Properties Questions in English

(B) Both aldehydes and ketones react with $Grignard$ reagent $(RMgX)$ to form alcohols.
$Fehling's$ solution,$Schiff's$ reagent,and $Tollens'$ reagent are specific tests used to distinguish aldehydes from ketones,as they react only with aldehydes.

(C) The conversion of a carbonyl group $(R_2CO)$ to a methylene group $(R_2CH_2)$ is known as reduction of carbonyl compounds.
Clemmensen reduction $(Zn-Hg/HCl)$,Wolff-Kishner reduction $(NH_2NH_2/KOH)$,and reduction with red phosphorus and $HI$ are standard methods for this transformation.
Wurtz reaction is used for the preparation of higher alkanes from alkyl halides,not for the reduction of carbonyl compounds.

(D) $Wolff-Kishner$ reduction is used to reduce carbonyl groups $(C=O)$ of aldehydes and ketones to methylene groups $(CH_2)$.
In this reaction,$Benzophenone$ $(C_6H_5-CO-C_6H_5)$ is reduced to $Diphenylmethane$ $(C_6H_5-CH_2-C_6H_5)$.

(A) $SeO_2$ is an oxidizing agent that specifically oxidizes the $\alpha$-methylene or $\alpha$-methyl group of a carbonyl compound to a carbonyl group,forming a $1,2-$dicarbonyl compound.
In the reaction of acetone $(CH_3-CO-CH_3)$ with $SeO_2$,one of the $\alpha$-methyl groups is oxidized to an aldehyde group.
The reaction is: $CH_3-CO-CH_3 + SeO_2 \rightarrow CH_3-CO-CHO + Se + H_2O$.
Thus,$P$ is $CH_3-CO-CHO$ (methylglyoxal or pyruvaldehyde).

(C) Compounds containing the $CH_3-CO-$ group or the $CH_3-CH(OH)-$ group give the iodoform test. Diethyl ketone $(CH_3CH_2-CO-CH_2CH_3)$ does not contain either of these groups,therefore it does not give the iodoform test.

(B) The mechanism of the Cannizzaro reaction involves the nucleophilic attack of $OH^-$ on the carbonyl carbon to form a dihydroxyalkoxide intermediate.
This is followed by the transfer of a hydride ion $(H^-)$ from the intermediate to another molecule of aldehyde.
The hydride shift is the rate-determining step (slow step) of the reaction.
Finally,a rapid proton exchange occurs to form the alcohol and carboxylate ion.

(B) The reaction of carbonyl compounds with ammonia derivatives (like hydroxylamine) is acid-catalyzed.
It proceeds most efficiently in a weakly acidic medium,typically in the range of $pH = 3 - 4$.

(B) Chloral $(CCl_3CHO)$ reacts with water to form a stable gem-diol known as chloral hydrate. The reaction is: $CCl_3CHO + H_2O \rightarrow CCl_3CH(OH)_2$. The stability of the gem-diol is due to the strong electron-withdrawing effect of the three chlorine atoms,which makes the carbonyl carbon highly electrophilic and facilitates the formation of the hydrate.

(B) The transformation involves the reduction of a ketone group $(-COCH_3)$ to an alkyl group $(-CH_2CH_3)$.
Both Clemmensen reduction $(Zn(Hg)/HCl)$ and Wolff-Kishner reduction $(NH_2NH_2/OH^-)$ can reduce ketones to alkanes.
However,the substrate contains a hydroxyl $(-OH)$ group. Under acidic conditions $(Zn(Hg)/HCl)$,the $-OH$ group can undergo dehydration or other side reactions.
Therefore,the Wolff-Kishner reduction $(NH_2NH_2/OH^-)$ is the appropriate reagent as it operates under basic conditions,which will not affect the $-OH$ group.

(B) $2,4-$Hexanedione contains an active methylene group $(-CH_2-)$ situated between two electron-withdrawing carbonyl groups. This makes the hydrogen atoms on this carbon highly acidic.
The structure is: $CH_3-CO-CH_2-CO-CH_2CH_3$.
Upon removal of a proton,the resulting carbanion is stabilized by resonance across both carbonyl groups,significantly increasing its stability compared to the other options.

(A) This reaction is a crossed Cannizzaro reaction.
In a crossed Cannizzaro reaction involving formaldehyde $(HCHO)$ and another aldehyde (like benzaldehyde,$C_6H_5CHO$),formaldehyde is more reactive towards nucleophilic attack by $OH^-$.
Therefore,formaldehyde undergoes oxidation to form a formate ion $(HCOO^-)$,which forms sodium formate $(HCOONa)$ in the presence of $NaOH$.
The other aldehyde (benzaldehyde) undergoes reduction to form the corresponding alcohol,which is benzyl alcohol $(C_6H_5CH_2OH)$.
The overall reaction is: $C_6H_5CHO + HCHO + NaOH \rightarrow C_6H_5CH_2OH + HCOONa$.

(A) Since product '$Y$' gives a silver mirror test,it must be an aldehyde. '$Y$' is formed by the oxidation of '$X$',which indicates that '$X$' is a primary alcohol $(CH_3CH_2CH_2OH)$.
$CH_3CH_2CH_2OH (X) \xrightarrow{K_2Cr_2O_7 + H_2SO_4} CH_3CH_2CHO (Y)$
When '$Y$' $(CH_3CH_2CHO)$ reacts with semicarbazide $(NH_2CONHNH_2)$ in the presence of sodium acetate,it forms a semicarbazone derivative.
$CH_3CH_2CHO + NH_2CONHNH_2 \rightarrow CH_3CH_2CH=NNHCONH_2 + H_2O$
The product '$Z$' is $CH_3CH_2CH=NNHCONH_2$.

(C) The given reaction is the Perkin reaction.
In the Perkin reaction,an aromatic aldehyde reacts with an acid anhydride (containing at least two $\alpha$-hydrogen atoms) in the presence of the corresponding sodium salt of the acid to form an $\alpha,\beta$-unsaturated carboxylic acid.
Here,$p$-methoxybenzaldehyde reacts with acetic anhydride $(CH_3CO)_2O$ in the presence of sodium acetate to yield $p$-methoxycinnamic acid.
Therefore,the compound $X$ is acetic anhydride $(CH_3CO)_2O$.

(C) Acetaldehyde reacts with $HCN$ to form a cyanohydrin,which upon hydrolysis yields lactic acid.
$CH_3CHO + HCN \to CH_3CH(OH)CN$
$CH_3CH(OH)CN + 2H_2O \xrightarrow{H^+} CH_3CH(OH)COOH + NH_3$
Here,$CH_3CH(OH)COOH$ is lactic acid.

(C) Mesityl oxide is $(CH_3)_2C=CH-CO-CH_3$.
It contains a methyl ketone group $(-COCH_3)$,which undergoes the iodoform reaction with $I_2$ and $NaOH$.
The reaction is: $(CH_3)_2C=CH-CO-CH_3 + 3I_2 + 4NaOH \rightarrow (CH_3)_2C=CH-COONa + CHI_3 + 3NaI + 3H_2O$.
The organic acid salt formed is sodium salt of $3-$methylbut$-2-$enoic acid.
Therefore,the corresponding acid is $(CH_3)_2C=CH-COOH$.

(A) The given reaction is a haloform reaction followed by decarboxylation.
Step $1$: Treatment with $I_2 + NaOH$ (iodoform test) converts the acetyl group $(-COCH_3)$ into a carboxylate group $(-COO^-)$ and produces $CHI_3$ (iodoform).
Step $2$: Acidification $(H^+)$ gives the corresponding $\beta$-keto acid.
Step $3$: Heating the $\beta$-keto acid causes decarboxylation (loss of $CO_2$) to yield cyclohexanone as the final product.

(B) The oxidation of $Butan-2-one$ $(CH_3CH_2COCH_3)$ to $Propionic \ acid$ $(CH_3CH_2COOH)$ is achieved via the $Haloform$ reaction.
$Butan-2-one$ contains a $CH_3CO-$ group,which reacts with $NaOH$ and $I_2$ (iodine) to form $Propionic \ acid$ and $Iodoform$ $(CHI_3)$.
Reaction: $CH_3CH_2COCH_3 + 3I_2 + 4NaOH \rightarrow CH_3CH_2COONa + CHI_3 + 3NaI + 3H_2O$.
Acidification of the resulting salt yields $Propionic \ acid$.

(D) The reaction of an aldehyde in the presence of aluminum ethoxide $(Al(OC_2H_5)_3)$ to form an ester is known as the Tishchenko reaction.
The general reaction is: $2RCHO \xrightarrow{Al(OC_2H_5)_3} RCOOCH_2R$.

(A) The reaction of a carbonyl compound $(R-CO-R')$ with $HCN/KCN$ is a nucleophilic addition reaction that forms a cyanohydrin.
Thus,$X$ is the cyanohydrin: $RR'C(OH)(CN)$.
Next,the reduction of the cyano group $(-CN)$ to a primary amine $(-CH_2NH_2)$ is achieved using a strong reducing agent like $LiAlH_4$.
Therefore,$Y$ is $LiAlH_4$.

(C) The conversion of an oxime into an amide in the presence of a strong acid (like $Conc. H_2SO_4$) is known as the $Beckmann$ rearrangement.
The reaction is represented as:
$R-C(=NOH)-R' \xrightarrow{H^+} R'-CO-NHR$ (or $R-CO-NHR'$ depending on the configuration).

(A) The reactivity of carbonyl compounds towards nucleophilic attack depends on the electrophilicity of the carbonyl carbon and the leaving group ability.
Acid chlorides $(RCOCl)$ are the most reactive because the $Cl^-$ ion is an excellent leaving group and the electron-withdrawing effect of the chlorine atom increases the electrophilicity of the carbonyl carbon.
Therefore,$MeCOCl$ is the most reactive among the given options.

(C) The correct answer is $(C)$.
Acetone $(CH_3COCH_3)$ shows keto-enol tautomerism because it contains $\alpha$-hydrogen atoms adjacent to the carbonyl group.
The keto form exists in equilibrium with its enol form:
$CH_3COCH_3 \rightleftharpoons CH_3-C(OH)=CH_2$.
$HCHO$ and $HCOOH$ do not have $\alpha$-carbons,and while $CH_3CN$ can show nitrile-ketenimine tautomerism,$CH_3COCH_3$ is the standard example for keto-enol tautomerism.

(D) Butanal $(CH_3CH_2CH_2CHO)$ contains $\alpha$-hydrogen atoms.
In the presence of dilute $NaOH$,it undergoes aldol condensation.
The $\alpha$-carbon of one butanal molecule attacks the carbonyl carbon of another butanal molecule to form a $\beta$-hydroxy aldehyde.
The product formed is $2$-ethyl-$3$-hydroxyhexanal,which is $CH_3-CH_2-CH_2-CH(OH)-CH(C_2H_5)-CHO$.

(A) Aldols ($\beta$-hydroxy aldehydes or $\beta$-hydroxy ketones) readily undergo dehydration in acidic or basic conditions to form $\alpha,\beta$-unsaturated carbonyl compounds.
This is because the resulting double bond is conjugated with the carbonyl group,which provides significant stability to the product.
Among the given options,$4$-hydroxy-$2$-pentanone is a $\beta$-hydroxy ketone.
Upon protonation of the hydroxyl group followed by the loss of water,it forms a stable conjugated system.
Other options are either simple alcohols or $\gamma$-hydroxy ketones,which do not form conjugated systems as readily as $\beta$-hydroxy carbonyl compounds.

(B) $p$-cresol reacts with $CHCl_3$ in an alkaline medium (Reimer-Tiemann reaction) to form $2$-hydroxy-$5$-methylbenzaldehyde (compound $A$).
Compound $A$ reacts with $HCN$ to form a cyanohydrin,$2$-hydroxy-$5$-methylmandelonitrile (compound $B$).
Acidic hydrolysis of the cyanohydrin $(B)$ yields $2$-hydroxy-$5$-methylmandelic acid,which is a chiral carboxylic acid due to the presence of a chiral carbon atom at the $\alpha$-position.
The structure is $2$-hydroxy-$5$-methylmandelic acid,which corresponds to the structure shown in option $B$.

(A) The conversion of benzophenone to benzene is achieved by fusion with a strong alkali like $KOH$. This reaction is a type of deacylation or cleavage reaction.
Step $1$: $C_6H_5COC_6H_5 + KOH \xrightarrow{\text{Fusion}} C_6H_6 + C_6H_5COOK$
Step $2$: The resulting potassium benzoate can further react with $KOH$ upon heating to produce benzene: $C_6H_5COOK + KOH \xrightarrow{\Delta} K_2CO_3 + C_6H_6$.
Thus,the correct reagent is fused alkali.

(D) Acetophenone $(C_6H_5COCH_3)$ contains a methyl ketone group $(-COCH_3)$,which gives a positive iodoform test with $I_2$ and $Na_2CO_3$ (or $NaOH$),resulting in a yellow precipitate of iodoform $(CHI_3)$.
Benzophenone $(C_6H_5COC_6H_5)$ does not contain a methyl ketone group and therefore does not give the iodoform test.
Thus,$I_2$ and $Na_2CO_3$ can be used to distinguish between them.

(C) Acetaldehyde $(CH_3CHO)$ is an aliphatic aldehyde that reduces Fehling solution to a red precipitate of cuprous oxide $(Cu_2O)$.
The chemical reaction is:
$CH_3CHO + 2Cu^{2+} + 5OH^- \to CH_3COO^- + Cu_2O \text{ (Red ppt.)} + 3H_2O$

(A) . The reactivity of carbonyl compounds towards nucleophilic addition is governed by two factors: steric hindrance and electronic effects.
$1$. Steric hindrance: As the number and size of alkyl or aryl groups attached to the carbonyl carbon increase,the approach of the nucleophile becomes more difficult.
$2$. Electronic effects: Alkyl groups are electron-donating ($+I$ effect),which decreases the electrophilicity of the carbonyl carbon.
Formaldehyde $(H_2C = O)$ has the least steric hindrance and no electron-donating groups,making it the most reactive.
Aldehydes $(RCHO)$ are more reactive than ketones $(R_2C = O)$ due to less steric hindrance.
Aryl groups $(Ar)$ provide resonance stabilization to the carbonyl carbon,further reducing its electrophilicity compared to alkyl groups.
Thus,the correct order is $H_2C = O > RCHO > ArCHO > R_2C = O > Ar_2C = O$.

(B) The reaction of an aldehyde with concentrated $NaOH$ to produce an alcohol and a salt of a carboxylic acid is known as the Cannizzaro reaction.
This reaction occurs in aldehydes that do not possess any $\alpha$-hydrogen atoms.
Among the given options,$C_6H_5CHO$ (Benzaldehyde) lacks $\alpha$-hydrogen atoms.
Therefore,it undergoes the Cannizzaro reaction:
$2C_6H_5CHO \xrightarrow{conc. NaOH} C_6H_5CH_2OH + C_6H_5COONa$

(B) The correct answer is $(B)$.
Cannizzaro's reaction is given by aldehydes that do not contain any $\alpha$-hydrogen atom.
$C_6H_5CHO$ (Benzaldehyde) does not have an $\alpha$-hydrogen atom,so it undergoes the Cannizzaro reaction.
Propionaldehyde $(CH_3CH_2CHO)$ and Acetaldehyde $(CH_3CHO)$ both contain $\alpha$-hydrogen atoms,while Bromobenzene is an aryl halide and does not undergo this reaction.

(C) $CH_3COCH_3$ (Acetone) contains $\alpha$-hydrogens.
In the presence of a base like $NaOH$,it undergoes an Aldol condensation reaction.
$H^{+}$ is typically used in the workup step to neutralize the product or facilitate dehydration.

(C) The reaction involves the catalytic hydrogenation of cyclohexanone using $H_2$ in the presence of a $Pt$ catalyst.
$Cyclohexanone + H_2 \xrightarrow{Pt} Cyclohexanol$
The carbonyl group $(C=O)$ is reduced to a hydroxyl group $(-OH)$,resulting in the formation of cyclohexanol.

(A) The reactivity of carbonyl compounds towards nucleophilic addition depends on the steric hindrance and electronic effects around the carbonyl carbon.
$HCHO$ (formaldehyde) has two hydrogen atoms attached to the carbonyl carbon,which provides minimum steric hindrance.
Alkyl groups (like $-CH_3$ or $-C_2H_5$) are electron-donating due to the $+I$ (inductive) effect,which decreases the electrophilicity of the carbonyl carbon.
Since $HCHO$ has no alkyl groups,it is the most electrophilic and thus the most reactive towards nucleophilic addition.
$CH_3COCH_3$ is a ketone,which is generally less reactive than aldehydes due to both steric and electronic factors.

(D) The Tollen's reagent is an ammoniacal silver nitrate solution.
The active species in Tollen's reagent is the diamminesilver$(I)$ complex ion,represented as $[Ag(NH_3)_2]^+$.
During the reaction with aldehydes,the $Ag^+$ ion in the complex is reduced to metallic silver $(Ag)$,which deposits on the inner walls of the test tube to form a silver mirror.
Therefore,the correct option is $(D)$.

(D) Acetone $(CH_3COCH_3)$ reacts with primary amines $(R-NH_2)$ to form imines $(>C=N-)$ and with hydrazines $(R-NHNH_2)$ to form hydrazones $(>C=N-NH-R)$.
$C_6H_5NH_2$ (aniline) is a primary amine,which reacts with acetone to form an imine $(CH_3C(CH_3)=NC_6H_5)$.
$C_6H_5NHNH_2$ (phenylhydrazine) is a hydrazine derivative,which reacts with acetone to form a hydrazone $(CH_3C(CH_3)=NNHC_6H_5)$.
Both products contain the $>C=N-$ linkage.
Therefore,the correct option is $D$.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$C_2H_5OH$ contains the $CH_3CH(OH)-$ group,$CH_3CHO$ contains the $CH_3CO-$ group,and $CH_3COCH_3$ contains the $CH_3CO-$ group.
$HCHO$ (formaldehyde) does not contain either of these groups,so it does not give a yellow precipitate of iodoform $(CHI_3)$ with $I_2$ and $NaOH$.
Therefore,the correct option is $(D)$.

(D) The reaction of acetaldehyde $(CH_3CHO)$ with $HCN$ is a nucleophilic addition reaction.
Since the carbonyl carbon in $CH_3CHO$ is $sp^2$ hybridized and planar,the cyanide ion $(CN^-)$ can attack from either side of the plane with equal probability.
This leads to the formation of a racemic mixture of the cyanohydrin intermediate.
Subsequent hydrolysis of the cyanohydrin to the carboxylic acid ($CH_3CH(OH)COOH$,lactic acid) preserves this racemic nature.
Therefore,the final product is a $50\% \ D$ and $50\% \ L$ mixture,which is a racemic mixture.

(B) Tollen's reagent is an ammoniacal silver nitrate solution. Its active oxidizing species is $Ag^+$. It oxidizes both aliphatic and aromatic aldehydes to their corresponding carboxylic acids,while $Ag^+$ is reduced to metallic silver $(Ag)$,forming a silver mirror.The reaction is: $R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag(s) + 4NH_3 + 2H_2O$.Among the given options,$Acetaldehyde$ $(CH_3CHO)$ is an aldehyde and therefore gives a positive Tollen's test.

(A) The iodoform test is given by compounds containing the $CH_3-C(=O)-$ group or the $CH_3-CH(OH)-$ group.
$A$. Pentan-$1$-one (which is actually $Pentanal$) does not contain the $CH_3-C(=O)-$ group,as the carbonyl group is at the terminal position.
$B$. Pentan-$2$-one contains the $CH_3-C(=O)-$ group and gives a positive iodoform test.
$C$. Propan-$2$-one (acetone) contains the $CH_3-C(=O)-$ group and gives a positive iodoform test.
$D$. Ethanol contains the $CH_3-CH(OH)-$ group and gives a positive iodoform test.
Therefore,the correct option is $A$.

(A) $NaHSO_3$ forms a crystalline addition product with acetaldehyde (an aliphatic aldehyde),whereas acetophenone (an aromatic ketone) does not react with $NaHSO_3$ due to steric hindrance.
$CH_3CHO + NaHSO_3 \to CH_3CH(OH)SO_3Na$ (White crystalline solid)
$C_6H_5COCH_3 + NaHSO_3 \to \text{No reaction}$
The addition product can be separated by filtration and then decomposed by dilute acid or base to recover the pure acetaldehyde.

(B) Ethyl acetate $(CH_3COOC_2H_5)$ reacts with two equivalents of Grignard reagent $(CH_3MgBr)$.
In the first step,the ester reacts with one equivalent of $CH_3MgBr$ to form a ketone (acetone,$CH_3COCH_3$).
In the second step,the ketone reacts with the second equivalent of $CH_3MgBr$ to form an alkoxide intermediate,which upon hydrolysis yields a tertiary alcohol (tert-butyl alcohol,$(CH_3)_3COH$).

(C) Methyl lithium $(CH_{3}Li)$ acts as a strong base.
It abstracts an acidic $\alpha$-hydrogen from the cyclopentanone molecule.
This process results in the formation of methane $(CH_{4})$ and a resonance-stabilized enolate intermediate,which is a cyclopentanonyl anion associated with the lithium cation.

(C) The reaction of ethyne $(HC \equiv CH)$ with $H_2SO_4$ in the presence of $Hg^{2+}$ is a hydration reaction (Kucherov reaction).
$HC \equiv CH + H_2O \xrightarrow[{Hg^{2+}}]{{H_2SO_4}} CH_3CHO$ (Acetaldehyde).
Thus,product $'P'$ is acetaldehyde $(CH_3CHO)$.
$1$. Acetaldehyde gives a positive Tollen's reagent test (silver mirror test).
$2$. Acetaldehyde gives a positive Brady's reagent test ($2$,$4$-$DNP$ test) as it is an aldehyde.
$3$. Acetaldehyde gives a positive Iodoform test because it contains the $CH_3CO-$ group.
$4$. The Victor Meyer test is used to distinguish between primary,secondary,and tertiary alcohols,not aldehydes. Therefore,acetaldehyde will not give the Victor Meyer test.

(C) The iodoform test is given by compounds containing either a methyl keto group $(CH_3-CO-)$ or a methyl hydroxy group $(CH_3-CH(OH)-)$ which can be oxidized to a methyl keto group.
$(i) CH_3-CH_2-OH$ (Ethanol) oxidizes to $CH_3-CHO$,which gives the iodoform test.
$(ii) CH_3-CO-CH_3$ (Acetone) contains the $CH_3-CO-$ group.
$(iii) CH_3-CH(OH)-CH_3$ (Propan$-2-$ol) oxidizes to $CH_3-CO-CH_3$.
$(iv) CH_3-OH$ (Methanol) does not contain the required group.
Therefore,$(i), (ii),$ and $(iii)$ will give a positive iodoform test.

(B) $X$ is $C_2H_6O$. Since it reacts with $Cu$ at $573 \ K$ to give $A$,and $A$ gives a silver mirror test,$X$ must be a primary alcohol and $A$ must be an aldehyde.
$X = CH_3CH_2OH$ (Ethanol).
$A = CH_3CHO$ (Ethanal).
$2CH_3CHO \xrightarrow{^-OH, \Delta} CH_3CH=CHCHO$ ($Y$,But$-2-$enal) via Aldol condensation.
$CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2$ ($Z$,Semicarbazone).
Thus,$A = \text{Ethanal}, X = \text{Ethanol}, Y = \text{But-2-enal}, Z = \text{Semicarbazone}$.

(A) Cyclohexanone undergoes self-aldol condensation in the presence of a base $(OH^-)$ to form a $\beta$-hydroxy ketone intermediate. Upon heating $(\Delta)$,this intermediate undergoes dehydration (elimination of $H_2O$) to form an $\alpha,\beta$-unsaturated ketone. The final product is $2$-cyclohexylidenecyclohexanone.

(B) Carbonyl compounds containing at least one $\alpha-$hydrogen atom exhibit tautomerism due to the acidic nature of the $\alpha-$hydrogen. The $\alpha-$hydrogen atom migrates to the carbonyl oxygen atom,resulting in an equilibrium between the keto form and the enol form. This phenomenon is specifically known as keto-enol tautomerism.

(A) The reaction of carbonyl compounds with ammonia derivatives (like hydrazine,$NH_2NH_2$) involves nucleophilic addition to the carbonyl group followed by the elimination of a water molecule $(H_2O)$ to form a product containing a $C=N$ bond (e.g.,hydrazone).
This is a characteristic nucleophilic addition-elimination reaction.