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

IUPAC nomenclature prioritizes the longest carbon chain containing the double bond, which is identified here as a pentene backbone with the double bond starting at position 2, determining the base name as 2-chloro-3-iodo-2-pentene. The geometric descriptor "trans" is applied because the higher priority groups, specifically the chlorine atom on the second carbon and the iodine atom on the third carbon, are situated on opposite sides of the $C=C$ double bond.

Nucleophilic aromatic substitution proceeds most efficiently when the benzene ring is made electron-deficient, as this stabilizes the negatively charged intermediate formed during the reaction. Electron-withdrawing groups, such as the $-NO_2$ group, are particularly effective because they pull electron density away from the aromatic ring through both inductive and resonance effects, significantly lowering the activation energy. Consequently, a chlorobenzene derivative containing such a powerful electron-withdrawing substituent undergoes nucleophilic attack much more readily than an unsubstituted chlorobenzene molecule.

Dumas' method relies on the measurement of nitrogen gas liberated from an organic compound, which is captured over water. Since the gas is collected under these conditions, it contains water vapor, meaning the observed pressure must be corrected by subtracting the aqueous tension to find the pressure of the dry nitrogen gas. The fundamental principle involved here is the Ideal Gas Law, which allows for the conversion of the gas volume from the experimental temperature and pressure to Standard Temperature and Pressure (STP).

To begin, the pressure of the dry nitrogen gas is obtained by subtracting the aqueous tension from the total pressure, resulting in 700 mm Hg. We then convert the observed gas volume to STP using the combined gas law, which states that the ratio of the product of pressure and volume to temperature remains constant. By applying the formula

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

and substituting the values $P_1 = 700$, $V_1 = 55$, $T_1 = 300$, $P_2 = 760$, and $T_2 = 273$ into the equation, the calculated volume of nitrogen at STP is approximately 46.099 mL.

The final percentage of nitrogen is determined by relating the mass of the gas at STP to the mass of the organic sample. Since 22,400 mL of nitrogen gas at STP corresponds to 28 g of nitrogen, the percentage can be found using the simplified expression $\frac{V_2}{8 \times W}$, where $W$ is the mass of the organic compound. Substituting the calculated STP volume of 46.099 mL and the given sample mass of 0.35 g into this ratio yields a value of approximately 16.46%.

Establishing the longest carbon chain that includes the double bond is the fundamental priority in IUPAC nomenclature, which in this structure results in a six-carbon parent chain defined as hex-1-ene. By numbering the chain from the end nearest the double bond to ensure the alkene receives the lowest positional index, we can systematically identify the attached side chains. Alphabetizing the substituents while noting their specific locations confirms that the ethyl group is at the fourth position and the propyl group is at the third, leading to the name 4-ethyl-3-propyl hex-1-ene.

The geometric structure of a molecule is dictated by the hybridization state of its atoms, where $sp$ hybridization specifically necessitates a linear configuration with a bond angle of $180^\circ$. For a molecule like $\mathrm{CH_3-CH=CH-CH_3}$, the carbons involved in the double bond are $sp^2$ hybridized, resulting in a planar or bent orientation rather than a straight line. Conversely, $\mathrm{CH_3-C\equiv C-CH_3}$ contains two carbon atoms joined by a triple bond, and these $sp$ hybridized carbons impose the required $180^\circ$ bond angle that renders the central carbon skeleton linear.

Other structures like $\mathrm{CH_2=CH-CH_2-C\equiv CH}$ fail to be linear because they incorporate $sp^3$ hybridized carbons that introduce kinks in the chain, disrupting any potential for a single straight line. Similarly, in a fully saturated alkane such as $\mathrm{CH_3-CH_2-CH_2-CH_3}$, all carbons are $sp^3$ hybridized, which gives rise to a tetrahedral geometry with bond angles of $109.5^\circ$, forcing the molecule to adopt a non-linear, zigzag shape.

An electrophile is defined by its electron deficiency, characterized by either an incomplete octet or a positive charge, which drives it to seek and accept electron pairs. Within the provided options, $\overset{\oplus}{\mathrm{Cl}}$, $\mathrm{BH}_3$, and $\overset{\oplus}{\mathrm{N}\mathrm{O}_2}$ all possess this electron-demanding nature, either due to a formal positive charge or, in the case of boron, an incomplete valence shell. Conversely, $\overset{\oplus}{\mathrm{H}_3\mathrm{O}}$ does not act as an electrophile in this context because the oxygen atom maintains a complete octet and possesses lone pairs, leading it to function primarily as a Brønsted acid by donating a proton rather than acting as an electron-pair acceptor.

The rate of an $S_N1$ reaction is dictated by the stability of the intermediate carbocation, where a more stable carbocation results in higher reactivity. In the case of $C_6H_5C(CH_3)(C_6H_5)Br$, the formed carbocation is tertiary and experiences significant resonance stabilization from two phenyl rings in addition to the inductive effect of a methyl group. This makes it the most stable intermediate among the choices, directly resulting in the highest reactivity towards the reaction.

When naming unsaturated hydrocarbons that possess both a double and a triple bond, the parent carbon chain must be numbered to assign the lowest possible locants to these functional groups as a combined set. In the molecule $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHC}\equiv\mathrm{CH}$, numbering from the right assigns the triple bond to carbon-1 and the double bond to carbon-3, resulting in the locant set {1, 3}. If the chain were numbered from the left, the double bond would begin at carbon-2 and the triple bond at carbon-4, yielding the higher set {2, 4}. Because the set {1, 3} provides a lower combination than {2, 4}, this numbering is preferred, defining the structure as Pent-3-en-1-yne.

The stability of ethylene glycol conformers is primarily determined by the presence of hydrogen bonding, which acts to lower the potential energy of the molecular structure. Conformation D is the most stable arrangement because it facilitates intermolecular hydrogen bonding, providing a crucial stabilizing interaction that is absent or less favorable in the other conformers. This attractive force allows the molecule to exist in a more energetically favorable state compared to the alternative spatial orientations.

Reactivity towards electrophilic reagents in aromatic systems is primarily governed by the electron-donating strength of the substituents attached to the ring, where a stronger resonance donation increases the electron density available for electrophilic attack. Because the $+R$ effect of the hydroxyl group ($-\text{OH}$) is inherently greater than that of the methoxy group ($-\text{OCH}_3$), the presence of the hydroxyl group creates a more electron-rich ring structure. Consequently, the derivative substituted with the hydroxyl group exhibits the highest reactivity, facilitating a faster interaction with electrophilic species.

Hybridization is determined by the number of sigma bonds surrounding a carbon atom, with four sigma bonds indicating $sp^3$ hybridization, three indicating $sp^2$, and two indicating $sp$ hybridization. Starting with $C_2$, this atom is part of a triple bond and is thus involved in only two sigma bonds, characterizing it as $sp$ hybridized. The carbon $C_3$ is bonded to a hydrogen, a methyl group, $C_2$, and $C_4$ using four single bonds, which results in $sp^3$ hybridization. For $C_5$, the participation in a carbon-carbon double bond limits the sigma bond count to three, designating it as $sp^2$ hybridized, whereas $C_6$ is bonded to two methyl groups, $C_5$, and $C_7$ solely through single bonds, placing it in an $sp^3$ hybridized state.

When naming organic compounds that contain both double and triple bonds, the carbon chain must be numbered such that the unsaturated bonds receive the lowest possible locants. In the hydrocarbon $CH\equiv C-CH=CH_2$, the chain consists of four carbons, giving it the parent name butane. Numbering the chain from either end assigns the positions 1 and 3 to the double and triple bonds, creating an equivalent locant set in both directions. In cases where the locant sets are identical, IUPAC priority rules dictate that the double bond is assigned the lower number, so the chain is numbered starting from the end closer to the double bond. This results in the double bond at the first position and the triple bond at the third position, forming the name 1-butene-3-yne.

Geometrical isomerism requires restricted rotation about a double bond, which prevents groups from rotating freely and enables the existence of different spatial arrangements. For the molecule $CH_3CH=CHCH_3$, this rigidity ensures that the hydrogen and methyl groups on each carbon of the double bond are locked in place, allowing for both cis and trans configurations. Compounds lacking this specific structural requirement, such as those with triple bonds or different bonding patterns, cannot generate these unique geometrical isomers.

The basic strength of carbanions is fundamentally dictated by the hybridization of the carbon atom carrying the negative charge, as an increase in s-character enhances the electronegativity of that carbon and stabilizes the lone pair. As the s-character increases from $\mathrm{sp}^3$ in $\mathrm{H_3C-CH_2^-}$ to $\mathrm{sp}^2$ in $\mathrm{H_2C=CH^-}$ and finally $\mathrm{sp}$ in $\mathrm{HC \equiv C^-}$, the negative charge is held more tightly to the nucleus, making the species significantly less willing to donate it and thus reducing its basicity. Consequently, the basic strength decreases as the hybridization shifts from $\mathrm{sp}^3$ to $\mathrm{sp}^2$ to $\mathrm{sp}$, resulting in an overall order of $\mathrm{H_3C-CH_2^-} > \mathrm{H_2C=CH^-} > \mathrm{HC \equiv C^-}$.

The stability of a carbanion is fundamentally governed by the hybridization of the carbon atom bearing the negative charge, as higher $s$-character increases electronegativity and allows the atom to hold the lone pair more effectively. The $sp$-hybridized carbanion in (i) provides the greatest stability due to its high electronegativity, whereas the $sp^3$-hybridized carbanion in (iv) is the least stable because the electron-donating inductive effect of the three $R$ groups increases electron density at the charge center, further destabilizing it. Following this trend, where increased electron-releasing groups and lower electronegativity reduce stability, the relative order is correctly represented as (i) > (ii) > (iii) > (iv).

Electrophilic aromatic substitution relies on the availability of electron density within the benzene ring to attract electron-seeking species. Substituents such as $-Cl$ and $-NO_2$ withdraw electrons via the $-I$ effect, which depletes the ring and lowers its reactivity. In contrast, while the $-OH$ group exerts an inductive effect, its strong $+M$ effect predominates to significantly increase the electron density of the ring, rendering it the most reactive towards electrophilic attack.

The total number of stereoisomers for the molecule $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_2\mathrm{CHBrCH}_3$ is determined by the number of independent stereogenic elements, specifically the geometrical and optical centers. The carbon-carbon double bond acts as a site for geometrical isomerism because the different groups attached to each carbon atom allow for two distinct spatial configurations. Additionally, the single chiral carbon atom bonded to four unique groups enables optical isomerism, which contributes two enantiomeric forms. Since these features are independent, we calculate the total number of stereoisomers by multiplying the two geometrical isomers by the two optical isomers, resulting in $2 \times 2 = 4$ stereoisomers.

The capacity to rotate plane-polarized light depends on the overall chirality of a molecule rather than the mere presence of individual stereocenters. Even when a substance contains tetrahedral carbon atoms bonded to four distinct groups, it remains optically inactive if the molecule as a whole possesses an internal plane of symmetry. This specific arrangement constitutes a meso compound, where one part of the structure induces a rotation of polarized light to the right while an equivalent part induces an equal rotation to the left. Because this internal compensation effectively cancels out the optical activity of the individual segments, the molecule as a whole exhibits no net rotation.

Nucleophilicity for ions in the same periodic group is governed by polarizability, where an increase in atomic size and mass facilitates a more effective donation of electron pairs. As we move down the group from chlorine to bromine to iodine, the ions become larger and more polarizable, thereby enhancing their nucleophilic strength. This direct relationship means that the smallest ion is the weakest nucleophile, resulting in the increasing order of $\mathrm{Cl}^{-} < \mathrm{Br}^{-} < \mathrm{I}^{-}$.

The reactivity of an aromatic ring towards electrophilic reagents depends on the electron density of the ring, which is increased by electron-donating groups and decreased by electron-withdrawing groups. Phenol possesses an $-OH$ group that strongly donates electrons via resonance, while toluene contains a $-CH_3$ group that provides milder activation, causing both to be more reactive than unsubstituted benzene. Conversely, the chlorine atom in chlorobenzene exerts an inductive $-I$ effect that withdraws electron density from the ring, rendering it the least reactive compound in this set.