Basics of Organic Chemistry (GOC)

The correct statement for paper chromatography is: (B) Water present in the pores of the paper forms the stationary phase.

In paper chromatography, the stationary phase is the water that is absorbed and trapped in the pores of the paper, while the mobile phase is a liquid solvent that moves through the paper carrying the sample with it.

The different components of the sample have different affinities for the stationary phase and the mobile phase, causing them to move at different rates and separate from each other on the paper.

The term meta-directing refers to substituents on a benzene ring that direct incoming electrophiles to the meta position (relative to themselves) during an electrophilic aromatic substitution reaction.

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.

Let's examine the options: 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.

7-Hydroxyheptan-2-one is correct IUPAC name

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.

(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.

(a) Charcoal treatment removes coloured impurity through adsorption. (b) Steam distillation separates the mixture of o-nitrophenol and p-nitrophenol.

The o-nitrophenol is steam volatile (due to intramolecular hydrogen bonding), and the para isomer is not volatile. (c) Fractional distillation separates crude naphtha.

Naphtha is a flammable liquid hydrocarbon mixture. (d) Distillation under reduced pressure separates mixture of glycerol and sugars.

Vacuum distillation lowers the boiling point and prevents decomposition.