Properties Questions in English

(D) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetaldehyde $(CH_3CHO)$ in the presence of a dilute base is a cross-aldol condensation reaction.
Acetaldehyde acts as the nucleophile because it has $\alpha$-hydrogens,forming an enolate ion.
This enolate attacks the carbonyl carbon of benzaldehyde to form a $\beta$-hydroxy aldehyde $(C_6H_5CH(OH)CH_2CHO)$.
Upon heating or dehydration,this intermediate loses a water molecule to form an $\alpha,\beta$-unsaturated aldehyde,which is cinnamaldehyde $(C_6H_5CH=CHCHO)$.

(D) $Grignard$ reagent $(RMgX)$ reacts with aldehydes (except formaldehyde) to form secondary alcohols and with ketones to form tertiary alcohols.
$R^{-}MgX + CH_3CHO$ $\rightarrow CH_3-CH(OMgX)-R$ $\xrightarrow{H_2O} CH_3-CH(OH)-R$ (Secondary alcohol).
$R^{-}MgX + CH_3COCH_3$ $\rightarrow CH_3-C(OMgX)(CH_3)-R$ $\xrightarrow{H_2O} CH_3-C(OH)(CH_3)-R$ (Tertiary alcohol).

(C) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the aldehyde or ketone molecule.
$1$. Benzaldehyde $(C_6H_5CHO)$ has no $\alpha$-hydrogen.
$2$. $2, 2-$dimethylpropionaldehyde $((CH_3)_3CCHO)$ has no $\alpha$-hydrogen.
$3$. Acetaldehyde $(CH_3CHO)$ has three $\alpha$-hydrogen atoms.
$4$. Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen.
Therefore,only acetaldehyde can undergo base-catalyzed aldol condensation.

(C) Fehling solution is a mixture of two solutions:
$1$. Fehling solution $A$: Aqueous copper sulphate $(CuSO_4 \cdot 5H_2O)$.
$2$. Fehling solution $B$: Alkaline sodium potassium tartrate (Rochelle salt,$KNaC_4H_4O_6 \cdot 4H_2O$ in $NaOH$).
Therefore,the correct composition is copper sulphate,sodium hydroxide,and Rochelle salt.

(A) The reduction of an aldehyde $(R-CHO)$ using reducing agents like $NaBH_4$ or $LiAlH_4$ results in the formation of a primary alcohol $(R-CH_2OH)$.
$R-CHO + 2[H] \xrightarrow{\text{Reduction}} R-CH_2OH$ (Primary alcohol).

(A) Aldehydes that do not contain $\alpha$-hydrogen atoms undergo the Cannizzaro reaction when treated with concentrated $NaOH$.
Methanal $(HCHO)$ lacks $\alpha$-hydrogen atoms.
The reaction is: $HCHO + HCHO \xrightarrow{Conc. NaOH} CH_3OH + HCOONa$.
Here,$CH_3OH$ (methanol) is an alcohol.

(A) Schiff's reagent is an aqueous solution of magenta or pink coloured Rosaniline Hydrochloride which gets decolourised when $SO_{2}$ is passed through it.
It is used as a diagnostic test for the presence of aldehydes.

(C) : The pyrolysis of acetone at high temperatures leads to the formation of ketene and methane.
$CH_3COCH_3 \xrightarrow{\Delta} CH_2=C=O + CH_4$
The product $CH_2=C=O$ is known as ketene.

(A) . $CH_3-CO-CH_3$ (acetone) is a ketone.
Ketones,upon vigorous oxidation,undergo $C-C$ bond cleavage,resulting in carboxylic acids with a fewer number of carbon atoms than the parent ketone.
In contrast,primary alcohols $(CH_3-CH_2-CH_2-OH)$ and aldehydes ($CH_3-CH_2-CHO$ and $CCl_3-CH_2-CHO$) yield carboxylic acids with the same number of carbon atoms upon oxidation.

(A) Acetal is obtained by the reaction of an aldehyde with two equivalents of an alcohol in the presence of dry $HCl$ gas.
Dry $HCl$ acts as an acid catalyst to protonate the carbonyl oxygen,making the carbonyl carbon more electrophilic.
For example,acetaldehyde reacts with $2$ molecules of ethanol to form $1,1-$diethoxyethane (an acetal).
The reaction is: $CH_3CHO + 2C_2H_5OH \xrightarrow{dry \ HCl} CH_3CH(OC_2H_5)_2 + H_2O$.

(B) Grignard reagents $(RMgX)$ react with both aldehydes and ketones to form addition products (adducts),which upon hydrolysis yield alcohols.
For aldehydes: $RCHO + R'MgX \to RCH(OMgX)R' \xrightarrow{H_2O} RCH(OH)R'$
For ketones: $RCOR' + R''MgX \to RC(OMgX)(R')R'' \xrightarrow{H_2O} RC(OH)(R')R''$
Fehling's reagent,Schiff's reagent,and Tollen's reagent are specific tests used to distinguish aldehydes from ketones,as they react with aldehydes but not with ketones.

(B) The iodoform reaction is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
Among the given options,$CH_3CHO$ (acetaldehyde) contains the $CH_3CO-$ group.
Therefore,$CH_3CHO$ undergoes the iodoform reaction to produce a yellow precipitate of iodoform $(CHI_3)$.
Reaction: $CH_3CHO + 3I_2 + 4NaOH \rightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$.

(B) Fehling solution is a mild oxidizing agent that reacts with aliphatic aldehydes but fails to oxidize aromatic aldehydes like $Benzaldehyde$ due to the resonance stabilization of the benzene ring. Therefore,$Benzaldehyde$ does not give a positive test with Fehling solution.

(D) Benzaldehyde undergoes benzoin condensation in the presence of ethanolic $KCN$ to form benzoin.
$2C_6H_5CHO \xrightarrow{Alc. KCN} C_6H_5CH(OH)COC_6H_5$ (Benzoin).
Therefore,the correct option is $(D)$.

(A) $Aldehydes$ react with $Schiff's \ reagent$ (which is colourless) to produce a characteristic pink colour.
$Ketones$ do not give this test.

(A) When acetaldehyde $(CH_3CHO)$ reacts with excess chlorine $(Cl_2)$,all three hydrogen atoms of the methyl group are replaced by chlorine atoms to form chloral $(CCl_3CHO)$.
The chemical reaction is: $CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO + 3HCl$.

(B) The reaction of an aldehyde with no $\alpha$-hydrogen with concentrated alkali $(50\% \, NaOH)$ is known as the Cannizzaro reaction.
In this reaction,one molecule of the aldehyde is oxidized to the corresponding carboxylic acid (as a salt),and another molecule is reduced to the corresponding alcohol.
Benzaldehyde $(C_6H_5CHO)$ does not contain any $\alpha$-hydrogen atom.
Therefore,it undergoes the Cannizzaro reaction to produce benzyl alcohol $(C_6H_5CH_2OH)$ and sodium benzoate $(C_6H_5COONa)$.
The reaction is:
$2C_6H_5CHO + NaOH (50\%) \rightarrow C_6H_5CH_2OH + C_6H_5COONa$
Thus,the correct option is $(B)$.

(D) Butan-$2$-one is a ketone. It undergoes Clemmensen reduction when treated with zinc amalgam and hydrochloric acid $(Zn-Hg/HCl)$ to give the corresponding hydrocarbon,butane.
$CH_3-CO-CH_2-CH_3 \xrightarrow{Zn-Hg/HCl} CH_3-CH_2-CH_2-CH_3$
Amides,carboxylic acids,and esters are not reduced to hydrocarbons by this reagent.

(B) When three molecules of acetone $(CH_3COCH_3)$ react in the presence of dry $HCl$ gas,they undergo an aldol condensation followed by dehydration to form Phorone.
The reaction is as follows:
$3CH_3COCH_3 \xrightarrow{Dry \ HCl} (CH_3)_2C=CH-CO-CH=C(CH_3)_2 + 2H_2O$
Thus,the correct product is Phorone.

(D) The reduction of aldehydes and ketones to their corresponding hydrocarbons using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$ is known as the Clemmensen reduction.
The general reaction is:
$R_2C=O + 4[H] \xrightarrow{Zn-Hg/HCl} R_2CH_2 + H_2O$
Thus,option $(D)$ is the correct answer.

(B) The reaction of acetone with iodine in the presence of a base like $NaOH$ is known as the iodoform test.
This reaction is specific to compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group.
The balanced chemical equation is:
$CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$
Thus,the correct reagent is $NaOH$.

(B) Acetaldehyde $(CH_3-CHO)$ reacts with $HCN$ to form acetaldehyde cyanohydrin $(CH_3-CH(OH)CN)$.
Upon acid hydrolysis,the cyano group $(-CN)$ is converted into a carboxylic acid group $(-COOH)$,yielding lactic acid $(CH_3-CH(OH)COOH)$.
$CH_3-CHO + HCN$ $\rightarrow CH_3-CH(OH)CN$ $\xrightarrow{H_3O^{+}} CH_3-CH(OH)COOH$

(B) Tollen's reagent $(AgNO_3 + NH_4OH)$ is used to detect aldehydes.
Aldehydes reduce Tollen's reagent to metallic silver,forming a silver mirror on the inner wall of the test tube.

(C) Only aliphatic aldehydes reduce Fehling solution.
Benzaldehyde and salicylaldehyde are aromatic aldehydes and do not respond to this test.
Acetaldehyde $(CH_3CHO)$ is an aliphatic aldehyde and gives a red precipitate of cuprous oxide $(Cu_2O)$ with Fehling solution.

(A) The product $(CH_3)_2C=CH-CO-CH_3$ (mesityl oxide) is formed by the self-aldol condensation of acetone $(CH_3-CO-CH_3)$ followed by dehydration.
$2CH_3-CO-CH_3$ $\xrightarrow{OH^-} (CH_3)_2C(OH)-CH_2-CO-CH_3$ $\xrightarrow{\Delta, -H_2O} (CH_3)_2C=CH-CO-CH_3$.

(A) The reaction described is the Cannizzaro reaction,which occurs in aldehydes that do not possess an $\alpha$-hydrogen atom.
$HCHO$ (Formaldehyde) has no $\alpha$-hydrogen atom.
When $HCHO$ reacts with concentrated $NaOH$,it undergoes self-oxidation and reduction (disproportionation) to form methanol and sodium formate.
$2HCHO + NaOH \xrightarrow{} CH_3OH + HCOONa$
Therefore,the correct option is $A$.

(C) Acetaldehyde is an aldehyde,while Acetone is a ketone.
Fehling's solution test is used to distinguish between aliphatic aldehydes and ketones.
Acetaldehyde reduces Fehling's solution to form a red precipitate of $Cu_2O$,whereas acetone does not give this test.
Therefore,the correct option is $(C)$.

(C) $(C)$. The given reaction is an intramolecular Cannizzaro reaction.
In this reaction,the glyoxal molecule $(OCH-CHO)$ undergoes disproportionation in the presence of a base $(OH^-)$,where one aldehyde group is reduced to an alcohol group $(-CH_2OH)$ and the other is oxidized to a carboxylate group $(-COO^-)$.
The overall reaction is: $OCH-CHO + OH^- \xrightarrow{} HOCH_2-COO^-$.

(B) The susceptibility of nucleophilic attack on the carbonyl carbon of an aldehyde depends on the steric hindrance and electronic effects of the alkyl groups attached to the carbonyl carbon.
As the number of alkyl groups increases,the steric hindrance increases and the electron-donating effect ($+I$ effect) increases,which reduces the electrophilicity of the carbonyl carbon.
Therefore,the order of reactivity towards nucleophilic attack is $1^o > 2^o > 3^o$.

(A) . Wolff-Kishner reduction: Aldehydes and ketones are treated with hydrazine $(NH_2NH_2)$ followed by heating with a strong base like potassium hydroxide $(KOH)$ in a high-boiling solvent.
The reaction proceeds via the formation of a hydrazone intermediate,which then undergoes base-catalyzed decomposition to yield the corresponding alkane.
Reaction: $> C = O + NH_2NH_2$ $\rightarrow > C = NNH_2 + H_2O$ $\xrightarrow{KOH, \text{heat}} > CH_2 + N_2$.

(D) Aldehydes and most ketones react with sodium bisulphite $(NaHSO_3)$ to form a crystalline bisulphite addition product.
The general reaction is:
$R_2C=O + NaHSO_3 \rightarrow R_2C(OH)(SO_3Na)$
Since $CH_3COCH_3$ (acetone),$CH_3CHO$ (acetaldehyde),and $HCHO$ (formaldehyde) are all carbonyl compounds,they all undergo this addition reaction.
Therefore,the correct option is $(d)$.

(D) . Fehling's solution is a mixture of two solutions: Fehling's solution $A$ ($CuSO_4$ solution) and Fehling's solution $B$ (alkaline solution of sodium potassium tartrate,also known as Rochelle salt).
Since the given options do not include Rochelle salt,the correct answer is $D$.

(C) The Wolff-Kishner reduction is a chemical reaction used to convert carbonyl groups (aldehydes or ketones) into alkanes.
Specifically,it reduces the $-CHO$ (aldehyde) or $C=O$ (ketone) group into a $-CH_2-$ (methylene) group.
Therefore,it reduces the $-CHO$ group.

(A) The molecular weight of the compound is calculated as $2 \times \text{Vapour density} = 2 \times 29 = 58$.
The molecular weights of the given compounds are:
$CH_3CH_2CHO = 58$
$CH_3CHOHCH_3 = 60$
$CH_3COCH_3 = 58$
$CH_3CH_2COOH = 74$
Both $CH_3CH_2CHO$ and $CH_3COCH_3$ have a molecular weight of $58$.
The iodoform test (yellow precipitate with alkali and iodine) is given by compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group.
Among the options,$CH_3COCH_3$ (acetone) contains the $CH_3CO-$ group and gives a yellow precipitate of iodoform. $CH_3CH_2CHO$ does not give this test.

(C) The reactivity of carbonyl compounds towards nucleophilic addition reactions (such as with Grignard reagents like $PhMgBr$) depends on two main factors:
$1.$ Steric hindrance: As the size and number of groups attached to the carbonyl carbon increase,the approach of the nucleophile becomes more difficult,decreasing reactivity.
$2.$ Electronic effect: Electron-donating groups (like alkyl groups via $+I$ effect) or resonance stabilization (like phenyl groups) decrease the positive charge on the carbonyl carbon,making it less electrophilic.
Comparing the given compounds:
$(II)$ Acetaldehyde $(CH_3-CHO)$ has the least steric hindrance and only one $+I$ group.
$(III)$ Acetone $(CH_3-CO-CH_3)$ has two $+I$ groups and more hindrance than acetaldehyde.
$(I)$ Benzophenone $(Ph-CO-Ph)$ has two bulky phenyl groups and significant resonance,making it the least reactive.
Therefore,the correct order is $(II) > (III) > (I)$.

(D) Acetaldehyde contains a $CH_3CO-$ group,which allows it to undergo the iodoform test in the presence of $I_2$ and $NaOH$ to form a yellow precipitate of iodoform $(CHI_3)$.
Formaldehyde does not contain a $CH_3CO-$ group and therefore does not give a positive iodoform test.
Both acetaldehyde and formaldehyde give positive tests with Fehling's solution,Tollen's reagent,and Schiff's reagent,making them unsuitable for distinguishing between the two.

(C) The silver mirror test (Tollens' test) is used to distinguish between aldehydes and other carbonyl compounds or acids.
Aldehydes are oxidized to carboxylate ions by Tollens' reagent,which reduces $Ag^+$ ions to metallic silver,forming a silver mirror on the inner surface of the test tube.
Carboxylic acids do not react with Tollens' reagent.
Therefore,the correct option is $(C)$.

(B) Paraldehyde is formed by the trimerization of acetaldehyde $(CH_3CHO)$ in the presence of concentrated $H_2SO_4$ at room temperature.
$3CH_3CHO \xrightarrow{\text{conc. } H_2SO_4, \text{ room temp.}} \text{Paraldehyde}$
Since three molecules of acetaldehyde combine to form one molecule of paraldehyde,it is a trimer of acetaldehyde.
Therefore,the correct option is $B$.

(A) The iodoform test is used to detect the presence of the $CH_3CO-$ group in an organic compound.
$\therefore$ Methyl ketones are identified by the iodoform test.
They produce a yellow precipitate of iodoform $(CHI_3)$ upon reaction with $I_2$ and $NaOH$.

(A) Fehling solution test is given by aliphatic aldehydes.
Ketones do not give Fehling solution test.
Acetone is a ketone,whereas $Propanal$,$Ethanal$,and $Butanal$ are aldehydes.

(C) The reaction shown is the iodoform reaction.
Methyl ketones like butan-$2$-one $(CH_3-CH_2-CO-CH_3)$ react with $I_2$ and $NaOH$ to form iodoform $(CHI_3)$ and the sodium salt of a carboxylic acid $(C_2H_5COONa)$.
Subsequent acidification with $H^+$ gives propanoic acid $(C_2H_5COOH)$.

(B) The reaction of ketones with $Mg-Hg$ (magnesium amalgam) in the presence of a solvent like benzene leads to the formation of a magnesium pinacolate complex.
Upon subsequent hydrolysis (treatment with water),this complex yields a pinacol (a vicinal diol).
The reaction is known as the pinacol reduction.
For example,$2(CH_3)_2C=O$ $\xrightarrow{Mg/Hg} [(CH_3)_2C(O)]_2Mg$ $\xrightarrow{H_2O} (CH_3)_2C(OH)-C(OH)(CH_3)_2 + Mg(OH)_2$.

(A) For a compound to exhibit geometrical isomerism ($E$ and $Z$ isomers) upon reaction with semicarbazide,the resulting semicarbazone must have two different groups attached to the carbon atom of the $C=N$ bond.
$Benzaldehyde$ $(C_6H_5CHO)$ reacts with semicarbazide to form benzaldehyde semicarbazone $(C_6H_5CH=NNHCONH_2)$.
In this product,the carbon atom is bonded to a phenyl group $(-C_6H_5)$ and a hydrogen atom $(-H)$,which are different.
Thus,it can exist in two geometrical isomeric forms ($E$ and $Z$).
Acetone,benzoquinone,and benzophenone do not show this specific type of isomerism with semicarbazide because they either have identical groups on the carbon atom or are constrained differently.

(D) Acetaldehyde reacts with $HCN$ to form acetaldehyde cyanohydrin $(A)$. The reaction is as follows:
$CH_3-CHO + HCN \to CH_3-CH(OH)-CN$ (Compound $A$)
On hydrolysis,the nitrile group $(-CN)$ is converted into a carboxylic acid group $(-COOH)$:
$CH_3-CH(OH)-CN \xrightarrow{H_3O^{+}} CH_3-CH(OH)-COOH$ (Lactic acid).

(B) Cannizzaro's reaction is given by aldehydes that do not have $\alpha$-hydrogen atoms.
$2$-methylpropanal $(CH_3-CH(CH_3)-CHO)$ contains an $\alpha$-hydrogen atom,hence it primarily undergoes Aldol condensation rather than Cannizzaro's reaction.
Benzaldehyde,$p$-methoxybenzaldehyde,and $2,2$-dimethylpropanal do not possess $\alpha$-hydrogen atoms and thus undergo Cannizzaro's reaction.

(D) Primary alcohols (like ethanol) and secondary alcohols (like $2$-propanol) are easily oxidized by $K_2Cr_2O_7$ and dilute $H_2SO_4$ to aldehydes/carboxylic acids and ketones,respectively.
Acetaldehyde is also easily oxidized to acetic acid.
Acetone (propanone) is a ketone and is resistant to oxidation under mild conditions like dilute $H_2SO_4$ and $K_2Cr_2O_7$ because it requires the breaking of a strong $C-C$ bond.
Therefore,acetone fails to react under these specific conditions.

(B) Acetaldehyde is an aldehyde and reacts with Tollen's reagent $(Ag(NH_3)_2^+)$ to form a silver mirror,whereas acetone is a ketone and does not react with Tollen's reagent.

(B) Chloral $(Cl_3C-CHO)$ reacts with water to form chloral hydrate $(Cl_3C-CH(OH)_2)$.
This is a stable gem-diol due to the strong electron-withdrawing inductive effect of the three chlorine atoms and intramolecular hydrogen bonding.

(C) The compound $A$ with formula $C_3H_6O$ gives the iodoform test,so it is propanone $(CH_3COCH_3)$.
Propanone undergoes self-aldol condensation in the presence of dilute $HCl$ (acid-catalyzed) to form mesityl oxide $(C_6H_{10}O)$ and further condensation leads to phorone $(C_9H_{14}O)$,which is $2,6-$dimethyl$-2,5-$heptadien$-4-$one.
Thus,$A$ is propanone and $B$ is $2,6-$dimethyl$-2,5-$heptadien$-4-$one.

(A) $(C_3H_8O)$ is a primary alcohol.
Since it oxidizes to an aldehyde $B$ $(C_3H_6O)$ that gives a positive Tollens' test (shining silver mirror),$B$ must be propanal.
$CH_3CH_2CH_2OH \xrightarrow{K_2Cr_2O_7/H^{+}} CH_3CH_2CHO$
$A$ (Propan$-1-$ol) $\rightarrow$ $B$ (Propanal)
Propanal reacts with semicarbazide $(NH_2NHCONH_2)$ in the presence of sodium acetate to form propanal semicarbazone $(C)$:
$CH_3CH_2CHO + NH_2NHCONH_2 \rightarrow CH_3CH_2CH=NNHCONH_2 + H_2O$
Thus,the structure of $C$ is $CH_3CH_2CH=NNHCONH_2$.