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All questions of Organic Named Reactions & Reagents for Chemistry Exam

 The major product formed in the following reaction is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Saurabh Chavan answered
OH group is stronger EDG than R(-CH3). No2 is EWG then it will give more electron density at OH group.
Clear all your doubts with EduRev

A carboxylic acid can best be converted into acid chloride by using:
  • a)
    PCl5
  • b)
    SOCl2
  • c)
    HCl
  • d)
    ClCOCOCl 
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered
Use of SOCl2 and CICOCOCI forms gaseous by products which can be easily removed, giving better yield of RCOCI. Further, oxalyl chIoride is particularly easy to use becasue any excess of it can be easily evaporated due to its low b.p. (62∘C)

Choose the reaction(s) that will not proceed as shown hereunder:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Pie Academy answered
Reduction of amide by LAH gives amine but hydrolysis of amine not give alcohol.
Hence B is the answer.

Which of the following statement is false:
  • a)
    Cannizzaro reaction is given by aldehydes in presence of alkali
  • b)
    Aldol condensation is given by aldehydes in presence of alkali
  • c)
    Aldol condensation is given by aldehydes and ketones in presence of acids
  • d)
    None of these.
Correct answer is option 'D'. Can you explain this answer?

Shivani Mehta answered
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.

Which of the following compounds does not give a tertiary alcohol upon reaction with methylmagnesium bromide/?
  • a)
    4-heptanone
  • b)
    Ethyl benzoate
  • c)
    4,4-dimethylcyclohexanone
  • d)
    3-methylpentanal
Correct answer is option 'D'. Can you explain this answer?

Ameya Reddy answered
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

Which of the following compounds is not formed in iodoform reaction of acetone?
  • a)
    CH3COCH2I
  • b)
    CH3COCHI2
  • c)
    CH3COCl3
  • d)
    ICH2COCH2I
Correct answer is option 'D'. Can you explain this answer?

Saranya Mehta answered
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.

1-Methylcyclopentene can be converted into 2-methylcyclopentanol by:
  • a)
    Acid-catalyzed hydration
  • b)
    Hydroboration oxidation
  • c)
    Epoxide formation followed by reduction with LiAlH4
  • d)
    Oxymercuration-demercuration
Correct answer is option 'B'. Can you explain this answer?

Rishabh Mehta answered
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.

A Grignard’s reagent may be made by reacting magnesium with which of the following compound?
  • a)
    Methyl amine
  • b)
    Diethyl ether
  • c)
    Ethyl iodide
  • d)
    Ethyl alcohol
Correct answer is option 'C'. Can you explain this answer?

Asf Institute answered
A Grignard’s reagent can be formed by reacting magnesium with Ethyl iodide, as shown below;
C2H5I + Mg + Dryether → C2H5 − Mg − I (Ethylmagnesium iodide)

 Which of the following structure is more stable:
 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Pooja Choudhury answered
The negative charge is delocalized and NO2 shows -I and -M effect and hence stabilizes -ve charge.

Which of the following is most reactive towards aqueous HBr:
a. 1- Phenyl-2-propanol
b. 1-Phenyl-1-propanol
c. 3-Phenyl-1-propanol
d. 2-Phenyl-1-propanol
Correct answer is option 'B'. Can you explain this answer?

Siddharth Majumdar answered
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.

Which of the following compounds gives a primary alcohol upon reaction with phenylmagnesium bromide?
  • a)
    2-methyloxirane
  • b)
    ethylene oxide
  • c)
    ethyl formate
  • d)
    carbon dioxide
Correct answer is option 'B'. Can you explain this answer?

Ishaan Sengupta answered
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.

Which of the following carbonyl compounds does not give 2-methylpentan-2-ol upon reaction with methylmagnesium iodide in ether followed by a workup with aqueous ammonium chloride?
  • a)
    pentan-2-one
  • b)
    b methyl propyl ketone
  • c)
    ethyl butanoate
  • d)
    ethyl 2-methylbutanoate
Correct answer is option 'D'. Can you explain this answer?

Pranavi Mishra answered
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).

Which of the following gives effervescences of CO2 with NaHCO3 solution:
  • a)
    HCOOH
  • b)
    2,4,6-trinitrophenol
  • c)
    Both (a) and (b)
  • d)
    None of these.
Correct answer is option 'C'. Can you explain this answer?

Amar Chawla answered
**Answer:**

The correct answer is option 'C', which means both HCOOH and 2,4,6-trinitrophenol give effervescences of CO2 with NaHCO3 solution.

**Explanation:**

When NaHCO3 (sodium bicarbonate) is added to a solution containing a compound that can release CO2, effervescence occurs. Let's analyze each option to see if they can release CO2 when reacted with NaHCO3.

a) HCOOH (formic acid):
Formic acid, HCOOH, is a weak acid. When it reacts with NaHCO3, it undergoes a neutralization reaction, resulting in the formation of sodium formate (NaHCOO) and the release of carbon dioxide gas (CO2).

HCOOH + NaHCO3 → NaHCOO + CO2 + H2O

b) 2,4,6-trinitrophenol (picric acid):
2,4,6-trinitrophenol, also known as picric acid, is a strong acid. Similar to formic acid, it can undergo a neutralization reaction with NaHCO3, resulting in the formation of sodium picrate (NaC6H2(NO2)3O7) and the release of carbon dioxide gas (CO2).

C6H2(NO2)3OH + NaHCO3 → NaC6H2(NO2)3O7 + CO2 + H2O

Therefore, both HCOOH and 2,4,6-trinitrophenol can release CO2 when reacted with NaHCO3 solution. Hence, option 'C' is the correct answer.

Which is not present in Grignard reagent?
  • a)
    Methyl group
  • b)
    Magnesium
  • c)
    Halogen
  • d)
    −COOH group
Correct answer is option 'D'. Can you explain this answer?

Vivek Khatri answered
Grignard reagents are made from halogenoalkanes (haloalkanes or alkyl halides) and introduces some of their reactions. A Grignard reagent has a formula RMgX where X is a halogen, and R is an alkyl or aryl (based on a benzene ring) group.

Comprehension Type
Direction (Q. Nos. 20-22) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Passage
Aldol condensation is an important reaction in organic chemistry, particularly for the formation of carbon-carbon bonds. This reaction typically involves the enolate ion derived from an aldehyde or ketone reacting with another carbonyl compound to form a β-hydroxy aldehyde or β-hydroxy ketone. This product can undergo dehydration to form an α,β-unsaturated carbonyl compound. The reaction is usually catalyzed by a base, although acid-catalyzed versions are also known. The presence of at least one α-hydrogen in the reactant is crucial for the aldol condensation to occur. Acetaldehyde, for example, can undergo aldol condensation to yield 3-hydroxybutanal, which can then dehydrate to form crotonaldehyde.
Ques: In the aldol condensation reaction of acetaldehyde (CH₃CHO), the initial product formed before dehydration is:
  • a)
    2-Hydroxybutanal
  • b)
    Crotonaldehyde
  • c)
    3-Hydroxybutanal
  • d)
    Acetone
Correct answer is option 'C'. Can you explain this answer?

Amar Choudhary answered
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.

Which can be deduced correctly regarding keto-enol tautomerism in general?
  • a)
    Increasing temperature increases the enol content at equilibrium
  • b)
    Mono-enols are usually more stable than dienols
  • c)
    Enols of ketones are generally more stable than enols of aliphatic aldehydes
  • d)
    Keto-enol taytomerism is catalysed by both acidic and basic catalys
Correct answer is option 'A,B,C,D'. Can you explain this answer?

Ishita Deshpande answered
Increasing temperature increases equilibrium content of less stable enol tautomers.
Enolisation decreases stability, hence introducing two or more enol groups are further more difficult.
Enols of ketones are more substituted at double bond, hence more stable.
Both acid and base catalyses keto-enol tautomerism.

Nitrobenzene can be reduced to aniline by:
(I) H2/Ni
(II) Sn/HCl
(III) Zn/NaOH
(IV) LiAlH4
  • a)
    I, II and III
  • b)
    I and II
  • c)
     I, II and IV
  • d)
    Only IV
Correct answer is option 'B'. Can you explain this answer?

Niharika Kulkarni answered
Correct Answer :- B
Explanation : When nitrobenzene is reduced with a metal and acid (Sn/HCl, Zn/HCl etc.), aniline is obtained. The intermediate products nitrosobenzene and phenylhydroxylamine are never isolated because these are reduced more rapidly than nitrobenzene.
 
However, the catalytic hydrogenation of nitrobenzene gives aniline.
 PhNO2  (H2+Ni)------------->PhNH2
​           
Azobenzene is formed when nitrobenzene is reduced by LiAlH4.

 Fehling’s solution can be used for distinguishing between:
  • a)
    CH3CHO and C6H5CHO
  • b)
    CH3CHO and CH3COCH2OH
  • c)
    Both (a) and (b)
  • d)
    None of these.
Correct answer is option 'A'. Can you explain this answer?

Anisha Banerjee answered
Fehling refers to a chemical reagent known as Fehling's solution, which is used to test for the presence of reducing sugars in a solution. It consists of two separate solutions, Fehling's A and Fehling's B, which are mixed together before use. Fehling's A contains copper(II) sulfate, while Fehling's B contains sodium potassium tartrate and sodium hydroxide.

When heated with a reducing sugar, such as glucose or fructose, in the presence of Fehling's solution, a redox reaction occurs. The reducing sugar is oxidized, while the copper(II) ions in Fehling's A are reduced to copper(I) oxide, which forms a brick-red precipitate. This precipitate indicates the presence of a reducing sugar in the solution.

Fehling's solution is commonly used in laboratory settings to identify and quantify reducing sugars in various samples, such as food products or biological fluids. It is particularly useful for detecting sugars like glucose, fructose, and lactose.

Which is the mildest reducing agent which reduces only carbonyl group in presence of nitro, carboxyl, double bond and ester groups?
  • a)
    LiAIH4
  • b)
    Na-NH3
  • c)
    NaBH4
  • d)
    H2-Ni
Correct answer is option 'C'. Can you explain this answer?

Vivek Khatri answered
NaBH4 is mild reducing agent for this conversion. It is very effective for the reduction of aldehydes and ketones to alcohols. By itself, it will generally not reduce esters, carboxylic acids, or amides (although it will reduce acyl chlorides to alcohols).

The ethereal linkage (—C—O—C—) is cleaved by: 
  • a)
    HBr
  • b)
    HNO3
  • c)
    Both (a) and (b)
  • d)
    None 
Correct answer is option 'A'. Can you explain this answer?

Akash Kulkarni answered
Also known as the ethereal connection or ethereal bond) is a metaphysical concept that describes a connection between two or more individuals or entities on a spiritual or energetic level.

According to this concept, the ethereal linkage is not dependent on physical proximity or communication, but rather on the alignment of energy, consciousness, or souls. It is believed to transcend time and space, allowing for a profound connection between individuals even when they are physically separated.

The ethereal linkage is often associated with deep emotional, mental, or spiritual connections. It is believed to enable individuals to understand and connect with each other on a profound level, often leading to a sense of unity, empathy, and shared experiences.

This concept is prevalent in various spiritual and esoteric traditions, such as certain forms of mysticism, New Age philosophies, and some belief systems focused on energy work or psychic abilities. It is often used to explain the bond between soulmates, twin flames, or spiritual allies.

The ethereal linkage is considered to be a powerful force that can facilitate growth, healing, and transformation. It is believed that individuals connected through this ethereal bond can support each other's spiritual journeys, provide guidance, and contribute to each other's personal and collective evolution.

However, it is important to note that the ethereal linkage is a concept that is not scientifically proven and may be interpreted differently by different individuals and belief systems. It is a subjective and personal experience that relies on intuition, spirituality, and metaphysical understandings.

What is the correct order of equilibrium enol content of the following compounds?
I. CH3COCH3
II. CH3COCH2COOC2H5
III. CH3COCH2COCH3
IV. CH3COCH2COH
  • a)
    I > II > III > IV
  • b)
    I > IV > III > II
  • c)
    IV > II > III > I
  • d)
    III > IV > II > I
Correct answer is option 'D'. Can you explain this answer?

Asha Nair answered
A 1,3-diketo compound forms more stable enol than a monocarbonyls. Also ester group forms less stable enol than carbonyls. Hence, III, a 1 , 3-diketo ne form s highest enol content while I (monocarbonyl) forms least enol content at equilibrium.

Which of the following reagents, when treated with phenylmagnesiuim bromide followed by acid workup, will yield 2-phenylethanol?
  • a)
    Ethanol
  • b)
    Diethyl ether
  • c)
    Ethanal
  • d)
    Oxirane
Correct answer is option 'D'. Can you explain this answer?

Nandini Das answered
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.

Which of the following compounds would not give tert-butyl alcohol when treated with excess methylmagnesium bromide?
  • a)
    acetyl chloride
  • b)
    acetaldehyde
  • c)
    methyl acetate
  • d)
    acetic anhydride
Correct answer is option 'B'. Can you explain this answer?

Athul Menon answered
The compound that would not give tert-butyl alcohol when treated with excess methylmagnesium bromide is acetaldehyde (option B).

Explanation:
When an alkyl halide, such as methylmagnesium bromide (CH3MgBr), reacts with a carbonyl compound, a nucleophilic addition reaction takes place. This reaction is known as a Grignard reaction. The alkyl magnesium halide acts as a nucleophile and adds to the carbonyl carbon, forming an alcohol.

In the case of acetaldehyde, the nucleophilic addition of methylmagnesium bromide would result in the formation of a secondary alcohol called tert-butanol. The reaction can be represented as follows:

CH3MgBr + CH3CHO → CH3CH(OH)CH3

However, in this question, we are specifically looking for the compound that would not give tert-butyl alcohol. Let's analyze the other options:

a) Acetyl chloride:
Acetyl chloride (CH3COCl) is an acyl chloride. When it reacts with methylmagnesium bromide, it forms a ketone called ethyl methyl ketone. The reaction can be represented as follows:

CH3MgBr + CH3COCl → CH3COCH2CH3

This reaction does not result in tert-butyl alcohol.

c) Methyl acetate:
Methyl acetate (CH3COOCH3) is an ester. When it reacts with methylmagnesium bromide, it undergoes a nucleophilic acyl substitution reaction, resulting in the formation of a tertiary alcohol called tert-butyl alcohol. The reaction can be represented as follows:

CH3MgBr + CH3COOCH3 → (CH3)3COH + MgBrOCH3

This reaction results in the formation of tert-butyl alcohol.

d) Acetic anhydride:
Acetic anhydride (CH3CO)2O is an anhydride. When it reacts with methylmagnesium bromide, it forms a ketone called methyl isobutyl ketone. The reaction can be represented as follows:

CH3MgBr + (CH3CO)2O → CH3COCH(CH3)2

This reaction does not result in tert-butyl alcohol.

Therefore, the correct answer is option B, acetaldehyde, as it does not give tert-butyl alcohol when treated with excess methylmagnesium bromide.

Which of the following compounds gives a secondary alcohol upon reaction with methylmagnesium bromide?
  • a)
    Butyl formate
  • b)
    3- pentanone
  • c)
    Pentanal
  • d)
    Methyl butanoate
Correct answer is option 'C'. Can you explain this answer?

Vivek Khatri answered
Pentanal gives a secondary alcohol upon reaction with methylmagnesium bromide, as the Grignard Reaction is the addition of an organomagnesium halide (Grignard reagent) to a ketone or aldehyde, to form a tertiary or secondary alcohol, respectively.

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