Organic Chemistry Some Basic Principles And Techniques – NEET UG ChemistryPractice Questions & PYQs

In covalent structures, a single bond consists solely of one $\sigma$ bond, whereas a double bond is formed by one $\sigma$ bond and one $\pi$ bond. In the tetrahedral molecule $XeO_4$, the central Xenon atom is bonded to four Oxygen atoms via four double bonds. This specific arrangement results in a total of 4 $\sigma$ bonds and 4 $\pi$ bonds, meaning the number of $\sigma$ and $\pi$ bonds is exactly equal.

Nucleophiles are defined by their ability to donate electron pairs, which categorizes them chemically as Lewis bases. Because these species are inherently electron-rich, they are attracted to and attack sites with low $e^-$ density rather than seeking electrons themselves. Describing a nucleophile as a Lewis acid contradicts this fundamental behavior, as Lewis acids are specifically defined as electron-deficient species that accept electron pairs.

The stability of a carbocation is primarily determined by the degree to which its positive charge can be delocalized through resonance, which offers significantly greater stabilization than inductive effects alone. In benzyl chloride, ionization of the $C-Cl$ bond generates the benzyl carbocation ($C_6H_5CH_2^+$), in which the positive charge is conjugated with the $\pi$-electrons of the aromatic ring. This extensive delocalization across the benzene ring confers superior stability to the intermediate compared to aliphatic or vinyl carbocations, making the parent compound the most favorable for generating a stable carbocation.

In Dumas's method for nitrogen estimation, the measured total pressure includes both the nitrogen gas and the aqueous vapor pressure. To isolate the pressure of the dry nitrogen gas, subtract the aqueous tension from the observed pressure, which is 725 mm minus 25 mm, yielding 700 mm. Convert this volume to standard conditions, where $P_2 = 760$ mm and $T_2 = 273$ K, using the combined gas law:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Substituting the experimental values $P_1 = 700$ mm, $V_1 = 40$ mL, and $T_1 = 300$ K into the equation leads to a standard volume of 33.52 mL, as calculated by $V_2 = \frac{700 \times 40 \times 273}{300 \times 760}$. Given that 22400 mL of nitrogen gas corresponds to 28 g at standard conditions, use this relationship to find the mass of the nitrogen collected. Calculating the percentage requires dividing this mass by the initial sample mass of 0.25 g and multiplying by 100, which yields the final result of 16.76%.

Methyl lithium ($CH_3Li$) acts as a strong nucleophile because the carbon-lithium bond is highly polarized, placing a significant partial negative charge on the methyl carbon. Upon reaction, this nucleophilic methyl group attacks the electrophilic carbonyl carbon of cyclopentanone, causing the electrons of the $C=O$ double bond to shift completely onto the oxygen atom. This nucleophilic addition results in the formation of a negatively charged alkoxide intermediate, which constitutes the cyclopentanonyl anion, $C_5H_8O(CH_3)^-$.

Kjeldahl's method determines nitrogen content by evolving ammonia gas and neutralizing it with sulfuric acid. Because sulfuric acid is diprotic, it reacts with ammonia in a 1:2 molar ratio. Calculating the moles of acid using $n = M \times V$ with a concentration of 1 M and a volume of 0.01 L gives 0.01 moles, which effectively neutralizes 0.02 moles of ammonia.

Since each mole of ammonia contains one mole of nitrogen, the sample contains 0.02 moles of nitrogen. Multiplying this amount by the molar mass of 14 grams per mole results in 0.28 grams of nitrogen. The percentage of nitrogen is determined by the following formula:

$\text{Percentage} = \frac{\text{Mass of Nitrogen}}{\text{Mass of Sample}} \times 100$

Substituting the values of 0.28 grams for the nitrogen and 0.75 grams for the sample into this formula yields 37.33%.

Hückel's rule states that an aromatic system must contain $(4n+2) \pi$ electrons, where $n$ is an integer. In this case, the radical exhibits aromatic character because it possesses 6 p-orbitals and 6 unpaired electrons, providing a total count that satisfies the requirement for a stable aromatic sextet. This specific configuration of 6 p-orbitals and 6 unpaired electrons ensures the continuous cyclic conjugation necessary to sustain aromaticity.

Carbocation stability is governed by the positive inductive effect, where alkyl groups donate electron density to help disperse the positive charge at the cationic center. The stability increases with more alkyl substituents and more alpha-hydrogens, moving sequentially from methyl to primary, secondary, and finally tertiary carbocations. Within the tertiary category, the specific number of alpha-hydrogens determines the extent of hyperconjugation, which is why the stability of $(\mathrm{CH}_{3})_{2}-\overset{+}{\mathrm{C}}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$ is lower than that of $(\mathrm{CH}_{3})_{3}-\overset{+}{\mathrm{C}}$, resulting in the increasing order $5 < 4 < 3 < 1 < 2$.

The "iso" prefix in organic nomenclature designates a structure containing a $(CH_3)_2CH-$ group at the end of a carbon chain, and an isobutyl group is specifically defined by the four-carbon arrangement $(CH_3)_2CH-CH_2-$. This configuration features a methyl group branching off the second-to-last carbon, which leaves a terminal methylene group available for bonding to the parent molecule. Option C accurately depicts this connectivity, showing two methyl groups attached to a central methine carbon that is subsequently bonded to the CH₂ group.

In Lassaigne's test, nitrogen present in an organic compound is first converted into sodium cyanide, which subsequently reacts with ferrous salts to produce sodium ferrocyanide. Upon the addition of ferric salts, the ferric ions interact with the ferrocyanide ions to precipitate a deep blue coordination complex known as Prussian blue. The composition of this specific pigment is represented by the formula ${Fe}_{4}[ {Fe}( {CN})_{6}]_{3}$, which arises from the chemical reaction:
$3 {Na}_{4}[ {Fe}( {CN})_{6}]+2 {Fe}_{2}( {SO}_{4})_{3} \longrightarrow {Fe}_{4}[ {Fe}( {CN})_{6}]_{3}+6 {Na}_{2} {SO}_{4}$

Nomenclature rules prioritize the carboxylic acid as the terminal functional group, which establishes the six-carbon parent chain starting from the carboxyl carbon. The structure builds sequentially from this point, placing a hydroxyl group at position two and a carbon-carbon double bond at position three, where an ethyl group is also attached. Carbon four anchors a methyl group, and the carbon-carbon triple bond initiates at position five to complete the six-carbon backbone. This specific sequence of substituents and multiple bonds defines the unique connectivity required to satisfy the criteria for the compound.

The reactivity of a benzene ring toward electrophilic aromatic substitution is governed by the net electron-withdrawing capacity of the attached functional group, which operates through both inductive ($-I$) and mesomeric ($-M$) effects. While all the listed groups are meta-directing and reduce ring reactivity, the nitro group ($-\mathrm{NO}_2$) exerts a significantly stronger pull on electron density compared to groups like $-\mathrm{COOH}$, $-\mathrm{C}\equiv \mathrm{N}$, and $-\mathrm{SO}_3\mathrm{H}$. This occurs because $-\mathrm{NO}_2$ benefits from a particularly potent combination of these inductive and mesomeric effects, making it the most effective deactivator among the options provided.

The stability of alkyl free radicals is primarily governed by the extent of hyperconjugation, where a higher number of alpha-hydrogens creates more resonance structures and stabilizes the electron-deficient center. Because alkyl substitution increases the availability of these hydrogens, primary radicals are less stable than secondary ones, which in turn are less stable than tertiary radicals. Within the tertiary category, the radical with nine alpha-hydrogens is more stable than the one with eight, resulting in the increasing stability order: $\mathrm{CH}_3-\overset{\bullet}{\mathrm{CH}}_2 < \mathrm{CH}_3-\overset{\bullet}{\mathrm{CH}}-\mathrm{CH}_3 < (\mathrm{CH}_3)_{2}\overset{\bullet}{\mathrm{C}}-\mathrm{CH}_2-\mathrm{CH}_3 < (\mathrm{CH}_3)_{3}\overset{\bullet}{\mathrm{C}}$.

The benzyl carbocation, represented as $C_6H_5CH_2^+$, features a benzylic carbon atom bonded to two hydrogen atoms and one carbon atom of the benzene ring through three sigma bonds. With no lone pairs and three bonding domains, the carbon atom adopts a planar geometry and exhibits $sp^2$ hybridisation to facilitate conjugation with the adjacent aromatic pi system, providing stability to the carbocation.

IUPAC rules mandate that the principal functional group, such as a carbon-carbon double bond, takes precedence over substituents and must be assigned the lowest possible locant. For the structure $\mathrm{Br}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2$, the parent chain is numbered starting from the right to place the double bond at the first position, which results in the correct name 3-bromopropene. Consequently, any naming convention that incorrectly prioritizes the halogen atom for a lower position over the double bond violates these fundamental guidelines.

Optical isomerism relies on the presence of a chiral center, which is a carbon atom attached to four distinct groups, ensuring the molecule is non-superimposable on its mirror image. Maleic acid contains a carbon-carbon double bond where the involved carbons are $sp^2$ hybridized, resulting in a planar geometry that lacks the necessary tetrahedral asymmetry for optical activity. Due to this rigid configuration and the restricted rotation around the double bond, the molecule exhibits geometrical isomerism rather than optical isomerism.

Tertiary alkyl halides undergo substitution via an $S_N1$ mechanism, which involves the formation of a carbocation intermediate that must subsequently react with the incoming nucleophile. For this pathway to proceed most effectively, the metal fluoride reagent should exhibit high ionic character to facilitate the necessary ionisation. As the ionic character of the metal-fluoride bond increases with the size of the alkali metal, the Rb-F bond possesses the highest ionic character among the given examples, making it the most favourable choice for the reaction.

Electrophilic nitration depends on the electron density of the benzene ring, where the reaction rate increases with the presence of electron-donating groups and decreases with electron-withdrawing substituents. The methyl group in toluene exerts a $+I$ effect that donates electron density to the ring, thereby enhancing its reactivity towards the attacking $NO_2^+$ electrophile compared to unsubstituted benzene. In contrast, the nitro group in nitrobenzene and the carboxyl group in benzoic acid exert strong $-I$ and $-M$ effects, which withdraw electron density and deactivate the ring toward substitution. Consequently, toluene emerges as the most reactive compound because it is the only derivative listed that effectively activates the aromatic system.

A nucleophilic substitution reaction occurs when a nucleophile attacks a molecule, displacing a leaving group that departs with an electron pair. In the reaction $\mathrm{CH}_3-\mathrm{CH}_2-\overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{C}}}\mathrm{H}-\mathrm{CH}_2\mathrm{Br}+\mathrm{NH}_3\rightarrow\mathrm{CH}_3-\mathrm{CH}_2-\overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{C}}}\mathrm{H}-\mathrm{CH}_2\mathrm{NH}_2$, the bromine atom functions as the leaving group and is replaced by the incoming ammonia nucleophile, which perfectly matches this mechanism. Other options represent addition reactions where species are added across double or carbonyl bonds rather than replacing an existing group.

The Lassaigne's test is designed to identify halogens, but if nitrogen and sulphur are present in the sample, the fusion process generates sodium cyanide and sodium sulphide, which interfere with the detection of halide ions. Boiling the sodium extract with concentrated $\mathrm{HNO}_{3}$ decomposes these unwanted compounds and drives them out of the solution as volatile gases, as shown by the reactions $\mathrm{NaCN}+\mathrm{HNO}_{3} \rightarrow \mathrm{NaNO}_{3}+\mathrm{HCN} \uparrow$ and $\mathrm{Na}_{2} \mathrm{S}+2 \mathrm{HNO}_{3} \rightarrow 2 \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{S} \uparrow$. By removing these contaminants, this essential step ensures that any observed precipitation is due solely to the presence of halogens.