Organic Chemistry

The correct answer is A displacement of electron

The inductive effect helps spread the unequal sharing of the bonding electron through the atoms in a chain of the molecule. This leads to the formation of a dipole in a bond. This distribution of electrons is also called displacement of electrons.

In an inductive effect, the sigma (σ) electrons are getting displaced resulting in the polarization of the bond. The inductive effect is the polarization of a σ bond due to the presence of an electron-withdrawing or an electron-donating group or atoms in a molecule. During the inductive effect, shifting of electrons takes place due to which partial charges occur on the atom.

The mesomeric effect involves the complete transfer or release of or lone pair of electrons to the adjacent atom or covalent bond. Therefore, it involves the delocalization of pi electrons.

Mesomeric effect involves the delocalization of π electrons. Hence option A is correct.

In the photoelectric effect, the number of electrons ejected from a material increases with the intensity of the incident light, as intensity is directly related to the number of photons striking the material. However, the energy of the ejected electrons is dependent on the frequency of the incident light, not the intensity. The work function (the minimum energy required to eject an electron) is a characteristic of the material and does not affect the number of electrons ejected.

The correct option is (A): Number of e⁻ ejected increases with the increase in the intensity of incident ligh

Looking at the reaction steps:

1. The reaction of CH₃COOH with SOCl₂ will result in the conversion of the carboxyl group (–COOH) into an acyl chloride, forming A (C₆H₅COCl).
2. This acyl chloride then undergoes Friedel–Crafts acylation with benzene in the presence of AlCl₃ to form B (C₆H₅COCH₂), where the acyl group is added to the benzene ring.
3. Next, compound B reacts with HCN, resulting in the formation of a nitrile group (-CN) at the benzylic position, forming compound C (C₆H₅CH(C₂CH₃)CN).
4. Finally, hydrolysis of C with water leads to the formation of D, where the nitrile group is converted to a carboxyl group, yielding compound D (C₆H₅CH(C₂CH₃)COOH).

The correct answer is (D) : .

The reactivity of alkyl halides in S_N² reactions is primarily influenced by the steric hindrance around the carbon atom to which the halogen is attached. The order of reactivity follows from the fact that less sterically hindered compounds react more readily in S_N² reactions. Hence, the reactivity decreases as the size of the substituents (such as phenyl groups or alkyl groups) increases.

• The compound with a simple alkyl group (C₆H₅CH₂Br) will react the fastest due to minimal steric hindrance.
• Adding an additional alkyl group (C₆H₅CH₂CH₂Br) increases the steric hindrance slightly, reducing the reactivity compared to (A).
• When substituting a phenyl group (C₆H₅) onto the carbon attached to the bromine, as in C₆H₅CH(C₆H₅)Br, the reactivity decreases further because of steric hindrance from the bulky phenyl group.

The correct order of reactivity is (A) C₆H₅CH₂Br < C₆H₅CH₂CH₂Br < C₆H₅CH(C₆H₅)Br < C₆H₅CH(C₆H₅)Br < C₆H₅CH₂CH₃Br.

The correct answer is (A) : .

To solve the problem, we need to determine the major product formed when the given compound undergoes a reaction with under UV light. The reaction involves the substitution of a hydrogen atom in the alkyl group by bromine ( ).

1. Understanding the Reaction:
The given compound is:
$ \text{NO}_2 \text{CH}_2\text{CH}_3 \text{Br}_2 \text{Br}_2 \text{Br}^\bullet \text{Br}^\bullet \text{CH}_2\text{CH}_3 \text{CH}_3\text{CH}_2^\bullet \text{HBr} \text{CH}_3\text{CH}_2^\bullet \text{Br}^\bullet \text{CH}_2\text{CH}_3 \text{Br} \text{CH}_2\text{CH}_2\text{Br} \text{CH}_2\text{CHBrCH}_3 \text{CH}_2\text{CHBrCH}_3 \text{O}_{2}\text{N}-\underset{\text{CH}_{2}-\text{CH}_{2}\text{-Br}}{\underbrace{\text{C}_{6}\text{H}_{4}}} $

1. Identify the Compounds

The compounds are:
- HF (Hydrogen Fluoride)
- (Water)
- (Ammonia)

2. Recognize the Dominant Intermolecular Force
For these compounds, the dominant intermolecular force is hydrogen bonding.

3. Recall Factors Affecting Hydrogen Bond Strength
Factors that affect the strength of hydrogen bonds are:
- Electronegativity of the atom bonded to hydrogen. Higher electronegativity leads to stronger hydrogen bonds.
- Number of hydrogen bonds a molecule can form.

4. Consider Electronegativity
The electronegativity of F > O > N. Thus, if electronegativity was the only factor, the order of boiling points would be HF > > .

5. Consider the Number of Hydrogen Bonds in Water
Water ( ) can form up to four hydrogen bonds per molecule because it has two H atoms and two lone pairs of electrons on the oxygen.

6. Consider the Number of Hydrogen Bonds in Hydrogen Fluoride
Hydrogen Fluoride (HF) can form hydrogen bonds. However, due to steric factors, and the fact that there is only one H, the hydrogen bonding is less extensive than that in water. It forms chains or rings, but not a 3D network.

7. Consider the Number of Hydrogen Bonds in Ammonia
Ammonia ( ) can form hydrogen bonds, though weaker than those in HF and water, and the network is less extensive than in water due to having only three hydrogen atoms and one lone pair.

8. Compare Water and Hydrogen Fluoride
has a more extensive, three-dimensional hydrogen bonding network compared to HF. Although F is more electronegative than O, the extent of hydrogen bonding in water is higher. This higher extent of hydrogen bonding makes water have a higher boiling point than HF.

Final Answer:
(B) > HF >

1. Understand Acidity and Hybridization
The acidity of a C-H bond is related to the stability of the resulting carbanion after deprotonation (removal of the proton). The stability of the carbanion is, in turn, influenced by the hybridization of the carbon atom bearing the negative charge. A higher s-character in the hybrid orbital leads to greater electronegativity of the carbon, thus stabilizing the carbanion.

- sp hybridization: 50% s-character
- sp² hybridization: 33.3% s-character
- sp³ hybridization: 25% s-character

Therefore, the order of acidity based on hybridization is: sp > sp² > sp³.

2. Determine the hybridization of carbon atoms in the given compounds

- Ethyne (Acetylene, HC≡CH): The carbon atoms are sp hybridized. The hydrogen atoms are directly attached to sp hybridized carbons, making it acidic.

- Benzene (C₆H₆): The carbon atoms are sp² hybridized. Benzene is less acidic than ethyne because sp² hybridized carbon is less electronegative than sp hybridized carbon.

- n-Hexane (C₆H₁₄): The carbon atoms are sp³ hybridized. n-Hexane is the least acidic among the three because the sp³ hybridized carbon is the least electronegative.

3. Arrange in decreasing order of acidic behavior
Based on the hybridization, the decreasing order of acidity is:

Ethyne > Benzene > n-Hexane

4. Compare with the given options
Comparing the derived order with the options, we find that it matches option (D).

Final Answer:
(D) ethyne > Benzene > n-hexane

Reaction by which Benzaldehyde cannot be prepared

Benzaldehyde can be prepared by several reactions, such as:

• Partial oxidation of toluene
• Hydrolysis of benzal chloride ( )
• Gattermann–Koch reaction

However, Stephen’s reductiononly in the aliphatic series. It is not suitable for aromatic nitriles like benzonitrile because the reaction does not stop at the aldehyde stage—it tends to proceed further or fails due to ring stabilization.

Correct answer: Option (D) — the one involving Stephen’s reduction of benzonitrile.

A secondary amine is a compound in which two of the hydrogens of NH3 have been replaced by organic groups. In a secondary amine, two of the three hydrogen atoms of ammonia (NH3) are substituted by organic groups, resulting in a structure where two carbon atoms are directly bonded to the nitrogen atom, along with one remaining hydrogen. The general formula for a secondary amine is R2NH, where R represents an organic group or hydrogen. Secondary amines are characterized by the presence of an amino group (NH2) attached to a carbon atom, which is further attached to two other carbon atoms or organic groups.

The iodoform test is used to identify compounds containing a methyl ketone (CH3CO-) group or a methyl carbonyl group. In this test, the compound is treated with iodine (I2) in the presence of a base, typically sodium hydroxide (NaOH) or potassium hydroxide (KOH). If the compound contains a methyl ketone group, such as pentan-3-one, it will react with iodine to produce a yellow precipitate of iodoform (CHI3). On the other hand, if the compound does not contain a methyl ketone group, such as pentan-2-one, no yellow precipitate will form.

Therefore, the correct answer is (C) Iodoform test.

Correct answer: The compound that undergoes an SN¹ reaction is the one that can form the most stable carbocation intermediate.

The SN¹ reaction mechanism involves two key steps:

1. Formation of a carbocation intermediate by the departure of the leaving group.
2. Nucleophilic attack on the carbocation to form the final product.

The rate-determining step is the formation of the carbocation; therefore, the stability of this intermediate is crucial. The order of carbocation stability is:

Benzylic ≈ Allylic > Tertiary > Secondary > Primary > Methyl

In the given pairs of halogen compounds, the compound that can form the most stable carbocation will undergo the SN¹ reaction more readily. Specifically:

- Benzylic halides (where the halogen is attached to a carbon adjacent to a benzene ring) can form a **benzylic carbocation**, which is highly stabilized by resonance with the aromatic ring.

- Allylic halides (where the halogen is attached to a carbon adjacent to a double bond) can form an **allylic carbocation**, also stabilized by resonance.

- Tertiary alkyl halides form **tertiary carbocations**, stabilized by inductive and hyperconjugative effects from surrounding alkyl groups.

Therefore, when comparing pairs:

- A benzylic halide will undergo SN¹ more readily than an allylic or tertiary halide due to the exceptional stability of the benzylic carbocation.

- An allylic halide will undergo SN¹ more readily than a tertiary halide, but less readily than a benzylic halide.

- A tertiary alkyl halide will undergo SN¹ more readily than secondary or primary alkyl halides.

In summary, the compound in each pair that can form the most stable carbocation will be the one that undergoes the SN¹ reaction.

In the given reaction sequence, the steps can be broken down as follows:

The starting compound has an amine group (NH₂) attached to a benzene ring.

In the first step, the amine group undergoes diazotization, where it reacts with NaNO₂ and dilute HCl to form the diazonium salt (P).

In the second step, the diazonium salt reacts with BF₃, leading to the formation of Q.

In the third step, the diazonium compound (Q) is reacted with Cu and NaNO₂, resulting in the formation of R, which is a nitro group (NO₂) attached to the benzene ring.

Conclusion:

The correct sequence of reactions corresponds to Option C:

P is the diazonium salt formed with NaNO₂ and dilute HCl.

Q is the intermediate after reacting with BF₃.

R is the final product after reacting with Cu and NaNO₂, which is a nitrobenzene.

Thus, the answer is Option C.

Aromatic compounds follow the rules defined by Hückel's rule, which states that a compound is aromatic if it is cyclic, planar, and has a total of delocalized pi electrons, where is an integer.

- Option (1) is a benzene ring ( ), which satisfies the criteria for aromaticity, as it has 6 pi electrons, a multiple of .
- Option (2) is naphthalene, which is also aromatic as it has 10 pi electrons, satisfying Hückel's rule.
- Option (3) is another aromatic compound, specifically a 5-membered ring with conjugated double bonds.
- Option (4), however, is a non-aromatic compound. It does not satisfy the aromaticity condition because it lacks a fully conjugated system, making it non-aromatic.

Thus, the correct answer is:

- When glucose reacts with acetic anhydride, it results in an esterification reaction, where glucose forms an ester with the anhydride, producing a product that has a straight chain of six carbon atoms (prediction ii). Hence, a-i is incorrect.
- When glucose reacts with bromine water, it leads to oxidation of the aldehyde group in glucose. The resulting product has an aldehyde group, which corresponds to prediction i. Hence, b-i is correct.
- Hydroiodic acid reaction with glucose results in the replacement of the hydroxyl groups by iodine, giving the product with five hydroxyl groups, which matches prediction iii. Hence, c-ii is correct.
- Hydrogen cyanide reacts with glucose, resulting in the formation of a cyano group, which corresponds to prediction iv (a carbonyl group). Hence, d-iv is correct.

Thus, the correct match is: