Properties Questions in English

(A) The reaction of acetone $(CH_3COCH_3)$ with magnesium amalgam $(Mg/Hg)$ followed by hydrolysis $(H^{+})$ is a classic example of bimolecular reduction.
This reaction is known as the pinacol reduction.
Two molecules of acetone undergo coupling to form pinacol ($2,3$-dimethyl-$2,3$-butanediol).
The reaction is: $2CH_3COCH_3 \xrightarrow{Mg/Hg, H^{+}} (CH_3)_2C(OH)-C(OH)(CH_3)_2$.
Thus,the correct product is $CH_3-C(OH)(CH_3)-C(OH)(CH_3)-CH_3$.

(A) The reaction between benzaldehyde $(C_6H_5CHO)$ and formaldehyde $(HCHO)$ in the presence of aqueous $NaOH$ is a $Crossed \ Cannizzaro \ reaction$.
In this reaction,the more reactive aldehyde (formaldehyde) undergoes oxidation to form a formate salt,while the less reactive aldehyde (benzaldehyde) undergoes reduction to form an alcohol.
$C_6H_5CHO + HCHO \xrightarrow{NaOH_{(aq)}} C_6H_5CH_2OH + HCOONa$
Thus,the products are benzyl alcohol $(C_6H_5CH_2OH)$ and sodium formate $(HCOONa)$.

(A) The reaction between formaldehyde $(HCHO)$ and ammonia $(NH_3)$ produces hexamethylenetetramine,commonly known as urotropine.
The balanced chemical equation is:
$6HCHO + 4NH_3 \to (CH_2)_6N_4 + 6H_2O$
Thus,the formula of urotropine is $(CH_2)_6N_4$.

(A) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the aldehyde or ketone.
$HCHO$ (Formaldehyde) does not contain any $\alpha$-carbon atom,and therefore,it lacks $\alpha$-hydrogen atoms.
Consequently,$HCHO$ cannot undergo aldol condensation.

(C) $1$. Propanone $(CH_3COCH_3)$ is a ketone. It does not react with either Tollen's reagent or Fehling's solution. Thus,it corresponds to $(i)$.
$2$. Benzaldehyde $(C_6H_5CHO)$ is an aromatic aldehyde. It reacts with Tollen's reagent to form a silver mirror but does not react with Fehling's solution. Thus,it corresponds to $(ii)$.
$3$. Ethanal $(CH_3CHO)$ is an aliphatic aldehyde. It reacts with both Tollen's reagent and Fehling's solution. Thus,it corresponds to $(iii)$.

(A) The reaction of aldehydes or ketones with hydrazine $(NH_2NH_2)$ involves the nucleophilic addition of the amino group to the carbonyl carbon followed by the elimination of a water molecule $(H_2O)$.
This results in the formation of a hydrazone,which has the general structure $>C=N-NH_2$.

(C) The correct answer is $(C)$.
In the Benzoin condensation reaction,two molecules of benzaldehyde react in the presence of an alcoholic solution of sodium cyanide $(NaCN)$ to form benzoin.
The reaction is: $2C_6H_5CHO \xrightarrow{\text{alc. } NaCN} C_6H_5-CH(OH)-CO-C_6H_5$.

(D) The $Wolf-Kishner$ reduction is a method used to convert carbonyl compounds (aldehydes and ketones) into their corresponding alkanes (hydrocarbons).
In this reaction, the ketone is treated with hydrazine $(NH_2NH_2)$ followed by heating with a strong base like $KOH$ in a high-boiling solvent such as ethylene glycol.
The reaction for acetone is:
$CH_3COCH_3 \xrightarrow{NH_2NH_2 / KOH, \text{glycol}, \Delta} CH_3CH_2CH_3 + N_2 + H_2O$

(A) Clemmensen reduction is a chemical reaction used to reduce carbonyl groups $( >C=O )$ of aldehydes and ketones to methylene groups $( >CH_2 )$.
This reaction is carried out using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.

(A) The aldol condensation of two molecules of acetaldehyde $(CH_3-CHO)$ in the presence of a dilute base (like $dil. NaOH$) results in the formation of $3$-hydroxybutanal $(CH_3-CH(OH)-CH_2-CHO)$.
$2CH_3-CHO \xrightarrow{dil. NaOH} CH_3-CH(OH)-CH_2-CHO$

(C) $1$. $HCN$ reacts with all aldehydes and ketones to form cyanohydrins.
$2$. Tollen's reagent is a mild oxidizing agent that oxidizes all aldehydes (both aliphatic and aromatic) to their corresponding carboxylate ions.
$3$. Fehling's solution is a milder oxidizing agent than Tollen's reagent and is capable of oxidizing only aliphatic aldehydes,but not aromatic aldehydes (like benzaldehyde) or ketones.
$4$. Benzaldehyde $(C_6H_5CHO)$ reacts with $HCN$ and Tollen's reagent,but it is not oxidized by Fehling's solution because it is an aromatic aldehyde.

(B) Benzaldehyde $(C_6H_5CHO)$ lacks $\alpha$-hydrogen atoms. When treated with a concentrated alkali (like $NaOH$),it undergoes the Cannizzaro reaction,which is a disproportionation reaction.
In this reaction,one molecule of benzaldehyde is oxidized to sodium benzoate $(C_6H_5COONa)$ and another molecule is reduced to benzyl alcohol $(C_6H_5CH_2OH)$.
The reaction is:
$2C_6H_5CHO + NaOH \rightarrow C_6H_5COONa + C_6H_5CH_2OH$
Thus,benzyl alcohol is one of the products.

(A) The Cannizzaro reaction is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking an $\alpha$-hydrogen atom.
$HCHO$ (formaldehyde) does not have any $\alpha$-hydrogen atoms.
Therefore,it undergoes the Cannizzaro reaction to form methanol and sodium formate in the presence of a concentrated base.

(B) The given reaction is between benzaldehyde $(C_6H_5CHO)$ and acetaldehyde $(CH_3CHO)$ in the presence of a dilute base to form cinnamaldehyde $(C_6H_5CH=CHCHO)$.
This is a type of cross-aldol condensation,specifically known as the Claisen-Schmidt condensation.

(C) The reaction of two molecules of acetaldehyde $(CH_3CHO)$ in the presence of a dilute alkali (like $NaOH$) is known as the Aldol condensation reaction.
In this reaction,the $\alpha$-hydrogen of one acetaldehyde molecule is abstracted by the base to form an enolate ion,which then attacks the carbonyl carbon of another acetaldehyde molecule.
The product formed is $3$-hydroxybutanal,which is commonly known as Aldol $(CH_3CH(OH)CH_2CHO)$.

(C) Acetaldehyde $(CH_3CHO)$ contains $\alpha$-hydrogens and undergoes an aldol condensation reaction in the presence of dilute $NaOH$.
Step $1$: Two molecules of acetaldehyde react to form $3$-hydroxybutanal (aldol).
$2CH_3CHO \xrightarrow{dil. NaOH} CH_3-CH(OH)-CH_2-CHO$
Step $2$: Upon heating,the aldol undergoes dehydration to form an $\alpha,\beta$-unsaturated aldehyde,which is but-$2$-enal.
$CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} CH_3-CH=CH-CHO + H_2O$
Thus,the final product is $CH_3-CH=CHCHO$.

(C) The reaction of a carbonyl compound $(R_2C=O)$ with hydrogen cyanide $(HCN)$ involves the attack of the nucleophile $(CN^-)$ on the electrophilic carbonyl carbon atom.
This is a characteristic reaction of aldehydes and ketones known as nucleophilic addition reaction.

(A) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetophenone $(CH_3COC_6H_5)$ in the presence of dilute $NaOH$ is a Claisen-Schmidt condensation reaction.
Benzaldehyde acts as the electrophile,and acetophenone acts as the nucleophile (enolate donor).
The reaction proceeds as follows:
$C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{NaOH} C_6H_5CH=CHCOC_6H_5 + H_2O$.
The final product is benzylidene acetophenone (also known as chalcone).

(D) The reduction of the carbonyl group $(-C(=O)-)$ to a methylene group $(-CH_2-)$ is known as Clemmensen reduction or Wolff-Kishner reduction.
$1$. Clemmensen reduction uses $Zn-Hg/conc. HCl$.
$2$. Wolff-Kishner reduction uses $NH_2-NH_2/C_2H_5ONa$ (or $KOH$ in ethylene glycol).
$3$. Reduction with $HI/P$ also converts carbonyls to hydrocarbons.
Therefore,the correct reagent among the options is $NH_2-NH_2/C_2H_5ONa$.

(A) When acetone $(CH_3COCH_3)$ is distilled with concentrated $H_2SO_4$,it undergoes autocondensation to form $1,3,5$-trimethylbenzene,which is commonly known as Mesitylene.
The reaction is as follows:
$3CH_3COCH_3 \xrightarrow{Conc. H_2SO_4} \text{Mesitylene} + 3H_2O$

(D) The reaction described is the Cannizzaro reaction,which occurs in aldehydes that do not possess an $\alpha$-hydrogen atom.
$2C_6H_5CHO + NaOH \xrightarrow{50\% \ NaOH} C_6H_5COONa + C_6H_5CH_2OH$
In this reaction,$Benzaldehyde$ $(C_6H_5CHO)$ undergoes self-oxidation and reduction (disproportionation) to form sodium benzoate $(C_6H_5COONa)$ and benzyl alcohol $(C_6H_5CH_2OH)$.

(A) The correct answer is $A$.
Fehling solution is a mild oxidizing agent used to distinguish between aldehydes and ketones.
Aldehydes (like acetaldehyde and formalin) and reducing sugars (like $D^{-}$-glucose) are oxidized by Fehling solution to give a brick red precipitate of $Cu_2O$.
Ketones,such as acetone,are not easily oxidized by Fehling solution because they require a stronger oxidizing agent to break the $C-C$ bond.
Therefore,acetone does not give a brick red precipitate.

(D) $CH_3CHO$ gives a positive Tollen's test (silver mirror test) because it is an aldehyde,forming a silver mirror: $CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag_{\downarrow} + 4NH_3 + 2H_2O$.
Acetone is a ketone and does not react with Tollen's reagent.

(D) $HCHO$,$CH_3CHO$,and $CH_3COCH_3$ are all capable of forming hydrogen bonds with water molecules.
Due to the formation of these $H$-bonds,all the given compounds are soluble in water.

(B) Acetaldehyde $(CH_3-CHO)$ reacts with methylmagnesium bromide $(CH_3-MgBr)$ to form an addition product,which upon hydrolysis yields propan-$2$-ol $(CH_3-CH(OH)-CH_3)$.
Since the $-OH$ group is attached to a carbon atom which is further attached to two other carbon atoms,it is a secondary alcohol.
The reaction is: $CH_3-CHO + CH_3-MgBr$ $\rightarrow CH_3-CH(OMgBr)-CH_3$ $\xrightarrow{H_2O} CH_3-CH(OH)-CH_3 + Mg(OH)Br$.

(B) The reaction proceeds as follows:
$1$. Acetaldehyde $(CH_3CHO)$ reacts with $HCN$ to form a cyanohydrin: $CH_3CHO + HCN \rightarrow CH_3CH(OH)CN$.
$2$. Subsequent acid hydrolysis of the cyanohydrin yields lactic acid ($2$-hydroxypropanoic acid): $CH_3CH(OH)CN + 2H_2O \xrightarrow{H^+} CH_3CH(OH)COOH + NH_3$.

(A) This reaction is an aldol condensation reaction.
$2CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO$ ($3-$hydroxybutanal).

(C) Fehling solution is an oxidizing agent that reacts with aldehydes to form a carboxylic acid salt and a red precipitate of cuprous oxide $(Cu_2O)$.
The chemical reaction is: $HCHO + 2Cu^{2+} + 5OH^{-} \xrightarrow{\text{Fehling solution}} HCOO^{-} + Cu_2O(s) + 3H_2O$.
Thus,the precipitate formed is red in color.

(B) The presence of at least one $\alpha$ $H$ atom in an aldehyde or ketone is the necessary condition for undergoing the aldol condensation reaction.
This is because the $\alpha$ $H$ atoms are acidic in nature and can be removed by a base to form a nucleophilic enolate ion.

(D) Clemmensen reduction of a ketone is carried out in the presence of $Zn-Hg$ with $HCl$.
This reduction converts the carbonyl group to a methylene group.
It is suitable for compounds that are sensitive to alkali.
$R-CO-CH_3 \xrightarrow{Zn-Hg, \text{ conc. } HCl} R-CH_2-CH_3$

(A) The reaction involves the reduction of a carbonyl group $(C=O)$ to a methylene group $(CH_2)$ using hydrazine $(NH_2NH_2)$ in the presence of a strong base like $KOH$ and a high-boiling solvent like ethylene glycol.
This specific reduction of aldehydes and ketones to alkanes is known as the Wolff-Kishner reduction.

(C) Acetaldehyde $(CH_3-CHO)$ reacts with methyl magnesium iodide $(CH_3-MgI)$ to form an addition product $X$,which on hydrolysis gives Propan$-2-$ol $(Y)$.
$CH_3-CHO + CH_3-MgI \xrightarrow{\text{Ether}} CH_3-CH(OMgI)-CH_3 (X)$
$CH_3-CH(OMgI)-CH_3 + H_2O/H^{+} \rightarrow CH_3-CH(OH)-CH_3 (Y) + Mg(OH)I$

(D) The reaction in which an aldehyde lacking an $\alpha$-hydrogen atom undergoes self-oxidation and reduction in the presence of a concentrated base is known as the Cannizzaro reaction.
Benzaldehyde $(C_6H_5CHO)$ does not have an $\alpha$-hydrogen atom,so it undergoes this reaction to form benzyl alcohol $(C_6H_5CH_2OH)$ and sodium benzoate $(C_6H_5COONa)$.

(A) $2,4-Dinitrophenylhydrazine$ $(2,4-DNP)$,also known as Brady's reagent,reacts with aldehydes and ketones to form $2,4-dinitrophenylhydrazones$.
These derivatives typically appear as orange,yellow,or red crystalline precipitates.
Acetaldehyde $(CH_3CHO)$ reacts with $2,4-DNP$ to form an orange crystalline solid,which is acetaldehyde $2,4-dinitrophenylhydrazone$.

(D) In the manufacture of mirrors,ammoniacal $AgNO_3$ (Tollens' reagent) and $HCHO$ (formaldehyde) are used.
The process involves the reduction of silver ions to metallic silver,which deposits on the glass surface to form a reflective mirror coating.
The chemical reaction is:
$HCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow HCOO^- + 2Ag(s) + 4NH_3 + 2H_2O$
Here,$HCHO$ acts as a reducing agent,reducing the silver complex to metallic silver $(Ag(s))$,which forms the mirror.

(A) The reaction of acetone $(CH_3COCH_3)$ with chlorine $(Cl_2)$ in the presence of a base $(NaOH)$ is a haloform reaction.
$CH_3COCH_3 + 3Cl_2 + 4NaOH \to CHCl_3 + CH_3COONa + 3NaCl + 3H_2O$.
Thus,chloroform $(CHCl_3)$ is formed as the main product.

(D) . Fehling's solution consists of alkaline $CuSO_4$ and sodium potassium tartrate (Rochelle salt).
Tollen's reagent consists of ammoniacal silver nitrate $(AgNO_3 + NH_4OH)$.
Schiff's reagent is a solution of $p$-rosaniline hydrochloride decolorized by sulfur dioxide.
All these reagents are specifically used to distinguish between aldehydes and ketones because aldehydes are easily oxidized to carboxylic acids by these mild oxidizing agents,whereas ketones are generally resistant to oxidation under these conditions.

(B) Schiff's reagent is a solution of rosaniline hydrochloride which is decolourised by $SO_2$. When an aldehyde is added to this decolourised solution,the pink colour is restored. Therefore,the correct option is $(B)$.

(C) Aldol condensation is exhibited by aldehydes and ketones that possess at least one $\alpha$-hydrogen atom.
In $CCl_3-CHO$,the $\alpha$-carbon is bonded to three chlorine atoms and has no hydrogen.
In $CH_3-C(CH_3)_2-CHO$,the $\alpha$-carbon is bonded to three methyl groups and has no hydrogen.
In $HCHO$ (formaldehyde),there is no $\alpha$-carbon.
In $CH_3-CH_2-CHO$ (propanal),the $\alpha$-carbon is bonded to two hydrogen atoms.
Therefore,$CH_3-CH_2-CHO$ undergoes aldol condensation.

(A) Acetaldehyde $(CH_3CHO)$ is an aldehyde.
When treated with a strong oxidizing agent like concentrated $KMnO_4$,it undergoes oxidation.
The aldehyde group $(-CHO)$ is oxidized to a carboxylic acid group $(-COOH)$.
Therefore,the reaction is: $CH_3CHO + [O] \xrightarrow{KMnO_4} CH_3COOH$ (Acetic acid).

(C) Acetaldehyde $(CH_3CHO)$ reacts with Tollen's reagent $([Ag(NH_3)_2]OH)$ to form an ammonium salt of the corresponding carboxylic acid and metallic silver.
The reaction is: $CH_3CHO + 2[Ag(NH_3)_2]OH \xrightarrow{\Delta} CH_3COONH_4 + 2Ag \downarrow (\text{Silver mirror}) + 3NH_3 + H_2O$.
Thus,a silver mirror is formed on the inner walls of the test tube.

(A) Urotropine is the common name for $C_6H_{12}N_4$,which is chemically known as $1,3,5,7$-tetraazatricyclo$[3.3.1.1^{3,7}]$decane or hexamethylene tetramine.
It is prepared by the reaction of formaldehyde with ammonia.
It is used as a medicine to treat urinary tract infections.

(D) The reaction of an aldehyde $(R-CHO)$ with hydroxylamine $(NH_2OH)$ involves the elimination of a water molecule to form a product known as an oxime.
The general reaction is: $R-CHO + NH_2OH \rightarrow R-CH=N-OH + H_2O$ (Oxime).
For example,acetaldehyde reacts with hydroxylamine to form acetaldoxime: $CH_3CHO + NH_2OH \rightarrow CH_3-CH=N-OH + H_2O$.

(C) The Cannizzaro reaction is given by aldehydes that do not possess an $\alpha-H$ atom.
$CH_3CHO$ (acetaldehyde) contains $3$ $\alpha-H$ atoms,therefore,it does not undergo the Cannizzaro reaction.

(B) The reaction of acetone $(CH_3COCH_3)$ with hydroxylamine $(NH_2OH)$ is a nucleophilic addition-elimination reaction.
Acetone reacts with hydroxylamine to form acetoxime and water.
The chemical equation is:
$(CH_3)_2C=O + NH_2OH \rightarrow (CH_3)_2C=N-OH + H_2O$
Since the product formed is an oxime,the correct option is $(B)$.

(A) Formaldehyde reacts with ammonia to form hexamethylenetetramine,which is commonly known as urotropine.
The balanced chemical equation for this reaction is:
$6HCHO + 4NH_3 \rightarrow (CH_2)_6N_4 + 6H_2O$

(C) The oxidation of an aldehyde yields a carboxylic acid with the same number of carbon atoms.
Propionaldehyde is $CH_3CH_2CHO$ (propanal).
Upon oxidation,it forms propanoic acid $(CH_3CH_2COOH)$.
Reaction: $CH_3CH_2CHO \xrightarrow{[O]} CH_3CH_2COOH$.

(C) The reaction of acetaldehyde $(CH_{3}CHO)$ with phosphorus pentachloride $(PCl_{5})$ is a nucleophilic substitution reaction where the carbonyl oxygen is replaced by two chlorine atoms.
$CH_{3}CHO + PCl_{5} \rightarrow CH_{3}CHCl_{2} + POCl_{3}$
The product $CH_{3}CHCl_{2}$ is known as $1,1$-dichloroethane or Ethylidene chloride.

(D) Benzaldehyde and acetaldehyde can be differentiated using $NaOH$ solution.
Benzaldehyde lacks $\alpha$-hydrogen atoms and undergoes the Cannizzaro reaction in the presence of concentrated $NaOH$ to form sodium benzoate and benzyl alcohol.
Acetaldehyde contains $\alpha$-hydrogen atoms and undergoes aldol condensation in the presence of dilute $NaOH$ to form $3-$hydroxybutanal.
$2CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO$ (Aldol condensation).