Step 1: The Function of Sodium Fusion Extract (Lassaigne's Test)

In the detection of halogens, the organic compound is fused with metallic sodium to convert covalently bonded halogens ($X$, where $X = Cl, Br, I$) into water-soluble ionic sodium halides ($NaX$). The aqueous extract prepared from this fusion is known as the Sodium Fusion Extract (SFE). This extract contains the following species which interfere with subsequent qualitative tests:

• Sodium halides ($NaX$)
• Sodium cyanide ($NaCN$)
• Sodium sulfide ($Na_2S$)

Step 2: Interference by Anions

If silver nitrate ($AgNO_3$) is added directly to the SFE, it will react not only with the halide ions ($X^-$) but also with the cyanide ($CN^-$) and sulfide ($S^{2-}$) ions present, as these also form insoluble precipitates with silver ions:

• $Ag^+ (aq) + CN^- (aq) \rightarrow AgCN (s)$ (White precipitate)
• $2Ag^+ (aq) + S^{2-} (aq) \rightarrow Ag_2S (s)$ (Black precipitate)

Step 3: Role of Nitric Acid ($HNO_3$)

Concentrated nitric acid ($HNO_3$) is added to the SFE and the solution is boiled before the addition of silver nitrate. This step serves two primary chemical purposes:

1. Decomposition of Interfering Ions: The acid decomposes the sodium cyanide and sodium sulfide into volatile gases, removing them from the reaction mixture:
   • $NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$
   • $Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$
2. Prevention of Hydrolysis: It prevents the formation of silver oxide or other silver salts in alkaline media, ensuring the precipitation reaction is specific to halide ions.

Step 4: Summary of Chemical Logic

Final Solution: Nitric acid is added to decompose any sodium cyanide ($NaCN$) and sodium sulfide ($Na_2S$) present in the sodium fusion extract into volatile hydrogen cyanide ($HCN$) and hydrogen sulfide ($H_2S$) gases, respectively. This prevents these ions from forming interfering precipitates (such as $AgCN$ or $Ag_2S$) with silver nitrate, thereby ensuring the test specifically identifies the presence of halide ions.

Theoretical Basis for Hybridization

The hybridization state of a carbon atom is determined by the number of sigma ($\sigma$) bonds and lone pairs attached to it, as defined by the Steric Number ($SN$):

$SN = \text{Number of } \sigma \text{ bonds} + \text{Number of lone pairs}$

• $SN = 4 \implies sp^3$ hybridization (Tetrahedral geometry)
• $SN = 3 \implies sp^2$ hybridization (Trigonal planar geometry)
• $SN = 2 \implies sp$ hybridization (Linear geometry)

Step 1: Analysis of $CH_2=C=O$ (Ketene)

Structure: $H_2C=C=O$

• Terminal $C$ (in $CH_2$): Attached to 2 $H$ atoms ($\sigma$) and 1 double bond ($1 \sigma, 1 \pi$). Total $\sigma = 3$. Hybridization: $sp^2$.
• Central $C$: Attached to 1 double bond on left ($1 \sigma$) and 1 double bond on right ($1 \sigma$). Total $\sigma = 2$. Hybridization: $sp$.

Step 2: Analysis of $CH_3CH=CH_2$ (Propene)

Structure: $H_3C^1-C^2H=C^3H_2$

• $C^1$ ($CH_3$): Attached to 3 $H$ atoms and 1 $C$. Total $\sigma = 4$. Hybridization: $sp^3$.
• $C^2$: Attached to 1 $C$, 1 $H$, and 1 double bond ($1 \sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.
• $C^3$ ($CH_2$): Attached to 2 $H$ atoms and 1 double bond ($1 \sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.

Step 3: Analysis of $(CH_3)_2CO$ (Acetone)

Structure: $CH_3-C(=O)-CH_3$

• Methyl carbons ($CH_3$): Each attached to 3 $H$ and 1 $C$. Total $\sigma = 4$. Hybridization: $sp^3$.
• Carbonyl carbon ($C=O$): Attached to 2 methyl groups and 1 oxygen ($1 \sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.

Step 4: Analysis of $CH_2=CHCN$ (Acrylonitrile)

Structure: $H_2C^1=C^2H-C^3 \equiv N$

• $C^1$: 2 $H$ and 1 $\sigma$ from double bond. Total $\sigma = 3$. Hybridization: $sp^2$.
• $C^2$: 1 $H$, 1 $C^1$ ($\sigma$), 1 $C^3$ ($\sigma$). Total $\sigma = 3$. Hybridization: $sp^2$.
• $C^3$: 1 $C^2$ ($\sigma$) and 1 triple bond to $N$ ($1 \sigma$). Total $\sigma = 2$. Hybridization: $sp$.

Step 5: Analysis of $C_6H_6$ (Benzene)

Benzene is a planar hexagonal ring where each carbon is bonded to two adjacent carbons and one hydrogen. Each carbon participates in one double bond ($1 \sigma + 1 \pi$).

• Each $C$: 2 $\sigma$ (C-C) + 1 $\sigma$ (C-H) = 3 $\sigma$. Hybridization: $sp^2$ for all six carbons.

Summary Table

Final Solution: The hybridization states are: $CH_2=C=O$ ($sp^2, sp$); $CH_3CH=CH_2$ ($sp^3, sp^2, sp^2$); $(CH_3)_2CO$ ($sp^3, sp^2, sp^3$); $CH_2=CHCN$ ($sp^2, sp^2, sp$); $C_6H_6$ (all $sp^2$).

Step 1: Determining the Mass Composition of the Organic Compound

Given the percentage composition of the compound:

• Carbon ($C$) = $69\%$
• Hydrogen ($H$) = $4.8\%$
• Oxygen ($O$) = $100\% - (69\% + 4.8\%) = 26.2\%$

For a sample mass of $m_{sample} = 0.20\text{ g}$, the mass of each element is:

• $m_{C} = 0.69 \times 0.20\text{ g} = 0.138\text{ g}$
• $m_{H} = 0.048 \times 0.20\text{ g} = 0.0096\text{ g}$

Step 2: Stoichiometric Analysis of Combustion

The complete combustion reaction is represented by the oxidation of elements into their respective oxides:

$C + O_2 \rightarrow CO_2$

$2H + \frac{1}{2}O_2 \rightarrow H_2O$

From the law of conservation of mass and stoichiometric proportions [Per Law of Definite Proportions]:

• Molar mass of $C = 12.01\text{ g/mol}$
• Molar mass of $CO_2 = 12.01 + (2 \times 16.00) = 44.01\text{ g/mol}$
• Molar mass of $H = 1.008\text{ g/mol}$
• Molar mass of $H_2O = (2 \times 1.008) + 16.00 = 18.016\text{ g/mol}$

Step 3: Calculating Masses of Products

The mass of $CO_2$ produced is determined by the ratio of the molar masses of $CO_2$ and $C$:

$m_{CO_2} = m_{C} \times \frac{M_{CO_2}}{M_{C}}$

$m_{CO_2} = 0.138\text{ g} \times \frac{44.01}{12.01} \approx 0.5056\text{ g}$

The mass of $H_2O$ produced is determined by the ratio of the molar masses of $H_2O$ and $2H$:

$m_{H_2O} = m_{H} \times \frac{M_{H_2O}}{2 \times M_{H}}$

$m_{H_2O} = 0.0096\text{ g} \times \frac{18.016}{2.016} \approx 0.0858\text{ g}$

Visual Representation of Stoichiometric Conversion

Final Solution: The combustion of 0.20 g of the substance produces 0.506 g of carbon dioxide ($CO_2$) and 0.086 g of water ($H_2O$).

1. Definitions and Operational Parameters

The selection of a separation technique depends on the thermal stability of the substances and the difference in their boiling points.

2. Comparative Analysis of Mechanisms

Step 1: Simple Distillation (Atmospheric Pressure)

This process relies on the principle that if two liquids have widely different boiling points, the more volatile component will vaporize first upon heating [Per Raoult's Law]. The vapor is then cooled to liquid form.

Step 2: Distillation Under Reduced Pressure (Vacuum Distillation)

Liquids boil when their vapor pressure equals the external atmospheric pressure. By employing a vacuum pump to decrease the pressure above the liquid, the vapor pressure required for boiling is reached at a significantly lower temperature [Clausius-Clapeyron Relation].
Equation: $ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Step 3: Steam Distillation

This technique is applied when the compound is steam-volatile and insoluble in water. The total vapor pressure ($P_{total}$) of the system is the sum of the vapor pressure of water ($P_{H_2O}$) and the organic compound ($P_{org}$):
$P_{total} = P_{H_2O} + P_{org}$
Since $P_{total} = \text{Atmospheric Pressure}$, the mixture boils at a temperature lower than the boiling point of either pure water or the pure organic liquid [Dalton's Law of Partial Pressures].

3. Visual Representation of Boiling Point Relationships

Summary of Differentiators

• Simple Distillation: Used for purification of liquids boiling at high temperatures without decomposition.
• Reduced Pressure: Vital for thermally sensitive substances (e.g., Glycerol boiling at $563\,K$ at $760\,mm$ vs. $453\,K$ at $12\,mm$).
• Steam Distillation: Essential for separating organic compounds that decompose before reaching their normal boiling point but are stable in the presence of steam (e.g., Aniline, Essential oils).

Final Solution: The core difference lies in the thermodynamic state of the boiling process: Simple distillation uses atmospheric pressure, reduced pressure distillation lowers the boiling point by decreasing ambient pressure to prevent thermal decomposition, and steam distillation uses the partial pressure of water vapor to distill substances at temperatures below $373\,K$.

Step 1: Structural Representation of the Molecules

To determine the correct IUPAC name, we must first construct the carbon chain for both names provided. Both names imply a pentane backbone (a chain of 5 carbon atoms).

Compound A: 2-Chloro-4-methylpentane

Compound B: 4-Chloro-2-methylpentane

Step 2: Application of IUPAC Lowest Locant Rule

According to the IUPAC nomenclature rules for aliphatic hydrocarbons [Per IUPAC Rule P-14.4], when numbering the parent chain, the set of locants must be the lowest possible at the first point of difference.

Step 3: Application of Alphabetical Priority Rule

When two different numbering sequences yield the same set of locants, the priority is given to the substituent that appears first in alphabetical order [Per IUPAC Rule P-14.5].

• Substituents: Chloro vs. Methyl
• Alphabetical Order: 'C' (Chloro) comes before 'M' (Methyl).
• Therefore, the substituent 'Chloro' must be assigned the lower locant number.

Step 4: Comparative Analysis

• In 2-chloro-4-methylpentane, Chloro is at position 2.
• In 4-chloro-2-methylpentane, Chloro is at position 4.
• Comparing position 2 and position 4, position 2 is lower.

Final Solution: The correct IUPAC name is 2-Chloro-4-methylpentane, as it assigns the lower locant to the substituent that appears first alphabetically (Chloro < Methyl).

Given Variables & Initial Setup

The chemical reaction provided is:

$CH_3CH_2Br + HS^- \rightarrow CH_3CH_2SH + Br^-$

Here, the substrate is bromoethane ($CH_3CH_2Br$) and the attacking species is the hydrosulfide ion ($HS^-$), which acts as a nucleophile. The bromide ion ($Br^-$) acts as the leaving group.

Step 1: Identifying the Reaction Mechanism

To classify this reaction, we observe the fundamental transformation occurring at the electrophilic carbon center ($C_1$ of the ethyl group). The nucleophile ($HS^-$) possesses a lone pair of electrons and is electron-rich. It attacks the carbon atom bonded to the electronegative bromine atom, resulting in the displacement of the bromide ion. [Per the definition of Nucleophilic Substitution reactions].

Step 2: Analytical Breakdown of Atomic Movement

• Nucleophile: $HS^-$ (electron donor)
• Electrophile: $CH_3CH_2^\delta^+ - Br^{\delta^-}$ (electron acceptor at the $\alpha$-carbon)
• Leaving Group: $Br^-$ (anionic species that departs with the electron pair from the $C-Br$ bond)

The reaction follows the general form:

$Nu^- + R-LG \rightarrow R-Nu + LG^-$

where $Nu = HS^-$, $R = CH_3CH_2-$, and $LG = Br^-$. This mechanism is classically defined as a Nucleophilic Substitution reaction.

Step 3: Visual Representation

Step 4: Theoretical Justification

The reaction involves the replacement of a bromine atom by a hydrosulfide group. Since the incoming group is a nucleophile and the departing group is a stable anion, the reaction is categorized under Nucleophilic Substitution ($S_N$ type). Specifically, as bromoethane is a primary alkyl halide, this reaction typically follows an $S_N2$ mechanism where the attack and departure occur in a concerted step [By IUPAC nomenclature standards and reaction kinetics theory].

Final Solution: The reaction is a Nucleophilic Substitution reaction.

Step 1: Analyzing the IUPAC Nomenclature

The name 2,2,4-trimethylpentane signifies the following structural parameters:

• Parent Chain: "Pentane" indicates a longest continuous carbon chain of $n=5$ atoms.
• Substituents: "Trimethyl" indicates three methyl ($-CH_3$) groups attached to the parent chain.
• Positions: The substituents are located at carbons $2, 2,$ and $4$ of the pentane chain.

Step 2: Deriving the Condensed Structural Formula

To construct the molecule, place the five carbon atoms in a row and assign the methyl branches accordingly:

1. Carbon-1: $CH_3$
2. Carbon-2: Quaternary carbon bonded to two methyl groups ($C(CH_3)_2$)
3. Carbon-3: $CH_2$ group
4. Carbon-4: Methine group bonded to one methyl group ($CH(CH_3)$)
5. Carbon-5: $CH_3$

Combining these, the condensed formula is: $(CH_3)_3CCH_2CH(CH_3)_2$

Step 3: Generating the Bond-Line Structural Formula

In bond-line notation, carbon atoms are represented by line endings and vertices. Hydrogen atoms attached to carbons are omitted [Per IUPAC Convention for Bond-Line Structures].

Step 4: Identification of Functional Groups

Examine the molecular structure for heteroatoms or characteristic unsaturation:

• The molecule consists entirely of carbon-carbon and carbon-hydrogen single bonds ($C-C$ and $C-H$).
• There are no double bonds, triple bonds, or heteroatoms (e.g., $O, N, S, X$).
• Conclusion: The compound belongs to the class of Alkanes.
• Functional Group Status: Alkanes do not possess a specific functional group; they represent the hydrocarbon skeleton.

Final Solution: The condensed formula is $(CH_3)_3CCH_2CH(CH_3)_2$. There are no functional groups present, as the compound is an alkane.

Step 1: Chemical Basis of Lassaigne’s Test

The Lassaigne’s test (or sodium fusion test) is utilized to detect the presence of nitrogen, sulfur, and halogens in organic compounds. In the case of nitrogen-containing organic compounds, the fusion of the substance with metallic sodium converts covalently bonded nitrogen into ionic sodium cyanide ($NaCN$).

The chemical reaction is represented as: $Na + C + N \xrightarrow{\Delta} NaCN$ [Ref: Chemical fusion principle]

Step 2: Conversion to Ferrous Cyanide

The resulting $NaCN$ extract is treated with freshly prepared ferrous sulfate ($FeSO_4$). This reacts to form sodium ferrocyanide, chemically known as sodium hexacyanoferrate(II).

The reaction is: $FeSO_4 + 6NaCN \rightarrow Na_4[Fe(CN)_6] + Na_2SO_4$ [Ref: Ligand substitution in coordination chemistry]

Step 3: Formation of Ferri-Ferrocyanide (Prussian Blue)

Upon addition of a small amount of dilute sulfuric acid or hydrochloric acid, some $FeSO_4$ is oxidized to $Fe^{3+}$ ions (ferric ions) by the atmosphere or the acidic medium. These ferric ions react with the sodium ferrocyanide to form the complex iron(III) hexacyanoferrate(II).

The balanced stoichiometric equation is: $3Na_4[Fe(CN)_6] + 4Fe^{3+} \rightarrow Fe_4[Fe(CN)_6]_3 \cdot xH_2O$ [Ref: Coordination complex precipitation]

Step 4: Molecular Structure Analysis

The deep blue precipitate, known as Prussian blue, is mathematically and chemically defined as ferric ferrocyanide. The coordination entity consists of an octahedral $[Fe(CN)_6]^{4-}$ core coordinating with $Fe^{3+}$ cations.

Step 5: Tabular Summary of Reagents

Final Solution: In the Lassaigne’s test, the Prussian blue colour is obtained due to the formation of Ferric ferrocyanide, chemically represented as $Fe_4[Fe(CN)_6]_3$.

Step 1: Identifying the Parent Carbon Chain

The chemical structure provided contains a benzene ring as a substituent attached to a six-carbon aliphatic chain. Following IUPAC nomenclature rules for hydrocarbons [Per IUPAC Blue Book P-44.4.1], the parent chain is determined by the length of the aliphatic backbone. Here, the longest carbon chain consists of 6 carbon atoms, which corresponds to the parent alkane hexane.

Step 2: Locating the Substituent

We must number the chain to give the substituent the lowest possible locant.

• Numbering from left to right: The phenyl group is attached to carbon-2.
• Numbering from right to left: The phenyl group would be attached to carbon-5.

Step 3: Determining Substituent Identity

A benzene ring attached to an alkyl chain is named as a phenyl group [Per IUPAC nomenclature for cyclic substituents].

Step 4: Assembling the IUPAC Name

Combining the substituent locant (2-), the substituent name (phenyl), and the parent chain name (hexane):

Resulting Name: 2-phenylhexane

Step 5: Visual Representation

Final Solution: 2-phenylhexane

Theoretical Foundation: Inductive and Hyperconjugative Effects

Alkyl groups attached to $\pi$ systems (such as alkenes, alkynes, or benzene rings) exhibit electron-donating properties primarily through two distinct electronic phenomena: the Inductive Effect and the Hyperconjugation Effect (also known as the "no-bond resonance" or Baker-Nathan effect).

Step 1: The Inductive Effect (+I Effect)

The alkyl group (e.g., $-CH_3$, $-CH_2CH_3$) consists of carbon atoms with $sp^3$ hybridization. The electronegativity of $sp^3$ hybridized carbon (approx. 2.5 on the Pauling scale) is slightly higher than that of hydrogen (approx. 2.1). However, when an alkyl group is attached to an $sp^2$ hybridized carbon of a $\pi$ system, the difference in electronegativity creates a polarity gradient [Per the Inductive Effect Principle]. The alkyl group effectively releases electron density through the $\sigma$-bond framework toward the more electronegative $sp^2$ carbon, acting as an electron-donating group.

Step 2: The Hyperconjugation Effect (Resonance)

The primary reason for the significant electron donation from an alkyl group is the phenomenon of hyperconjugation. This involves the delocalization of $\sigma$-electrons of the $C-H$ bond of an alkyl group into the adjacent empty or partially filled $\pi$-orbital or $p$-orbital of the $\pi$ system.

For an alkyl group to exhibit hyperconjugation, it must possess at least one $\alpha$-hydrogen atom attached to the $\alpha$-carbon, which is directly linked to the unsaturated system. The overlap occurs between the $C-H$ $\sigma$-bonding orbital and the adjacent $\pi$-orbital.

Step 3: Visual Representation of Hyperconjugation

Step 4: Analytical Summary of Electron Donation

The total electron-donating capacity of the alkyl group is the sum of these effects. The following table summarizes the mechanisms:

Because the $\pi$ system typically has an electron-deficient nature or a lower energy $p$-orbital compared to the $\sigma$-bonding electrons of the alkyl group, the electrons flow into the $\pi$ system, increasing the electron density on the attached unsaturated carbon. This effectively stabilizes the $\pi$ system (e.g., increasing the stability of alkenes or activating the benzene ring toward electrophilic substitution).

Final Solution: Alkyl groups act as electron donors because they exert a positive inductive effect (+I) due to the polarizability of $\sigma$-bonds and, more significantly, they undergo hyperconjugation where the $C-H$ $\sigma$-electrons are delocalized into the adjacent $\pi$ system, thereby increasing the electron density of the $\pi$ system.

Step 1: Analyzing the Reaction Components

The given chemical equation is:

$(CH_3)_2C = CH_2 + HCl \rightarrow (CH_3)_2ClC - CH_3$

In this transformation, we observe two distinct chemical phenomena:

• Substrate: 2-Methylpropene, an unsaturated hydrocarbon containing a carbon-carbon double bond ($C=C$). [Per IUPAC nomenclature for alkenes].
• Reagent: Hydrogen chloride ($HCl$), a polar electrophilic reagent.
• Product: 2-Chloro-2-methylpropane, a saturated haloalkane.

Step 2: Identifying the Reaction Mechanism

The reaction involves the breaking of the $\pi$-bond of the alkene and the addition of the $H^+$ and $Cl^-$ species across the double bond. This is a characteristic Electrophilic Addition Reaction [Per the mechanism of electrophilic addition to alkenes].

The process follows these sub-stages:

1. Formation of Electrophile: The double bond attacks the electrophilic proton ($H^+$) from $HCl$.
2. Carbocation Formation: The proton adds to the terminal carbon to form a tertiary carbocation $(CH_3)_3C^+$, which is highly stable due to the inductive effect and hyperconjugation.
3. Nucleophilic Attack: The chloride ion ($Cl^-$) acts as a nucleophile and attacks the tertiary carbocation to form the final product.

Step 3: Visualizing the Molecular Geometry

Step 4: Theoretical Justification

According to Markovnikov's Rule, in the addition of a protic acid to an asymmetric alkene, the acid hydrogen ($H$) becomes attached to the carbon with fewer alkyl substituents (or the carbon with more hydrogens), and the halide group ($Cl$) attaches to the carbon with more alkyl substituents [Per Markovnikov’s Regioselectivity Principle].

Final Solution: The reaction is classified as an Electrophilic Addition Reaction.

1. Fundamental Principle of Crystallisation

The principle of crystallisation is based upon the difference in the solubility of the compound and its impurities in a suitable solvent. The technique relies on the fact that a substance becomes more soluble in a solvent at higher temperatures and less soluble at lower temperatures [Per the principle of solubility-temperature dependence].

2. Operational Methodology

• Dissolution: The crude solid is dissolved in the minimum amount of a suitable solvent at or near its boiling point [Thermodynamic saturation].
• Hot Filtration: The hot solution is filtered to remove any insoluble impurities.
• Cooling/Nucleation: The filtrate is allowed to cool slowly. As the temperature decreases, the solubility of the compound drops, forcing it to crystallize out of the solution.
• Isolation: The pure crystals are separated from the "mother liquor" (which contains the dissolved impurities) via filtration under reduced pressure.

3. Visual Representation of Solubility Curve

4. Exemplification: Purification of Benzoic Acid

To purify crude benzoic acid containing sand or charcoal as an impurity:

1. Dissolution: Dissolve the crude benzoic acid in minimum boiling water [Solvent selection based on polarity matching].
2. Filtration: Pass the hot solution through filter paper; sand/charcoal remains as the residue [Separation by physical state].
3. Crystallization: Allow the filtrate to cool to room temperature. Pure, white needle-like crystals of benzoic acid precipitate out [Per the principle of super-saturation].
4. Verification: The purity is confirmed by checking the melting point, which should match the standard value ($122.4 ^\circ C$) [Analytical validation].

Final Solution: Crystallisation is a process of purifying a solid substance based on the differential solubility of the solute and impurities at varying temperatures. By dissolving the impure solid in a minimum volume of hot solvent and subsequently cooling the solution, the pure compound crystallizes out while the impurities remain in the mother liquor.

Theoretical Foundation: Lassaigne’s Test

The elemental analysis of organic compounds requires the conversion of elements chemically bound in covalent form into their corresponding ionic forms. Because organic compounds typically contain elements like Nitrogen ($N$), Sulphur ($S$), and Halogens ($X$) bonded covalently to a carbon skeleton, they do not ionize in water to yield characteristic tests. The sodium fusion process, also known as the Lassaigne’s test, is used to facilitate this conversion.

Step 1: Chemical Transformation via Sodium Fusion

When an organic compound is fused with metallic sodium ($Na$), the strong reducing nature of sodium at high temperatures breaks the covalent bonds between carbon and the heteroatoms ($N, S, X$). This results in the formation of water-soluble sodium salts.

Step 2: Chemical Reactions for Specific Elements

The transformation reactions are represented by the following chemical equations:

• For Nitrogen: The carbon and nitrogen present in the compound react with sodium to form sodium cyanide.
  $Na + C + N \rightarrow NaCN$
• For Sulphur: The sulphur in the compound reacts with sodium to form sodium sulphide.
  $2Na + S \rightarrow Na_2S$
• For Halogens: The halogens (where $X = Cl, Br, I$) react with sodium to form sodium halides.
  $Na + X \rightarrow NaX$
• For Nitrogen and Sulphur simultaneously: If both elements are present, they react to form sodium thiocyanate.
  $Na + C + N + S \rightarrow NaSCN$

Step 3: Justification for Fusion

The fusion is essential for the following reasons:

1. Ionic Solubility: Elements in covalent compounds are held by shared electron pairs. Fusion with $Na$ [per the principle of chemical reduction] converts these into ionic salts ($NaCN, Na_2S, NaX$), which are highly soluble in water, allowing them to dissociate into their constituent ions ($CN^-$, $S^{2-}$, $X^-$).
2. Reactive Accessibility: Once the extract (Lassaigne’s extract) is prepared by boiling the fused mass in distilled water, the dissociated ions become readily available for identification through qualitative inorganic precipitation or color-change reactions.

Visual Representation of Fusion Process

Final Solution: The fusion of an organic compound with metallic sodium is necessary to convert the covalently bonded elements (Nitrogen, Sulphur, and Halogens) into their corresponding water-soluble ionic salts ($NaCN, Na_2S, NaX$). This process renders these elements detectable through standard aqueous qualitative analysis, as they would otherwise remain unreactive in their original covalent states.

Step 1: Structural Analysis of the Given Pair

The provided pair consists of two structures:

• Structure A: $CH_3-CH=CH-CH_3$ (But-2-ene)
• Structure B: $CH_3-CH=CH-CH_3$ (But-2-ene)

Upon analysis of the connectivity, both molecules possess the same molecular formula ($C_4H_8$) and the same connectivity of atoms (the double bond is positioned between the second and third carbon atoms in both). [Per IUPAC Nomenclature rules for constitutional isomers].

Step 2: Evaluating Isomerism

Since the connectivity (constitution) is identical, these are not structural isomers. To determine if they are stereoisomers, we examine the spatial arrangement of the substituents around the $C=C$ double bond.

Step 3: Justification of Geometric Isomerism

The restriction of rotation around the carbon-carbon double bond leads to different spatial orientations. In the cis-form, similar groups ($CH_3$) are on the same side of the double bond, while in the trans-form, they are on opposite sides. [Per the Cahn-Ingold-Prelog priority rules and the definition of diastereomers].

Conclusion

Since the structures differ only in the spatial arrangement of atoms/groups across a rigid double bond without altering the structural connectivity, they are categorized as geometrical isomers (a subset of stereoisomers).

Final Solution: The members of the pair are Geometrical Isomers.

Given Variables & Analysis of Carbocation Stability

Stability of a carbocation is primarily governed by three electronic effects: Inductive Effect ($+I/-I$), Hyperconjugation, and Resonance. The stability order generally follows: Resonance > Hyperconjugation > Inductive Effect.

The options typically presented for this problem are: $(CH_3)_3C^+$ (tert-butyl), $(CH_3)_2CH^+$ (isopropyl), $CH_3CH_2^+$ (ethyl), and $CH_3^+$ (methyl). Below is the analytical breakdown of their stability:

Step 1: Evaluation of Hyperconjugation

Hyperconjugation involves the delocalization of $\sigma$-electrons of a $C-H$ bond attached to an $\alpha$-carbon into the empty $p$-orbital of the carbocation. The number of hyperconjugative structures is equal to the number of $\alpha$-hydrogens.

Step 2: Analysis of Inductive Effect ($+I$)

Alkyl groups are electron-releasing in nature [$+I$ effect]. In $(CH_3)_3C^+$, three methyl groups donate electron density toward the positively charged central carbon, effectively dispersing the positive charge and lowering the internal energy of the system [Per the principle of Charge Delocalization].

Step 3: Comparative Graphical Representation

Step 4: Logical Conclusion

By comparing the hyperconjugative structures and the inductive effects, $(CH_3)_3C^+$ possesses 9 $\alpha$-hydrogens [Most effective stabilization] and the highest degree of $+I$ donor capability. Therefore, it is the most stable among simple alkyl carbocations.

Final Solution: The most stable carbocation is $(CH_3)_3C^+$ (tert-butyl carbocation) due to the maximum number of hyperconjugative structures (9) and the cumulative $+I$ inductive effect of three methyl groups.

1. Analysis of the Chemical Equation

The given reaction is: $C_6H_6 + \mathbf{CH_3C^+O} \rightarrow C_6H_5COCH_3$

To identify the nature of the species $CH_3C^+O$ (the acetylium ion), we must examine its electronic configuration and its role in the reaction mechanism [Per the theory of electrophilic aromatic substitution].

2. Electronic Configuration of the Acetylium Ion ($CH_3C^+O$)

The species $\mathbf{CH_3C^+O}$ can be represented by resonance structures:

• Structure A: $CH_3 - C \equiv O^+$
• Structure B: $CH_3 - C^+ = O$

In structure B, the carbon atom bears a formal positive charge ($C^+$) and has an incomplete octet (only 6 electrons in its valence shell). Even in structure A, the carbon atom is electron-deficient due to the high electronegativity of the oxygen atom pulling electron density away via the triple bond [By Inductive and Mesomeric effects].

3. Classification Logic

• Nucleophile: A chemical species that donates an electron pair to form a chemical bond in a reaction [Definition by Lewis Acid-Base Theory].
• Electrophile: A chemical species that accepts an electron pair to form a chemical bond, characterized by an electron-deficient center [Definition by Lewis Acid-Base Theory].

In this reaction, $C_6H_6$ (Benzene) acts as a nucleophile, providing a pair of $\pi$-electrons from its aromatic ring to attack the positively charged carbon atom of the acetylium ion ($CH_3C^+O$). Therefore, $CH_3C^+O$ functions as the electron acceptor.

4. Visual Representation of the Electrophilic Attack

5. Conclusion

Since $\mathbf{CH_3C^+O}$ possesses a vacant orbital or a positive charge center that seeks electron density to stabilize its valence shell, it fits the definition of an electrophile.

Final Solution: The reagent $\mathbf{CH_3C^+O}$ is an electrophile.

Step 1: Identification of Chemical Species

The two species under comparison are:

• Species A: $O_2NCH_2CH_2O^–$ (2-nitroethoxide ion)
• Species B: $CH_3CH_2O^–$ (ethoxide ion)

Both species are alkoxide ions characterized by a negatively charged oxygen atom bonded to a carbon chain. The stability of an anion is primarily dictated by the dispersal of its negative charge [Per the Principle of Charge Delocalization].

Step 2: Analysis of Inductive Effects

The stability of the conjugate base is enhanced by the presence of electron-withdrawing groups (EWG) which decrease the electron density on the oxygen atom through the Inductive Effect ($-I$ effect).

• In Species B ($CH_3CH_2O^–$): The ethyl group ($-CH_2CH_3$) acts as an electron-donating group ($+I$ effect). This increases the electron density on the already negatively charged oxygen atom, destabilizing the species [Per the Inductive Effect Theory].
• In Species A ($O_2NCH_2CH_2O^–$): The nitro group ($-NO_2$) is a powerful electron-withdrawing group ($-I$ effect). It pulls electron density away from the oxygen atom through the carbon chain.

Step 3: Comparative Analysis of Charge Density

The stability of an ion is inversely proportional to the concentration of charge on the central atom. The $-I$ effect of the $-NO_2$ group reduces the negative charge density on the oxygen atom in $O_2NCH_2CH_2O^–$.

Step 4: Conclusion on Thermodynamic Stability

Because the nitro group significantly stabilizes the negative charge via the inductive effect, $O_2NCH_2CH_2O^–$ is significantly more stable than the ethoxide ion.

Final Solution: The $O_2NCH_2CH_2O^–$ ion is more stable than the $CH_3CH_2O^–$ ion because the nitro group ($-NO_2$) exerts a strong electron-withdrawing inductive effect ($-I$ effect), which disperses the negative charge on the oxygen atom, whereas the ethyl group in $CH_3CH_2O^–$ exerts an electron-donating inductive effect ($+I$ effect), which increases the electron density and destabilizes the ion.