Aldehydes And Ketones Test - 17

Hint: -I effect of oxygen atom

Acid strength can be determined by the l-effect of oxygen atom and resonance effect. Here stability of conjugate base is important. Stability of conjugate base is directly proportional to the acidity of an acid.

The structures of conjugate base are as follows:

The stability of conjugate base is b > c > a. The negative charge in oxygen is stable by resonance effect and -I effect of

oxygen atom in b and c.  As the distance between oxygen and –COO^- group increases -I effect of oxygen decreases.

Hence corresponding carboxylic acid will become less acidic. Thus, acidity order is II>III>I

Hint: Hydrogen bonding

Due to intermolecular H-bonding between molecules boiling point increases as the intermolecular force of attraction increases. The structure is as follows:

Hint: The last reaction is an example of clemmensen Reduction

Step 1:

The compound is pentan-3-one.

The formation of phenyl hydrazone indicates the presence of ketone or an aldehyde group. The compound gives a negative tollen's test which indicates that the aldehyde group is not present. The -ve iodoform test means the group is missing.

Step 2:

As the given compound does not answer Tollen's test, it is a ketone. The ketone is not a methyl ketone, as it fails to answer the iodoform reaction.  Ketones on reaction with phenylhydrazine give phenyl hydrazone.

Thus, the ketone is 3-pentanone, which on reduction using Zn/Hg and HCl gives hydrocarbon pentane.

Hint: LiAlH₄ reduce both an ester and ketone group due to strong ionic hydride character

Explanation:

Follow the chart given below:

NaBH₄ reduces the ketone functional group only, it will not reduce ester. Hence, the correct answer is 3

Hint: X is a geminal chloride

Side chain chlorination of toluene gives benzal chloride, which on hydrolysis gives benzaldehyde. This is a commercial method
of manufacture of benzaldehyde.

The reaction is as follows:

Hint: Hydrogen bonding containing molecules has a higher boiling point than those compounds in which hydrogen bonding did not occur.

Step 1:

Which factors decide the Boiling point?

Boiling point 

∝ Molar mass

∝ Polarity

∝ Hydrogen Bond

Step 2:

Analyze those factors w.r.t. given options.

Acid > Alcohol > Aldehyde/Ketone

Hint: NaBH₄ will reduce aldehyde group not double bond.

Hydrogen gas with Pt is a strong reducing agent hence, in the case of α, β unsaturated aldehyde, both double bond and CHO group will reduce.

The reaction is as follows:

The NaBH₄ is a mild reducing agent. It will only reduce CHO group not a double bond.

The reaction is as follows:

Hint: Use of analytical tests for aldehyde and ketone

Explanation:

Step 1:

It is given that the compound (with molecular formula C₉H₁₀O) forms 2, 4-DNP derivative and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde. Again, the compound undergoes cannizzaro reaction and on oxidation gives 1, 2benzenedicarboxylic acid. Therefore, the −CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2ethylbenzaldehyde.

Step 2:

The given reactions can be explained by the following equations.

Hint: Only carbon carbon bond reduce

Pd-catalyst shows selective reduction of α,β–unsaturated carbonyl compound. Only carbon-carbon double bond gets reduce not carbonyl group. It is an example of selective reduction. The reaction is as follows:

Hint: Nucleophilic addition reaction

cis-Cyclopenta-1,2-diol when reacts with acetone cyclic ketal is formed whereas its trans isomer cannot do so. 

Selectively only cis diol reacts with acetone not trans diol. The reaction is as follows: