleton, so we consider R P for molar mass estimate. Thus:

Step 3: Calculate mass of S

Final Answer: