Class 11 Chemistry (Chemistry Part-II) Chapter 8 Organic Chemistry-Some Basic Principles And Technique

Key Concept: Calculation Formula

c) 72.73%

[Solution Description]

To find the percentage of oxygen, first calculate the amount of carbon present: $\text{Molar mass of } \text{CO}_2 = 44 \, \text{g/mol}$

The number of moles of $\text{CO}_2$ formed is: $\frac{88 \, \text{g}}{44 \, \text{g/mol}} = 2 \, \text{mol}$

Since one mole of $\text{CO}_2$ contains one mole of C, 2 mol of $\text{CO}_2$ contains 2 mol of C, which is: $2 \times 12 \, \text{g/mol} = 24 \, \text{g}$

Total mass of organic compound = Mass of C + Mass of O

$\text{Mass of O} = 88 \, \text{g} – 24 \, \text{g} = 64 \, \text{g}$

Percentage of oxygen is: $\left( \frac{64}{88} \right) \times 100 = 72.73\%$

Key Concept: Complex Reaction Analysis, Comparative Analysis

a) Ammonium phosphomolybdate: 0.413%, $Mg_2P_2O_7$: 4.19%

[Solution Description]

Using the formula for ammonium phosphomolybdate: $\frac{0.25 \times 31 \times 100}{1877 \times 1} = \frac{7.75}{1877} = 0.413\%$

For $Mg_2P_2O_7$: $\frac{0.15 \times 62 \times 100}{222 \times 1} = \frac{930}{222} = 4.19\%$

Therefore, the $Mg_2P_2O_7$ method indicates a higher percentage of phosphorus at 4.19%.

Key Concept: Application of Formula

b) 41.16%

[Solution Description]

To calculate the percentage of sulphur, we use the formula:

$\text{Percentage of Sulphur} = \left( \frac{\text{Mass of S in BaSO}_4}{\text{Mass of Organic Compound}} \right) \times 100$

The molar mass of $BaSO_4$ is about 233 g/mol and that of S is 32 g/mol. So, the ratio of S in $BaSO_4$ is:

$\frac{32}{233}$

Therefore, the mass of S in 0.750 g of $BaSO_4$ is: $\frac{32}{233} \times 0.750 = 0.1029 \, \text{g}$

Now, substituting in the first equation: $\text{Percentage of Sulphur} = \left( \frac{0.1029}{0.250} \right) \times 100 = 41.16\%$

Key Concept: Advanced Conceptual Understanding, Critical Analysis

c) Iodine

[Solution Description]

In the qualitative test for halogens, the behavior of the silver halides with ammonium hydroxide and concentrated nitric acid can help determine which halogen is present.

– Silver chloride (AgCl) dissolves in dilute ammonium hydroxide due to the formation of a complex ion.

– Silver bromide (AgBr) is sparingly soluble.

– Silver iodide (AgI) is insoluble in ammonium hydroxide.

Since the precipitate is insoluble in ammonium hydroxide but soluble in concentrated nitric acid, it indicates that it is likely silver iodide (AgI).

Therefore, iodine is the possible halogen present.

Key Concept: Method Suitability, Multi-step Problem Solving

c) Aromatic nitrogen compounds resistant to digestion

[Solution Description]

If the Kjeldahl method provides significantly lower nitrogen values compared to Dumas, it might indicate the presence of nitrogen in forms that are not amenable to Kjeldahl analysis. For instance, the presence of nitrogen in aromatic heterocyclic compounds, like pyridines, can result in incomplete conversion to ammonium sulfate because such structures resist breakdown even after prolonged boiling in sulfuric acid. Thus, these nitrogen atoms will not be accounted for in the Kjeldahl method, leading to lower measured nitrogen content than what is actually present and detectable by Dumas method, where all nitrogen is converted to gaseous form.

Key Concept: Real-world Application, Combustion Analysis, Free Radicals

b) Unsaturated hydrocarbons

[Solution Description] Unsaturated hydrocarbons typically burn less completely than saturated ones, often leading to the emission of unburnt hydrocarbons and CO due to incomplete combustion. Their multiple bonds require specific conditions for complete oxidation. Thus, the likely culprit behind such pollution is unsaturated hydrocarbons.

Key Concept: Real-World Application, Problem Solving

c) 45.8%

[Solution Description] To convert phosphorus content to phosphorus pentoxide ($P_2O_5$), utilize the relation between their molar masses.

First, recognize that the molar mass of P is 30.97 g/mol and that of $P_2O_5$ is $(2\times30.97 + 5\times16) = 141.94$ g/mol.

Knowing that 1 mole of P yields half a mole of $P_2O_5$, calculate the conversion factor:

$\text{Conversion factor} = \frac{141.94 \text{ g/mol}}{61.94 \text{ g/mol}} = 2.29$

Therefore, if the phosphorus content is 20%, the equivalence in terms of $P_2O_5$ will be: $\text{Equivalent P}_2\text{O}_5\% = 20\% \times 2.29 = 45.8%$

Key Concept: Comparative Analysis, Real-world Application

a) Insufficient ammonium molybdate leads to false negatives in phosphorus detection, similar to lacking lead acetate in sulphur tests.

[Solution Description] Ammonium molybdate is crucial in forming the yellow precipitate indicative of phosphorus presence. Using inadequate amounts may not provide enough reagent to react completely with all available phosphate ions, potentially leading to undetected phosphorus. Conversely, using an excessive amount might not worsen the test result since it provides an abundance of reagent ensuring full reaction, unlike some elemental tests where reagent excess could cause side reactions affecting results.

Key Concept: Reaction Mechanism, Advanced Solubility

b) Chlorine; dissolves due to complex ion formation

[Solution Description] The formation of a white precipitate that dissolves in excess ammonium hydroxide indicates the presence of chlorine. This is due to the formation of $AgCl$, which is more soluble in ammonia compared to other silver halides such as $AgBr$ or $AgI$. The dissolution is explained by the equilibrium: $AgCl_{(s)} + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-_{(aq)}$, where AgCl forms a complex ion with ammonia, making it soluble.

Key Concept: Reaction Mechanism, Advanced Comparative Analysis

c) Formation of ferric thiocyanate

[Solution Description] The blood-red color arises from the formation of the ferric thiocyanate complex. Initially, sodium thiocyanate ($NaSCN$) is formed due to the presence of both nitrogen and sulphur. When $Fe^{3+}$ ions are subsequently added, they react with $SCN^{-}$ ions to form the blood-red colored complex $[Fe(SCN)]^{2+}$. The reaction sequence is: $Na + C + N + S \rightarrow NaSCN$

followed by $Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$

Key Concept: Detailed Process Analysis, Real-world Application

d) Ferrous ions participate in forming Prussian blue complex.

[Solution Description]

In the confirmation test involving Prussian blue formation, ferrous ions ($Fe^{2+}$) react with hexacyanoferrate(III) ions to form insoluble ferric ferrocyanide, commonly known as Prussian blue. The reaction mechanism can be described as follows:

$Fe^{2+} + [Fe(CN)_6]^{3-} \rightarrow Fe^{3+}_{4}[Fe(CN)_6]_{3}$ (Prussian Blue)

This indicates the conversion of ferrous ions into ferric state by combining with cyanide complexes, reflecting their key role in the distinctive blue color development.

Key Concept: Element Conversion

c) $Na_2$S

[Solution Description] During the Lassaigne’s test, when an organic compound containing sulphur is fused with sodium, it forms sodium sulphide ($Na_2$S). This process involves converting covalently bonded sulphur atoms into ionic form as part of the detection procedure.

Key Concept: Comparative Analysis, Complex Reaction Analysis

c) Copper(II) oxide method due to water formation

[Solution Description] Evaluating the effectiveness of different methods requires analyzing their sensitivity and reliability in detecting hydrogen:

– Method using copper(II) oxide: This approach involves heating the compound, allowing hydrogen to oxidize into water vapor. Anhydrous copper sulfate then absorbs this water, turning blue, visibly indicating hydrogen presence.

While both methods can work, the copper(II) oxide method directly transforms hydrogen into measurable water, offering clear qualitative results through color change when anhydrous copper sulfate turns from white to blue. Hence, it’s generally considered more direct and reliable for simple hydrogen detection.

Key Concept: Complex Reaction, Multi-step Process

a) Chlorine

[Solution Description] The formation of a white precipitate upon treatment with silver nitrate, which is soluble in aqueous ammonia, confirms the presence of chloride ions ($\text{Cl}^-$). Silver chloride ($\text{AgCl}$), a white precipitate, forms under these conditions and dissolves in ammonia.

Key Concept: Real-world Application, Critical Analysis

c) Use a more polar solvent.

[Solution Description] If the pigments are not separating well and remain close to the baseline, it indicates they haven’t traveled much with the solvent front. This scenario often requires changing the solvent’s polarity. To improve separation among closely packed pigments, increasing the solvent polarity can help dissolve more components and carry them further up the stationary phase, thus improving their separation on the chromatogram.

Key Concept: Comparative Chromatography, Advanced Reasoning

c) TLC is faster and cheaper; ideal for quick screening

[Solution Description] TLC is generally faster and less expensive than column chromatography, allowing for quick screening of multiple samples simultaneously, which is beneficial in the initial stages of pharmaceutical quality control. However, TLC may lack the resolution and separation capacity of column chromatography, limiting its effectiveness for complex mixtures or detailed analyses. Despite these disadvantages, TLC’s simplicity, speed, and cost-effectiveness make it ideal for preliminary checks and routine monitoring.

Key Concept: Advanced Rf Analysis, Conceptual Integration

c) Compound B moves faster; separation is efficient.

[Solution Description] In thin layer chromatography, a higher Rf value implies that compound B moved further up the stationary phase compared to compound A. Translating this into column chromatography with a polar stationary phase, compound B interacts less with the polar stationary phase and thus moves faster through the column than compound A. This suggests better separation efficiency and faster elution for compound B.

Key Concept: TLC vs Column, Calculation of Retardation Factor

b) It would be strongly adsorbed and elute last.

[Solution Description] The retardation factor (Rf value) indicates the proportion of the distance travelled by the compound relative to the solvent front in TLC. An Rf value of 0.3 implies that the compound has significant interaction with the stationary phase, showing moderate progression compared to the mobile phase. In column chromatography, which shares the same stationary phase, this suggests it would be retained longer within the column due to stronger adsorption interactions but will eventually be eluted depending on the polarity of the eluant used.

Key Concept: Mechanism Exploration, Real-world Application

b) 0.5; indicating balanced interaction with both phases

[Solution Description] The Rf value in paper chromatography is calculated as the ratio of the distance moved by the compound to the distance moved by the solvent front. Given distances: the spot (4 cm) and the solvent front (8 cm), the Rf value is $Rf = \frac{4\, \text{cm}}{8\, \text{cm}} = 0.5$. An Rf of 0.5 indicates moderate solubility in the mobile phase compared to the stationary phase, suggesting balanced interactions between the pigment and both phases.

Key Concept: Advanced Technique Analysis, Numerical and Analytical Difficulty

a) Component 1 with an Rf value of 0.25 has the highest affinity for the stationary phase

[Solution Description] The Rf value is calculated using the formula $Rf = \frac{\text{Distance moved by the compound}}{\text{Distance moved by the solvent front}}$. The Rf values for the compounds are:

Component 1: $Rf_1 = \frac{1.5}{6} = 0.25$

Component 2: $Rf_2 = \frac{3}{6} = 0.5$

Component 3: $Rf_3 = \frac{4.5}{6} = 0.75$

The lower the Rf value, the higher the affinity for the stationary phase. Component 1 has the lowest Rf value and therefore the highest affinity for the stationary phase.

Key Concept: Equipment Usage

c) Separatory funnel

[Solution Description] In differential extraction, a separatory funnel is used to separate the layers formed by the immiscible organic solvent and aqueous solution. It allows the denser layer to be removed from the bottom while retaining the other layer.

Key Concept: Process Optimization, Advanced Reasoning

d) Using a lower boiling point organic liquid

[Solution Description] Optimizing steam distillation involves ensuring efficient separation while maintaining energy efficacy. Using a lower boiling point liquid facilitates easier vaporization by reducing the energy consumption needed for boiling.

Key Concept: Concept Application

b) 0.75

[Solution Description] The retardation factor (Rf) value in thin layer chromatography is calculated using the formula $Rf = \frac{\text{distance travelled by the compound}}{\text{distance travelled by the solvent front}}$. Here, the distance travelled by the solvent front is 10 cm, and the distance travelled by the compound is 7.5 cm. Thus, the Rf value is: $Rf = \frac{7.5 \, \text{cm}}{10 \, \text{cm}} = 0.75$

Key Concept: Theoretical Plates

b) They increase efficiency by providing more surface area for separation

[Solution Description] Theoretical plates provide multiple stages of condensation and re-vaporization within the column, enhancing the separation efficiency by allowing more surface area for interaction between ascending vapours and condensed liquid. More theoretical plates mean better separation of components with close boiling points.

Key Concept: Advanced Application, Conceptual Analysis

b) Fractional distillation

[Solution Description] For liquids with boiling points close together and low relative volatility, fractional distillation is preferred. The fractionating column allows for multiple condensation and vaporization cycles, increasing the efficiency of separation.

Key Concept: Continuous Extraction Technique, Differential Extraction

d) Change the pH of the aqueous phase

[Solution Description] For an organic compound with poor solubility in both water and typical organic solvents, enhancing the solvent’s ability to dissolve the target compound is crucial. This requires selecting a solvent that better mimics the polarity or structure of the compound, or modifying the pH of the aqueous phase to alter the compound’s ionization state, hence improving its solubility in the chosen solvent for more efficient extraction.

Key Concept: Solvent Mixture Strategy, Crystallisation Efficiency

c) To achieve controlled crystallisation and maximize yield

[Solution Description] The selection of solvents for crystallization can significantly affect the process’s efficiency. In this scenario, the compound dissolves well in ethanol and poorly in water. When these two solvents are mixed, the solubility properties help control how the compound crystallizes as the solution cools. Ethanol serves to dissolve the compound at higher temperatures, while water precipitates it upon cooling due to reduced solubility. This combination offers controlled crystallisation, maximizing yield and purity by minimizing the amount of compound left dissolved when cooled.

Key Concept: Theoretical Analysis, Real-World Application

b) Below the boiling point of pure water due to reduced total pressure

[Solution Description]

In steam distillation, the boiling point of the mixture occurs when the total vapor pressures of both substances equal the atmospheric pressure. Since the total vapor pressure of aniline and water (10 kPa + 20 kPa = 30 kPa) is less than the atmospheric pressure (normally around 101 kPa), the boiling point of this mixture will be lower than that of pure water, 373 K (100°C).

Thus, statement b) is correct; the boiling point of the aniline-water mixture is below that of pure water.

Key Concept: Pressure Concept

c) The vapor pressure of the mixture equals the atmospheric pressure.

[Solution Description] In steam distillation, boiling occurs when the sum of the vapor pressures of the organic liquid and water equals atmospheric pressure. Mathematically, this is represented as $p = p_1 + p_2$, where $p$ is atmospheric pressure, $p_1$ is the vapor pressure of the organic liquid, and $p_2$ is the vapor pressure of water.

Key Concept: Comparative Analysis, Critical Thinking

b) The atmospheric pressure prevents camphor’s sublimation.

[Solution Description] Camphor is a sublimable substance meaning it changes from solid to gas without becoming liquid. Salt, however, does not sublime under standard conditions. Sublimation works effectively by using reduced pressure or elevated temperatures to help camphor transition directly into vapor while salt remains. However, at standard atmospheric pressure, the sublimation process would be inefficient and much slower due to lack of favorable conditions for sublimation.

Key Concept: Distillation Under Reduced Pressure, Real-world Application

c) Distillation under reduced pressure

[Solution Description] Distillation under reduced pressure lowers the boiling point of a liquid by reducing the external pressure. This prevents thermal decomposition by allowing the liquid to boil at a lower temperature than its normal boiling point. Therefore, this method is suitable for purifying a high boiling point liquid prone to decomposition.

Key Concept: Real-world Application, Reaction Pathway Analysis

c) Provides $H^+$ ions for generating a stable carbocation.

[Solution Description]

Dilute sulfuric acid acts as a catalyst providing $H^+$ ions. It facilitates the initial protonation of isobutylene, generating a highly stable tertiary carbocation $((CH_3)_3C^+)$. Water then attacks this carbocation to form an oxonium ion, which subsequently loses a proton to yield tert-butanol.

Key Concept: Predictive Analysis, Concept Integration

a) $(CH_3)_3C^+$

[Solution Description] To evaluate where hyperconjugation exerts more influence than inductive effects:

– Hyperconjugation is more influential in cases where multiple C-H bonds are present adjacent to the positive charge.

– Inductive effects dominate when fewer alkyl groups surround the charged center.

– Comparing $(CH_3)_3C^+$, $CH_3CH_2^+$, $CH_3^+$, and $(CH_3)_2CH^+$, for $(CH_3)_3C^+$, there are 9 C-H bonds providing significant hyperconjugative stability, overshadowing any inductive effect.

Thus, in $(CH_3)_3C^+$, hyperconjugation prevails over inductive effects.

Key Concept: Effect on Molecular Polarity, Real-world Application

a) Increases polarity

[Solution Description] When HCl adds to an unsymmetrical alkene, the electrophile $(H^+)$ initially provokes an electromeric shift where $\pi$-electrons are displaced toward the more substituted carbon, increasing temporary electron density there. This shifting leads to a polarized transition state, thus enhancing molecular polarity. The resultant positive charge on the less substituted carbon and negative charge induced on the chlorine create a dipole, demonstrating increased polarity during the reaction.

Key Concept: Comparative Analysis, Real-World Application

b) -R effect of -CN causing higher acidity increase

[Solution Description]

Analyze and compare the electronic effects imparted by -CN and $-OCH_3$:

1. The -CN group provides a strong -R effect, pulling electron density from the benzene ring and increasing acidity by stabilizing the conjugate base through resonance.

2. The $-OCH_3$ group exhibits a +M (mesomeric) effect, supplying electron density into the ring, generally lowering acidity.

3. For determining dominance over \$K_a\$, focus on which effect alters the electron density balance more significantly in favor of dissociation.

4. Given the strong electron-withdrawing nature of -CN versus the electron-donating character of $-OCH_3$, the former has a more pronounced impact on increasing \$K_a\$.

Therefore, the -CN group’s -R effect dominates the scenario.

Key Concept: Theoretical Predictions, Structural Influence

b) Single C-C bonds will shorten and double C-C bonds will lengthen

[Solution Description] Introducing a -OR group leads to resonance donation of electrons into the conjugated system, altering the bond characteristics. The single bonds adjacent to the -OR group gain partial double-bond character while the double bonds become shorter due to increased electron density facilitating greater overlap. As a result, the single C-C bonds near the -OR group become shorter, whereas the original double bonds may extend slightly but remain shorter than typical single bonds.

Key Concept: Complex Stability Analysis, Resonance Energy

c) Completion of the octet

[Solution Description] To determine the stability of resonance structures, we consider several factors: completion of the octet, fewer charges, and placing negative charges on more electronegative atoms when applicable. When comparing a structure that violates the octet rule to one that maintains it, the octet-complete structure is generally favored even if it involves separated charges. Thus, octet completion takes precedence over charge separation in determining stability, as it maximizes bonding interactions and minimizes potential energy.

Key Concept: Resonance Effect

c) $-NO_2$

[Solution Description] A negative resonance effect involves electron-withdrawing groups that pull electrons away from the rest of the molecule. Groups such as nitro $(-NO_2)$, cyano (-CN), or carbonyl (>C=O) exhibit this effect.

Key Concept: Substituent Interaction, Predictive Reactivity

a) Compound (i)

[Solution Description] To determine which compound better stabilizes a carbocation, analyze the inductive effects of substituents:

– In compound (i), the $-CH_3$ groups are electron-donating via the +I effect, providing some stabilization to a carbocation.

– In compound (ii), the $-NO_2$ groups are strongly electron-withdrawing, destabilizing a carbocation due to the -I effect.

– A stable carbocation typically requires electron-donating groups to offset the positive charge.

As such, compound (i) with electron-donating $-CH_3$ groups is better able to stabilize a carbocation.

Key Concept: Electromeric Effect Types

b) The atom to which the reagent does not attach

[Solution Description] The electromeric effect occurs when electrons are transferred within a molecule upon interaction with a reagent. In a negative electromeric effect, $$\pi$$-electrons shift away from the atom that does not form a new bond with the incoming electrophilic reagent, stabilizing the positive charge on the attacking species.

Key Concept: Detailed Mechanism, Real World Problem Solving

b) C-H bond

[Solution Description]

In propene, hyperconjugation involves the overlap between the $\sigma$ bond electrons of C-H adjacent to a double bond and the $\pi$ system or empty p orbital, leading to electron delocalization. Thus:

– The C=C bond itself does not directly participate in hyperconjugation.

– The C-H bond adjacent to the C=C bond participates by providing its $\sigma$ electron for stabilization through overlap.

– The C-C bond remains unaffected in terms of hyperconjugation.

– There is no C-O bond present in propene.

Therefore, the C-H bond is involved in the delocalization process.

Key Concept: Complex Reaction Analysis, Advanced Concept Integration

c) Towards the carbon that forms a bond with the hydrogen atom.

[Solution Description] The positive electromeric effect (+E effect) involves the transfer of $\pi$-electrons to the atom where the reagent attaches. In 1,3-butadiene, upon approaching HCl, the $\pi$-electrons will be directed towards the carbon atom that will form a bond with the hydrogen atom, as the hydrogen ion $(H^+)$ acts as the electrophile attaching to the diene system, directing the electron density towards itself.

Key Concept: Comparative Stability, Advanced Resonance Structures

b) Phenoxide ion

[Solution Description]

Nitrobenzene has several resonance structures where negative charge can be delocalized onto the oxygen atoms, but it involves charge separation leading to lesser stability.

Phenoxide ion, however, achieves increased stability through effective delocalization of negative charge across the aromatic ring, contributing to complete octet fulfillment without significant charge separation.

Overall, the phenoxide ion is more stable compared to nitrobenzene because its resonance structures exhibit less charge separation and more optimal electron distribution.

Key Concept: Complex Analysis, Predicting Product Formation

d) 2-methyl-2-propanol

[Solution Description] The reaction involves nucleophilic substitution where hydroxide ion $(OH^-)$ attacks the carbon bonded to bromine. The methyl group’s electron-donating inductive effect stabilizes the carbocation intermediate, favoring a rearrangement to form tertiary alcohol over a direct attack on the primary carbon. Therefore, the main product will be 2-methyl-2-propanol due to this stabilization by the inductive effect.

Key Concept: Real-World Applications, Advanced Stability Concepts

c) 3-bromo-cyclohexene

[Solution Description]

– The reaction under UV light promotes homolytic cleavage of the $Br_{2}$ bond, generating bromine radicals.

– These radicals abstract hydrogen atoms from cyclohexene, forming allylic radicals.

– The resonance stabilization of these radicals favors the formation of 3-bromo-cyclohexene as the major product through radical substitution.

Hence, the major product is 3-bromo-cyclohexene.

Key Concept: Stability Order

a) Methyl carbocation

[Solution Description] The stability of carbocations increases with the number of alkyl groups attached due to the +I inductive effect. Tert-butyl carbocation is most stable due to three alkyl groups, while methyl carbocation has no alkyl group, making it the least stable.

Key Concept: Real-World Application, Electrophile-Nucleophile Interaction

d) Hydrolysis of ethyl chloride using $OH^-$.

[Solution Description]

A classic example of nucleophilic substitution is the hydrolysis of an alkyl halide. Here, $OH^-$ acts as a nucleophile and attacks the electrophilic carbon bonded to the halogen, displacing the halide ion.

Hydrolysis of ethyl chloride $(C_2H_5Cl)$:

$\text{C}_2\text{H}_5\text{Cl (aq)} + \text{OH}^- \rightarrow \text{C}_2\text{H}_5\text{OH (aq)} + \text{Cl}^-$

This process occurs in various industrial applications, particularly in the synthesis of alcohols.

Key Concept: Substrate and Reaction Mechanism, Substrate and Reaction Kinetics

b) Its electronic configuration

[Solution Description] The electronic configuration of a substrate often dictates its reactivity and interactions during a reaction. Analyzing the electronic structure can reveal potential sites for bond formation or breaking, thereby helping to deduce the reaction pathway.

Key Concept: Electrophile/Nucleophile Pairing, Mechanism Analysis

b) $NH_2^{-}$ acts as a nucleophile; Product: Aniline

[Solution Description] In the given reaction:

$NH_2^{-}$ acts as a strong nucleophile. When it reacts with $C_6H_5-Br$, it substitutes the bromine atom on the benzene ring through an SNAr (Nucleophilic Aromatic Substitution) mechanism. This results in the formation of aniline $(C_6H_5-NH_2)$, where the amino group takes the place of the bromine atom, forming a more stable carbon-nitrogen bond.

Thus, $NH_2^{-}$ is acting as a nucleophile, leading to the substitution of bromine by an amine group.

Key Concept: Comparative Stability, Reaction Condition Influence

d) $(CH_3)_3C\cdot$

[Solution Description]

To determine the stability of radicals under increased temperature conditions, we must consider both the intrinsic stability and how this stability might change with temperature. Tertiary radicals have greater hyperconjugation and dispersion compared to primary radicals, leading to more stabilization as temperature increases due to entropy factors.

Considering these elements:

a) $CH_3\cdot$: Primary radical with no alpha hydrogen for hyperconjugation.

b) $C_2H_5\cdot$: Secondary radical with limited hyperconjugation.

c) $(CH_3)_2CH\cdot$: Tertiary radical, higher hyperconjugation potential.

d) $(CH_3)_3C\cdot$: Highly tertiary radical, maximum hyperconjugation and therefore highest stability.

Under increased temperature conditions favoring entropy-driven processes, the highly substituted radical (d) would show enhanced stability.

Key Concept: Homolytic vs Heterolytic

b) A covalent CH bond undergoing homolytic cleavage

[Solution Description] In homolytic cleavage, each atom takes one electron from the bond pair, resulting in free radicals. Thus, when a covalent CH bond undergoes homolytic cleavage, it forms two neutral radicals.

Key Concept: Hybridization and Shape

b) $sp^3$, Trigonal Pyramidal

[Solution Description] A carbanion like $CH_3^-$ has one extra pair of electrons compared to its neutral counterpart $(CH_4)$. This results in $sp^3$ hybridization because one s orbital and three p orbitals combine to form four $sp^3$ orbitals. The geometry is tetrahedral, but the presence of a lone pair distorts the shape into a trigonal pyramidal structure.

Key Concept: Advanced Conceptual Understanding, Analytical Techniques

d) Polarimetry helps in determining the optical purity of a mixture containing enantiomers.

[Solution Description] Optical isomers, or enantiomers, are known to rotate plane-polarized light in opposite directions. To distinguish between them, polarimetry is a common technique used, as it measures the angle of rotation caused by passing polarized light through a substance. Enantiomers will rotate light at equal angles but in opposite directions. Therefore, they have opposite specific rotations when measured under identical conditions.

Key Concept: Complex Structural Analysis, Reaction Mechanisms

b) maleic acid

[Solution Description] Geometrical isomers can have different reactivities based on their spatial arrangement. Consider the influence of sterics and electronic factors in different environments for cis and trans configurations.

For compound 1, cis-2-butene, the bulky groups are on the same side which could cause steric hindrance affecting the reaction path with certain reagents compared to trans-2-butene.

Compound 2, maleic acid (cis-butenedioic acid), has carboxylic groups positioned such that they can form an intramolecular anhydride upon heating, unlike fumaric acid (trans-butenedioic acid).

Compounds 3 and 4 do not exhibit significant differences in reaction mechanisms caused by their geometry due to either lacking sufficient complexity or having similar spatial arrangements that do not affect the reaction significantly.

Thus, the correct answer is option b).

Key Concept: Optical Isomers

b) Optical isomerism

[Solution Description] Optical isomerism involves chiral molecules, which are non-superimposable on their mirror images and can rotate plane-polarized light.

Key Concept: Alkyl Chain Variation

c) A different isomer is formed.

[Solution Description]

Changing the alkyl chains will result in different metamers, as metamerism involves rearranging the groups around a functional group while maintaining the same molecular formula.

Key Concept: Functional Group Comparison

a) Butanal and Butanone

[Solution Description]

The molecular formula $C_4H_8O$ can represent compounds with different functional groups. One must identify the separate functional groups present and see if they form a possible pair of isomers.

– Butanal: An aldehyde with the structure $CH_3(CH_2)_2CHO$

– Butanone: A ketone with the structure $CH_3COCH_2CH_3$

– Butanol: An alcohol with the structure $CH_3(CH_2)_3OH$

– Methyl Propyl Ether: An ether with the structure $CH_3OCH_2CH_2CH_3$

Comparing:

– Butanal and Butanone are functional group isomers because one is an aldehyde, and the other is a ketone, sharing the same formula but different groups.

Therefore, the correct answer is Butanal and Butanone, option a).

Key Concept: Functional Group Shift

a) 3-methylbutan-2-ol

[Solution Description]

To form a position isomer by shifting the hydroxyl group in 3-methylbutan-1-ol, move the OH group from the first carbon to another position. The possible shift can be to the second carbon, leading to 3-methylbutan-2-ol:

– Original: $CH_3CH(CH_3)CH_2CH_2OH$ (3-methylbutan-1-ol)

– Isomer: $CH_3CH(CH_3)CHOHCH_3$ (3-methylbutan-2-ol)

So, the correct answer is 3-methylbutan-2-ol.

Key Concept: Comparative Analysis

b) Chain isomerism

[Solution Description] Chain isomerism refers to compounds with the same molecular formula but different arrangements or branches in their carbon skeletons. This contrasts with other types of isomerism like position or functional group isomerism, where the differences lie in the positions of substituents or in functional groups themselves.

Key Concept: Identify Isomer Type

a) Chain isomerism

[Solution Description] Butan-1-ol and 2-methylpropan-1-ol have the same molecular formula of $C_4H_{10}O$. However, they differ in their carbon skeletons. Butan-1-ol has a straight chain, while 2-methylpropan-1-ol has a branched chain with a methyl group attached at the second carbon. This structural difference indicates that these compounds exhibit chain isomerism.

Key Concept: Functional Group Isomerism

d) Propanal and Propanone

[Solution Description] Functional group isomerism occurs when compounds have the same molecular formula but possess different functional groups. For instance, Propanal (an aldehyde) and Propanone (a ketone) are functional group isomers.

Key Concept: Branched Chain Naming

b) 2-Ethyl-4-methylpentane

[Solution Description]

The longest carbon chain contains 5 carbon atoms, making it pentane. The substituents are an ethyl group at position 2 and a methyl group at position 4. According to the alphabetical order rule, the ethyl group is named before the methyl group. Thus, the correct name is 2-Ethyl-4-methylpentane.

Key Concept: Equivalent Positions

a) Lowest locant rule

[Solution Description] When two substituents are in equivalent positions around a benzene ring, the lowest locant rule is used to assign the smallest possible numbers, ensuring clarity in the structure’s naming.

Key Concept: Equivalent Position

c) 4-ethyl-2-methylheptane

[Solution Description]

To determine the correct numbering, we first look for the longest carbon chain. The longest chain has 7 carbons: $CH_3-CH(CH_2CH_3)-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$. Now, we number the chain from left to right to get the lowest locants: 1-ethyl-4-methylheptane or 4-ethyl-2-methylheptane. By numbering so that substituents receive the lowest possible numbers, the correct name is 4-ethyl-2-methylheptane.

Key Concept: Chain Selection

a) Chain A, which has more side chains

[Solution Description] The IUPAC naming rule states that if there are two chains of equal size, the chain with more substituents should be chosen as the parent chain.

Key Concept: Chain Selection

c) Hexane

[Solution Description] The parent chain of an alkane is the longest continuous chain of carbon atoms. In 3-(tert-butyl)hexane, hexane indicates that the main chain consists of six carbon atoms. Therefore, despite having a tert-butyl substituent, the longest chain remains hexane.

Key Concept: Branched Alkyl Group Naming

b) 3-(3,3-Dimethylbutyl)hexane

[Solution Description] When naming branched alkyl groups, the carbon of attachment is considered as number 1. For a 3,3-dimethylbutyl group attached to hexane:

– Butyl implies four carbons in the side chain.

– Methyl groups are at positions 3 and 3.

– The main chain is hexane.

So, the correct name is 3-(3,3-dimethylbutyl)hexane.

Key Concept: Complex Branching, Functional Group Combination

b) 2-Hydroxyhexan-4-one

[Solution Description] The highest priority functional group here is the ketone. Therefore, the root name will be based on the ketone, which is a hexanone since there are six carbons. The ketone at position 4 gives us “hexan-4-one”. Since the alcohol group is at position 2, it becomes a substituent as “2-hydroxy”.

The correct IUPAC name is “2-Hydroxyhexan-4-one”.

Key Concept: Advanced Locant Rule, Complex Substitution

c) 3-nitro-1,4-dimethylbenzene

[Solution Description] To determine the correct IUPAC name, we follow these steps:

– Identify the principal functional group: ‘Nitro’ has higher priority over ‘Methyl’.

– Prioritize numbering to give the lowest possible numbers to substituents.

By placing the nitro group at position 3, the methyl groups will be at positions 1 and 4.

Thus, the compound is named: 3-nitro-1,4-dimethylbenzene.

Key Concept: Disubstituted Benzene

c) p-Dichlorobenzene

[Solution Description] For disubstituted benzene compounds, the term para (p) is used when substituents are at the 1,4 positions. Thus, the compound is named p-Dichlorobenzene.

Key Concept: Polyfunctional Naming

d) Butanedioic acid

[Solution Description] The compound $HOOC-(CH_2)_2$-COOH consists of two carboxylic acid groups attached to a butane chain. According to IUPAC naming conventions, the suffix “dioic acid” is used when there are two identical carboxylic acid groups. Therefore, the correct IUPAC name for this compound is Butanedioic acid.

Key Concept: Alphabetical Order

c) 3,3-Diethyl-2-methylpentane

[Solution Description] The correct name must list substituents alphabetically as per IUPAC rules. Therefore, the two ethyl groups are prioritized over the methyl group when writing the name: 3,3-Diethyl-2-methylpentane.

Key Concept: Real-World Application, Branched Alkyl Group Naming

a) 2-Isopropyl-4-methylpentane

[Solution Description] The longest carbon chain consists of 5 carbons, making it a pentane. The molecule features a branched alkyl group at carbon 2, which is an isopropyl group. Additionally, there is a methyl group at carbon 4. Alphabetically arranging the substituents leads to the name 2-Isopropyl-4-methylpentane.

Key Concept: Incorrect Naming Correction, Branched Chain Hydrocarbons

a) Incorrect chain length; Correct name: 2,3-dimethylhexane

[Solution Description] Analyze the given structure: 2,3-dimethyl groups suggest additional branches on a pentane chain. However, checking the parent chain reveals that the longest continuous chain contains six carbon atoms, not five. Therefore, the base name should be hexane. Re-evaluating substituent positioning gives us “2,3-dimethylhexane” as the correct name since both methyl groups are still correctly placed at positions 2 and 3.

Key Concept: Structural Formula Analysis, Series and Nomenclature Integration

c) $CH_3-C(CH_3)-C≡CH$

[Solution Description]

In the name 3-methylbut-1-yne, “but” indicates four carbon atoms in the main chain, and the “yne” suffix identifies it as an alkyne with a triple bond starting at the first carbon position. “3-methyl” signifies a methyl group attached to the third carbon. Hence, the structure is represented as $CH_3-C(CH_3)-C≡CH$.

Key Concept: Isomerism Identification, Real-world Application

b) Functional isomerism; ester group

[Solution Description]

The molecular formula $C_4H_8O_2$ corresponds to esters that can exhibit functional isomerism. Methyl propanoate has the structure $CH_3CH_2COOCH_3$, while ethyl acetate has $CH_3COOCH_2CH_3$. These are different compounds having the same molecular formula but differing in connectivity and functional group placement, hence showing functional isomerism.

The ester functional group $(-COOR’)$ undergoes hydrolysis resulting in alcohol and carboxylic acid formation. For instance, hydrolysis of methyl propanoate yields methanol and propanoic acid, whereas ethyl acetate yields ethanol and acetic acid.

Therefore, the type of isomerism is functional isomerism, and the ester functional group undergoes hydrolysis.

Key Concept: Complex Series Identification, Functional Group Impact

b) Alkanoic acids

[Solution Description] Compounds with a carboxyl group $(COOH)$ are typically known as carboxylic acids or alkanoic acids. These acids are characterized by their ability to donate a hydrogen ion (H+), giving them strong acidic properties compared to other homologous series without this reactive group.

Key Concept: Polyfunctional Compound Naming

b) 2-hydroxypropanoic acid

[Solution Description] To determine the IUPAC name, identify the principal functional group. Here, the carboxylic acid $(\text{-COOH})$ has higher priority than both alcohol and aldehyde groups. Numbering starts from the carboxyl end, giving 1-hydroxypropan-2-al as part of the structure, leading to the name: 2-hydroxypropanoic acid.

Key Concept: Electronegativity Effects, Hybridization

a) sp hybridized

[Solution Description] Electronegativity increases with increasing s-character because s-orbitals are closer to the nucleus than p-orbitals. Thus, an sp-hybridized carbon, having 50% s-character, is more electronegative than $sp^2$ (33% s-character) and $sp^3$ (25% s-character). Hence, sp-hybridized carbon is the most electronegative.

Key Concept: Nomenclature Basics

b) 1-methylcyclopentene

[Solution Description] The base structure is cyclopentene due to the presence of a double bond in a five-membered ring. With a methyl substituent, it becomes 1-methylcyclopentene.

Key Concept: Basic Nomenclature

a) Oxacyclopentadiene

[Solution Description] Furan is a five-membered aromatic ring consisting of four carbon atoms and one oxygen atom. According to IUPAC nomenclature rules for naming heterocyclic compounds, this specific structure is named Oxacyclopentadiene.

Key Concept: Substituent Effects

c) Nitro

[Solution Description] Substituents that withdraw electrons from the benzene ring generally decrease the electron density. The nitro group $(NO_2)$ is a strong electron-withdrawing group due to its resonance and inductive effects, thus decreasing the electron density of the benzene ring it is attached to.

Key Concept: Isomer Identification

a) Hex-1-ene

[Solution Description] Chain isomers have the same molecular formula but differ in the arrangement of the carbon skeleton. For cyclohexane $(C_6H_12)$, hex-1-ene is a potential chain isomer where the cyclic structure is opened to form a linear one with a double bond.

Key Concept: Position Isomerism

b) 2-methylcyclopentanol

[Solution Description] Position isomers differ in the location of the functional group or substituent on the same carbon skeleton. In 1-methylcyclopentanol, the hydroxyl group is at position 1. Therefore, moving the methyl group to another position on the cyclopentanol ring will create a position isomer, such as 2-methylcyclopentanol.

Key Concept: Functional Group Impact

b) Amine

[Solution Description] Ethanolamine contains an amino group $(NH_2)$, which acts as a weak base because it can accept protons, thereby giving ethanolamine its basic character.

Key Concept: Chain Isomerism

b) Pentane and Isopentane

[Solution Description] Chain isomers have the same molecular formula but different arrangements of the carbon skeleton. Pentane $(C_5H_{12})$ and isopentane $(C_5H_{12})$ exhibit chain isomerism as they have different structures but the same molecular formula.

Key Concept: Functional Group Role

c) $–NH_2$

[Solution Description] Amines are basic in nature due to the presence of the $-NH_2$ functional group. This group acts as a base because the nitrogen atom has a lone pair of electrons which can accept protons $(H^+)$, thus exhibiting basicity.

Key Concept: Model Characteristics

c) Highlights van der Waals radii

[Solution Description]

The space-filling model highlights the van der Waals radii of atoms, showing how much physical space each atom occupies. This characteristic makes it distinct from other models like the ball-and-stick model, which focuses more on bonds and angles.

Key Concept: Advanced Application, Model Analysis

a) It provides a detailed representation of bond angles.

[Solution Description]

The ball-and-stick model helps visualize the bond angles and spatial arrangement of atoms, which is crucial for understanding the bent shape of the water molecule due to its $104.5^\circ$ bond angle. In contrast, the space-filling model gives a more accurate depiction of the volume occupied by each atom but makes it harder to see bond angles precisely, which is essential in analyzing molecular geometry.

Key Concept: 3D Visualization

d) Framework model

[Solution Description] The visualization technique that focuses on the structure of molecules by emphasizing bond connections rather than the details of each atom’s size or presence is called the framework model.

Key Concept: Bond Characteristics

b) Approximately 120 degrees

[Solution Description] Molecules with $sp^2$ hybridization around the central atom typically have bond angles of approximately 120 degrees due to the trigonal planar geometry.

Key Concept: Application in Reactions

c) In-plane carbon-oxygen double bond

[Solution Description] In acetaldehyde, the carbonyl group $(\ce{C=O})$ is planar with the oxygen having partial negative charge. Nucleophilic attacks usually occur at the electrophilic carbon atom of the carbonyl group. Therefore, the in-plane bond representing the carbon-oxygen double bond is most likely involved in the reaction.

Key Concept: Advanced Model Application, Real-World Model Application

c) Space filling model

[Solution Description] Steric hindrance refers to the restriction of chemical reactivity due to the spatial arrangement of atoms within a molecule. The space-filling model is most effective for visualizing this because it provides a realistic representation of the molecule’s volume and surface area, allowing chemists to assess potential clashes at the chiral center.

Key Concept: Hybridization Influence

a) $109.5^\circ, 120^\circ, 180^\circ$

[Solution Description] Methane $(CH_4)$ has a central carbon atom that is $sp^3$ hybridized, leading to a tetrahedral geometry and a bond angle of approximately 109.5°. Ethylene $(C_2H_4)$ involves $sp^2$ hybridized carbons, resulting in a planar trigonal geometry with bond angles around 120°. Acetylene $(C_2H_2)$ has $sp$ hybridized carbons, forming a linear structure with a bond angle of 180°. Thus, the order according to increasing hybridization would be:

– Methane: 109.5° $(sp^3)$,

– Ethylene: 120° $(sp^2)$,

– Acetylene: 180° (sp).

Hence, the influence of increasing s-character leads to decreasing bond angles from methane to acetylene.

Key Concept: Structural Formula Expansion, Multi-functional Group Analysis

b) $H_2N-CH(CH_3)$-COOH

[Solution Description] The given formula represents alanine, an amino acid. The complete structural formula is $H_2N-CH(CH_3)$-COOH. It contains an amino group $(NH_2)$, a carboxylic acid group (COOH), and a side chain represented by a methyl group $(CH_3)$.

Key Concept: Resonance Energy Calculation, Hybridization and Electronegativity

c) Resonance energy is 5 kJ/mol; sp hybridization is expected.

[Solution Description] The resonance energy is calculated as the difference between the lowest energy canonical structure and the actual energy of the molecule. Here, the lowest canonical energy is -320 kJ/mol, and the actual energy is -325 kJ/mol. Thus, the resonance energy is 5 kJ/mol ($-320 – (-325)$). Given that maximizing resonance energy often involves extensive pi-bonding, sp or sp2 hybridizations are common; however, without specific details on bonding, further hybridization analysis is speculative.

Key Concept: Molecular Shape Prediction, Structural Formula Conversion

c) The structural formula can be represented as CH4 for methane.

[Solution Description] A tetrahedral molecular shape indicates that the carbon atom is \$sp^3\$ hybridized, forming four sigma bonds arranged symmetrically. The complete structural formula would show these four covalent bonds emanating from the central carbon atom in a three-dimensional arrangement. However, in condensed structural formulas or bond-line structures, these details may be omitted for simplicity, focusing instead on connectivity.

Key Concept: Structural Formula Derivation, Advanced Functional Group Recognition

b) $C_6H_4(OH)_2$

[Solution Description] The description provided fits the structure of resorcinol, which is 1,3-dihydroxybenzene:

– The core structure here is benzene, a six-membered ring with alternating double bonds.

– Two hydroxyl groups are attached to the ring, specifically at positions 1 and 3.

The condensed formula can be written by recognizing that the phenolic ring remains unchanged while the $\text{-OH}$ groups are specified by their position: $C_6H_4(OH)_2$. This captures the essential features of the aromatic structure along with the placement of the substituents.

Key Concept: Functional Group Influence, Real-world Application

c) Carboxyl group due to hydrogen bonding ability

[Solution Description] Functional groups that can form hydrogen bonds with water molecules significantly enhance water solubility. Hydroxyl $(-OH)$ groups are highly polar and can both donate and accept hydrogen bonds, leading to increased solubility in water compared to other groups like ketones or aldehydes which primarily act as hydrogen bond acceptors but not donors.

Key Concept: Pi Bond Reactivity, Inductive and Resonance Effects

c) Resonance stabilization potential of resulting cations.

[Solution Description] In a conjugated diene, resonance stabilization plays a crucial role. The most reactive pi bond will be where resonance can best stabilize positive charge formation during the transition state. Typically, this is affected by inductive effects from substituents but primarily by the potential resonance structures.

Key Concept: π Bond Orientation and Molecular Reactivity, π Bond and Electrophilic Attack

c) At the terminal $\pi$ bond due to resonance stabilization.

[Solution Description]

Polyenes typically exhibit conjugation due to alternating single and double bonds. The orientation of $\pi$ bonds allows for delocalization of electrons across multiple bonds, which stabilizes the molecule but simultaneously makes it susceptible to attack by electrophiles. An electrophile tends to attack at sites where $\pi$ bond electrons are most readily available.

The terminal $\pi$ bonds in a conjugated system often bear a higher electron density due to resonance structures, making them preferable sites for electrophilic attacks compared to internal $\pi$ bonds.

Hence, electrophiles preferentially attack at the terminal $\pi$ bonds in conjugated systems.

Key Concept: Hybridization and Molecular Properties, Pi Bond Electron Cloud

c) The electron cloud localizes above and below, affecting polarity and reactivity.

[Solution Description]

In a carbon-carbon double bond, the $\pi$ bond results from the sideways overlap of p orbitals, creating an electron cloud above and below the plane of the atoms. This electron cloud leads to areas of negative charge that can interact with other molecules, influencing properties like polarity and reactivity. Thus, the location of the $\pi$ electron cloud directly impacts intermolecular interactions and molecular behavior in reactions.

Key Concept: Structural Representation Challenges, Conjugated Systems

c) $CH_2=CH-CH=CH-CH_3$

[Solution Description] The given bond-line structure represents a pentadiene system where each vertex corresponds to a carbon atom. Assuming typical valency for carbon, there are five carbons, making up a conjugated diene chain with alternating single and double bonds. Thus, the complete structural formula is $CH_2=CH-CH=CH-CH_3$.

Key Concept: Resonance Energy, Conjugated Systems

a) Benzene has higher resonance energy due to aromaticity

[Solution Description] Benzene exhibits a conjugated pi system allowing for complete delocalization of electrons across all six carbon atoms, forming a highly stable aromatic ring structure. This results in significant resonance energy, contributing to benzene’s stability. In contrast, 1,3-butadiene also has a conjugated system but lacks the aromaticity found in benzene, resulting in lower resonance energy when compared to benzene.

Key Concept: Comparative Reactivity, Hybridization Influence

b) Acetylene is more reactive because of multiple $\pi$ bonds in sp hybridization.

[Solution Description]

Ethylene consists of two $sp^2$-hybridized carbons forming a C=C double bond with one $\pi$ bond resulting from sideways overlap of p orbitals above and below the sigma bond plane. Acetylene involves sp-hybridized carbons forming a $C$\equiv$C$ triple bond containing two $\pi$ bonds, leading to a linear molecular shape. The additional $\pi$ bond in acetylene provides more reactive sites for nucleophilic attack than ethylene. Moreover, the sp hybridization leads to a higher percentage of s character, increasing the electronegativity and causing the $\pi$ bonds in acetylene to have significant electron-withdrawing properties, making it generally less reactive toward electrophiles but quite susceptible in nucleophilic additions. Thus, acetylene exhibits distinct reactivity differences compared to ethylene attributable to the nature of its $\pi$ bond and hybrid state.

Key Concept: π Bond Formation Conditions, Restricted Rotation Implications

b) Parallel orientation of p orbitals forming the π bond

[Solution Description] The restricted rotation around a C=C bond is primarily due to the presence of π bonds, which require parallel orientation of p orbitals. These p orbitals overlap to form the π bond above and below the plane formed by the σ bond.

If these p orbitals rotate, the overlap would be lost, breaking the π bond. Therefore, the critical condition for restricted rotation is having parallel p orbitals forming the π bond.

Key Concept: Hybridization and Reactivity, Electronegativity and Bond Strength

c) sp hybridization increases bond strength due to increased electronegativity.

[Solution Description] In sp hybridization, carbon has greater electronegativity due to 50% s character, pulling electrons more strongly towards itself. This results in shorter and stronger C-H bonds compared to $sp^3$ hybridized ones where there is only 25% s character. Therefore, the bond strength in C-H bonds for sp hybridized carbon is higher.

Key Concept: Hybridization and Bond Length

c) It results in shorter bond lengths due to increased s-character.

[Solution Description]

Hybridization affects bond lengths significantly. In acetylene (C$_2$H$_2$), each carbon atom is sp-hybridized. Sp hybridization means that the carbon forms two $\pi$ bonds and one $\sigma$ bond, creating a linear geometry.

Due to the triple bond and the high s character (50%) in sp hybridization, the bond length is shorter compared to sp$^2$ or sp$^3$ hybridizations. Shorter bonds are stronger and result from greater overlap of orbitals. Thus, the C-C bond length in acetylene is shorter than that in alkenes or alkanes.

Key Concept: Molecular Planarity

b) C$_2$H$_4$

[Solution Description] Planarity in molecules due to $\pi$ bonds generally arises when there is an arrangement that allows for delocalization across the plane. Ethene (C$_2$H$_4$) is planar because the double bond restricts rotation and maintains planarity through the presence of one $\sigma$ and one $\pi$ bond.

Key Concept: Bond Length and Strength

c) Ethyne \[
(C \equiv C)
\]

[Solution Description] The strength and length of a carbon-carbon bond depend on its type: single (C-C), double (C=C), or triple \[
(C \equiv C)
\]
. A triple bond involves more shared electrons than either single or double bonds, resulting in a stronger and shorter bond. Therefore, among ethane (C-C), ethene (C=C), and ethyne \[
(C \equiv C)
\]
, the bond in ethyne will be the shortest because it is a triple bond formed by $sp$ hybridized carbon atoms.

Key Concept: 3D Representation

c) Bond projecting away from the observer

[Solution Description] In the wedge-and-dash notation used to represent 3D structures, a dashed line indicates that the bond is projecting away from the observer into the plane of the paper.

Key Concept: Hybridization and Electronegativity, Reaction Pathways

c) A secondary carbocation rearranges to give a tertiary carbocation via a hydride shift.

[Solution Description] In this reaction pathway, during hydrohalogenation, a hydrogen halide adds across a double bond, forming a secondary carbocation intermediate. Carbocation rearrangements such as hydride or alkyl shifts occur because they lead to more stable tertiary carbocations due to hyperconjugation and inductive effects. The hybridization state of carbon changes from $sp^2$ in the double-bonded carbon to $sp^3$ in the carbocation.

Key Concept: Structural Analysis, IUPAC Nomenclature

a) $HOCH_2C(CH_3)(OH)CH_3$

[Solution Description]

2-methylpropane-1,2-diol indicates a propane backbone with two hydroxyl groups on the first and second carbon, and a methyl group on the second carbon. The correct structural representation should have three carbon atoms with OH groups attached at C1 and C2, and a $CH_3$ group also on C2.

Key Concept: Homologous Series Properties

b) Boiling point

[Solution Description] In a homologous series, as the molecular size or weight increases, there is a corresponding increase in the boiling point. This is largely due to enhanced van der Waals forces between molecules as their size becomes larger and more significant.

Key Concept: Structural Analysis, Polyfunctional Compounds

c) Carboxyl group

[Solution Description] Acetic acid has a chemical structure represented as $CH_3COOH$. In this structure, the carboxyl $(-COOH)$ group is present. This group consists of a carbonyl $(>C=O)$ and a hydroxyl $(-OH)$ group attached to the same carbon atom. The hydrogen atom in the hydroxyl part of the carboxyl group can dissociate to release $H\^+$ ions, making the compound acidic. Therefore, it is the carboxyl group that imparts acidic properties to acetic acid.

Key Concept: Chromatography Techniques, Complex Reaction Types

b) Adsorption chromatography

[Solution Description] Chromatography techniques rely on differential interactions between compounds and the stationary/mobile phases:

– Adsorption chromatography relies on the adhesion of molecules on surfaces (solid stationary phase), affecting their travel speeds through the column.

– Partition chromatography involves solute partitioning between two liquid phases.

– Distillation separates based on boiling points, not adsorption.

– Sublimation separates based on changes from solid to gas without liquefying.

Adsorption chromatography fits the description of using differential adsorption.