Aldehydes Ketones And Carboxylic Acids

Step 1: Reaction analysis. In this reaction, the organic product involves an alcohol group with a carboxyl group.
Step 2: Conclusion. Thus, the correct product is H C–C OH + COOH.

Step 1: Reaction of Calcium acetate. When calcium acetate is heated, it decomposes to form acetone and calcium carbonate.
Step 2: Conclusion. Thus, calcium acetate on heating gives acetone and calcium carbonate.

Step 1: Reactivity of Aldehydes and Ketones. Aldehydes are reduced to primary alcohols, while ketones are reduced to secondary alcohols. Ketones do not react with Fehling’s solution.
Step 2: Conclusion. Thus, ketones do not react with alcohols.

Step 1: Aldol condensation.
Aldol condensation typically involves the reaction of aldehydes or ketones with at least one alpha-hydrogen. Acetaldehyde, being a simple aldehyde with alpha-hydrogens, undergoes aldol condensation.

Step 2: Conclusion.
Thus, the correct answer is option (B).

Final Answer:

Step 1: Understanding the reaction.
This is a reaction of an ester with a Grignard reagent, followed by hydrolysis. The product is a beta-hydroxy acid.

Step 2: Conclusion.
Thus, the correct product is option (B).

Final Answer:

Step 1: Recall Stephen reduction.
Stephen reduction is the partial reduction of nitriles ( ) using .
Step 2: Intermediate formation.
Nitrile is reduced to iminium salt:

Step 3: Hydrolysis gives aldehyde.
On hydrolysis:

Step 4: Conclusion.
Hence alkyl cyanides give aldehydes in Stephen reduction.
Final Answer:

Step 1: Recall Wittig reaction.
Wittig reaction involves reaction of an aldehyde/ketone with a phosphonium ylide to form an alkene (olefin).
Step 2: General reaction.

Step 3: Product type.
The carbonyl oxygen is removed and replaced by a C=C bond, so an olefin (alkene) is formed.
Final Answer:

Step 1: Condition for Cannizzaro reaction.
Cannizzaro reaction is shown by aldehydes that do not have alpha hydrogen.
Step 2: Check each option.
(A) HCHO (formaldehyde) has no alpha hydrogen because it has no carbon chain.
So it undergoes Cannizzaro reaction.
(C) CH CHO has alpha hydrogens, so it undergoes aldol reaction instead.
Others are not aldehydes.
Step 3: Conclusion.
Hence HCHO gives Cannizzaro reaction.
Final Answer:

Step 1: Identify the reaction.
Crotonaldehyde is formed by aldol condensation of acetaldehyde.

Step 2: Explain mechanism.
Aldol condensation involves:
- nucleophilic addition of enolate ion to carbonyl carbon
- followed by elimination of water to form -unsaturated aldehyde
Step 3: Conclusion.
Thus it is nucleophilic addition followed by elimination.
Final Answer:

Step 1: Recall Gattermann aldehyde synthesis.
Gattermann reaction converts aromatic compounds to aromatic aldehydes using:
followed by hydrolysis.
Step 2: Identify correct option.
Option (C) matches the exact reagents: benzene + HCN + HCl with and then hydrolysis to benzaldehyde.
Step 3: Why other options are not Gattermann.
(A) is Rosenmund reduction.
(B) is Gattermann-Koch reaction (CO + HCl).
(D) is Etard reaction.
Final Answer:

Step 1: Meaning of aldol.
Aldol is the product of aldol condensation of acetaldehyde.
Step 2: Aldol from acetaldehyde.
Two molecules of acetaldehyde combine to form:
This is 3-hydroxybutanal.
Step 3: Classification.
In this compound, is on -carbon with respect to aldehyde group.
Hence it is -hydroxybutyraldehyde.
Final Answer:

Step 1: Identify reaction type.
Propanone (acetone) in presence of dilute base undergoes aldol addition (self-condensation).
Step 2: Product of aldol addition of acetone.
Two acetone molecules combine to form diacetone alcohol:
Step 3: IUPAC name.
Diacetone alcohol is:
Final Answer:

Step 1: Reaction involved.
Ketone reacts with HCN to form cyanohydrin:
Step 2: Identify attacking species.
In this reaction, attacks the carbonyl carbon.
Carbonyl carbon is electrophilic due to polar bond.
Step 3: Type of reaction.
Since a nucleophile adds to carbonyl carbon, this is nucleophilic addition.
Final Answer:

In the presence of , propanone (acetone) undergoes aldol type condensation as

Step 1: Understand the reaction.
The reaction shown corresponds to the Lederer-Manasse reaction, which involves the halogenation of aromatic compounds.
Step 2: Conclusion.
Thus, the given reaction is identified as the Lederer-Manasse reaction.
Final Answer:

The strength of the C—O bond in carbonyl complexes is influenced by the metal's ability to back-donate electrons. The larger the metal's atomic number, the weaker the C—O bond typically is due to stronger back-donation. Thus, will have the strongest C—O bond.

Step 2: Conclusion.
The correct answer is , corresponding to option (a).

Iodoform reaction occurs with compounds that have a methyl group adjacent to a carbonyl group. The compounds in (i), (ii), and (iv) will give iodoform on reaction with iodine and NaOH.

Step 2: Conclusion.
The compounds that will give iodoform are (i), (ii), and (iv), corresponding to option (a).

Dehydration of alcohols involve formation of carbocation intermediate. More the stability of carbocation, higher is the ease of dehydration. The order of stability of carbocation, given by the given compounds, is. Thus, compound given in option readily undergoes dehydration.

Oxidation of Ketones, yield secondary alcohol

When two molecules of an ester undergo a condensation reaction, the reaction is called as Claisen condensation and product is a - keto ester.


Mechanism
Step I

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Step II



Step III

Step 1: Identifying ketone formation.
Ketones are formed by oxidation of secondary alcohols, where the alcohol is converted to a carbonyl group (C=O).

Step 2: Conclusion.
The correct answer is oxidation of secondary alcohols, which corresponds to option (3).

Step 1: Understanding benzoin condensation.
Benzoin condensation is a reaction between aldehydes in the presence of a catalyst. Certain functional groups on aldehydes, like the presence of , prevent them from undergoing benzoin condensation.

Step 2: Conclusion.
does not undergo benzoin condensation, which corresponds to option (1).

The common name of Pentanoic acid is valeric acid.

Compounds containing group give iodoform test. Acetaldehyde satisfies this condition.

Aldehydes without -hydrogen undergo Cannizzaro reaction.