MHT CET 2026 Chemistry Question Paper with Answer and Solution

(B) $1$. Oxidation of alcohols depends on the type of alcohol.
$2$. Primary alcohols $(R-CH_2OH)$ oxidize to aldehydes and then to carboxylic acids.
$3$. Secondary alcohols $(R_2CHOH)$ oxidize to ketones.
$4$. $2-Propanol$ $(CH_3-CH(OH)-CH_3)$ is a secondary alcohol. Upon oxidation, it forms $propanone$ $(CH_3-CO-CH_3)$:
$CH_3-CH(OH)-CH_3 \xrightarrow{[O]} CH_3-CO-CH_3 + H_2O$.

(C) Alcohols $(R-OH)$ have a $pK_a$ value comparable to water (approximately $16-18$). In an aqueous medium, they do not donate a proton to water to a significant extent, nor do they accept a proton from water to act as a base. Therefore, alcohols are considered neutral in an aqueous medium.

(A) Step $1$: Primary alcohols $(R-CH_2OH)$ are oxidized to aldehydes $(R-CHO)$ and further to carboxylic acids $(R-COOH)$ by strong oxidizing agents.
Step $2$: $K_2Cr_2O_7 / \text{dil. } H_2SO_4$ and acidic $KMnO_4$ are strong oxidizing agents that oxidize primary alcohols directly to carboxylic acids.
Step $3$: $CrO_3$ in anhydrous conditions can stop at the aldehyde stage, but $PCC$ (Pyridinium chlorochromate) is specifically designed to oxidize primary alcohols to aldehydes without further oxidation to carboxylic acids.
Step $4$: Therefore, $PCC$ is the correct reagent to limit the oxidation to the aldehyde stage.

(B) $1$. $DIBAL-H$ (Diisobutylaluminium hydride) is a selective reducing agent.
$2$. When an ester $(R - CO - O - R')$ is treated with $DIBAL-H$ at low temperature followed by acidic workup $(H_3O^+)$, it is reduced to an aldehyde $(R - CHO)$ and an alcohol $(R'OH)$.
$3$. The reaction stops at the aldehyde stage because the tetrahedral intermediate formed is stable at low temperatures and does not undergo further reduction to the primary alcohol.
$4$. Therefore, the main organic product containing the carbonyl group is $R - CHO$.

(B) Step $1$: The chemical formula $HCOOH$ represents the simplest carboxylic acid, which contains only one carbon atom.
Step $2$: The $IUPAC$ name for $HCOOH$ is methanoic acid.
Step $3$: The common name for methanoic acid is derived from the Latin word 'formica', meaning ant, as it was first isolated by the distillation of ants. Thus, it is known as formic acid.

(C) The reaction is: $C_6H_5COOH + PCl_5 \xrightarrow{\Delta} C_6H_5COCl + POCl_3 + HCl$.
$PCl_5$ (Phosphorus pentachloride) reacts with carboxylic acids to form acid chlorides, phosphorus oxychloride $(POCl_3)$, and hydrogen chloride $(HCl)$.

(A) $1$. The acidity of carboxylic acids depends on the stability of the conjugate base (carboxylate ion).
$2$. Electron-withdrawing groups (like $-Cl$) increase acidity by stabilizing the negative charge on the carboxylate ion through the inductive effect ($-I$ effect).
$3$. Electron-donating groups (like alkyl groups $-CH_3$) decrease acidity by destabilizing the carboxylate ion through the $+I$ effect.
$4$. Comparing the options:
- $(CH_3)_2CHCOOH$ (Isobutyric acid) has two methyl groups donating electrons ($+I$ effect).
- $CH_3CH_2COOH$ (Propanoic acid) has one ethyl group donating electrons ($+I$ effect).
- $ClCH_2COOH$ (Chloroacetic acid) has one $-Cl$ atom ($-I$ effect).
- $Cl_3CCOOH$ (Trichloroacetic acid) has three $-Cl$ atoms (strong $-I$ effect).
$5$. Since $(CH_3)_2CHCOOH$ has the strongest electron-donating effect among the choices, it destabilizes the conjugate base the most, making it the weakest acid.

(D) $1$. The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion) formed after the loss of a proton.
$2$. Electron-withdrawing groups (EWGs) stabilize the carboxylate ion through the inductive effect ($-I$ effect), thereby increasing the acidity of the parent acid.
$3$. In $ClCH_2COOH$, the chlorine atom is a strong electron-withdrawing group due to its high electronegativity.
$4$. In $CH_3COOH$ and $CH_3CH_2COOH$, the alkyl groups ($CH_3-$ and $CH_3CH_2-$) are electron-donating groups ($+I$ effect), which destabilize the carboxylate ion and decrease acidity.
$5$. $HCOOH$ has only a hydrogen atom, which has no inductive effect.
$6$. Therefore, $ClCH_2COOH$ is the strongest acid among the given options.

(D) $1$. Sodium bicarbonate $(NaHCO_3)$ is a weak base.
$2$. Carboxylic acids $(-COOH)$ are acidic enough to react with $NaHCO_3$ to release carbon dioxide $(CO_2)$ gas.
$3$. The evolution of $CO_2$ gas causes brisk effervescence.
$4$. The reaction is: $R-COOH + NaHCO_3 \rightarrow R-COONa + H_2O + CO_2 \uparrow$.
$5$. Therefore, the presence of the carboxylic acid group is indicated.

(B) Step $1$: The reaction between a carboxylic acid $(RCOOH)$ and an alcohol $(R'OH)$ in the presence of an acid catalyst $(H^+)$ is an esterification reaction, forming an ester $(RCOOR')$ and water $(H_2O)$.
Step $2$: The intermediate is the ester $RCOOR'$.
Step $3$: Catalytic hydrogenation of an ester using $Ni, H_2$ at high temperature $(\Delta)$ leads to the cleavage of the ester bond, reducing the carboxylic acid part to a primary alcohol $(RCH_2OH)$ and releasing the original alcohol $(R'OH)$.
Step $4$: The final products are $RCH_2OH + R'OH$.

(C) $1$. The acidic strength of a carboxylic acid is directly proportional to its acid dissociation constant $(K_a)$.
$2$. The relationship between $pK_a$ and $K_a$ is defined as $pK_a = -\log(K_a)$.
$3$. Therefore, a higher $K_a$ value corresponds to a lower $pK_a$ value.
$4$. Consequently, a higher $K_a$ value and a lower $pK_a$ value indicate a stronger acid.

(C) Step $1$: Isopropyl magnesium chloride is $(CH_3)_2CHMgCl$.
Step $2$: Reaction with dry ice $(CO_2)$ forms a complex: $(CH_3)_2CHMgCl + CO_2 \rightarrow (CH_3)_2CHCOOMgCl$.
Step $3$: Acidic hydrolysis of the complex yields the carboxylic acid: $(CH_3)_2CHCOOMgCl + H_2O/H^+ \rightarrow (CH_3)_2CHCOOH + Mg(OH)Cl$.
Step $4$: The product $(CH_3)_2CHCOOH$ is $2-\text{Methylpropanoic acid}$ (also known as isobutyric acid).

(C) Step $1$: $A$ monocarboxylic acid contains only one $-COOH$ group in its structure.
Step $2$: Toluic acid $(CH_3C_6H_4COOH)$ has one $-COOH$ group.
Step $3$: Acrylic acid $(CH_2=CHCOOH)$ has one $-COOH$ group.
Step $4$: Phthalic acid $(C_6H_4(COOH)_2)$ has two $-COOH$ groups attached to the benzene ring at ortho positions, making it a dicarboxylic acid.
Step $5$: Isobutyric acid $((CH_3)_2CHCOOH)$ has one $-COOH$ group.
Step $6$: Therefore, Phthalic acid is not a monocarboxylic acid.

(D) $1$. Carboxylic acids $(R-COOH)$ exhibit higher boiling points than alcohols, aldehydes, ketones, and alkanes of comparable molecular masses.
$2$. This is due to the presence of extensive intermolecular hydrogen bonding between the carboxylic acid molecules.
$3$. In the vapor phase, carboxylic acids often exist as stable dimers held together by two intermolecular hydrogen bonds.
$4$. This strong association requires more energy to break, resulting in a higher boiling point.

(B) $1$. The Gattermann reaction is a variation of the Sandmeyer reaction.
$2$. In the Sandmeyer reaction, $Cu_2Cl_2$ or $Cu_2Br_2$ is used as a catalyst.
$3$. In the Gattermann reaction, copper powder $(Cu)$ in the presence of the corresponding halogen acid ($HCl$ or $HBr$) is used to replace the diazonium group with a halogen atom.
$4$. Therefore, the correct catalyst system is copper powder and $HCl$ or $HBr$.

(A) The given reaction involves the oxidation of toluene $(C_6H_5CH_3)$ to benzaldehyde $(C_6H_5CHO)$ using chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ as a solvent, followed by hydrolysis. This specific chemical transformation is known as the Etard reaction.

(A) Acetaldehyde $(CH_3CHO)$ undergoes the iodoform reaction when treated with $I_2$ and $NaOH$.
Step $1$: The reaction is: $CH_3CHO + 3I_2 + 4NaOH \rightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$.
Step $2$: The products formed are iodoform $(CHI_3)$ and sodium formate $(HCOONa)$.

(D) $1$. The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$2$. Ethyl methyl ketone $(CH_3COCH_2CH_3)$ contains the $CH_3CO-$ group, so it gives the test.
$3$. Isopropyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group, so it gives the test.
$4$. $3-\text{Methyl}-2-\text{butanone}$ $(CH_3COCH(CH_3)_2)$ contains the $CH_3CO-$ group, so it gives the test.
$5$. Isobutyl alcohol $((CH_3)_2CHCH_2OH)$ does not contain either the $CH_3CO-$ or the $CH_3CH(OH)-$ group. Therefore, it cannot form iodoform.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$(1)$ Ethyl methyl ketone $(CH_3COCH_2CH_3)$ contains the $CH_3CO-$ group, so it gives a positive test.
$(2)$ Isopropyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group, so it gives a positive test.
$(3)$ $3-\text{Methyl} - 2 - \text{butanone}$ $(CH_3COCH(CH_3)_2)$ contains the $CH_3CO-$ group, so it gives a positive test.
$(4)$ Isobutyl alcohol $((CH_3)_2CHCH_2OH)$ does not contain either the $CH_3CO-$ or the $CH_3CH(OH)-$ group. Therefore, it does not respond to the iodoform test.

(B) Step $1$: Tollens' reagent is an ammoniacal silver nitrate solution, $[Ag(NH_3)_2]^+ OH^-$.
Step $2$: When an aldehyde is heated with Tollens' reagent, the aldehyde is oxidized to a carboxylate ion, and the silver ion is reduced to metallic silver.
Step $3$: The metallic silver deposits on the inner walls of the test tube, forming a characteristic 'silver mirror'.
Step $4$: Therefore, the correct observation is the formation of an $Ag$ layer on the sides of the test tube.

(C) Step $1$: The Schiff test is a chemical test used to detect the presence of an aldehyde group $(-CHO)$.
Step $2$: Schiff's reagent is a magenta-colored solution of $p$-rosaniline hydrochloride that has been decolorized by sulfur dioxide.
Step $3$: When an aldehyde is added to Schiff's reagent, the magenta color is restored, indicating a positive test.
Step $4$: Ketones, carboxylic acids, and ethers do not react with Schiff's reagent under standard conditions to restore the color.
Step $5$: Therefore, aldehydes are the compounds that exhibit the Schiff test.

(D) $1$. $A$ disproportionation reaction is a type of redox reaction in which the same element is simultaneously oxidized and reduced.
$2$. In the $Cannizzaro$ reaction, aldehydes that do not have an $\alpha$-hydrogen atom undergo self-oxidation and reduction in the presence of a concentrated base.
$3$. For example, two molecules of $benzaldehyde$ $(C_6H_5CHO)$ react to form $benzyl$ $alcohol$ $(C_6H_5CH_2OH)$ and $sodium$ $benzoate$ $(C_6H_5COONa)$.
$4$. Here, the carbon atom in the aldehyde group is oxidized to a carboxylic acid salt and reduced to an alcohol simultaneously.
$5$. Therefore, the $Cannizzaro$ reaction is a classic example of a disproportionation reaction.

(D) In cross-aldol condensation between ethanal $(CH_3CHO)$ and propanal $(CH_3CH_2CHO)$, four products are possible: two self-aldol and two cross-aldol products.
$1$. Cross-aldol $1$: $CH_3CHO$ acts as the nucleophile (enolate) and $CH_3CH_2CHO$ acts as the electrophile. The product is $CH_3CH_2CH(OH)CH(CH_3)CHO$, which dehydrates to form $CH_3CH_2CH=C(CH_3)CHO$ ($2$-methylpent$-2-$enal).
$2$. Cross-aldol $2$: $CH_3CH_2CHO$ acts as the nucleophile (enolate) and $CH_3CHO$ acts as the electrophile. The product is $CH_3CH_2CH(OH)CH_2CHO$, which dehydrates to form $CH_3CH=C(CH_3)CHO$ ($2$-methylbut$-2-$enal).
Thus, the cross-aldol products are $2-$methylbut$-2-$enal and $2-$methylpent$-2-$enal.

(B) The reaction of a ketone $(R_2C=O)$ with hydrazine $(NH_2-NH_2)$ is a nucleophilic addition-elimination reaction.
Step $1$: The lone pair on the nitrogen of hydrazine attacks the electrophilic carbonyl carbon of the ketone.
Step $2$: $A$ molecule of water $(H_2O)$ is eliminated from the intermediate.
Step $3$: The final product formed is a hydrazone with the general formula $R_2C=N-NH_2$.

(C) $1$. The reaction of a ketone with a derivative of ammonia $(NH_2-Z)$ is a nucleophilic addition-elimination reaction.
$2$. Semicarbazide is the reagent used to form a semicarbazone.
$3$. The chemical formula for semicarbazide is $NH_2-NH-CONH_2$.
$4$. Therefore, the correct reagent is $NH_2-NH-CONH_2$.

(B) $1$. Nucleophilic addition reactions in carbonyl compounds depend on the electrophilicity of the carbonyl carbon and steric hindrance.
$2$. Aldehydes have one alkyl group and one hydrogen atom attached to the carbonyl carbon, whereas ketones have two alkyl groups.
$3$. Alkyl groups are electron-donating ($+I$ effect), which reduces the positive charge (electrophilicity) on the carbonyl carbon.
$4$. Ketones have two such groups, making them less electrophilic than aldehydes.
$5$. Additionally, two alkyl groups in ketones create more steric hindrance compared to one alkyl group and one small hydrogen atom in aldehydes, making the approach of the nucleophile more difficult in ketones.

(A) $1$. The Cannizzaro reaction is a disproportionation reaction involving aldehydes that do not have an $\alpha$-hydrogen atom.
$2$. In the presence of a concentrated base, one molecule of the aldehyde is reduced to the corresponding alcohol, while another molecule is oxidized to the salt of the corresponding carboxylic acid.
$3$. Therefore, the correct statement is that some aldehydes are converted to the respective alcohols and the salt of the respective carboxylic acid.

(D) Step $1$: Reaction in container $A$ with $H_2 / Ni$. $H_2 / Ni$ is a strong reducing agent that reduces both the alkene double bond and the aldehyde group. Thus, $CH_3-CH=CH-CH_2-CHO + 2H_2 \xrightarrow{Ni} CH_3-CH_2-CH_2-CH_2-CH_2OH$ (or $CH_3(CH_2)_3CH_2OH$). So, $P = CH_3(CH_2)_3CH_2OH$.
Step $2$: Reaction in container $B$ with $LiAlH_4$. $LiAlH_4$ is a selective reducing agent that reduces the aldehyde group to a primary alcohol but does not reduce the isolated alkene double bond. Thus, $CH_3-CH=CH-CH_2-CHO \xrightarrow{LiAlH_4 / H_3O^+} CH_3-CH=CH-CH_2-CH_2OH$. So, $Q = CH_3-CH=CH-CH_2-CH_2OH$.
Step $3$: Comparing with options, $P = CH_3(CH_2)_3CH_2OH$ and $Q = CH_3-CH=CH-CH_2-CH_2OH$ matches option $D$.

(A) Step $1$: Acetone $(CH_3COCH_3)$ is a ketone.
Step $2$: $CrO_3$ is a strong oxidizing agent.
Step $3$: Oxidation of ketones with strong oxidizing agents like $CrO_3$ leads to the cleavage of $C-C$ bonds, resulting in a mixture of carboxylic acids with fewer carbon atoms than the original ketone.
Step $4$: The oxidation of acetone $(CH_3COCH_3)$ yields acetic acid $(CH_3COOH)$ and formic acid $(HCOOH)$, where acetic acid is the major product.
Therefore, the correct option is $A$.

(A) $1$. The reduction of alkanones (ketones) to alkanes can be achieved using Clemmensen reduction.
$2$. In Clemmensen reduction, the carbonyl group $(>C=O)$ is reduced to a methylene group $(-CH_2-)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
$3$. The reaction is: $R-CO-R' + 4[H] \xrightarrow{Zn-Hg/HCl} R-CH_2-R' + H_2O$.
$4$. Etard reaction is used for the oxidation of toluene to benzaldehyde. Rosenmund reduction is used for the reduction of acid chlorides to aldehydes. Haloform reaction is used for the detection of methyl ketones.

(B) Step $1$: The Grignard reagent $(R'MgX)$ acts as a nucleophile and attacks the electrophilic carbon atom of the alkyl cyanide $(R-C \equiv N)$.
Step $2$: This forms an intermediate imine salt $(R-C(R')=NMgX)$.
Step $3$: Acid hydrolysis of the imine salt yields an imine $(R-C(R')=NH)$, which is unstable and undergoes further hydrolysis to form a ketone $(R-C(=O)R')$.
Step $4$: Therefore, the final product is a ketone.

(A) Step $1$: Reaction of $\text{Benzonitrile}$ $(C_6H_5CN)$ with $\text{Phenylmagnesium bromide}$ $(C_6H_5MgBr)$ leads to the formation of an imine intermediate $(X)$, which is $\text{N-magnesiobenzophenone imine}$ $(C_6H_5-C(=NMgBr)-C_6H_5)$.
Step $2$: Acidic hydrolysis $(H_3O^+)$ of the intermediate $X$ converts the imine group into a carbonyl group, yielding $\text{Benzophenone}$ $(C_6H_5-CO-C_6H_5)$ as the final product '$Y$', along with $MgBr(OH)$ and $NH_3$.

(D) $1$. The reaction of benzonitrile $(C_6H_5CN)$ with a Grignard reagent $(RMgX)$ followed by acid hydrolysis yields a ketone.
$2$. The reaction proceeds as follows: $C_6H_5CN + C_6H_5MgBr \rightarrow C_6H_5C(NMgBr)C_6H_5$.
$3$. Upon hydrolysis: $C_6H_5C(NMgBr)C_6H_5 + 2H_2O \rightarrow C_6H_5COC_6H_5 + NH_3 + Mg(OH)Br$.
$4$. Since the product is benzophenone $(C_6H_5COC_6H_5)$, the Grignard reagent must be phenylmagnesium bromide $(C_6H_5MgBr)$.

(C) $1$. The reaction is the Stephen reduction of a nitrile. Isopropyl cyanide is $(CH_3)_2CH-CN$.
$2$. In the first step, the nitrile is reduced by $SnCl_2/HCl$ to form an imine intermediate $(A)$, which is $(CH_3)_2CH-CH=NH$.
$3$. In the second step, the acid hydrolysis of the imine $(A)$ yields the corresponding aldehyde $(B)$.
$4$. The reaction is: $(CH_3)_2CH-CN \xrightarrow{SnCl_2, HCl} (CH_3)_2CH-CH=NH \xrightarrow{H_3O^+} (CH_3)_2CH-CHO + NH_4Cl$.
$5$. The product '$B$' is $(CH_3)_2CH-CHO$, which is $2-\text{Methylpropanal}$.

(B) The given reaction is the Rosenmund reduction, where an acid chloride is reduced to an aldehyde using hydrogen gas in the presence of a poisoned palladium catalyst.
Step $1$: The reagent used is $H_2$ gas in the presence of $Pd$ supported on $BaSO_4$.
Step $2$: $BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.
Step $3$: The reaction is $C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$.

(C) Step $1$: $\text{DIBAL-H}$ (Diisobutylaluminium hydride) is a selective reducing agent.
Step $2$: It reduces nitriles $(-CN)$ to imines, which upon acidic hydrolysis $(H_3O^+)$ yield aldehydes.
Step $3$: The reaction is: $CH_3-CH=CH-CH_2-CN \xrightarrow[2. H_3O^+]{1. DIBAL-H} CH_3-CH=CH-CH_2-CHO$.
Step $4$: The product formed is $\text{Pent-3-enal}$.

(C) Rosenmund reduction involves the catalytic hydrogenation of acid chlorides to aldehydes using $Pd$ supported on $BaSO_4$ (poisoned with sulfur or quinoline).
Reaction: $C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$.
Benzoyl chloride $(C_6H_5COCl)$ is reduced to Benzaldehyde $(C_6H_5CHO)$.

(A) Step $1$: Rosenmund reduction is the catalytic hydrogenation of acid chlorides to aldehydes using $H_2$ gas in the presence of $Pd$ catalyst poisoned with $BaSO_4$ (Lindlar's catalyst).
Step $2$: The reaction is $R - CO - Cl + H_2 \xrightarrow{Pd-BaSO_4} R - CHO + HCl$.
Step $3$: Option $(A)$ represents this reaction. Option $(B)$ is Stephen reduction, $(C)$ is Clemmensen reduction, and $(D)$ is Wolff-Kishner reduction.

(A) $1$. The reaction of alkyl halides $(RX)$ with $AgCN$ is a nucleophilic substitution reaction.
$2$. $AgCN$ is a covalent compound. The carbon atom in $AgCN$ is not free to act as a nucleophile, but the nitrogen atom has a lone pair of electrons available for bonding.
$3$. Therefore, the nitrogen atom attacks the alkyl group, leading to the formation of alkyl isocyanides $(R-NC)$.
$4$. In contrast, $KCN$ is an ionic compound that provides $CN^-$ ions, where the carbon atom acts as the nucleophile, leading to the formation of alkyl cyanides $(R-CN)$.

(C) $1$. The reaction described is the Carbylamine reaction.
$2$. Primary amines $(R-NH_2)$ react with chloroform $(CHCl_3)$ and alcoholic $KOH$ to form alkyl isocyanides (carbylamines), which have a foul smell.
$3$. Secondary $(R_2NH)$ and tertiary $(R_3N)$ amines do not undergo the Carbylamine reaction.
$4$. Methanamine $(CH_3NH_2)$, Propan$-2-$amine $(CH_3CH(NH_2)CH_3)$, and Ethanamine $(CH_3CH_2NH_2)$ are primary amines.
$5$. $N, N-Diethylethanamine$ $((C_2H_5)_3N)$ is a tertiary amine and therefore does not give this test.

(B) $1$. The reaction of primary amines with ethanolic $KOH$ and $CHCl_3$ is known as the Carbylamine reaction.
$2$. This reaction produces isocyanides (carbylamines), which have a characteristic foul smell.
$3$. Secondary and tertiary amines do not undergo the Carbylamine reaction because they lack the necessary hydrogen atoms on the nitrogen to form the isocyanide group.
$4$. Benzenamine $(C_6H_5NH_2)$, Propan$-2-$amine $(CH_3CH(NH_2)CH_3)$, and Phenylmethanamine $(C_6H_5CH_2NH_2)$ are primary amines and will give the test.
$5$. $N, N - \text{Dimethylaniline}$ is a tertiary amine $(C_6H_5N(CH_3)_2)$ and will not produce a foul smell.

(C) Step $1$: Reaction of $CH_3-I$ with $KCN$ is a nucleophilic substitution reaction. $CH_3-I + KCN \rightarrow CH_3-CN + KI$. Thus, '$A$' is $CH_3-CN$ (methyl cyanide or ethanenitrile).
Step $2$: Reduction of $CH_3-CN$ with $Na/C_2H_5OH$ is known as Mendius reduction. $CH_3-CN + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3-CH_2-NH_2$. Thus, '$B$' is $CH_3-CH_2-NH_2$ (ethylamine).

(C) Step $1$: Ethyl cyanide is $\text{CH}_3\text{CH}_2\text{CN}$ (Propanenitrile).
Step $2$: Partial hydrolysis of $\text{CH}_3\text{CH}_2\text{CN}$ gives $\text{CH}_3\text{CH}_2\text{CONH}_2$ (Propanamide) as '$A$'.
Step $3$: Complete hydrolysis of '$A$' $(\text{CH}_3\text{CH}_2\text{CONH}_2)$ gives $\text{CH}_3\text{CH}_2\text{COOH}$ (Propanoic acid) as '$B$' and $\text{NH}_3$ as a byproduct.
Step $4$: Therefore, '$B$' is Propanoic acid.

(A) The chemical reaction for the reduction of acetamide $(CH_3CONH_2)$ to ethanamine $(CH_3CH_2NH_2)$ using $LiAlH_4$ is:
$CH_3CONH_2 + 4[H] \xrightarrow{LiAlH_4 / \text{ether}} CH_3CH_2NH_2 + H_2O$
In this reaction, $4$ moles of hydrogen atoms (or $2$ moles of $H_2$ molecules) are required to reduce one mole of acetamide to ethanamine.
Therefore, the number of moles of hydrogen atoms needed is $4$.

(C) $1$. The $Hinsberg$ test uses $Benzenesulfonyl \ chloride$ $(C_6H_5SO_2Cl)$ as the reagent.
$2$. Primary amines react with $C_6H_5SO_2Cl$ to form an $N$-alkylbenzenesulfonamide, which is soluble in alkali due to the acidic hydrogen on the nitrogen atom.
$3$. Secondary amines react to form an $N,N$-dialkylbenzenesulfonamide, which lacks an acidic hydrogen and is insoluble in alkali.
$4$. Tertiary amines do not react with $C_6H_5SO_2Cl$.

(C) The reaction of alkyl nitriles $(RCN)$ with sodium $(Na)$ and ethanol $(C_2H_5OH)$ is known as the Mendius reduction.
In this reaction, the nitrile group $(-C \equiv N)$ is reduced to a primary amine $(-CH_2NH_2)$.
The chemical equation is: $RCN + 4[H] \xrightarrow{Na/C_2H_5OH} RCH_2NH_2$.
Therefore, the product obtained is $RCH_2NH_2$.

(C) The carbylamine test is a chemical test used to detect the presence of primary amines $(R-NH_2)$.
In this reaction, a primary amine is heated with chloroform $(CHCl_3)$ and an alcoholic base $(KOH)$, resulting in the formation of an isocyanide (carbylamine), which has a characteristic foul smell.
Secondary and tertiary amines do not undergo this reaction because they lack the necessary hydrogen atoms on the nitrogen atom to form the isocyanide linkage.
Therefore, only primary amines exhibit the carbylamine test.

(D) Step $1$: Primary amines $(R-NH_2)$ react with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form isocyanides (carbylamines), which have a foul smell. This is known as the carbylamine test.
Step $2$: Secondary and tertiary amines do not give this test because they lack the necessary hydrogen atoms on the nitrogen to form the isocyanide structure.
Step $3$: Acetylation, alkylation, and reaction with nitrous acid are exhibited by primary, secondary, and tertiary amines to varying degrees or with different products, but the carbylamine test is specific to primary amines.

(B) The conversion of a nitrile $(-CN)$ group to a primary amine $(-CH_2NH_2)$ is a reduction reaction.
This reaction is known as the Mendius reduction.
Sodium in ethanol $(Na + C_2H_5OH)$ acts as a strong reducing agent capable of reducing the $-CN$ group to a $-CH_2NH_2$ group.
The reaction is: $R-CN + 4[H] \xrightarrow{Na/C_2H_5OH} R-CH_2NH_2$.

(A) $1$. The structure of isopropylamine is $(CH_3)_2CHNH_2$.
$2$. In this molecule, the nitrogen atom is attached to only one carbon atom (the central carbon of the isopropyl group).
$3$. An amine where the nitrogen atom is bonded to only one alkyl or aryl group is classified as a primary amine ($1^\circ$ amine).
$4$. Therefore, isopropylamine is a primary amine.