Aldehydes Ketones and Carboxylic Acids Test - 10

Higher aldehydes are soluble in Benzene. Therefore, statement (B) is correct.

Lower Aldehydes are pungent but higher Aldehydes are fragrant. Therefore, statement (A) is incorrect.

Acetone is more polar than n-Butane and weight is comparable, so acetone has higher boiling point. Therefore, statement (C) is incorrect.

Methanal is soluble in water. Therefore, statement (D) is incorrect.

Hence, the correct option is (B).

Carboxylic acids form dimers by hydrogen bonding of the acidic hydrogen and the carbonyl oxygen when anhydrous.

Due to the small size and strong hydrogen bonding between the molecule (intermolecular), they overlap and form a dimer in aqueous condition.

For example, acetic acid forms a dimer in the gas phase, where the monomer units are held together by hydrogen bonds.

Hence, the correct option is (C).

For aldol condensation to occur, \(\alpha\)-hydrogen atom is required for carbonyl compound.

For formaldehyde, no \(\alpha\)-carbon and no \(\alpha\)-hydrogen is present. Therefore, \(H C H O\) will not undergo aldol condensation.

An alpha \((\alpha)\) hydrogen is the hydrogen which is bonded to alpha carbon. Alpha carbon is the first carbon which is bonded to the functional group. 

Hence, the correct option is (C).

Maleic acid is stronger than fumaric acid.

• Maleic acid is cis-butenedioic acid whereas fumaric acid is trans-butenedioic acid.
• Maleic acid is able to lose \(H^{+}\)ion and it results in the formation of intra hydrogen bond. Whereas fumaric acid being a trans isomer has strong interaction with both the oxygen atoms of each carboxylic group. Therefore, fumaric acid is unable to give hydrogen ions as compared to maleic acid.
• That is why, maleic acid is stronger than fumaric acid.

Hence, the correct option is (B).

Aldehydes with no \(\alpha\)-H atom undergo Cannizzaro reaction on heating with conc. alkali solution.

• Aldehydes undergo self -oxidation and reduction reaction on heating with concentrated alkali.
• In Cannizzaro reaction, one molecule of an aldehyde is reduced to alcohol and at the same time the second molecule is oxidized to carboxylic acid salt. 
• Thus, the reaction is an example of disproportionation reaction.

Therefore, only \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) will undergo the reaction. The reaction is as follows:

Hence, the correct option is (C).

An alpha \((\alpha)\) hydrogen is the hydrogen which is bonded to alpha carbon. Alpha carbon is the first carbon which is bonded to the functional group. 

As propanal contains alpha hydrogen, it undergoes self aldol condensation to give beta hydroxy (aldol) compound.

The reaction is as follows:

\(CH_3CH_2CHO+{\underset{CH_3}{\underset{|}{C}H_2}}CHO\xrightarrow{dil.NaOH}CH_3-CH_2-{\overset{OH}{\overset{|}{C}H}}-{\underset{CH_3}{\underset{|}{C}}HCHO}\)

Hence, the correct option is (C).

Presence of one vinyl (or ethenyl) group gives formaldehyde as one of the product in ozonolysis.

CH₂=CH- on ozonolysis will give HCHO.

• Ozonolysis is an organic reaction where the unsaturated bonds of alkenes, alkynes, or azo compounds are cleaved with ozone. 
• Alkenes and alkynes form organic compounds in which the multiple carbon–carbon bond has been replaced by a carbonyl group while azo compounds form nitrosamines.

Hence, the correct option is (B).

Order of reactivity of different alcohols towards esterification is \(1^{\circ}>2^{\circ}>3^{\circ}\).

Thus, as the steric hinderance (or bulkiness) increases from primary to secondary to tertiary alcohol, the order of esterification decreases.

Hence, the correct option is (C).

Methyl salicylate has a smell of oil of wintergreen.

• Methyl salicylate (oil of wintergreen or wintergreen oil) is an organic compound with the formula \(\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})\left(\mathrm{CO}_{2} \mathrm{CH}_{3}\right)\). 
• It is produced by many species of plants, particularly wintergreens. 
• It is also produced synthetically. 
• It is used as a fragrance, in foods and beverages, and in liniments.

Hence, the correct option is (B).

The oxidation of toluene \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CH}_{3}\right)\) with chromyl chloride \(\left(\mathrm{CRO}_{2} \mathrm{Cl}_{2}\right)\) in \(\mathrm{CCl}_{4}\) or \(C S_{2}\) to give benzaldehyde is called Etard reaction. 

In this reaction, the chromyl chloride first forms a brown complex, which is seperated and then decomposed with \(\mathrm{H}_{2} \mathrm{O}\) to give benzaldehyde \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\right)\).

The solvent used is carbon disulfide. Chromium chloride converts the methyl group to a chromium complex. This complex on acid hydrolysis gives the corresponding aldehyde.

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3} \stackrel{\mathrm{CrO}_{2} \mathrm{Cl}_{2}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{OCrOHCl}_{2}\right)_{2} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\)

Hence, the correct option is (A).

\({HOCl} \longrightarrow {OH}^{-}+{Cl}^{+}\)

\(C l^{+}\)shall attack acetylene as an electrophile and \(O H^{-}\)will end the addition with a nucleophilic addition.

Therefore, the final reaction will be:

\({CH} \equiv {CH}+2 {HOCl} \longrightarrow \underset{\text { Unstable }}{{CH}({OH})_{2}-{CHCl}_{2} \stackrel{-{H}_{2} {O}}{\longrightarrow} {CHCl}_{2}-{CHO}}\)

Hence, the correct option is (A).

Given compound:

\(\mathrm{\underset { \overset { | }{ CHO }  }{ { CH }_{ 2 } } -\underset { \overset { | }{ CHO }  }{ CH } -\underset { \overset { | }{ CHO }  }{ { CH }_{ 2 } }}\)

Numbering the carbon atoms, we get:

\(\mathrm{\underset { \overset { | }{ CHO }  }{ { \overset {3}{C}H }_{ 2 } } -\underset { \overset { | }{ CHO }  }{\overset {2}{C}H } -\underset { \overset { | }{ CHO }  }{ { \overset {1}{C}H }_{ 2 } }}\)

• Since there are three carbon atoms in long chain and it contains 3 -CHO groups (one on each carbon atom). 
• IUPAC suffix "aldehyde" will added due to presence of functional group -CHO.

Therefore, the correct IUPAC name will be propane-1, 2, 3-tricarbaldehyde.

Hence, the correct option is (D).

Given reaction:

The above reaction is known as Etard reaction.

• The Etard Reaction is a chemical reaction that involves the direct oxidation of an aromatic or heterocyclic bound methyl group to an aldehyde using chromyl chloride. 
• When toluene is reacted with Chromyl Chloride, then a chromium complex is formed (Etard Complex) whose hydrolysis gives Benzaldehyde.

Hence, the correct option is (C).

The reaction of HCN with carbonyl compounds is an example of the nucleophilic addition reaction. The mechanism involved is discussed below:

Step – 1: Attack of nucleophile from HCN to the slightly positive carbon atom.

The base is used as a catalyst and it deprotonates the HCN molecule to produce a stronger nucleophile – cyanide ion.

Step – 2: The negative charge in the tetrahedral intermediate is then neutralized by accepting a proton from a water molecule. This results in the formation of cyanohydrin – an additional product and hydroxyl ion. This hydroxyl ion can further be used up in the formation of the nucleophile.

HCN is not directly used in this reaction as it is highly poisonous. The reagent is prepared by adding dilute acid to sodium or potassium cyanide.

Hence, the correct option is (A).

Benzyl chloride gives benzoic acid on oxidation.

The reaction takes place as follows:

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}+2 \mathrm{KOH}+2[\mathrm{O}] \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOK}+\mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}\)

Hence, the correct option is (D).

Toluene reacts with halogen in presence of iron(III) chloride giving ortho and para halo compounds, the reaction is electrophilic substitution reaction.

The \(\mathrm{H}\) atom or aromatic nucleus is replaced with \(\mathrm{Cl}\) atom. The electrophile in this reaction is \(\mathrm{C l^{+}}\) ion obtained by the reaction of \(\mathrm{Cl}_{2}\) with \(\mathrm{AlCl}_{3}\).

Hence, the correct option is (D).

The correct order of increasing acidic strength is Ethanol < Phenol < Acetic acid < Chloroacetic acid.

• Phenol is more acidic than ethanol because in phenol, the phenoxide ion obtained on deprotonation is stabilized by resonance which is not possible in case of ethanol.
• Also carboxylic acids are more acidic than alcohols and phenols as the carboxylate ion is stabilized by resonance. Chloroacetic acid is more acidic than acetic acid due to inductive effect of chlorine atom which stabilizes the carboxylate anion.

Hence, the correct option is (C).

Baeyer's reagent (alk. \(\mathrm{KMnO}_{4}\) ) which is pink in colour decolourises due to the presence of unsaturation.

Thus, it shows the presence of unsaturation in an organic compound.

Hence, the correct option is (C).

Ionic species are stabilized by the dispersal of charge.

• The carboxylate ion \( \left(\mathrm{F}_{2} \mathrm{CHCOO}^{-}\right)\)is the most stable.
• The negative charge on \(\mathrm{O}\) is dispersed through resonance in the carboxylate group.
• The negative charge is further dispersed due to negative inductive effect (-I effect) of two \(\mathrm{F}\) atoms present on alpha carbon atoms. 

Hence, the correct option is (D).

\(\mathrm{FCH}_{2} \mathrm{CHO}\) is most reactive towards nucleophilic addition since presence of most electronegative \(F\) withdraws electron from carbon of carbonyl group making it more polar.

Hence, the correct option is (A).

Higher carboxylic acids are insoluble in water due to the increased hydrophobic interaction of the hydrocarbon part. 

• Thus, decanoic acid is highly insoluble in water, as it contains longer hydrocarbon chain.
• Decanoic acid is a ten-carbon, saturated fatty acid. It is present in palm kernel, coconut fat and in milk fat.

Hence, the correct option is (D).

Acetic acid is obtained when acetaldehyde is oxidised with potassium dichromate and sulphuric acid.

The reaction takes place as follows:

\(\ce{ CH3CHO->[K2Cr2O7][H2SO4]CH3COOH}\)

Hence, the correct option is (C).

Boiling points of carbonyl compounds are higher than those of alkanes due to dipole-dipole interactions.

• The boiling points of aldehydes and ketones (carbonyl compounds) are higher than boiling point of non-polar alkanes due to the presence of dipole-dipole interactions between molecules of carbonyl compounds which are much stronger than Van der Waals forces between molecules of alkanes. 
• So, energy required to break the interaction between carbonyl compound molecules is more than that of forces between molecules of alkane and therefore, boiling point is higher.

Hence, the correct option is (B).

Hydrolysis of \(\mathrm{CH _{3} CH _{2} NO _{2}}\) with \(\mathrm{85 \% ~H _{2} SO _{4}}\) gives carboxylic acid as shown in below chemical equation:

\(\ce{ CH3CH2NO2$+\mathrm{H}_2\mathrm{O}$->[H2SO4]CH3COOH$+$NH2OH}\)

Hence, the correct option is (D).

"Acidic hydrogen means it has a tendency to be released as \(H^{+}\) ion. So, if any H-atom is attached to another atom or group of atoms with higher electronegativity, that \(\mathrm{H}\)-atom can be released very easily as \(H^{+}\) ion."

(A) 

The hydrogen marked with red color is an acidic hydrogen, due to the presence of hydrogens in between electron withdrawing groups.

(B) 

The hydrogens marked with red color are acidic hydrogens, due to the presence of hydrogens in between electron withdrawing groups.

(C) 

The hydrogen marked with red color is acidic hydrogen, due to the presence of hydrogens in between electron withdrawing groups.

(D) 

The hydrogen marked with red color is acidic hydrogen, due to the attachment of hydrogen directly to the oxygen.

Oxygen atoms have higher electronegativity than carbon.

So, the hydrogen which is attached to oxygen is more acidic than the hydrogen atoms that are attached to carbon atoms (carbon atoms later attached to electron withdrawing groups).

Hence, the correct option is (D).

\(CH_3 - \underset{OH}{\underset{|}{C}H} - CH = \underset{CH_3}{\underset{|}{C} -} CHO\)

Numbering of parent chain will occur from right hand side because functional group like aldehyde is given lowest numbering.

\(\overset{5}{C}H_3 - \underset{OH}{\underset{|}{\overset{4}{C}}H} - \overset{3}{C}H = \underset{CH_3}{\underset{|}{\overset{2}{C}} -} \overset{1}{C}HO\)

4-hydroxy-2-methylpent-2-en-1-al

Hence, the correct option is (A).

We know that, \(-F\) is a strong electron withdrawing group.

Electron withdrawing group present nearer to the \(-\mathrm{COOH}\) group has more acidic strength. Therefore, will be having low \(p K_{a}\) value.

We can see that, in option (B), \(-F\) is placed at the adjacent to \(-\mathrm{COOH}\) group. Therefore, the compound will have the smallest value for \(p K_{a}\).

Hence, the correct option is (B).

Given: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl}\)

Numbering the carbon atoms as:

\(\mathrm{\stackrel{3}{C}H_{3}\stackrel{2}{C}H_{2}\stackrel{1}{C}OCl}\)

• This is an acid chloride.
• It consists of three carbon atoms in the parent chain and chloroformyl group \(\mathrm{(-COCl)}\).
• In the IUPAC name, the suffix "oyl chloride" is added. Therefore, the IUPAC name will be propanoyl chloride.

Hence, the correct option is (B).

There is a large difference in the boiling points of butanal and butan-1-ol due to intermolecular hydrogen bonding in butan-1-ol. 

Butan-1-ol has polar O-H bond due to which it shows intermolecular H-bonding which is not possible in case of butanal due to absence of polar bond. 

Hence, the correct option is (A).

As the compound after oxidation, contains only one oxygen this means alcohol is not primary alcohol. Oxidation of secondary alcohol gives ketone as a product.

As the ketone formed does not give iodoform test, so, the ketone formed is:

\(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CO}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\)

Therefore, the structure of the compound 'A' i.e., alcohol will be:

\(\mathrm{CH_3-CH_2-{\underset{OH}{\underset{|}C}H}-CH_2-CH_3}\)

Hence, the correct option is (C).