Properties Questions in English

(D) The $Wolff-Kishner$ reduction is used to reduce the carbonyl group $(C=O)$ of aldehydes or ketones to a methylene group $(CH_2)$.
It specifically reduces the ketone group to an alkyl group while leaving the carbon-carbon double bond $(C=C)$ intact.
Therefore,an unsaturated ketone is converted into an unsaturated hydrocarbon.

(D) Aldehydes that do not contain an $\alpha$-hydrogen atom undergo the Cannizzaro reaction in the presence of a base.
In the given options,$(CH_3)_3C-CHO$ (pivalaldehyde) has no $\alpha$-hydrogen atom attached to the carbonyl carbon,therefore it will undergo the Cannizzaro reaction.

(B) The addition of $NaHSO_3$ (sodium bisulphite) to carbonyl compounds is a nucleophilic addition reaction.
Steric hindrance plays a significant role in this reaction.
$HCHO$,acetaldehyde,and acetone are small enough to allow the nucleophilic attack of $HSO_3^-$.
Acetophenone is an aromatic ketone where the bulky phenyl group causes significant steric hindrance,preventing the formation of the bisulphite addition product.

(B) The Meerwein-Ponndorf-Verley $(MPV)$ reduction is a highly selective method for the reduction of aldehydes and ketones to their corresponding alcohols.
In this reaction,aluminum isopropoxide acts as a catalyst,and isopropanol serves as the hydride donor.
The reaction is chemoselective,meaning it reduces the carbonyl group $(C=O)$ while leaving other functional groups like carbon-carbon double bonds $(C=C)$ intact.
Therefore,when cyclohex$-2-$enone is subjected to $MPV$ reduction,the ketone group is reduced to an alcohol group,while the double bond remains unaffected.
The product formed is cyclohex$-2-$enol.

(C) The $Wolff-Kishner$ reduction is a chemical reaction used to convert carbonyl groups $(C=O)$ into methylene groups $(CH_2)$.
The general reaction is:
$R_2C=O + NH_2NH_2 \xrightarrow{KOH, \text{heat}} R_2CH_2 + N_2 + H_2O$
This reaction is specifically used for the reduction of aldehydes and ketones (carbonyl compounds).

(D) The Clemmensen reduction involves the reduction of a carbonyl group $(C=O)$ to a methylene group $(CH_2)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
Reaction: $R-CO-R + 4[H] \xrightarrow{Zn-Hg, HCl} R-CH_2-R + H_2O$

(A) The reaction of acetone $(CH_3)_2CO$ with $NaCN/HCl$ is a nucleophilic addition reaction forming a cyanohydrin.
$A = (CH_3)_2C(OH)CN$ (Acetone cyanohydrin).
Subsequent acid-catalyzed hydrolysis of the cyanohydrin $(A)$ converts the $-CN$ group into a $-COOH$ group.
$B = (CH_3)_2C(OH)COOH$ ($2$-hydroxy$-2-$methylpropanoic acid).
Therefore,the correct sequence is $A = (CH_3)_2C(OH)CN$ and $B = (CH_3)_2C(OH)COOH$.

(C) Tollens' reagent is used for the detection of the aldehyde group,not for detecting unsaturation. Therefore,the statement in option $C$ is incorrect.

(B) Paraldehyde is formed by the trimerization of acetaldehyde $(CH_3CHO)$ in the presence of an acid catalyst like concentrated $H_2SO_4$ at room temperature. The reaction involves three molecules of acetaldehyde combining to form a cyclic trimer,$2,4,6-trimethyl-1,3,5-trioxane$,commonly known as paraldehyde.

(C) $m$-Chlorobenzaldehyde does not have any $\alpha$-hydrogen atoms. Therefore,it undergoes the Cannizzaro reaction when treated with a concentrated base like $50\% KOH$. In this reaction,one molecule of the aldehyde is oxidized to a carboxylate ion ($m$-chlorobenzoate),and another molecule is reduced to an alcohol ($m$-chlorobenzyl alcohol). The correct product is shown in option $C$.

(A) Benzaldehyde $(C_6H_5CHO)$ does not contain $\alpha$-hydrogen atoms. When treated with concentrated $NaOH$,it undergoes the Cannizzaro reaction,which is a disproportionation reaction.
In this reaction,one molecule of benzaldehyde is reduced to benzyl alcohol $(C_6H_5CH_2OH)$ and another molecule is oxidized to sodium benzoate $(C_6H_5COONa)$.
The reaction is: $2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$.

(D) Carbonyl compounds react with hydroxylamine to form oximes,with hydrazine to form hydrazones,and with phenylhydrazine to form phenylhydrazones. All these reactions are characteristic tests used to detect the presence of the carbonyl group.

(B) The reduction of acetaldehyde $(CH_3CHO)$ with sodium $(Na)$ and alcohol $(C_2H_5OH)$ is a classic reduction reaction.
$CH_3CHO + 2[H] \xrightarrow{Na/C_2H_5OH} CH_3CH_2OH$
Therefore,the product formed is ethyl alcohol (ethanol).

(D) Acetaldehyde reacts with Tollen's reagent (ammoniacal silver nitrate) to form a silver mirror.
The chemical equation is:
$CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \xrightarrow{\Delta} CH_3COO^- + 2Ag \downarrow + 4NH_3 + 2H_2O$
Here,$2Ag$ represents the formation of a silver mirror.

(B) The reaction described is the $Clemmensen$ reduction,which reduces aldehydes and ketones to alkanes using amalgamated zinc $(Zn-Hg)$ and concentrated $HCl$.
The chemical equation is:
$CH_3CH_2CHO + 4[H] \xrightarrow{Zn-Hg, HCl} CH_3CH_2CH_3 + H_2O$
Thus,propionaldehyde $(CH_3CH_2CHO)$ is reduced to propane $(CH_3CH_2CH_3)$.

(D) The reaction described is the $Wolff-Kishner$ reduction.
In the first step,the carbonyl compound reacts with hydrazine $(H_2NNH_2)$ to form a hydrazone intermediate.
The reaction is: $R_2C=O + H_2NNH_2 \rightarrow R_2C=N-NH_2 + H_2O$.
Thus,the first intermediate formed is a hydrazone,$R_2C=N-NH_2$.

(A) Phenylhydrazine is a nucleophilic reagent that reacts with carbonyl compounds (aldehydes and ketones) to form phenylhydrazones. $Ethanol$ is an alcohol and does not contain a carbonyl group $(C=O)$,therefore it does not undergo this reaction.

(A) Aldehydes containing $\alpha$-hydrogen atoms undergo Aldol condensation in the presence of dilute alkali. However,some specific aldehydes,particularly those with a single $\alpha$-hydrogen atom (like isobutyraldehyde,$(CH_3)_2CHCHO$),can undergo both Aldol condensation and Cannizzaro reaction under specific conditions (e.g.,concentrated alkali at high temperatures).
$1$. Isobutyraldehyde $((CH_3)_2CHCHO)$ has one $\alpha$-hydrogen atom,allowing it to form an aldol product.
$2$. Due to the steric hindrance and the specific nature of the $\alpha$-carbon,it can also undergo a cross-Cannizzaro-like reaction or disproportionation under harsh conditions.
$3$. $HCHO$ and $C_6H_5CHO$ lack $\alpha$-hydrogens and only undergo Cannizzaro reaction.
$4$. $CH_3CH_2CHO$ has two $\alpha$-hydrogens and primarily undergoes Aldol condensation.

(D) The reduction of an aldehyde to an alcohol can be achieved using any of the given transition metal catalysts ($Ni$,$Pd$,or $Pt$) in the presence of hydrogen gas $(H_2)$.

(A) The reaction described is the Aldol condensation of acetaldehyde $(CH_3CHO)$.
Step $1$: Two molecules of acetaldehyde undergo aldol condensation in the presence of dilute $NaOH$ to form $\beta$-hydroxybutanal.
$CH_3CHO + CH_3CHO \xrightarrow{dil. NaOH} CH_3-CH(OH)-CH_2-CHO$ ($\beta$-hydroxybutanal).
Step $2$: Upon heating with dilute acid,$\beta$-hydroxybutanal undergoes dehydration to form crotonaldehyde $(CH_3CH=CH-CHO)$.
$CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, H^+} CH_3CH=CH-CHO + H_2O$ (Crotonaldehyde).
Therefore,the starting compound is $CH_3CHO$.

(C) Benzaldehyde $(C_6H_5CHO)$ reacts with methylamine $(CH_3NH_2)$ to form a Schiff base (imine) through a nucleophilic addition-elimination reaction.
The reaction is: $C_6H_5CHO + CH_3NH_2 \rightarrow C_6H_5CH=NCH_3 + H_2O$.
The product formed is $N$-benzylidenemethylamine $(C_6H_5CH=NCH_3)$.

(B) The reaction described is the $Perkin$ reaction. In this reaction,an aromatic aldehyde (e.g.,$Benzaldehyde$) reacts with an acid anhydride (e.g.,$Acetic$ $anhydride$) in the presence of the sodium salt of the corresponding acid (e.g.,$CH_3COONa$) to yield an $\alpha,\beta$-unsaturated aromatic acid (e.g.,$Cinnamic$ $acid$ after hydrolysis). The general reaction is: $ArCHO + (RCH_2CO)_2O \xrightarrow{RCH_2COONa} ArCH=C(R)COOH + RCH_2COOH$.

(C) Propanal $(CH_3CH_2CHO)$ contains $\alpha$-hydrogens and undergoes an aldol condensation reaction in the presence of dilute $NaOH$.
Two molecules of propanal react as follows:
$CH_3CH_2CHO + CH_3CH_2CHO \xrightarrow{dil. NaOH} CH_3CH_2CH(OH)CH(CH_3)CHO$
This product is $3-hydroxy-2-methylpentanal$.

(C) The given reaction is the catalytic hydrogenation of cyclohexanone using $H_2/Pt$.
Catalytic hydrogenation of a ketone reduces the carbonyl group $(C=O)$ to a secondary alcohol group $(-CHOH)$.
Thus,cyclohexanone is reduced to cyclohexanol.

(A) Acetaldehyde is an aliphatic aldehyde that reacts with $NaHSO_3$ to form a crystalline bisulfite addition product.
Acetophenone,being a bulky aromatic ketone,does not form this addition product due to steric hindrance.
Therefore,$NaHSO_3$ can be used to distinguish between them.

(C) The reaction where two molecules of aromatic aldehydes (like benzaldehyde) condense in the presence of cyanide ions $(CN^-)$ to form an $\alpha$-hydroxy ketone (acyloin),specifically benzoin,is known as the Benzoin condensation. The reaction is represented as: $2C_6H_5CHO \xrightarrow{KCN} C_6H_5CH(OH)COC_6H_5$.

(A) The reaction shown is an aldol condensation of cyclohexanone in the presence of a base $(NaOH)$ and heat $(\Delta)$.
$1$. Two molecules of cyclohexanone undergo aldol addition to form $\beta$-hydroxy ketone.
$2$. Upon heating,the $\beta$-hydroxy ketone undergoes dehydration (elimination of $H_2O$) to form an $\alpha,\beta$-unsaturated ketone.
$3$. The final product $A$ is $2-$cyclohexylidenecyclohexanone.

(A) The reaction is a Cannizzaro reaction involving formaldehyde-$d_2$ $(CD_2=O)$.
In the presence of a strong base $(OH^-)$,aldehydes without $\alpha$-hydrogens undergo disproportionation.
One molecule of $CD_2=O$ is oxidized to the formate ion $(D-COO^-)$ and the other is reduced to the corresponding alcohol $(CD_3OH)$.
Thus,$X = D-COO^-$ and $Y = CD_3OH$.

(A) The reaction sequence is as follows:
$1$. The starting material is $2$-acetyl-$2$-methylcyclohexanone. Treatment with $KBrO$ and $\Delta$ (haloform reaction) converts the acetyl group $(-COCH_3)$ into a carboxylate group $(-COOK)$ while releasing $CHBr_3$.
$2$. Acidification with $H^+$ converts the carboxylate salt into a $\beta$-keto acid,$1$-methyl-$2$-oxocyclohexanecarboxylic acid.
$3$. Heating $(\Delta)$ the $\beta$-keto acid causes decarboxylation,releasing $CO_2$ and forming $2$-methylcyclohexanone as the final product $A$.

(D) Tollens' reagent is a mild oxidizing agent that selectively oxidizes the aldehyde group $(-CHO)$ to a carboxylic acid group $(-COOH)$ without affecting the carbon-carbon double bond $(C=C)$.
$CH_3CH=CHCHO \xrightarrow{[Ag(NH_3)_2]^+} CH_3CH=CHCOOH$
Strong oxidizing agents like acidic $KMnO_4$ or acidic $K_2Cr_2O_7$ would oxidize both the aldehyde group and the double bond,leading to cleavage of the molecule.

(C) The given reactant is a tetra-aldehyde compound,specifically $2,2',6,6'$-tetrabenzenedicarbaldehyde. When treated with concentrated $NaOH$ at $100^{\circ}C$,it undergoes an intramolecular Cannizzaro reaction. In the Cannizzaro reaction,aldehydes lacking $\alpha$-hydrogen atoms undergo self-oxidation and reduction in the presence of a strong base. Here,the aldehyde groups are reduced to primary alcohol groups $(-CH_2OH)$ and oxidized to carboxylate groups $(-COO^-)$,which upon acidification with $H^+/H_2O$ yield carboxylic acid groups $(-COOH)$. The final product is a compound containing both alcohol and carboxylic acid functional groups,as shown in option $C$.

(C) mixed ketone is a ketone where the two groups attached to the carbonyl carbon are different.
$1$. Pentanone (e.g.,$CH_3CH_2COCH_2CH_3$ or $CH_3COCH_2CH_2CH_3$) can be simple or mixed.
$2$. Benzophenone $(C_6H_5COC_6H_5)$ is a simple (symmetrical) ketone.
$3$. Acetophenone $(C_6H_5COCH_3)$ has a phenyl group and a methyl group attached to the carbonyl carbon,making it a mixed (unsymmetrical) ketone.

(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 $NaOH$),producing 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.

(B) Acetaldehyde $(CH_3CHO)$ undergoes self-condensation in the presence of dilute alkali $(NaOH)$ to form $3$-hydroxybutanal,which is commonly known as aldol.
The reaction is: $2CH_3CHO \xrightarrow{dil. NaOH} CH_3-CH(OH)-CH_2-CHO$ (Aldol).

(C) The reduction of aldehydes and ketones using $NaBH_4$ or $LiAlH_4$ yields primary or secondary alcohols.
$1.$ Aldehydes $(RCHO)$ are reduced to primary alcohols $(RCH_2OH)$.
$2.$ Ketones $(RCOR')$ are reduced to secondary alcohols $(RCH(OH)R')$.
$3.$ Tertiary alcohols cannot be obtained by the reduction of aldehydes or ketones because they require a carbon atom bonded to three other carbon atoms,which is not possible through simple reduction of the carbonyl group.
$4.$ $2$-Methylpropan-$2$-ol is a tertiary alcohol,hence it cannot be obtained by this method.

(D) Fehling's solution is a mild oxidizing agent used to distinguish between aldehydes and ketones.
Aldehydes (except benzaldehyde) and reducing sugars like glucose give a brick-red precipitate of $Cu_2O$ with Fehling's solution.
Acetone is a ketone and does not undergo oxidation under these mild conditions; therefore,it does not give a brick-red precipitate.

(A) Aldehydes react with two equivalents of monohydric alcohol in the presence of dry $HCl$ gas to form gem-dialkoxy compounds known as acetals.
$CH_3CHO + R-OH \xrightarrow{dry \ HCl} CH_3CH(OH)(OR)$ (Hemiacetal)
$CH_3CH(OH)(OR) + R-OH \xrightarrow{dry \ HCl} CH_3CH(OR)_2 + H_2O$ (Acetal)
Thus,acetaldehyde reacts with alcohol to form an acetal.

(D) Ammonia $(NH_3)$ reacts differently with formaldehyde,acetaldehyde,and acetone.
$1$. Formaldehyde reacts with $NH_3$ to form hexamethylenetetramine (urotropine):
$6HCHO + 4NH_3 \to (CH_2)_6N_4 + 6H_2O$
$2$. Acetaldehyde reacts with $NH_3$ to form acetaldehyde ammonia (an addition product):
$CH_3CHO + NH_3 \to CH_3CH(OH)NH_2$
$3$. Acetone reacts with $NH_3$ to form diacetone amine:
$2CH_3COCH_3 + NH_3 \to (CH_3)_2C(NH_2)CH_2COCH_3 + H_2O$
Other reagents like $HCN$,$NH_2NH_2$,and $NH_2OH$ react with all three carbonyl compounds to form similar types of products (cyanohydrins,hydrazones,and oximes,respectively).

(C) The reaction of a Grignard reagent $(CH_3MgI)$ with acetone $(CH_3COCH_3)$ followed by hydrolysis yields tert-butyl alcohol $(2-methylpropan-2-ol)$.
$CH_3COCH_3 + CH_3MgI \xrightarrow{dry \, ether} (CH_3)_3C-OMgI$
$(CH_3)_3C-OMgI + H_2O \xrightarrow{H^+} (CH_3)_3C-OH + Mg(OH)I$
Thus,acetone is the correct reactant.

(A) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on the electrophilicity of the carbonyl carbon.
Presence of electron-withdrawing groups (like $-Cl$) increases the positive charge on the carbonyl carbon,making it more susceptible to nucleophilic attack.
In $CCl_3CHO$,the three chlorine atoms exert a strong $-I$ effect,making the carbonyl carbon highly electrophilic.
Thus,$CCl_3CHO$ is the most reactive among the given options.

(C) Carbonyl compounds undergo nucleophilic addition reactions.
The reaction is:
$ >C=O + HCN \to >C(OH)(CN) $
This product is known as a cyanohydrin.

(B) Ethanal $(CH_3CHO)$ is an aldehyde,so it gives the Fehling's test.
It also contains the $CH_3-C(=O)-$ group,which allows it to give the iodoform test.

(A) Benzophenone $(C_6H_5COC_6H_5)$ can be converted into benzene $(C_6H_6)$ by fusion with molten potassium hydroxide $(KOH)$.
This reaction is a type of cleavage reaction where the carbon-carbon bond between the carbonyl group and one of the phenyl rings is broken.
The reaction proceeds as follows:
$C_6H_5COC_6H_5 + KOH \xrightarrow{\text{fusion}} C_6H_6 + C_6H_5COOK$
Here,benzophenone reacts with fused $KOH$ to yield benzene and potassium benzoate.

(C) Acetophenone $(C_6H_5COCH_3)$ undergoes standard electrophilic aromatic substitution reactions because the acetyl group is deactivating and meta-directing.
Therefore,statement $III$ is incorrect.
Acetophenone contains a methyl ketone group $(-COCH_3)$,so it gives a positive iodoform test with iodine and alkali.
Therefore,statement $IV$ is incorrect.
Statement $I$ is correct: Reduction of acetophenone with $Na/C_2H_5OH$ yields $1$-phenylethanol (methyl phenyl carbinol).
Statement $II$ is correct: Oxidation of acetophenone with acidic $KMnO_4$ yields benzoic acid.
Thus,statements $III$ and $IV$ are incorrect.

(C) When benzaldehyde $(C_6H_5CHO)$ reacts with chlorine $(Cl_2)$ in the absence of a catalyst (like $FeCl_3$ or $AlCl_3$),the reaction is a substitution of the aldehydic hydrogen atom.
This reaction produces benzoyl chloride $(C_6H_5COCl)$ and hydrogen chloride $(HCl)$ as a byproduct.
The chemical equation is: $C_6H_5CHO + Cl_2 \rightarrow C_6H_5COCl + HCl$.

(C) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on two factors: steric hindrance and electronic effects.
Aldehydes are generally more reactive than ketones because they have less steric hindrance and less electron-donating alkyl groups attached to the carbonyl carbon.
In $HCHO$,there are two hydrogen atoms attached to the carbonyl carbon,providing the least steric hindrance.
In $CH_3CHO$,one methyl group is present,which increases steric hindrance and provides a $+I$ effect.
In $CH_3COCH_3$,two methyl groups are present,providing the maximum steric hindrance and the strongest $+I$ effect,which decreases the electrophilicity of the carbonyl carbon.
Therefore,the order of reactivity is $I > II > III$.

(C) The reduction of a carbonyl group $( > C = O)$ to a methylene group $( > CH_2)$ is known as the Wolff-Kishner reduction.
This reaction uses hydrazine $(NH_2NH_2)$ in the presence of a strong base like $KOH$ (potassium hydroxide) at high temperatures.
The chemical equation is: $ > C = O + NH_2NH_2 \xrightarrow{KOH} > CH_2 + N_2 + H_2O$.

(D) The reaction described is the Cannizzaro reaction. Aldehydes that do not contain an $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation) when treated with a concentrated alkali solution like $50\% \ NaOH$. Benzaldehyde $(C_6H_5CHO)$ lacks $\alpha$-hydrogen atoms and thus undergoes the Cannizzaro reaction to produce benzyl alcohol and sodium benzoate. The reaction is: $2C_6H_5CHO + NaOH (50\%) \rightarrow C_6H_5CH_2OH + C_6H_5COONa$.

(B) The reaction of a ketone (like cyclohexanone) with a secondary amine (like dimethylamine) in the presence of an acid catalyst leads to the formation of an enamine.
During this process,the initial addition product loses a molecule of water $(H_2O)$ to form a carbon-carbon double bond adjacent to the nitrogen atom.
The reaction is: $Cyclohexanone + HN(CH_3)_2 \xrightarrow{H^+} Enamine + H_2O$.
Thus,the product formed is an enamine.