All questions of Organic Named Reactions & Reagents for Chemistry Exam

OH group is stronger EDG than R(-CH3). No2 is EWG then it will give more electron density at OH group.

False Statement in Organic Chemistry

The false statement in organic chemistry is:

Cannizzaro reaction is given by aldehydes in the presence of alkali.

Explanation

Cannizzaro reaction is a redox reaction that occurs in aldehydes having no alpha-hydrogen atoms. In this reaction, one molecule of aldehyde is oxidized to a carboxylic acid, and the other molecule of the same aldehyde is reduced to an alcohol in the presence of a strong base. This reaction is not given by all aldehydes, only those without alpha-hydrogen atoms.

Aldol Condensation

Aldol condensation is a reaction between two carbonyl compounds, such as aldehydes or ketones, in the presence of a base or an acid catalyst. The reaction leads to the formation of a β-hydroxy carbonyl compound. Aldol condensation can be carried out using aldehydes or ketones or a combination of both.

Aldol Condensation with Aldehydes and Ketones

Aldol condensation can also occur between aldehydes and ketones in the presence of an acid catalyst. In this case, the reaction is called crossed aldol condensation. The acid catalyst promotes the formation of an enol intermediate, which then reacts with the carbonyl group of the other reactant to form a β-hydroxy carbonyl compound.

Conclusion

The false statement in organic chemistry is Cannizzaro reaction is given by aldehydes in the presence of alkali. The correct statement is that Cannizzaro reaction is only given by aldehydes without alpha-hydrogen atoms. Aldol condensation is a reaction between two carbonyl compounds, such as aldehydes or ketones, in the presence of a base or an acid catalyst. Aldol condensation can also occur between aldehydes and ketones in the presence of an acid catalyst, called crossed aldol condensation.

Reaction of Methylmagnesium bromide with Carbonyl Compounds

Carbonyl compounds contain a carbonyl group (C=O) and can undergo nucleophilic addition reactions with nucleophiles such as methylmagnesium bromide (CH3MgBr) to form alcohols. The reaction proceeds through the formation of an unstable intermediate called an alkoxy magnesium halide, which is then hydrolyzed to form the alcohol.

Reaction of Methylmagnesium bromide with 4-Heptanone

4-heptanone is a ketone with the formula C7H14O. When it is treated with methylmagnesium bromide, a Grignard reagent, it undergoes a nucleophilic addition reaction to form an alkoxy magnesium halide intermediate. The intermediate then undergoes hydrolysis to form the corresponding alcohol.

C7H14O + CH3MgBr → C7H13OMgBr + CH4
C7H13OMgBr + H2O → C7H16O + MgBrOH

The alcohol formed is a tertiary alcohol, 2,2,3,3-tetramethylbutan-2-ol.

Reaction of Methylmagnesium bromide with 3-Methylpentanal

3-methylpentanal is an aldehyde with the formula C6H12O. When it is treated with methylmagnesium bromide, it undergoes a nucleophilic addition reaction to form an alkoxy magnesium halide intermediate. The intermediate then undergoes hydrolysis to form the corresponding alcohol.

C6H12O + CH3MgBr → C6H11OMgBr + CH4
C6H11OMgBr + H2O → C6H14O + MgBrOH

The alcohol formed is a tertiary alcohol, 2,2-dimethyl-3-pentanol.

Reaction of Methylmagnesium bromide with Ethyl benzoate

Ethyl benzoate is an ester with the formula C9H10O2. When it is treated with methylmagnesium bromide, it undergoes a nucleophilic addition reaction to form an alkoxy magnesium halide intermediate. The intermediate then undergoes hydrolysis to form the corresponding alcohol.

C9H10O2 + CH3MgBr → C9H9OMgBr + C2H5Br
C9H9OMgBr + H2O → C9H12O2 + MgBrOH

The alcohol formed is a tertiary alcohol, 2-phenylethanol.

Reaction of Methylmagnesium bromide with 4,4-Dimethylcyclohexanone

4,4-dimethylcyclohexanone is a ketone with the formula C10H18O. When it is treated with methylmagnesium bromide, it undergoes a nucleophilic addition reaction to form an alkoxy magnesium halide intermediate. The intermediate then undergoes hydrolysis to form the corresponding alcohol.

C10H18O + CH3MgBr → C10H17OMgBr + CH4
C10H17OMgBr + H2O → C10H20O + MgBrOH

The alcohol formed is a tertiary alcohol, 4

Correct answer is LiAlH4

Cr2O7 2- strong oxidizing agent it can oxidized alkyl gr to carboxyl

The answer will be D not A

In the iodoform reaction, acetone (CH3COCH3) is treated with iodine and a base to form iodoform (CHI3) and an alkoxide ion. The reaction proceeds through a series of steps, starting with the formation of an enolate ion and followed by halogenation and deprotonation. Let's analyze each option to determine which compound is not formed in the iodoform reaction of acetone.

a) CH3COCH2I
In the iodoform reaction, the enolate ion of acetone is halogenated. In this case, the iodine atom replaces one of the hydrogen atoms, resulting in the formation of CH3COCH2I. This compound can be formed in the iodoform reaction, so option a) is not the correct answer.

b) CH3COCHI2
Similar to option a), in the iodoform reaction, the iodine atom can replace two hydrogen atoms on acetone, resulting in the formation of CH3COCHI2. This compound can also be formed in the iodoform reaction, so option b) is not the correct answer.

c) CH3COCl3
In the iodoform reaction, the iodine atom is required for the halogenation step. Chlorine atoms cannot replace hydrogen atoms in acetone during the reaction. Therefore, CH3COCl3 is not formed in the iodoform reaction. This compound is the correct answer.

d) ICH2COCH2I
This compound is not directly formed in the iodoform reaction of acetone. The iodoform reaction involves the substitution of iodine atoms for hydrogen atoms. In this case, there are no hydrogen atoms adjacent to the carbonyl group. Therefore, this compound is not formed in the iodoform reaction.

It gives iodoform test successfully becoz it has active methylene group

Conversion of 1-Methylcyclopentene into 2-methylcyclopentanol

Hydroboration-oxidation is a reaction in which an alkene reacts with borane (BH3) to form a trialkylborane. This trialkylborane then reacts with hydrogen peroxide (H2O2) and a base to form an alcohol. This reaction is used to convert alkenes into alcohols.

Steps involved in the conversion of 1-Methylcyclopentene into 2-methylcyclopentanol are as follows:

Step 1: Hydroboration of 1-Methylcyclopentene
1-Methylcyclopentene reacts with borane (BH3) to form trialkylborane.

Step 2: Oxidation of trialkylborane
Trialkylborane reacts with hydrogen peroxide (H2O2) and a base to form an alcohol.
2-methylcyclopentanol is obtained as a product of the reaction.

Thus, the correct answer is option 'B' that is hydroboration oxidation.

Reactivity towards Aqueous HBr

To determine which compound is most reactive towards aqueous HBr, we need to consider the stability of the carbocation intermediates that are formed during the reaction. The more stable the carbocation, the more reactive the compound will be towards HBr.

Stability of Carbocation Intermediates

To determine the stability of the carbocation intermediates, we need to consider the following factors:

- Resonance: Compounds that can form resonance structures with the positive charge are more stable.
- Inductive effects: Compounds with electron-donating groups near the positive charge are more stable.
- Steric hindrance: Compounds with bulky groups near the positive charge are less stable.

Based on these factors, we can rank the stability of the carbocation intermediates in the following order:

3-Phenyl-1-propanol > 2-Phenyl-1-propanol > 1-Phenyl-1-propanol > 1-Phenyl-2-propanol

Therefore, 1-Phenyl-1-propanol is the most reactive towards aqueous HBr as it forms the least stable carbocation intermediate.

Conclusion

In conclusion, the stability of the carbocation intermediates determines the reactivity of compounds towards aqueous HBr. 1-Phenyl-1-propanol is the most reactive towards HBr as it forms the least stable carbocation intermediate.

Primary alcohols are formed when a Grignard reagent reacts with an epoxide. In this case, the compound that gives a primary alcohol upon reaction with phenylmagnesium bromide is ethylene oxide.

- Epoxides are cyclic ethers with a three-membered ring containing an oxygen atom.
- Phenylmagnesium bromide (C6H5MgBr) is a Grignard reagent that contains a carbon-magnesium bond.
- Grignard reagents are highly reactive and can undergo nucleophilic addition reactions with a variety of electrophilic compounds.
- When phenylmagnesium bromide reacts with ethylene oxide, the oxygen atom of the epoxide is nucleophilically attacked by the carbon atom of the Grignard reagent.
- This results in the opening of the epoxide ring and the formation of a new carbon-oxygen bond.
- The product of this reaction is a primary alcohol, which has a hydroxyl group (-OH) directly attached to a carbon atom that is bonded to only one other carbon atom.
- In the case of ethylene oxide, the reaction with phenylmagnesium bromide would result in the formation of 1-phenylethanol.
- The hydroxyl group (-OH) is directly attached to the carbon atom that was previously part of the epoxide ring, which makes it a primary alcohol.
- The other compounds listed, 2-methyloxirane, ethyl formate, and carbon dioxide, do not have a carbon atom that can react with the Grignard reagent to form a primary alcohol.
- Therefore, only ethylene oxide gives a primary alcohol upon reaction with phenylmagnesium bromide.

The incoming h- attack from backside

Carbonyl compounds are organic compounds that contain a carbonyl functional group (-C=O). When reacted with methylmagnesium iodide in ether followed by a workup with aqueous ammonium chloride, they form alcohols.

Explanation:

The reaction of carbonyl compounds with methylmagnesium iodide in ether is a nucleophilic addition reaction. The nucleophile attacks the electrophilic carbon of the carbonyl group, forming an alkoxide intermediate. This intermediate then reacts with a proton from the solvent (ether), forming an alcohol.

The reaction can be represented as follows:

Carbonyl compound + MeMgI in ether → alkoxide intermediate → alcohol

In this specific case, the carbonyl compounds given are:

a) pentan-2-one
b) 3-methylbutan-2-one
c) ethyl butanoate
d) ethyl 2-methylbutanoate

Out of these, only option d) ethyl 2-methylbutanoate will not give 2-methylpentan-2-ol upon reaction with methylmagnesium iodide in ether followed by a workup with aqueous ammonium chloride.

The reason for this is that ethyl 2-methylbutanoate has a bulky ethyl group attached to the carbonyl carbon, which makes it less reactive towards nucleophilic attack. This results in a slower reaction rate and a lower yield of the desired alcohol product.

In contrast, the other three carbonyl compounds have simpler alkyl groups attached to the carbonyl carbon, which makes them more reactive towards nucleophilic attack. This results in a faster reaction rate and a higher yield of the desired alcohol product (2-methylpentan-2-ol).

First hydroxide ion will abstract alpha hydrogen from the given molecule of aldehyde. In second step anion produced in first step will attack on carbonyl carbon of second molecule of given aldehyde and oxygen of it get negative charge. Third step negatively charged oxygen will attack on hydrogen ion and then water will be removed.

D

D is the correct answer.aldehyde is more reactive than ketone

Answer:

When phenylmagnesium bromide (C6H5MgBr) reacts with a reagent followed by acid workup, it results in the formation of an alcohol. In this case, the reagent that will yield 2-phenylethanol is Oxirane (also known as ethylene oxide or epoxyethane).

Explanation:

When phenylmagnesium bromide reacts with oxirane, the phenyl group (C6H5) from the Grignard reagent adds to one of the carbon atoms of the oxirane ring. This leads to the formation of a cyclic intermediate known as a magnesium alkoxide.

The magnesium alkoxide intermediate then undergoes acid workup, which involves the addition of an acid such as dilute hydrochloric acid (HCl). This step protonates the alkoxide group, leading to the formation of an alcohol.

In the case of oxirane, the addition of phenylmagnesium bromide followed by acid workup results in the formation of 2-phenylethanol. The phenyl group adds to one of the carbon atoms of the oxirane ring, and the resulting product is a secondary alcohol with the phenyl group attached to the second carbon.

Summary:

- When phenylmagnesium bromide reacts with oxirane, it forms a cyclic intermediate called a magnesium alkoxide.
- Acid workup of the magnesium alkoxide leads to the formation of an alcohol.
- In this case, the reaction of phenylmagnesium bromide with oxirane yields 2-phenylethanol.

B)Na/NH3
c)LiAlH4
d)H2/Pd-C

Hydrates of Aldehyde and Ketone:

1. Thermodynamic Stability:
- Usually, hydrates have lower thermodynamic stability than the anhydrous form.
- The presence of water in the structure of aldehydes and ketones affects their stability and reactivity.
- The formation of hydrates is an equilibrium process, and the stability of the hydrate depends on the balance between the hydration energy and the energy required to break the water-aldehyde or water-ketone interactions.
- In most cases, the anhydrous form is more stable than the corresponding hydrate.

2. Hydrate Content of Acetone:
- The hydrate content of acetone is greater in water than in hexane.
- Acetone (CH3COCH3) is a ketone that readily forms hydrates in the presence of water.
- In an aqueous environment, water molecules can interact with the carbonyl oxygen of acetone through hydrogen bonding, leading to the formation of hydrates.
- Hexane, on the other hand, is a non-polar solvent and does not readily form hydrates with acetone.

3. Isotopic Labeling:
- When CH3CHO (acetaldehyde) is treated with H2O18 (water containing the stable isotope oxygen-18) in acidic medium, it gets converted into CH3CHO18.
- This reaction is an example of isotopic labeling, where a specific atom or group of atoms is replaced with its isotopic counterpart.
- The acidic medium helps in the protonation of the carbonyl oxygen, making it more susceptible to nucleophilic attack by water molecules containing oxygen-18.
- The resulting product, CH3CHO18, contains the oxygen-18 isotope, which can be used for various studies and analyses.

4. Hydrate Content of C6H5CHO and p-nitrobenzaldehyde:
- C6H5CHO (benzaldehyde) and p-nitrobenzaldehyde both can form hydrates, but C6H5CHO has a greater hydrate content than p-nitrobenzaldehyde.
- The hydrate content depends on the ability of the aldehyde or ketone to form hydrogen bonds with water molecules.
- Benzaldehyde (C6H5CHO) has a phenyl group attached to the carbonyl carbon, which enhances its ability to form hydrogen bonds with water, leading to a higher hydrate content.
- In contrast, p-nitrobenzaldehyde has a nitro group (-NO2) attached to the carbonyl carbon, which reduces its ability to form hydrogen bonds with water, resulting in a lower hydrate content compared to benzaldehyde.

Therefore, the correct statements regarding hydrates of aldehyde and ketone are A) Usually hydrates have lower thermodynamic stability than the anhydrous form, B) Hydrate content of acetone is greater in water than in hexane, and C) CH3CHO when treated with H2O18 in acidic medium gets converted into CH3CHO18.

• This would likely deprotonate the electrophilic carbonyl component, making it a poor electrophile and favoring side reactions. This is not favorable for a successful mixed aldol reaction.
  (Incorrect)

• This could result in deprotonation on both components, leading to a mixture of products and possibly side reactions.
  (Incorrect)

• Removing the aldol product shifts the equilibrium towards product formation, which is a beneficial condition for successful mixed aldol condensation.
  (Correct)

• This ensures that the pre-nucleophilic component forms an enolate, which can then attack the electrophilic carbonyl component without side reactions, favoring a successful aldol condensation.
  (Correct)

Understanding Aldol Condensation
Aldol condensation is a significant reaction in organic chemistry that allows for the formation of carbon-carbon bonds. When acetaldehyde is involved, the process can be broken down into key steps.
Formation of Enolate Ion
- The reaction begins with the formation of an enolate ion from acetaldehyde, which has at least one α-hydrogen.
- This enolate ion acts as a nucleophile, attacking another molecule of acetaldehyde.
Initial Product Formation
- The nucleophilic attack leads to the formation of a β-hydroxy aldehyde.
- In the case of acetaldehyde, the product formed is specifically 3-hydroxybutanal.
- This product has a hydroxyl group (-OH) and a carbonyl group (C=O) in its structure.
Dehydration Step
- The β-hydroxy aldehyde (3-hydroxybutanal) can undergo dehydration, resulting in the formation of an α,β-unsaturated carbonyl compound, which is crotonaldehyde.
- However, the question specifically asks for the product formed before this dehydration occurs.
Conclusion
- Therefore, the correct answer to the question is option 'C' - 3-hydroxybutanal, which is the initial product prior to dehydration in the aldol condensation of acetaldehyde.

Isomers of C4H8O when reacts with CH3MgBr followed by acidification to alcohol (only consider carbonyl isomers and Including stereoisomers)

There are different isomers possible for C4H8O when reacts with CH3MgBr followed by acidification to alcohol. Let's consider them one by one:

1. But-2-enal

- But-2-enal has a double bond between C2 and C3 and an aldehyde functional group (CHO) at C2.
- When reacted with CH3MgBr, it forms a Grignard reagent, which can then be protonated to form an alcohol.
- The resulting alcohol is (Z)-but-2-en-1-ol, which has a stereoisomer (E)-but-2-en-1-ol.
- Therefore, the total number of isomers for but-2-enal is 2.

2. But-2-yn-1-ol

- But-2-yn-1-ol has a triple bond between C2 and C3 and an alcohol functional group (OH) at C1.
- When reacted with CH3MgBr, it forms a Grignard reagent, which can then be protonated to form an alcohol.
- The resulting alcohol is (E)-but-2-en-1-ol, which has a stereoisomer (Z)-but-2-en-1-ol.
- Therefore, the total number of isomers for but-2-yn-1-ol is 2.

3. 2-Methylpropanal

- 2-Methylpropanal has an aldehyde functional group (CHO) at C2 and a methyl group (CH3) at C3.
- When reacted with CH3MgBr, it forms a Grignard reagent, which can then be protonated to form an alcohol.
- The resulting alcohol is (S)-2-methylpropan-1-ol, which has a stereoisomer (R)-2-methylpropan-1-ol.
- Therefore, the total number of isomers for 2-methylpropanal is 2.

4. 3-Methylbutanal

- 3-Methylbutanal has an aldehyde functional group (CHO) at C3 and a methyl group (CH3) at C2.
- When reacted with CH3MgBr, it forms a Grignard reagent, which can then be protonated to form an alcohol.
- The resulting alcohol is (R)-3-methylbutan-1-ol, which has a stereoisomer (S)-3-methylbutan-1-ol.
- Therefore, the total number of isomers for 3-methylbutanal is 2.

Therefore, the total number of isomers for C4H8O when reacts with CH3MgBr followed by acidification to alcohol is 2+2+2+2 = 8. However, the question only considers carbonyl isomers, which are but-2-enal and 2-methylpropanal. Therefore, the correct answer is option A, which is 2.

C) Wurtz reaction
d) Grignard reaction

Answer:

The correct answer is option 'B' - 2,4,6-Trinitrophenol can be prepared in good yield by the nitration of 2,4-dinitrophenol. Let's explain this answer in detail.

Nitration is a chemical reaction that involves the introduction of one or more nitro groups (-NO2) into a molecule. In this case, we are looking at the nitration of phenol compounds to produce 2,4,6-Trinitrophenol, also known as picric acid.

1. Nitration of 2,4-dinitrochlorobenzene:
2,4-dinitrochlorobenzene is a compound that contains both a nitro group (-NO2) and a chlorine group (-Cl) on a benzene ring. While it is possible to nitrate 2,4-dinitrochlorobenzene, the yield of 2,4,6-Trinitrophenol from this reaction is not as good as the alternative method.

2. Nitration of 2,4-dinitrophenol:
2,4-dinitrophenol is a compound that contains two nitro groups (-NO2) on a phenol ring. When 2,4-dinitrophenol is nitrated, one of the nitro groups is replaced by a nitric acid group (-NO2) to form 2,4,6-Trinitrophenol. This reaction occurs under acidic conditions and can be catalyzed by sulfuric acid.

The nitration of 2,4-dinitrophenol is the preferred method for the synthesis of 2,4,6-Trinitrophenol because it provides a higher yield of the desired compound. The presence of the hydroxyl group (-OH) on the phenol ring in 2,4-dinitrophenol makes it more reactive towards nitration compared to 2,4-dinitrochlorobenzene.

In addition, the nitration of 2,4-dinitrophenol can be controlled to produce the desired product with good selectivity. By carefully controlling the reaction conditions, such as the concentration of the reactants, temperature, and reaction time, a high yield of 2,4,6-Trinitrophenol can be obtained.

Overall, the nitration of 2,4-dinitrophenol is the most efficient and reliable method for the synthesis of 2,4,6-Trinitrophenol. It provides a good yield of the desired compound and allows for control over the reaction conditions to optimize the process.

Br has +m effect due to which ortho and para position available for No2

To determine which compound gives a secondary alcohol upon reaction with methylmagnesium bromide (CH3MgBr), we need to understand the mechanism of the Grignard reaction and how it leads to the formation of alcohols.

The Grignard reaction involves the nucleophilic addition of a Grignard reagent (an organomagnesium compound) to a carbonyl group, resulting in the formation of an alcohol. In this case, methylmagnesium bromide (CH3MgBr) will act as the nucleophile and attack the carbonyl group of the given compound.

Let's analyze each compound and see which one is most likely to give a secondary alcohol:

a) Butyl formate: This compound contains a formate ester group (R-COO-R'), where R is a butyl group. When the Grignard reagent reacts with the carbonyl group in butyl formate, it will form a tertiary alcohol. However, we are looking for a secondary alcohol, so this compound is not the answer.

b) 3-pentanone: This compound is a ketone, and ketones react with Grignard reagents to form secondary alcohols. Therefore, it is a potential candidate.

c) Pentanal: This compound is an aldehyde, and aldehydes also react with Grignard reagents to form secondary alcohols. Therefore, it is another potential candidate.

d) Methyl butanoate: This compound is an ester, and esters react with Grignard reagents to form tertiary alcohols. Therefore, this compound does not give a secondary alcohol.

Based on the analysis above, the correct answer is option 'c' (Pentanal). When pentanal reacts with methylmagnesium bromide, it will undergo nucleophilic addition, resulting in the formation of a secondary alcohol.

Overall, understanding the reactivity of different functional groups towards Grignard reagents allows us to predict the products of their reactions and determine which compound will give a secondary alcohol in this case.