Basics of Organic Chemistry

Carbocation intermediate is stabilised by $+\mathrm{I},+\mathrm{M} \&$ hyperconjugation effect. Since in option 2 carbocation is in conjugation with stronger +M group $-\mathrm{OCH}_3$ hence it will be most stable.

q is aromatic r is nonaromatic p is antiaromatic

Due to +M effect of -OMe at para position

To determine the number of different stereoisomers for a molecule, use the formula:

$ \text{Number of stereoisomers} = 2^n $

where $ n $ is the number of chiral centers in the molecule.

For the given molecule, there are 2 chiral centers, so:

$ n = 2 $

Therefore, the number of possible stereoisomers is:

$ 2^2 = 4 $

This means there are 4 different stereoisomers possible for the molecule in question.

(A) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$. GI is not possible

(B) Trans isomer has identical atoms/groups on the opposite side of double bond.

There are three stereo center

So No of stereoisomer $=2^3=8$

Firstly higher the number of bonds or say complete octet higher will be stability, then the negative charge situated over more electronegative element more will be stability. Hence, overall order of stability is I > II > III.

: Due to strength of I effect of substituents

Hence the correct order of basic strength

Higher the electron withdrawing strength of substituents higher will be acidic strength of phenols and lesser will be $\mathrm{pK}_{\mathrm{a}}$

The methods used for the purification of organic compounds are fundamentally based on both the nature of the compound and the presence of the impurity. This is because the selection of a suitable purification technique requires understanding not only what the organic compound is (i.e., its physical and chemical properties) but also the type of impurities that need to be removed.

For example, if an organic compound is a solid with a high melting point and the impurity is a low melting solid or a liquid, techniques like sublimation or crystallization could be used, which rely on the differences in physical properties between the compound and the impurity. On the other hand, if the compound needs to be purified from substances with similar boiling points, then a more refined method such as fractional distillation might be employed, relying on slight differences in boiling points. In cases where the compound is mixed with substances that have similar physical properties but different chemical reactivities, chemical methods such as extraction or chromatography could be used, which exploit these chemical differences.

Therefore, the correct answer is:

Option C: nature of compound and presence of impurity.

$\begin{aligned} & \mathrm{HCOOH}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}> \\ & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH} \end{aligned}$

Compound (C) is non planer due to trans annular hydrogen interaction and hence non aromatic.

Compound (B) and (D) are only aromatic compounds.

Correct name is 1-chloro-2,4-dinitro benzene

Statement-I is incorrect.

Statement-II is correct.

Methoxy group strongly activates benzene ring towards electrophilic aromatic substitution due to $+\mathrm{R}$ effect. Methyl group weakly activates benzene ring due to its $+\mathrm{I}$ and $+\mathrm{H}$ effects. $\mathrm{CF}_3$ group strongly deactivates the benzene ring due to $-\mathrm{I}$ and $-\mathrm{H}$ effects.

Therefore, the correct order of the given compounds towards electrophilic aromatic substitution is

Let's analyze each statement to determine which one is incorrect regarding the geometrical isomers of 2-butene.

Option A: "Cis-2-butene and trans-2-butene are not interconvertible at room temperature."

This statement is correct. Cis and trans isomers of 2-butene are not readily interconvertible at room temperature due to the energy barrier associated with the rotation around the double bond. The double bond ensures rigidity, making interconversion between these isomers difficult without sufficient energy input.

Option B: "Trans-2-butene is more stable than cis-2-butene."

This statement is correct. The trans isomer is more stable than the cis isomer due to reduced steric hindrance. In the cis isomer, the bulkier substituents (methyl groups) are on the same side, leading to increased steric repulsion.

Option C: "Cis-2-butene has less dipole moment than trans-2-butene."

This statement is incorrect. The cis isomer typically has a higher dipole moment than the trans isomer because the substituents on the same side of the double bond create a net dipole moment. In the trans isomer, the dipole moments of the substituents cancel each other out, resulting in a lower overall dipole moment.

Option D: "Cis-2-butene and trans-2-butene are stereoisomers."

This statement is correct. Cis-2-butene and trans-2-butene are indeed stereoisomers because they have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of the substituents around the double bond.

Therefore, the incorrect statement is:

Option C: Cis-2-butene has less dipole moment than trans-2-butene.

$-\mathrm{SO}_3 \mathrm{H}$ is functional group in sulphonic acids.

Compound given is option (2) is metamer of compound "$\mathrm{X}$".

$\begin{equation} \text { Dipole moment of cis }>\text { trans and } \mu_{\text {trans }}=0 \end{equation}$

(d) is aromatic. So it is most stable.

(a) is anti-aromatic. So it is least stable.

To determine the decreasing order of basic strength of the given conjugate bases, we need to understand the concept of basicity. A base is defined by its ability to accept protons (H⁺). The stronger the conjugate base, the more readily it accepts protons. The basicity of conjugate bases is also influenced by the stability of the anion (conjugate base); the more stable the anion, the weaker the base, and vice versa.

Let's analyze each conjugate base:

1. ${ }^{-} \mathrm{OH}$ (Hydroxide ion): This is a strong base because it is less stabilized by resonance or inductive effects.

2. $\mathrm{R} \overline{\mathrm{O}}$ (Alkoxide ion): This also is a strong base. However, alkyl groups (R) are electron donating, which can increase the electron density on oxygen, making it a stronger base compared to hydroxide ion.

3. $\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}$ (Acetate ion): This is a much weaker base because the negative charge is delocalized over the two oxygen atoms through resonance, which stabilizes the ion.

4. $\mathrm{Cl}$ (Chloride ion): Chloride ion is very weakly basic since it is highly stable due to its high electronegativity and full octet.

Based on these observations, the decreasing order of basic strength should be:

$\mathrm{R} \overline{\mathrm{O}}>^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{Cl}$

Therefore, the correct choice is:

Option D

$\mathrm{R} \overline{\mathrm{O}}>{ }^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{Cl}$

Let's analyze each statement to determine which is incorrect:

Option A: The electromeric effect is a temporary effect.

This statement is correct. The electromeric effect is indeed a temporary effect that occurs in the presence of a reagent and disappears when the reagent is removed.

Option B: Hydrogen ion $(\mathrm{H}^{+})$ shows negative electromeric effect.

To understand this, let's define the electromeric effect. The electromeric effect involves the transfer of electron pairs within a molecule in the presence of an attacking reagent. There are two types of electromeric effects: the positive electromeric effect ($E^+$) and the negative electromeric effect ($E^-$). A positive electromeric effect occurs when electrons are transferred towards the reagent, while a negative electromeric effect occurs when electrons are transferred away from the reagent.

Since a hydrogen ion $(\mathrm{H}^{+})$ is an electron-deficient species (it lacks electrons), it would generally induce the transfer of electrons towards itself, resulting in a positive electromeric effect. Hence, this statement is incorrect.

Option C: The organic compound shows electromeric effect in the presence of the reagent only.

This statement is correct. The electromeric effect is observed only in the presence of an attacking reagent. Once the reagent is removed, the effect ceases to exist.

Option D: Electromeric effect dominates over inductive effect.

This statement is correct. The electromeric effect generally has a stronger influence on the electronic distribution within a molecule compared to the inductive effect, especially when both effects are present simultaneously.

Therefore, the incorrect statement is:

Option B: Hydrogen ion $(\mathrm{H}^{+})$ shows negative electromeric effect.

Meta directors are typically deactivating groups with respect to the aromatic ring and they often posses an atom with a partial positive charge or a full formal positive charge directly attached to the ring, which tends to withdraw electron density from the ring through the inductive effect or resonance. This decreased electron density in the ring makes the ortho and para positions less reactive, thereby directing new substituents to the meta position. Furthermore, the presence of these withdrawing groups can stabilize the intermediate cation (whichever is formed) better when the substituent is at the meta position relative to the ortho or para positions.

Option A (Incorrect):

• $-\mathrm{CN}$ (cyanide) is a strong meta director due to its strong electron-withdrawing nature.
• $-\mathrm{NH}_2$ (amino), $-\mathrm{NHR}$ (alkylamino), and $-\mathrm{OCH}_3$ (methoxy) are strong ortho/para directors because they donate electron density to the ring through resonance.

Option B (Possibly correct):

• $-\mathrm{CN}$ (cyanide) is a strong meta director as stated before.
• $-\mathrm{CHO}$ (formyl) is a weak meta director because the carbonyl group withdraws electron density via resonance and induction.
• $-\mathrm{NHCOCH}_3$ (acetamido) is a bit tricky. While $-\mathrm{NH}_2$ usually directs to the ortho/para position, when it is part of the amide group, the resonance with the carbonyl can make it withdraw electron density, thus Acetamido become a meta-director.
• $-\mathrm{COOR}$ (ester) is a meta director due to the electron-withdrawing nature of the carbonyl group.

Option C (Incorrect):

• $-\mathrm{NO}_2$ (nitro) is a very strong meta director due to its strong electron-withdrawing nature both through induction and resonance.
• $-\mathrm{NH}_2$ (amino) and $-\mathrm{COOH}$ (carboxylic acid) are strong ortho/para directors. Carboxylic acid groups can be deactivating due to their overall electron-withdrawing nature, but they still direct to the ortho/para positions because the electronegative oxygen can donate electron density by resonance.
• $-\mathrm{COOR}$ (ester) is a meta director.

Option D (Correct):

• $-\mathrm{NO}_2$ (nitro) is a strong meta director as before.
• $-\mathrm{CHO}$ (formyl) is a meta director.
• $-\mathrm{SO}_3 \mathrm{H}$ (sulphonic acid) is a strong meta director because the sulphonic acid group is very electron-withdrawing due to the presence of several electronegative oxygen atoms.
• $-\mathrm{COR}$ (ketone) is a meta director because of the electron-withdrawing nature of the carbonyl group.

Based on the description above, the set of functional groups that are meta directing from the given options is Option D: $-\mathrm{NO}_2$, $-\mathrm{CHO}$, $-\mathrm{SO}_3 \mathrm{H}$, $-\mathrm{COR}$. These groups are all electron-withdrawing and would direct incoming substituents to the meta position during electrophilic aromatic substitution.

Lassaigne's test is a qualitative test in analytical chemistry used to detect the presence of certain elements, namely nitrogen, sulfur, and halogens in an organic compound. The test involves heating the organic compound with sodium metal to convert these elements to sodium salts, which can then react with specific reagents to yield visibly identifiable compounds.

Let's go through the options one by one:

Option A suggests that Lassaigne's test is used for the detection of phosphorous and halogens only. This is incorrect because the test is also used to detect nitrogen and sulfur.

Option B suggests that the test is for nitrogen, sulfur, and phosphorous only. However, Lassaigne's test does not typically involve a specific test for phosphorous.

Option C, which is the correct answer, indicates that this test is used to detect nitrogen, sulfur, phosphorous, and halogens. Even though phosphorous detection isn't as common as the detection of nitrogen, sulfur, and halogens in the typical Lassaigne's test, the option stating all four elements should be considered the most complete answer among those provided, given its inclusion of all elements that could potentially be detected with adaptations of the test.

Option D states that the test is for nitrogen and sulfur only, which is not entirely accurate because the test can also be used to detect halogens, and potentially, albeit less commonly, phosphorous.

Therefore, the most accurate answer is Option C: Nitrogen, Sulphur, phosphorous, and halogens.

The negative resonance effect, also called the -I effect, refers to the ability of a functional group to withdraw electron density through the p-orbitals via resonance. It is often the characteristic of groups that have atoms with high electronegativity or multiple bonds that allow for electron delocalization away from the conjugated system.

To determine which functional group shows a negative resonance effect from the options provided, let's review them one by one:

• Option A - $-\mathrm{OH}$: The hydroxyl group (-OH) is capable of donating electrons via resonance due to the lone pair of electrons on oxygen. Therefore, it has a positive resonance effect, not a negative one.


• Option B - $-\mathrm{OR}$: Similar to the hydroxyl group, alkoxides (-OR) can also donate electron density through resonance because of the lone pairs on the oxygen atom. This makes them electron-donating groups, showing a positive resonance effect.


• Option C - $-\mathrm{COOH}$: The carboxylic acid group (-COOH) contains a carbonyl group (C=O) which is highly electronegative and can withdraw electron density via resonance. The double bond between carbon and oxygen can participate in resonance, spreading the electron density away from other parts of the molecule. Thus, the carboxylic acid group shows a negative resonance effect.


• Option D - $-\mathrm{NH}_2$: The amino group (-NH₂) is an electron-donating group due to the lone pair of electrons on the nitrogen atom. It shows a positive resonance effect, similar to hydroxyl and alkoxy groups.

Given the choices, Option C - $-\mathrm{COOH}$ is the functional group that shows a negative resonance effect because of its ability to withdraw electron density through resonance with its carbonyl group which is highly electronegative and capable of delocalizing electrons.

The correct answer is Option B: (B) only.

In organic chemistry, ionic reactions typically involve heterolytic bond cleavage. This is when a bond between two atoms breaks and both electrons of the shared pair go to one of the atoms, creating ions. One atom becomes a positively charged ion (cation), as it loses an electron, and the other becomes a negatively charged ion (anion), as it gains an electron.

In contrast, homolytic bond cleavage occurs when each atom gets one electron from the shared pair, leading to the formation of two free radicals, which are highly reactive species with unpaired electrons. However, this is not typically how ionic reactions proceed in organic compounds.

The options (C), (D), and (E) mention free radicals. Since homolytic cleavage leads to free radicals, and we've established that ionic reactions in organic chemistry generally proceed through heterolytic cleavage, free radicals are not typically involved in these reactions. Therefore, these options are not correct in the context of ionic reactions with organic compounds.

Thus, the process of heterolytic bond cleavage (B) is the mechanism through which ionic reactions usually occur in organic chemistry.

$-\mathrm{CH}_3$ shows $+\mathrm{M}$ and $+\mathrm{I}$.

$-\mathrm{Cl}$ shows $+\mathrm{M}$ and $-\mathrm{I}$ but inductive effect dominates.

$-\mathrm{NO}_2$ shows $-\mathrm{M}$ and $-\mathrm{I}$.

Electrophilic substitution $\alpha \frac{1}{-\mathrm{M} \text { and }-\mathrm{I}}$

$\alpha+\mathrm{M} \text { and }+\mathrm{I}$

Hence, order is $\mathrm{B}>\mathrm{A}>\mathrm{C}>\mathrm{D}$.

The species described as having a carbon atom with a sextet of electrons—that is, six electrons in its valence shell—capable of acting as an electrophile (an electron-pair acceptor) is called a carbocation. Therefore, the correct answer is:

Here's a brief explanation of each option:

Option A: Pentavalent carbon - This term is not commonly used in organic chemistry, and it refers incorrectly to a carbon with five bonds, which contradicts the tetravalency of carbon.

Option B: Carbon free radical - This species has a carbon atom with an unpaired electron (a septet, or seven valence electrons). It is highly reactive but is not typically classified as an electrophile because it seeks to pair its single unpaired electron rather than accepting a pair of electrons.

Option C: Carbanion - This anion has a carbon atom with three bonds and a pair of electrons, which in total gives eight valence electrons. It is nucleophilic (electron-pair donor) rather than electrophilic.

Option D: Carbocation - A carbocation is indeed a carbon species with only six valence electrons and thus has an empty p-orbital that can accept a pair of electrons, making it electrophilic. Carbocations are usually trigonal planar in structure and carry a positive charge.

7-Hydroxyheptan-2-one is correct IUPAC name

More no. of hyperconjugable Hydrogens, more stable is the carbocations.

The purification method among the ones listed that is based on the principle of "Solubility" in two different solvents is Option D, Differential Extraction.

Differential extraction, also known as liquid-liquid extraction or solvent extraction, is a technique to separate compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent. It exploits the fact that the compound of interest has different solubility properties in two solvents which are not miscible with each other. A solution containing the desired compound and impurities is mixed with a second solvent in which the desired compound is more soluble. The two solvents form layers due to their immiscibility, and the compound partitions into the layer where it is more soluble. By separating the layers and repeating the process if necessary, a higher degree of purification can be achieved for the target compound.

Let's quickly review why the other options do not primarily use solubility in two different solvents:

• Distillation (Option A): Distillation is a process that separates the components of a mixture based on differences in their boiling points. It does not rely on differential solubility in two solvents.


• Sublimation (Option B): Sublimation is the phase transition of a substance from the solid to the gas phase without passing through the intermediate liquid phase. This process is based on the volatility of the substance, not on its solubility in solvents.


• Column Chromatography (Option C): Column chromatography separates compounds based on their differing affinities to a stationary phase (often a solid) and a mobile phase (often a liquid). The separation is achieved by partitioning between the phases, but it does not primarily involve the solubility of the compounds in two different immiscible solvents.

Cyclohex-2-en-1-ol

Stability of conjugate base $\alpha$ acidic strength

C < D < B < A

Due to unsymmetrical.

It is a type of conjugation responsible for resonance.

The statement regarding conformations of ethane that is incorrect is:

Option C - Eclipsed conformation is the most stable conformation.

To explain why Option C is incorrect, let's analyze all the options:

Option A: "The dihedral angle in staggered conformation is $60^{\circ}$." This statement is incorrect. In staggered conformation, the dihedral angle, or the angle between the planes containing the hydrogen atoms, is actually $60^{\circ}$. However, the staggered form is not unique; it refers to the form where any given pair of hydrogens on adjacent carbons has this dihedral angle, but rotating the molecule can yield a variety of staggered conformations with different specific hydrogen atoms at the $60^{\circ}$ dihedral angle. The full rotation of one ethyl group relative to the other through $360^{\circ}$ creates a cycle where the staggered conformation occurs every $120^{\circ}$.

Option B: "Ethane has infinite number of conformations." This statement is correct. The rotation around the carbon-carbon single bond (sigma bond) is relatively free, so ethane can theoretically adopt an infinite number of conformations between the two extremes of staggered and eclipsed. These conformations can have very slight differences in angles, resulting in finely varied overlap of electron clouds.

Option C: "Eclipsed conformation is the most stable conformation." This statement is incorrect, which makes it the answer to the question. The eclipsed conformation is actually the least stable conformation of ethane due to the increased torsional strain from electron repulsion. The electron clouds of the bonding pairs of hydrogens on adjacent carbons are aligned in such a way as to be as close to each other as possible, increasing repulsion. The most stable conformation is the staggered conformation, where the groups are as far apart as possible, reducing repulsion.

Option D: "The conformations of ethane are inter-convertible to one-another." This statement is correct. As you rotate around the carbon-carbon single bond, the conformations continuously change from one to another. This interconversion can be quite rapid and occurs without the need for a high energy input under normal conditions.

Thus, the incorrect statement is Option C.