Properties Questions in English

(C) The given transformation is the reduction of a carbonyl group $(C=O)$ to a methylene group $(CH_2)$.
Clemmensen reduction $(Zn-Hg/HCl)$ reduces aldehydes and ketones to alkanes.
Wolff-Kishner reduction $(NH_2NH_2/KOH, \Delta)$ also reduces aldehydes and ketones to alkanes.
Red $P$ and $HI$ at $200^{\circ}C$ is a strong reducing agent that reduces carbonyl compounds to alkanes.
Wurtz reaction is used for the preparation of higher alkanes from alkyl halides,not for the reduction of carbonyl compounds.
Therefore,Wurtz reduction does not effect this transformation.

(D) The reaction involving $Zn-Hg$ and concentrated $HCl$ is known as the $Clemmensen$ reduction.
This reaction specifically reduces carbonyl groups $(C=O)$ of aldehydes and ketones to methylene groups $(CH_2)$.
Among the given options,$Butan-2-one$ $(CH_3COCH_2CH_3)$ is a ketone,which undergoes $Clemmensen$ reduction to form $n-butane$ $(CH_3CH_2CH_2CH_3)$.
Other options like acetamide,acetic acid,and ethyl acetate do not undergo this reduction under these conditions.

(B) The given transformation involves the reduction of a ketone group $(C=O)$ to a methylene group $(-CH_2-)$ while keeping the carbon-carbon double bond $(C=C)$ and the hydroxyl group $(-OH)$ intact.
$NaBH_4$ reduces ketones to alcohols but does not reduce the $C=O$ group to $CH_2$.
$Zn-Hg / HCl$ (Clemmensen reduction) is acidic and would likely cause dehydration of the alcohol or other side reactions.
$NH_2NH_2, \mathop{OH}\limits^\Theta$ (Wolff-Kishner reduction) is a basic condition that specifically reduces the carbonyl group of aldehydes and ketones to a methylene group without affecting the $C=C$ double bond or the alcohol group.
Therefore,the correct reagent is $NH_2NH_2, \mathop{OH}\limits^\Theta$.

(A) The Clemmensen reduction involves the reduction of carbonyl groups (aldehydes and ketones) to methylene groups $(-CH_2-)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
The reaction shown in the image is:
$C_6H_5COCH_3 \xrightarrow{Zn-Hg/HCl} C_6H_5CH_2CH_3$
This is the classic Clemmensen reduction.
Option $B$ represents the Wolff-Kishner reduction.
Option $C$ represents reduction using red phosphorus and $HI$.
Therefore,only the reaction shown in the image is a Clemmensen reduction.

(B) The given transformation involves the reduction of a ketone group $(C=O)$ to a methylene group $(-CH_2-)$ while keeping the hydroxyl $(-OH)$ group intact.
$1.$ $Zn(Hg)/HCl$ is the reagent for Clemmensen reduction,which is acidic in nature. It would likely cause the dehydration of the alcohol group.
$2.$ $NH_2NH_2, OH^-$ is the reagent for Wolff-Kishner reduction,which is basic in nature. It reduces the carbonyl group to a methylene group without affecting the hydroxyl group.
$3.$ $H_2/Ni$ would reduce the ketone to an alcohol.
$4.$ $NaBH_4$ would reduce the ketone to an alcohol.
Therefore,the correct reagent is $NH_2NH_2, OH^-$.

(A) The reagent $Zn-Hg/HCl$ is used in the $Clemmensen$ reduction.
This reaction specifically reduces carbonyl groups $(C=O)$ present in aldehydes and ketones to methylene groups $(-CH_2-)$.
Among the given options,butan-$2$-one is a ketone,which undergoes $Clemmensen$ reduction to form butane.
Carboxylic acids,amides,and esters do not undergo $Clemmensen$ reduction under these conditions.

(D) The reduction of aldehydes and ketones to alkanes using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$ is known as the $Clemmensen$ reduction.
The general reaction is: $R-CO-R' + 4[H] \xrightarrow{Zn-Hg/HCl} R-CH_2-R' + H_2O$.

(D) The given reaction involves the reduction of an aldehyde $(CH_3CHO)$ to an alkane $(CH_3CH_3)$ using zinc amalgam $(Zn(Hg))$ and concentrated hydrochloric acid $(Conc. HCl)$.
This specific reagent system is used in the Clemmensen reduction,which reduces carbonyl groups $(C=O)$ to methylene groups $(CH_2)$.
Therefore,the correct option is $D$.

(A) The conversion of a carbonyl group $(C=O)$ to a methylene group $(CH_2)$ is achieved by the Clemmensen reduction.
Phenyl methyl ketone $(C_6H_5COCH_3)$ reacts with $Zn-Hg$ amalgam in the presence of concentrated $HCl$ to form ethylbenzene $(C_6H_5CH_2CH_3)$.
This reaction is a standard method for the reduction of ketones to alkanes.

(A) The reactivity of carbonyl compounds towards nucleophilic addition depends on two factors: steric hindrance and electronic effects.
$1$. Steric hindrance: As the number of alkyl groups attached to the carbonyl carbon increases,the attack of the nucleophile becomes more difficult due to crowding.
$2$. Electronic effect: Alkyl groups are electron-donating ($+I$ effect),which decreases the electrophilicity of the carbonyl carbon.
Comparing the given compounds:
- $HCHO$ (Formaldehyde): No alkyl groups,least steric hindrance,most electrophilic.
- $CH_3CHO$ (Acetaldehyde): One methyl group,moderate steric hindrance.
- $CH_3COCH_3$ (Acetone): Two methyl groups,highest steric hindrance,least electrophilic.
Therefore,the order of reactivity is $HCHO > CH_3CHO > CH_3COCH_3$.

(B) The reaction of ketones with hydrogen cyanide $(HCN)$ to form cyanohydrins is a classic example of a nucleophilic addition reaction.
In this process,the cyanide ion $(CN^-)$ acts as a nucleophile and attacks the electrophilic carbonyl carbon of the ketone.
This leads to the formation of an intermediate alkoxide ion,which is subsequently protonated to form the cyanohydrin product.

(B) The molecular formula $C_5H_{10}O$ represents an aldehyde with the functional group $-CHO$.
Subtracting one carbon for the $-CHO$ group,we have $4$ carbons remaining to form the alkyl chain.
The possible isomers for the butyl group $(-C_4H_9)$ are:
$1$. $CH_3-CH_2-CH_2-CH_2-CHO$ (pentanal)
$2$. $CH_3-CH_2-CH(CH_3)-CHO$ ($2$-methylbutanal)
$3$. $CH_3-CH(CH_3)-CH_2-CHO$ ($3$-methylbutanal)
$4$. $(CH_3)_3C-CHO$ ($2$,$2$-dimethylpropanal)
Therefore,the total number of possible aldehyde isomers is $4$.

(D) The enol content depends on the stability of the enol form,which is influenced by factors like conjugation and intramolecular hydrogen bonding.
$1. CH_3CHO$ (Acetaldehyde): Enol content is very low $(\approx 10^{-4} \%)$.
$2. CH_3COCH_3$ (Acetone): Enol content is very low $(\approx 10^{-6} \%)$.
$3. CH_3COCH_2CHO$: This is a $\beta$-dicarbonyl compound. The enol form is stabilized by conjugation and intramolecular hydrogen bonding. Enol content is $\approx 9 \%$.
$4. CH_3COCH_2COCH_3$ (Acetylacetone): This is a $\beta$-diketone. The enol form is highly stabilized by strong intramolecular hydrogen bonding and extended conjugation. Enol content is $\approx 76 \%$.
Comparing the values,the order of increasing enol content is $2 < 1 < 3 < 4$. However,among the given options,the sequence $1 < 2 < 3 < 4$ is the most appropriate representation of the increasing trend.

(A) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on two factors: steric hindrance and electronic effects.
$1.$ $HCHO$ has the least steric hindrance and no electron-donating groups,making it the most reactive.
$2.$ $CH_3COCH_3$ has two methyl groups,which provide slight steric hindrance and $+I$ effect.
$3.$ $PhCOCH_3$ has one phenyl group,which provides significant steric hindrance and resonance stabilization to the carbonyl carbon.
$4.$ $PhCOPh$ has two phenyl groups,providing the maximum steric hindrance and resonance stabilization,making it the least reactive.
Therefore,the increasing order of reactivity is $4 < 3 < 2 < 1$.

(D) The extent of enolization depends on the stability of the resulting enol form.
In the case of $1,3$-dicarbonyl compounds,the enol form is stabilized by intramolecular hydrogen bonding and conjugation.
Among the given options,the cyclic $1,3$-dicarbonyl compound (cyclohexane-$1,3$-dione) forms an enol that is highly stabilized due to the formation of a stable,conjugated system and intramolecular hydrogen bonding,leading to the highest degree of enolization.

(D) Acetone $(CH_3COCH_3)$ undergoes keto-enol tautomerism in the presence of $D_2O$ (a deuterium source).
In the presence of a base or acid catalyst,the $\alpha$-hydrogens are exchanged with deuterium atoms.
The keto form becomes $CD_3COCD_3$.
The enol form is formed by the tautomerization of this deuterated ketone.
The enol form of $CD_3COCD_3$ is $CD_2=C(OH)-CD_3$.

(A) The reaction of a ketone with a Grignard reagent $(R'MgX)$ proceeds via nucleophilic addition to the carbonyl group to form an adduct,which upon acidic hydrolysis yields a tertiary alcohol.
$R_2C=O + R'MgX \rightarrow R_2C(R')(OMgX)$
$R_2C(R')(OMgX) + H_2O/H^+ \rightarrow R_2C(R')OH + Mg(OH)X$
Thus,the product is a $3^o$ alcohol.

(C) The Victor Meyer test gives a blue color for secondary alcohols. This means compound $B$ must be a secondary alcohol. Only aldehydes (other than formaldehyde) react with Grignard reagents to form secondary alcohols. Since the Grignard reagent is methyl magnesium iodide $(CH_3MgI)$,the aldehyde $A$ must be acetaldehyde $(CH_3CHO)$ to produce isopropyl alcohol $(CH_3CH(OH)CH_3)$ as the secondary alcohol $B$.
Reaction:
$CH_3CHO + CH_3MgI \rightarrow CH_3CH(OMgI)CH_3$
$CH_3CH(OMgI)CH_3 + H_2O/H^+ \rightarrow CH_3CH(OH)CH_3 + Mg(OH)I$
Isopropyl alcohol $(CH_3CH(OH)CH_3)$ is a secondary alcohol,which gives a blue color in the Victor Meyer test.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$HCHO$ (formaldehyde) does not contain either of these groups.
Therefore,$HCHO$ will not give a yellow precipitate of iodoform $(CHI_3)$ with $I_2/NaOH$.

(A) The reaction of an ester with a Grignard reagent $(R'MgX)$ proceeds in two steps to form a tertiary alcohol.
First,the ester reacts with one equivalent of the Grignard reagent to form a ketone intermediate.
$RCOOR'' + R'MgX \rightarrow RCOR' + R''OMgX$.
Then,the ketone reacts with a second equivalent of the Grignard reagent to form the tertiary alcohol after acidic hydrolysis.
$RCOR' + R'MgX$ $\rightarrow R(R')_2COMgX$ $\xrightarrow{H_3O^+} R(R')_2COH$.
For ethyl acetate $(CH_3COOC_2H_5)$ reacting with methylmagnesium bromide $(CH_3MgBr)$:
$1$. $CH_3COOC_2H_5 + CH_3MgBr \rightarrow CH_3COCH_3 + C_2H_5OMgBr$ (Acetone is formed).
$2$. $CH_3COCH_3 + CH_3MgBr \rightarrow (CH_3)_3COMgBr$.
$3$. $(CH_3)_3COMgBr + H_2O/H^+ \rightarrow (CH_3)_3COH + Mg(OH)Br$.
The final product is tert-butyl alcohol,$(CH_3)_3COH$.

(C) $1$. $p$-Cresol reacts with $CHCl_3$ in the presence of $NaOH$ (Reimer-Tiemann reaction) to form $2$-hydroxy-$5$-methylbenzaldehyde (compound $A$ or $X$).
$2$. The aldehyde group $(-CHO)$ of $X$ reacts with $HCN$ to form a cyanohydrin $(Y)$,which is $2$-hydroxy-$5$-methylmandelonitrile.
$3$. Acidic hydrolysis of the cyanohydrin $(Y)$ converts the $-CN$ group into a $-COOH$ group,resulting in $2$-hydroxy-$5$-methylmandelic acid.
$4$. The structure of the final product is $2$-hydroxy-$5$-methylmandelic acid,which contains a chiral carbon atom at the alpha position relative to the carboxylic acid group.

(A) Aldehydes reduce $Tollens'$ reagent to form a silver mirror on the inner walls of the test tube. This reaction is specific to aldehydes.

(B) Ketones undergo bimolecular reduction in the presence of $Mg-Hg$ (magnesium amalgam) and water to form vicinal diols known as pinacols. The reaction is as follows:
$2(CH_3)_2C=O \xrightarrow{Mg-Hg/H_2O} (CH_3)_2C(OH)-C(OH)(CH_3)_2$ (Pinacol).

(C) Fehling's solution is a mild oxidizing agent used to distinguish between aldehydes and ketones.
Aldehydes (except benzaldehyde) and $\alpha$-hydroxy ketones (like glucose) give a positive test with Fehling's solution.
Ketones,such as $CH_3COCH_3$ (propanone),do not react with Fehling's solution because they cannot be easily oxidized.
Therefore,propanone is the correct answer.

(D) Compounds containing the $CHO$ group (aldehydes) reduce Tollen's reagent to form a silver mirror.
$1$. $CH_3(CHOH)_3CHO$ is an aldose (sugar),which contains an aldehyde group and gives the test.
$2$. $CH_3COCHOHCH_3$ is an $\alpha$-hydroxy ketone. $\alpha$-hydroxy ketones are oxidized by Tollen's reagent to dicarbonyl compounds,thus giving a positive silver mirror test.
$3$. $HCOOH$ (formic acid) contains an aldehyde-like hydrogen atom attached to a carbonyl group,which allows it to reduce Tollen's reagent.
Therefore,all of these substances give the silver mirror test.

(D) Benzaldehyde $(C_6H_5CHO)$ is an aldehyde. Aldehydes can be reduced to primary alcohols using strong reducing agents like lithium aluminium hydride $(LiAlH_4)$.
The reaction is as follows:
$C_6H_5CHO + 2[H] \xrightarrow{LiAlH_4} C_6H_5CH_2OH$
Thus,benzaldehyde is reduced to benzyl alcohol using $LiAlH_4$.

(C) Paraldehyde is a cyclic trimer of acetaldehyde. It has been historically used as a hypnotic and sedative agent in medicine.

(A) $1$. The reaction of $HCHO$ and $CH_3CHO$ with $dil. NaOH$ is a cross-aldol condensation. Since $HCHO$ has no $\alpha$-hydrogen,it acts as the electrophile,and $CH_3CHO$ acts as the nucleophile. The product $A$ is $CH_2=CH-CHO$ (after dehydration of the aldol product $HO-CH_2-CH_2-CHO$).
$2$. The reaction of $A$ $(CH_2=CH-CHO)$ with $HCN$ followed by hydrolysis $(H_3O^+)$ is a nucleophilic addition to the carbonyl group followed by hydrolysis of the nitrile group to a carboxylic acid.
$3$. The addition of $HCN$ to $CH_2=CH-CHO$ gives $CH_2=CH-CH(OH)CN$.
$4$. Hydrolysis of $CH_2=CH-CH(OH)CN$ gives $CH_2=CH-CH(OH)COOH$.

(C) Compound $A$ gives the iodoform test,which means it must contain the $CH_3CO-$ or $CH_3CH(OH)-$ group.
Given the molecular formula $C_3H_6O$,compound $A$ is propanone $(CH_3COCH_3)$.
When propanone is treated with $HCl$ (acid-catalyzed aldol condensation followed by dehydration),it undergoes self-condensation to form phorone $(2,6-dimethyl-2,5-heptadien-4-one)$.
The reaction is: $3CH_3COCH_3 \xrightarrow{HCl} (CH_3)_2C=CH-CO-CH=C(CH_3)_2 + 2H_2O$.
Thus,$A$ is propanone and $B$ is $2,6-dimethyl-2,5-heptadien-4-one$.

(B) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on steric hindrance and electronic effects.
Aldehydes are generally more reactive than ketones due to less steric hindrance and lower electronic stabilization of the carbonyl carbon.
Aliphatic carbonyl compounds are more reactive than their aromatic counterparts because the resonance effect in aromatic compounds stabilizes the carbonyl carbon,making it less electrophilic.
Thus,the correct order of reactivity is: $H_2C=O > RCHO > ArCHO > R_2C=O > Ar_2C=O$.

(C) Benzaldehyde does not contain $\alpha$-hydrogen atoms,so it does not undergo aldol condensation. It does not form an addition product with sodium hydrogen sulfite due to steric hindrance. However,it gives the Cannizzaro reaction and reduces Tollen's reagent. The question asks for the statement that is '$NOT$ false' (i.e.,true). Both $A$ and $C$ are true,but in the context of standard multiple-choice questions,$C$ is the most characteristic reaction for benzaldehyde.

(B) Ketones are resistant to oxidation under mild conditions but undergo oxidation with strong oxidizing agents like concentrated $HNO_3$ or $KMnO_4$ at high temperatures,leading to the cleavage of $C-C$ bonds.
$CH_3-CO-CH_2-CH_3 + 3[O] \xrightarrow{conc. HNO_3} 2CH_3COOH$ (Ethanoic acid).

(A) Acetone $(CH_3COCH_3)$ is an aliphatic ketone that reacts with saturated sodium bisulfite $(NaHSO_3)$ to form a white crystalline addition product.
Acetophenone $(C_6H_5COCH_3)$ is an aromatic ketone; due to steric hindrance and electronic effects,it does not form a bisulfite addition product with $NaHSO_3$.
Therefore,$NaHSO_3$ is used to distinguish between them.

(C) The formation of lactic acid $(2-hydroxypropanoic \ acid)$ involves the reaction of acetaldehyde with $HCN$ to form a cyanohydrin,followed by acid hydrolysis.
$CH_3CHO + HCN \to CH_3CH(OH)CN$ (Acetaldehyde cyanohydrin)
$CH_3CH(OH)CN + 2H_2O + H^{+} \to CH_3CH(OH)COOH + NH_4^{+}$ (Lactic acid)

(C) Phenacyl chloride $(C_6H_5COCH_2Cl)$ is commonly known as a tear gas or lachrymatory agent.
It is prepared by the chlorination of acetophenone $(C_6H_5COCH_3)$ in the presence of a catalyst.
The reaction is as follows:
$C_6H_5COCH_3 + Cl_2 \rightarrow C_6H_5COCH_2Cl + HCl$

(B) Acetaldehyde $(CH_3CHO)$ is an aldehyde.
It gives a positive iodoform test due to the presence of the $CH_3CO-$ group.
It gives positive Benedict's and Tollens' tests as it is an aldehyde.
Lucas test is specifically used to distinguish between primary,secondary,and tertiary alcohols,not aldehydes.

(A) Aldol condensation occurs in aldehydes or ketones that possess at least one $\alpha$-hydrogen atom.
$HCHO$ (formaldehyde) has no $\alpha$-hydrogen.
$CH_3CH_2CHO$ and $CH_3CHO$ possess $\alpha$-hydrogens.
$CH_3CH(OH)CH_3$ is an alcohol,not an aldehyde or ketone,and therefore does not undergo aldol condensation.

(A) The reaction of benzaldehyde $(C_6H_5CHO)$ with $HCN$ produces benzaldehyde cyanohydrin $(C_6H_5CH(OH)CN)$.
In this molecule,the carbon atom attached to the $-OH$,$-CN$,$-C_6H_5$,and $-H$ groups is a chiral center.
Since the starting material is achiral and the nucleophilic attack by $CN^-$ can occur from either side of the planar carbonyl group with equal probability,both enantiomers ($R$ and $S$) are formed in equal amounts.
Therefore,the resulting product is a racemic mixture.

(C) $C_2H_5CHO$ is an aldehyde,which reduces Fehling's solution to give a brick-red precipitate of $Cu_2O$.
$(CH_3)_2CO$ is a ketone,which does not react with Fehling's solution.
Therefore,Fehling's solution is used to distinguish between them.

(B) The compound shown in the image is $2,4-$dinitrophenylhydrazine $(2,4-DNP)$.
Only aldehydes and ketones react with $2,4-DNP$ to form characteristic colored (orange/red/yellow) crystalline precipitates,known as $2,4-$dinitrophenylhydrazones.
Among the given options,$CH_3COCH_3$ (acetone) is a ketone,while the others are acid derivatives (acid chloride,ester,and amide) which do not undergo this reaction.

(A) $HCHO$ undergoes the Cannizzaro reaction with $NaOH$ to produce an alcohol and a salt of a carboxylic acid.
$2HCHO + NaOH \rightarrow CH_3OH + HCOONa$
Here,$HCHO$ (formaldehyde) is converted into $CH_3OH$ (methanol) and $HCOONa$ (sodium formate).

(D) $1$. The hydration of ethyne $(HC \equiv CH)$ in the presence of $30\% \, H_2SO_4$ and $HgSO_4$ (Kucherov reaction) yields vinyl alcohol,which tautomerizes to form acetaldehyde $(A = CH_3CHO)$.
$2$. Acetaldehyde $(CH_3CHO)$ undergoes an aldol condensation reaction in the presence of dilute $NaOH$ to form $3-hydroxybutanal$ $(B = CH_3-CH(OH)-CH_2CHO)$.

(C) Aldol condensation is given by carbonyl compounds that possess at least one $\alpha$-hydrogen atom.
In $CH_3CH_2-CO-CH_3$ (Butan$-2-$one),the carbon atom adjacent to the carbonyl group (the $\alpha$-carbon) has two $\alpha$-hydrogen atoms,allowing it to undergo aldol condensation.
$C_6H_5OH$ is phenol,which does not contain a carbonyl group.
$C_6H_5-CO-C_6H_5$ (Benzophenone) has no $\alpha$-hydrogen atoms.
$(CH_3)_3C-CHO$ ($2$,$2$-Dimethylpropanal) has no $\alpha$-hydrogen atoms.

(B) $Acetophenone$ $(C_6H_5COCH_3)$ is a ketone.
Ketones do not reduce $Tollens'$ reagent,therefore they do not give a silver mirror test.
$Acetophenone$ contains a methyl ketone group $(-COCH_3)$,so it gives a positive iodoform test.
It forms $2,4-DNP$ derivatives as it is a carbonyl compound.
Oxidation with alkaline $KMnO_4$ followed by acidification yields benzoic acid.

(D) The reaction described is an aldol condensation reaction.
Two molecules of acetaldehyde $(CH_3CHO)$ undergo self-aldol condensation in the presence of a dilute base like $NaOH$ to form $3-$hydroxybutanal.
Reaction: $CH_3CHO + CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO$.
Here,$(X) = CH_3CHO$,$(Y) = CH_3CHO$,and $(Z) = NaOH$.

(D) Tollens' reagent is an alkaline solution of silver nitrate complexed with ammonia,specifically $[Ag(NH_3)_2]^+OH^-$. Therefore,ammoniacal silver nitrate is the correct answer.

(A) The molecular formula $C_2Cl_3OH$ corresponds to $CCl_3CHO$ (Chloral).
Since it is an aldehyde,it reduces Fehling's solution.
On oxidation,it gives trichloroacetic acid,which is a monocarboxylic acid.
$CCl_3CHO \xrightarrow{[O]} CCl_3COOH$
Chloral is also produced by the reaction of ethanol with chlorine:
$CH_3CH_2OH + 4Cl_2 \rightarrow CCl_3CHO + 5HCl$
Therefore,$A$ is Chloral.

(C) Schiff's reagent is a solution of $Rosaniline \ hydrochloride$ which has been decolorized by sulfur dioxide $(SO_2)$.

(D) Fehling's solution consists of two parts: Fehling's solution $A$ ($CuSO_4$ solution) and Fehling's solution $B$ (alkaline solution of sodium potassium tartrate,also known as Rochelle salt). The structure of sodium potassium tartrate is $CH(OH)COONa \mid CH(OH)COOK$.