Properties Questions in English

(C) The reduction of a carbonyl group $(>C=O)$ to a methylene group $(CH_2)$ is specifically achieved by Wolff-Kishner reduction (using hydrazine,$NH_2NH_2$,followed by a strong base like $KOH$ in ethylene glycol) or Clemmensen reduction (using $Zn(Hg)$ and $HCl$).
Therefore,the correct option is $(C)$.

(D) The correct option is $(D)$.
Benzaldehyde $(C_6H_5CHO)$ on reduction yields benzyl alcohol $(C_6H_5CH_2OH)$ and not phenol $(C_6H_5OH)$.
The reaction is: $C_6H_5CHO + 2[H] \rightarrow C_6H_5CH_2OH$.

(A) The reaction of a $Grignard$ reagent $(RMgX)$ with a ketone like acetone $(CH_3COCH_3)$ proceeds via nucleophilic addition to the carbonyl group.
The nucleophilic alkyl group $(R^-)$ from the $Grignard$ reagent attacks the electrophilic carbonyl carbon of the acetone.
This forms an intermediate magnesium alkoxide: $CH_3-C(R)(OMgX)-CH_3$.
Upon acidic hydrolysis $(H_3O^+)$,this intermediate is converted into a tertiary alcohol with the structure $CH_3-C(R)(OH)-CH_3$.

(C) $FeCl_3$ is used to detect phenols by forming a violet-colored complex.
Fehling solution is used to detect reducing sugars like glucose.
$NaHSO_3$ (sodium bisulfite) is used to detect carbonyl compounds (aldehydes and ketones) by forming a crystalline addition product.
Tollen’s reagent is used to detect aldehydes,not unsaturation. Unsaturation is typically detected using bromine water or Baeyer's reagent.
Therefore,the statement that Tollen’s reagent is used in the detection of unsaturation is incorrect.

(B) The Wolff-Kishner reduction is a chemical reaction used to convert carbonyl groups $(> C = O)$ into methylene groups $(> CH_2)$ using hydrazine $(NH_2NH_2)$ and a strong base.
It does not convert the carbonyl group into an alcohol $(- CHOH)$.
Therefore,the pair in option $(B)$ is incorrectly matched.

(C) The correct answer is $C$.
Aldol condensation occurs in aldehydes and ketones that possess at least one $\alpha$-hydrogen atom.
$C_6H_5OH$ (phenol) is not a carbonyl compound.
$C_6H_5-CO-C_6H_5$ (benzophenone) has no $\alpha$-hydrogen atoms.
$CH_3CH_2-CO-CH_3$ (butan$-2-$one) has $\alpha$-hydrogen atoms and undergoes aldol condensation.
$(CH_3)_3C-CO-CH_3$ ($3$,$3$-dimethylbutan$-2-$one) has $\alpha$-hydrogen atoms,but it does not undergo aldol condensation due to significant steric hindrance.

(A) When benzaldehyde $(C_6H_5CHO)$ reacts with a Grignard reagent $(CH_3MgBr)$,a nucleophilic addition occurs to form an intermediate alkoxide.
Upon subsequent acid hydrolysis $(H_3O^+)$,the intermediate is converted into $1$-phenylethanol,which is a secondary $(2^\circ)$ alcohol.
The reaction is: $C_6H_5CHO + CH_3MgBr$ $\rightarrow C_6H_5CH(OMgBr)CH_3$ $\xrightarrow{H_3O^+} C_6H_5CH(OH)CH_3$.

(A) Chloral,$CCl_3CHO$,has no $\alpha$-hydrogen atom and hence does not undergo aldol condensation.

(B) $2, 4-Dinitrophenylhydrazine$ $(2, 4-DNP)$ is a reagent used to identify aldehydes and ketones through a nucleophilic addition-elimination reaction,forming coloured crystalline $2, 4-dinitrophenylhydrazones$.
Among the given options:
$A$ $(CH_3COCl)$ is an acyl chloride.
$B$ $(CH_3COCH_3)$ is a ketone (acetone).
$C$ $(CH_3CO(OC_2H_5))$ is an ester (ethyl acetate).
$D$ $(CH_3CONH_2)$ is an amide (acetamide).
Only aldehydes and ketones react with $2, 4-DNP$ to form the characteristic coloured crystalline derivatives. Therefore,acetone $(CH_3COCH_3)$ is the correct answer.

(D) Ethanal $(CH_3CHO)$ contains an aldehyde group $(-CHO)$,which gives a positive Fehling test.
It also contains a methyl ketone group ($-COCH_3$ structure is present in the form of $CH_3-CH=O$),which gives a positive iodoform test.
Therefore,Ethanal is the correct answer.

(B) Aldol condensation occurs in aldehydes or ketones that possess at least one $\alpha$-hydrogen atom.
$C_6H_5CHO$ (benzaldehyde) and $HCHO$ (formaldehyde) do not have $\alpha$-hydrogen atoms.
$CH_3CH_2CHO$ (propanal) contains two $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,$CH_3CH_2CHO$ undergoes self-aldol condensation in the presence of cold dilute alkali.

(C) Acetaldehyde $(CH_3CHO)$ contains $\alpha$-hydrogen atoms.
When treated with dilute $NaOH$,it undergoes an aldol condensation reaction.
$2CH_3CHO \xrightarrow{dil. NaOH} CH_3-CH(OH)-CH_2-CHO$ ($3$-hydroxybutanal).

(C) The correct answer is $C$.
$C_2H_5CHO$ is an aldehyde,while $(CH_3)_2CO$ is a ketone.
Aldehydes reduce Fehling solution to give a red precipitate of cuprous oxide $(Cu_2O)$,whereas ketones do not react with Fehling solution.
The reaction for the aldehyde is: $C_2H_5CHO + 2Cu^{+2} + 5OH^{-} \to C_2H_5COO^{-} + Cu_2O (\text{red ppt}) + 3H_2O$.
Ketones like $(CH_3)_2CO$ do not show this reaction.

(D) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the carbonyl compound.
$A$ Acetaldehyde $(CH_3CHO)$ has three $\alpha$-hydrogens.
$B$ Propanaldehyde $(CH_3CH_2CHO)$ has two $\alpha$-hydrogens.
$C$ Trideuteroacetaldehyde $(CD_3CDO)$ has three $\alpha$-deuterium atoms. Since deuterium $(D)$ behaves chemically similar to hydrogen $(H)$,it can also undergo aldol condensation.
Therefore,all of these compounds will undergo aldol condensation.

(B) The $Lucas$ test is used to distinguish between $1^o$,$2^o$,and $3^o$ alcohols based on their reactivity with $Lucas$ reagent $(ZnCl_2 + conc. HCl)$.
$1^o$ Alcohols: No turbidity at room temperature.
$2^o$ Alcohols: Turbidity appears within $5-10$ minutes.
$3^o$ Alcohols: Immediate turbidity.
Acetaldehyde is an aldehyde,not an alcohol; therefore,it does not react with $Lucas$ reagent and cannot show the $Lucas$ test. It gives positive results for $Iodoform$,$Benedict's$,and $Tollen's$ tests.

(A) The reaction between $Benzaldehyde$ $(C_6H_5CHO)$ and concentrated $NaOH$ is a classic example of the $Cannizzaro$ reaction.
Since $Benzaldehyde$ lacks $\alpha$-hydrogen atoms,it undergoes self-oxidation and reduction (disproportionation) in the presence of a strong base.
The products formed are $Benzyl \ alcohol$ $(C_6H_5CH_2OH)$ and $Sodium \ benzoate$ $(C_6H_5COONa)$.

(A) The reaction of $C_6H_5COCHO$ (phenylglyoxal) with aqueous $NaOH$ is an example of an intramolecular Cannizzaro reaction.
In this reaction,the aldehyde group $(-CHO)$ is oxidized to a carboxylate group $(-COO^-)$ while the adjacent ketone group $(-CO-)$ is reduced to a hydroxyl group $(-CHOH-)$.
Thus,$C_6H_5COCHO + NaOH \rightarrow C_6H_5CHOHCOONa$.

(B) Benzaldehyde,on treatment with $50\%$ aqueous or ethanolic alkali solution,undergoes Cannizzaro's reaction because it lacks an $\alpha$-hydrogen atom.
In this reaction,one molecule of benzaldehyde is oxidized to sodium benzoate and another is reduced to benzyl alcohol.
$2C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$

(D) Formaldehyde $(HCHO)$ and acetaldehyde $(CH_3CHO)$ can be distinguished using caustic soda $(NaOH)$ solution.
Formaldehyde undergoes the Cannizzaro reaction because it lacks $\alpha$-hydrogen atoms:
$2HCHO + NaOH (conc.) \rightarrow CH_3OH + HCOONa$
Acetaldehyde undergoes aldol condensation because it possesses $\alpha$-hydrogen atoms:
$2CH_3CHO \xrightarrow{NaOH (dil.)} CH_3-CH(OH)-CH_2-CHO$
Thus,option $(D)$ is the correct choice.

(B) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$CH_3CH_2OH$ contains the $CH_3CH(OH)-$ group and gives a positive test.
$CH_3CHO$ contains the $CH_3CO-$ group and gives a positive test.
$PhCOCH_3$ (acetophenone) contains the $CH_3CO-$ group attached to a phenyl ring and gives a positive test.
$CH_3OH$ (methanol) does not contain either of these groups,so it does not give the iodoform test.

(D) The iodoform test is given by compounds containing the $CH_3-C(=O)-$ group or the $CH_3-CH(OH)-$ group.
$3-$pentanone $(CH_3-CH_2-C(=O)-CH_2-CH_3)$ does not contain the $CH_3-C(=O)-$ group,therefore it does not give the iodoform test.
Thus,the correct option is $(D)$.

(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test.
$2$. Ethanal $(CH_3CHO)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test.
$3$. Benzophenone $(C_6H_5COC_6H_5)$ does not contain either the $CH_3CO-$ group or the $CH_3CH(OH)-$ group,so it does not give the iodoform test.
$4$. Ethanol $(CH_3CH_2OH)$ contains the $CH_3CH(OH)-$ group,so it gives a positive iodoform test.
Therefore,the correct option is $C$.

(B) The haloform test is given by compounds containing the methyl ketone group $(CH_3CO-)$ or alcohols that can be oxidized to a methyl ketone group (like $CH_3CH(OH)-$).
Acetone $(CH_3COCH_3)$ contains the methyl ketone group and thus gives a positive haloform test.

(B) Dimethyl ketone $(CH_3-CO-CH_3)$ is a methyl ketone.
Methyl ketones react with $I_2$ in the presence of $NaOH$ to form a yellow precipitate of iodoform $(CHI_3)$.
This is known as the iodoform test.
Tollen's reagent,Schiff's reagent,and Benedict's reagent are primarily used to identify aldehydes.

(C) Acetone $(CH_3COCH_3)$ undergoes the iodoform reaction when treated with iodine $(I_2)$ and an alkali (like $NaOH$).
This reaction produces iodoform $(CHI_3)$,which is a light yellow crystalline solid.
The chemical equation is: $CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$.

(C) The reaction between formaldehyde $(HCHO)$ and concentrated $KOH$ is a Cannizzaro reaction.
In this reaction,formaldehyde undergoes self-oxidation and reduction.
$2HCHO + KOH \xrightarrow{\Delta} HCOOK + CH_3OH$
Thus,the products are potassium formate and methyl alcohol.

(D) Acetaldehyde $(CH_3CHO)$ reacts with the Grignard reagent $C_2H_5MgCl$ to form an addition product.
Upon acidic hydrolysis,this addition product yields a secondary alcohol.
The reaction is:
$CH_3CHO + C_2H_5MgCl \rightarrow CH_3CH(OMgCl)C_2H_5$
$CH_3CH(OMgCl)C_2H_5 + H_2O \rightarrow CH_3CH(OH)C_2H_5 + Mg(OH)Cl$
The product $CH_3CH(OH)C_2H_5$ is $butan-2-ol$,which is a secondary alcohol.

(B) Propionaldehyde $(CH_3CH_2CHO)$ contains $\alpha$-hydrogens,so it undergoes an aldol condensation reaction in the presence of a dilute base like $NaOH$.
The reaction proceeds as follows:
$CH_3CH_2CHO + CH_3CH_2CHO \xrightarrow{dil. NaOH} CH_3CH_2CH(OH)CH(CH_3)CHO$
The product formed is $3$-hydroxy-$2$-methylpentanal.

(B) The aldol condensation of an aldehyde involves the formation of a carbanion as the reaction intermediate in the first step.
This occurs by the removal of an $\alpha-H$ atom from the aldehyde by a base (e.g.,$OH^-$).
The resulting enolate ion is resonance-stabilized as shown below:
$OH^- + CH_3CHO \rightleftharpoons H_2O + [CH_2-CHO]^- \leftrightarrow [CH_2=CH-O]^-$

(A) When acetone $(CH_3COCH_3)$ is saturated with dry hydrogen chloride gas,it undergoes an aldol condensation followed by dehydration to form Phorone.
The reaction is:
$3 CH_3COCH_3 \xrightarrow{HCl} (CH_3)_2C = CH - CO - CH = C(CH_3)_2 + 2 H_2O$
The product formed is Phorone.

(B) The reaction described is the $Cannizzaro$ reaction.
In this reaction,aldehydes that do not contain an $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation) in the presence of a concentrated alkali solution.
For example,two molecules of benzaldehyde react with concentrated $NaOH$ to produce sodium benzoate and benzyl alcohol.

(D) $m$-chlorobenzaldehyde does not contain an $\alpha$-hydrogen atom. Therefore,it undergoes the Cannizzaro reaction in the presence of concentrated $KOH$.
In the Cannizzaro reaction,one molecule of the aldehyde is oxidized to the corresponding carboxylic acid salt,and another molecule is reduced to the corresponding alcohol.
Thus,$m$-chlorobenzaldehyde undergoes disproportionation to form potassium $m$-chlorobenzoate and $m$-chlorobenzyl alcohol.

(C) The reaction with $NaOH + I_2$ is the iodoform test,which is given by compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group.
$1$. Acetone $(CH_3COCH_3)$ contains the $CH_3CO-$ group and gives a positive iodoform test.
$2$. Acetaldehyde $(CH_3CHO)$ contains the $CH_3CO-$ group and gives a positive iodoform test.
$3$. Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group attached to a phenyl ring and gives a positive iodoform test.
$4$. Benzaldehyde $(C_6H_5CHO)$ does not contain the $CH_3CO-$ or $CH_3CH(OH)-$ group,so it does not give a yellow precipitate of iodoform $(CHI_3)$.

(D) The reagent described is Fehling's solution,which consists of $CuSO_4$ (Fehling's $A$) and sodium potassium tartrate (Rochelle salt) in $NaOH$ (Fehling's $B$).
Fehling's solution is a mild oxidizing agent used to distinguish aliphatic aldehydes from aromatic aldehydes.
Aliphatic aldehydes like $CH_3CHO$,$C_2H_5CHO$,and $C_6H_5CH_2CHO$ (phenylacetaldehyde) are oxidized by Fehling's solution to their corresponding carboxylate ions.
However,aromatic aldehydes like $C_6H_5CHO$ (benzaldehyde) do not react with Fehling's solution because the aldehyde group is directly attached to the benzene ring,making it resistant to oxidation by this mild reagent.
Therefore,the correct option is $D$.

(A) The reactivity of carbonyl compounds towards nucleophilic addition depends on two factors: steric hindrance and electronic effects.
$1$. Steric hindrance: As the size of the alkyl groups attached to the carbonyl carbon increases,the approach of the nucleophile becomes more difficult.
$2$. Electronic effect: Alkyl groups are electron-donating ($+I$ effect),which reduces the electrophilicity of the carbonyl carbon.
Comparing the compounds:
- $CH_3CHO$ (Acetaldehyde): One alkyl group $(CH_3)$ and one hydrogen atom.
- $CH_3COCH_3$ (Acetone): Two methyl groups.
- $C_2H_5COCH_3$ (Butan$-2-$one): One methyl group and one ethyl group.
Since $CH_3CHO$ has the least steric hindrance and the least electron-donating effect,it is the most reactive.
$C_2H_5COCH_3$ has more steric hindrance than $CH_3COCH_3$ due to the larger ethyl group.
Therefore,the order of reactivity is $CH_3CHO > CH_3COCH_3 > C_2H_5COCH_3$.

(D) The oxidation of an aldehyde to a carboxylic acid is represented by the equation:
$RCHO + 0.5 O_2 \to RCOOH$
From the stoichiometry of the reaction,$1$ mole of aldehyde requires $0.5$ mole of oxygen to produce $1$ mole of the corresponding carboxylic acid. Therefore,the compound is an aldehyde.

(D) Aldehydes are easily oxidised to carboxylic acids by mild oxidising agents like Tollen's reagent,Fehling solution,and Benedict solution,whereas ketones are resistant to oxidation by these reagents.
$1.$ Tollen's reagent: $RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \to RCOO^- + 2Ag + 4NH_3 + 2H_2O$
$2.$ Fehling solution: $RCHO + 2Cu^{2+} + 5OH^- \to RCOO^- + Cu_2O + 3H_2O$
$3.$ Benedict solution: Similar to Fehling solution,it contains $Cu^{2+}$ ions which oxidise aldehydes.
Therefore,the correct option is $D$.

(A) The silver mirror test,also known as the $Tollens'$ test,is a chemical test used to distinguish between aldehydes and ketones.
Aldehydes are oxidized to carboxylic acids by $Tollens'$ reagent,which is an ammoniacal silver nitrate solution,while the silver ions are reduced to metallic silver,forming a silver mirror on the inner surface of the reaction vessel.
Ketones do not give this test.

(C) $CH_3CH=CHCHO$ is an $\alpha,\beta$-unsaturated aldehyde.
Ammoniacal $AgNO_3$ (Tollens' reagent) is a mild oxidizing agent that selectively oxidizes the aldehyde group to a carboxylic acid group without affecting the carbon-carbon double bond.
The reaction is: $CH_3CH=CHCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \to CH_3CH=CHCOO^- + 2Ag + 4NH_3 + 2H_2O$.
Alkaline $KMnO_4$ is a strong oxidizing agent and would likely cleave the double bond.
Selenium dioxide is typically used for allylic oxidation.
Therefore,the correct reagent for this specific transformation is Ammoniacal $AgNO_3$.

(C) Schiff's reagent is a specific test used to detect the presence of aldehydes.
Aldehydes react with Schiff's reagent to produce a characteristic pink or magenta color.
Among the given options,$Formaldehyde$ $(HCHO)$,$Benzaldehyde$ $(C_6H_5CHO)$,and $Acetaldehyde$ $(CH_3CHO)$ are all aldehydes and will give a positive test.
$Acetone$ $(CH_3COCH_3)$ is a ketone,which does not react with Schiff's reagent to produce the pink color.

(A) Fehling's test is used to distinguish between aliphatic aldehydes and aromatic aldehydes or ketones.
Aliphatic aldehydes,such as $CH_3CHO$ (Acetaldehyde),give a positive Fehling's test by forming a reddish-brown precipitate of $Cu_2O$.
Benzaldehyde is an aromatic aldehyde and does not respond to Fehling's test.
Ethers and alcohols do not contain the carbonyl group required for this oxidation reaction.

(B) Acetaldehyde $(CH_3CHO)$ reacts with ammonia $(NH_3)$ to form an addition product known as acetaldehyde ammonia.
In contrast,acetone $(CH_3COCH_3)$ reacts with ammonia to form complex condensation products such as diacetone amine and triacetone amine.
Both aldehydes and ketones react similarly with sodium bisulphite,$PCl_5$,and phenyl hydrazine to form addition products,gem-dichlorides,and phenylhydrazones,respectively.

(B) The reduction of acetaldehyde $(CH_3CHO)$ with sodium and ethanol $(Na/C_2H_5OH)$ is a classic reduction reaction that yields a primary alcohol.
The chemical equation for the reaction is:
$CH_3CHO + 2[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2OH$
Thus,the final product is ethyl alcohol.

(B) The reaction of propionaldehyde with amalgamated zinc and concentrated $HCl$ is known as Clemmensen reduction.
In this reaction,the carbonyl group $(>C=O)$ is reduced to a methylene group $(-CH_2-)$.
$CH_3CH_2CHO + 4[H] \xrightarrow{Zn/Hg, HCl} CH_3CH_2CH_3 + H_2O$
Thus,propionaldehyde is reduced to propane.

(C) . In Cannizzaro reaction,an aldehyde lacking $\alpha$-hydrogen undergoes self-oxidation and reduction (disproportionation) in the presence of a concentrated base.
$2HCHO + KOH \xrightarrow{\text{conc.}} CH_3OH + HCOOK$
Here,one molecule of formaldehyde is reduced to methanol $(CH_3OH)$ and another is oxidized to potassium formate $(HCOOK)$.

(D) Carbonyl compounds $(>C=O)$ react with $HCN$ to form cyanohydrins via a nucleophilic addition reaction,which is a pure addition reaction.
Reaction: $>C=O + HCN \rightarrow >C(OH)(CN)$.
Conversely,reactions with hydrazine $(NH_2NH_2)$,phenyl hydrazine $(NH_2NHC_6H_5)$,and semicarbazide $(NH_2NHCONH_2)$ are addition-elimination reactions (nucleophilic addition followed by the elimination of a water molecule) to form derivatives like hydrazones,phenylhydrazones,and semicarbazones,respectively.
Therefore,$HCN$ is the correct choice for a pure addition reaction.

(D) Acetaldehyde contains a carbonyl group $(C=O)$.
Due to the difference in electronegativity between carbon and oxygen,the carbonyl carbon is electrophilic,making it susceptible to attack by nucleophiles (nucleophilic addition reactions).
Additionally,the oxygen atom has lone pairs and can act as a weak nucleophile,and the alpha-hydrogens can be involved in reactions with electrophiles (e.g.,in aldol condensation or halogenation).
Therefore,acetaldehyde can react with both electrophiles and nucleophiles.

(C) The typical reaction of an aldehyde is $Nucleophilic \text{ } addition$.
Due to the presence of a polar carbonyl group $(C=O)$,the carbon atom is electrophilic,making it susceptible to attack by nucleophiles.
Examples include the addition of $HCN$,$NaHSO_3$,and Grignard reagents.

(C) The addition of $HCN$ to carbonyl compounds is a classic example of nucleophilic addition reaction.
In this reaction,the cyanide ion $(CN^-)$ acts as a nucleophile and attacks the electrophilic carbonyl carbon atom.

(C) The reaction of benzaldehyde $(C_6H_5CHO)$ with chlorine $(Cl_2)$ in the absence of a catalyst proceeds via a free radical mechanism or direct substitution of the aldehydic hydrogen.
The chemical equation is:
$C_6H_5CHO + Cl_2 \to C_6H_5COCl + HCl$
Thus,the product formed is benzoyl chloride.