Test: General Organic Chemistry Level - 2 - Chemistry MCQ

To determine which of the compounds is most basic, we need to analyze the nitrogen atoms in each structure and their availability to donate a lone pair of electrons.

The basicity of nitrogen-containing compounds depends on the availability of the lone pair on nitrogen. Nitrogen atoms in aromatic systems, such as pyridine or quinoline derivatives, have lone pairs that participate in resonance, which can reduce their availability for protonation. Nitrogens that are less involved in resonance and conjugation will tend to be more basic.

Analyzing each structure:

1. Compound 1: This structure has nitrogen atoms in an aromatic system similar to quinoline. The lone pairs on these nitrogen atoms are somewhat delocalized due to resonance, reducing their basicity.

2. Compound 2: This structure is analogous to acridine, which has a nitrogen in an aromatic system. The lone pair is part of the conjugated system, reducing its basicity.

3. Compound 3: This structure has a nitrogen in a pyridine-like environment (quinazoline), but its lone pair is still somewhat available, making it relatively more basic compared to compounds where nitrogen is more deeply involved in conjugation.

4. Compound 4: This structure is a pyrimidine derivative. The nitrogen atoms are part of an aromatic system where their lone pairs are involved in resonance. Therefore, they are less basic.

Conclusion:

Among the given compounds, Compound 3 (quinazoline derivative) is the most basic because its nitrogen atoms are less involved in resonance compared to the other structures, making the lone pair more available for protonation.

The correct answer is:

Option 3.

It follows huckels rule. in the 1st case if it loses H+ ion then it's conjugate base is stabilized by all the CF3 groups . in 3rd case conjugate base is stabilized by electron withdrawing double bonds.hence stable. in 2nd case it becomes antiaromatic.

Aldehyde is more acidic then ester, as ther is +R effect of -OMe group in ester, it decreases the electron withdrawing power of carbonyl group.for second case it is ester.but for 3rd case, the (alfa H) is attached with two carbonyl group.
so the order of acidity is ¡>¡¡¡>¡¡

Correct Answer :- c

Explanation :

 CH₃Cl --------------> CH₃ + Cl

Methyl free radical sp² hybridized (with singly occupied p-orbital.

(sp3 hybridized carbon) tetrahedral bond angle 109^o28

The pKa of water is 14

The pKa of methanol (CH3OH) is 15.5

The pKa of Phenol (C₆H₅OH) is 9.88

As these petroleum and natural gas comprise of naturally occurring compounds such as carbon and hydrogen in abundance, these serve as the main source of arenes.

Negative charge over S atom gets additional stability by conjugation through vacent 3d orbital of S ,so HS^- is the most stable species and the reaction undergoes in the forward direction.

Naphthalene has fused ring of aromaticity and has the simplest structure when compared with other polycyclic aromatic hydrocarbons.

Picric acid is a representative arene compound but not picric chloride.

 An aromatic hydrocarbon always has a sigma as well as a delocalized pi bond found between the carbon atoms.

Electron donating groups increase the acidity of phenol when present as substituent on the benzene ring while electron withdrawing groups decrease the acidity of phenol.

Cl and NO₂ are electron withdrawing groups while -CH₃ and -OCH₃ are electron donating groups.

However, nitro undergoes resonance also which allows it to better stabilize the negative charge of the oxygen as compare to chlorine and thus acidic character of nitro phenol would be higher than that of chloro phenol.

So, Nitrophenol is the most acidic compound here followed by chloro phenol.

Hence, the correct option is 2.

Compound which gives an stable conjugate base deprotonation will deprotonate easily.

so, 1,3 and 6 will become aromatic after deprotonation.

To determine why a molecule is anti-aromatic, we must consider the following criteria:

1. **Planarity**: The molecule must be planar, allowing for p-orbitals to overlap and create a continuous cyclic conjugation.
2. **Cyclic structure**: The molecule must be cyclic.
3. **Conjugation**: The molecule must have a continuous overlap of p-orbitals (conjugation).
4. **Hückel's rule**: For a molecule to be aromatic, it must have 4n+2 π-electrons (where n is a non-negative integer). For a molecule to be anti-aromatic, it must have 4n π-electrons.

Based on the image provided:

- The structure appears to be a cyclic molecule.
- The structure appears to have conjugation (indicated by the bonds and charges).
- We need to count the number of π-electrons to apply Hückel's rule.

From the image, let's assume that the cyclic structure contains 4 π-electrons (commonly seen in anti-aromatic systems such as cyclobutadiene).

Since 4 π-electrons fit the 4n formula (with n = 1), this means the molecule has 4 π-electrons, making it anti-aromatic because it does not satisfy the 4n+2 rule for aromaticity.

Therefore, the molecule in the image is anti-aromatic because it is planar, cyclic, conjugated, and has 4n π-electrons (with n = 1, leading to 4 π-electrons), which fits the criteria for anti-aromaticity.

Both aromatic rings are preferred. Therefore, D option is correct.

In the 1st compound the double bond is at the bridgehead carbon , as a result this compound cannot exist in enol form because such conformation would give an impossibility strained ring .Such dis stabilizing effect is absent in 2 nd compound . Whereas the 3 rd compound exclusively exist in enol form due to the formation of intramolecular hydrogen bonding

According to britz's rule at bridge head position -ve charge is unstable so; 3>2>1

With electron withdrawing group, the keto-form dominates and with electron donating group, the enol form dominates. Since OCH3 is an electron donating group, (II) will have higher enol content.

In the first compound after formation of enol the ring becomes aromatic (when the lonepair over N atom involve in conjugation) and this is a stabilizing factor but no such stabilizing factor is present in the enol form of 2. So 1 has higher enol content and 2 has lower enol content.

To determine the order of the amount of enol form present at equilibrium for the given compounds, let's analyze each compound:

1. Compound I (CH₃)₃COOH: This compound is a carboxylic acid with three methyl groups attached to a carbonyl group. The presence of three methyl groups increases the stability of the keto form over the enol form because the enol form would involve forming a double bond next to these bulky groups, which is not favorable.

2. Compound II (Cyclohexanone): This is a cyclic ketone. The enol form here can be relatively stabilized by intramolecular hydrogen bonding, but the keto form is generally more stable. Cyclohexanone has moderate enol content.

3. Compound III (Phenol with carbonyl group): This is a phenol derivative where the enol form is highly stabilized due to the aromaticity of the phenol ring. The enol form is significantly stabilized by the resonance in the benzene ring, making it much more stable compared to the keto form.

Given these observations:

• Compound III will have the highest amount of enol form due to the resonance stabilization.
• Compound II will have a moderate amount of enol form.
• Compound I will have the least amount of enol form because of the stability of the keto form in the presence of the bulky methyl groups.

So, the correct order of enol content is III > II > I.

Therefore, the correct answer is (d) III > II > I.

• Option A shows a benzene ring with a methyl group (CH₃) attached, which is a correct representation of toluene.
• Option B shows a double bond outside the benzene ring, which is not a stable or common structure since benzene rings are known for their stability due to delocalized π electrons. This representation suggests a pentadienyl cation, which is highly unstable.
• Option C shows a methyl group attached to a benzene ring, similar to Option A, which is another correct representation of toluene.
• Option D shows a benzene ring with a methyl group attached, which is also a correct representation of toluene.

Based on this, Option B is the incorrect representation because it depicts a structure that is not chemically stable or common, especially in the context of benzene chemistry.

Correct Answer :- a

Halogen group have no mesomeric effect only inductive effect on benzene ring so correct answer is 2>3>4>1