MHT CET 2026 Chemistry Question Paper with Answer and Solution

(A) Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
In this reaction, the nucleophilic substitution of the phthalimide anion with an alkyl halide occurs.
Since aryl halides do not undergo nucleophilic substitution reactions easily under these conditions, $Aniline$ (an aromatic amine) cannot be prepared by this method.
Therefore, $Aniline$ is not obtained by Gabriel phthalimide synthesis.

(B) Hinsberg's reagent is chemically known as $benzene\ sulphonyl\ chloride$ $(C_6H_5SO_2Cl)$. It is used to distinguish between primary, secondary, and tertiary amines.
Therefore, the correct option is $B$.

(D) Step $1$: The general formula for an amine is $C_nH_{2n+3}N$. For $n=3$, the formula is $C_3H_{2(3)+3}N = C_3H_9N$.
Step $2$: $1^\circ$ amines (e.g., $CH_3CH_2CH_2NH_2$), $2^\circ$ amines (e.g., $CH_3NHCH_2CH_3$), and $3^\circ$ amines (e.g., $(CH_3)_3N$) all satisfy the formula $C_3H_9N$.
Step $3$: $A$ quaternary ammonium salt has the general formula $R_4N^+X^-$. For the simplest quaternary ammonium salt, the formula would be at least $C_4H_{12}N^+X^-$. Thus, it cannot be represented by $C_3H_9N$.

(D) Step $1$: Analyze the structure of amines. Primary amines $(R-NH_2)$ have two hydrogen atoms attached to the nitrogen, allowing for intermolecular hydrogen bonding.
Step $2$: Tertiary amines $(R_3N)$ have no hydrogen atoms attached to the nitrogen, so they cannot form intermolecular hydrogen bonds.
Step $3$: Tertiary butylamine is $(CH_3)_3C-NH_2$, which is a primary amine because the nitrogen is attached to only one carbon atom.
Step $4$: iso-propylamine is $(CH_3)_2CH-NH_2$, which is a primary amine because the nitrogen is attached to only one carbon atom. Therefore, the statement that iso-propylamine is a secondary amine is incorrect.

(A) Step $1$: The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts using copper salts as catalysts.
Step $2$: The reaction involves the replacement of the diazonium group $(-N_2^+)$ with a halogen atom (e.g., $-Cl, -Br, -CN$).
Step $3$: The Mendius reaction involves the reduction of nitriles to primary amines. The Hofmann rearrangement involves the conversion of an amide to a primary amine with one less carbon atom. The Carbylamine reaction is a test for primary amines using chloroform and base.
Step $4$: Therefore, the Sandmeyer reaction is the only one among the options that utilizes a diazonium salt.

(D) $1$. Primary $(R-NH_2)$ and secondary $(R_2NH)$ amines contain $N-H$ bonds, which allow for inter-molecular hydrogen bonding.
$2$. Tertiary amines $(R_3N)$ do not have any hydrogen atoms attached to the nitrogen atom.
$3$. Due to the absence of $N-H$ bonds, tertiary amines cannot form inter-molecular hydrogen bonds.
$4$. Consequently, they have weaker inter-molecular forces compared to primary and secondary amines, resulting in lower boiling points.

(B) Step $1$: In the aqueous phase, the basicity of aliphatic amines depends on the combined effect of the inductive effect $(+I)$, solvation effect, and steric hindrance.
Step $2$: For methyl-substituted amines, the order is $(CH_3)_2NH > CH_3NH_2 > NH_3$.
Step $3$: $C_6H_5NH_2$ (aniline) is the least basic because the lone pair on the nitrogen atom is involved in resonance with the benzene ring, making it less available for protonation.
Step $4$: Combining these, the decreasing order of basic strength is $(CH_3)_2NH > CH_3NH_2 > NH_3 > C_6H_5NH_2$.

(B) $1$. Ethyl amine $(CH_3CH_2NH_2)$ reacts with excess methyl iodide $(CH_3I)$ via nucleophilic substitution ($S_N2$ reactions).
$2$. The primary amine reacts to form a secondary amine: $CH_3CH_2NH_2 + CH_3I \rightarrow CH_3CH_2NH(CH_3) + HI$.
$3$. The secondary amine reacts further to form a tertiary amine: $CH_3CH_2NH(CH_3) + CH_3I \rightarrow CH_3CH_2N(CH_3)_2 + HI$.
$4$. The tertiary amine reacts with the remaining excess methyl iodide to form a quaternary ammonium salt: $CH_3CH_2N(CH_3)_2 + CH_3I \rightarrow [CH_3CH_2N(CH_3)_3]^+I^-$.
$5$. The final product is ethyltrimethylammonium iodide.

(B) Step $1$: The Hofmann bromamide degradation reaction involves the conversion of an amide $(RCONH_2)$ to a primary amine $(RNH_2)$ using $Br_2$ and $KOH$ (or $NaOH$).
Step $2$: In this reaction, the carbonyl group $(-CO-)$ is removed as a carbonate $(K_2CO_3)$, not reduced to a $-CH_2-$ group.
Step $3$: Since one carbon atom is lost during the reaction, the resulting amine has one carbon atom less than the parent amide, making its molar mass lower.
Step $4$: Therefore, the statement that the $-CO-$ group is reduced to a $-CH_2-$ group is false.

(A) $1$. The reaction described is the carbylamine reaction, which is a characteristic test for primary $(1^\circ)$ amines.
$2$. In this reaction, a primary amine reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine), which has a foul smell.
$3$. Among the given options, $\text{Benzenamine}$ (aniline, $C_6H_5NH_2$) is a primary aromatic amine.
$4$. $N-\text{methylaniline}$ and $N,N-\text{Dimethylaniline}$ are secondary and tertiary amines, respectively, and do not undergo the carbylamine reaction.
$5$. $\text{Ethylethanamine}$ (diethylamine) is a secondary aliphatic amine and also does not undergo this reaction.
$6$. Therefore, $\text{Benzenamine}$ is the correct answer.

(B) Step $1$: Hofmann's exhaustive alkylation involves the reaction of a primary amine with an excess of alkyl halide to form a mixture of secondary amine, tertiary amine, and finally a quaternary ammonium salt.
Step $2$: Option $(A)$ is correct as the final product is a quaternary ammonium salt.
Step $3$: Option $(B)$ is incorrect because the reaction of $1 \text{ mole}$ of primary amine with $2 \text{ moles}$ of alkyl halide produces a tertiary amine, not a tetraalkyl ammonium halide. Tetraalkyl ammonium halide requires $4 \text{ moles}$ of alkyl halide.
Step $4$: Option $(C)$ is correct as the product distribution depends on the molar ratio of reactants.
Step $5$: Option $(D)$ is correct as the reaction releases $HX$, which is neutralized by a base like $\text{NaHCO}_3$ or excess amine to drive the reaction forward.

(B) $1$. Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
$2$. It involves the reaction of potassium phthalimide with an alkyl halide followed by alkaline hydrolysis.
$3$. Aromatic amines $(C_6H_5NH_2)$ cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
$4$. Secondary and tertiary amines cannot be prepared because the reaction requires an $S_N2$ mechanism, which is hindered by steric factors in secondary and tertiary alkyl halides.
$5$. Among the options, $C_2H_5NH_2$ is a primary aliphatic amine, which can be prepared using this method.

(C) The reaction of an amide with $Br_2$ and aqueous $KOH$ is known as the $Hofmann$ bromamide degradation reaction.
In this reaction, the amide group $(-CONH_2)$ is converted into a primary amine $(-NH_2)$ with the loss of one carbon atom.
Propionamide is $CH_3CH_2CONH_2$.
Reaction: $CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$.
The product obtained is ethylamine $(CH_3CH_2NH_2)$.

(D) In the Hofmann bromamide degradation reaction, a primary amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide. This reaction results in the degradation of the amide, where the carbonyl group $(-CO-)$ is removed as a carbonate, yielding a primary amine with one carbon atom less than the original amide.
Reaction: $R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.

(A) $1$. The reaction is the Hofmann elimination of a quaternary ammonium salt.
$2$. Step $(i)$ involves the exchange of the halide ion with the hydroxide ion using moist $Ag_2O$ $(AgOH)$, forming the quaternary ammonium hydroxide: $A + AgOH \rightarrow [CH_3CH_2N^+(CH_3)_3]OH^- + AgX$.
$3$. Step (ii) is the thermal decomposition (Hofmann elimination) of the hydroxide. According to the Hofmann rule, the less substituted alkene is formed. Here, $CH_2=CH_2$ (ethene) is formed from the ethyl group, and the remaining part is $CH_3CH_2N(CH_3)_2$.
$4$. Thus, the substrate '$A$' must be $Ethyltrimethylammonium$ halide.

(A) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2 / HCl$ at $0^\circ C$ to form Benzenediazonium chloride $(A = C_6H_5N_2^+Cl^-)$.
Step $2$: $A$ reacts with $CuCN / KCN$ to form Benzonitrile $(B = C_6H_5CN)$.
Step $3$: $B$ undergoes catalytic hydrogenation with $H_2 / Ni$ to form Benzylamine $(C = C_6H_5CH_2NH_2)$.
Step $4$: Benzylamine $(C)$ reacts with nitrous acid $(HNO_2)$ to form Benzyl alcohol $(D = C_6H_5CH_2OH)$ along with $N_2$ gas and $H_2O$.

(B) Step $1$: $HDL$ (High-Density Lipoprotein) is often referred to as 'good cholesterol' because it helps remove other forms of cholesterol from the bloodstream.
Step $2$: Dietary intake of polyunsaturated fatty acids, specifically $\text{Omega-3 fatty acids}$ (found in fish oil, flaxseeds, and walnuts), is clinically shown to increase $HDL$ levels.
Step $3$: In contrast, $\text{Trans fats}$ are known to lower $HDL$ and raise $LDL$ (bad cholesterol), while $\text{Omega-6 fatty acids}$ have a neutral or slightly lower effect on $HDL$ compared to $\text{Omega-3}$.
Step $4$: Therefore, $\text{Omega-3 fatty acids}$ are the correct choice.

(A) $1$. Saturated fats contain no double bonds between carbon atoms in their fatty acid chains.
$2$. Coconut oil is unique among plant oils because it has a very high concentration of saturated fatty acids (approximately $80-90\%$), primarily lauric acid.
$3$. Olive oil, safflower oil, and fish oil are primarily composed of unsaturated fatty acids.
$4$. Therefore, coconut oil is the correct answer.

(A) Step $1$: Saturated fats are typically solid at room temperature and are commonly found in animal products and certain processed fats.
Step $2$: Olive oil, safflower oil, and fish oil are primarily composed of unsaturated fatty acids and are liquid at room temperature.
Step $3$: Margarine is produced by the hydrogenation of vegetable oils, which converts unsaturated fats into saturated fats, making it the correct choice among the options provided.

(A) Step $1$: Ascorbic acid is chemically known as Vitamin $C$.
Step $2$: Vitamin $C$ is abundantly found in citrus fruits such as oranges, lemons, and grapefruits.
Step $3$: Cinnamon, turmeric, and cloves are spices and are not primary sources of ascorbic acid.
Step $4$: Therefore, the correct option is $A$.

(B) Step $1$: Citrus fruits such as lemons, oranges, and grapefruits are well-known natural sources of ascorbic acid.
Step $2$: Ascorbic acid is commonly known as $\text{Vitamin C}$.
Step $3$: $\text{Vitamin C}$ is essential for human health, acting as an antioxidant and supporting immune function, making it a medicinally important compound.
Therefore, the correct option is $B$.

(C) Ascorbic acid is commonly known as $Vitamin \ C$. It is found in abundance in citrus fruits such as oranges, lemons, and grapefruits. Therefore, the correct option is $C$.

(C) The genetic information in $DNA$ is encoded in the specific order or sequence of nitrogenous bases ($Adenine$, $Thymine$, $Cytosine$, and $Guanine$) along the sugar-phosphate backbone. This sequence acts as a code that determines the synthesis of proteins and the traits of an organism.
Therefore, the correct option is $C$.

(D) $Teflon$ is a synthetic fluoropolymer of tetrafluoroethylene.
Its chemical formula is $(C_2F_4)_n$.
The monomer unit is tetrafluoroethylene, which is $CF_2 = CF_2$.
Therefore, the correct option is $D$.

(B) Step $1$: $Orlon$ is a synthetic polymer also known as polyacrylonitrile $(PAN)$.
Step $2$: The polymerization of acrylonitrile $(CH_2=CH-CN)$ leads to the formation of $Orlon$.
Step $3$: Therefore, the monomer used to prepare $Orlon$ is acrylonitrile.

(A) The manufacture of $HDPE$ (High-Density Polyethylene) is carried out by the polymerization of ethene in the presence of a Ziegler-Natta catalyst.
Ziegler-Natta catalyst is a mixture of triethyl aluminum $(Al(C_2H_5)_3)$ and titanium tetrachloride $(TiCl_4)$.
Therefore, the correct option is $A$.

(C) $1$. Vulcanization is a chemical process in which natural rubber is heated with sulfur.
$2$. This process creates cross-links between the polymer chains of rubber.
$3$. These cross-links significantly increase the mechanical properties of rubber, specifically its tensile strength, elasticity, and durability.
$4$. Therefore, vulcanization improves the tensile strength of rubber.

(B) $1$. $Dacron$ (also known as $Terylene$) is a polyester polymer.
$2$. It is formed by the condensation polymerization of ethylene glycol $(HO-CH_2-CH_2-OH)$ and terephthalic acid $(C_6H_4(COOH)_2)$.
$3$. During the formation of the polymer, the carboxylic acid group of the acid and the hydroxyl group of the alcohol react to form an ester linkage $(-COO-)$.
$4$. Therefore, $Dacron$ contains an ester linkage.

(C) $1$. $Bakelite$ is a thermosetting polymer formed by the condensation polymerization of $Phenol$ and $Formaldehyde$.
$2$. $Methanal$ is the $IUPAC$ name for $Formaldehyde$ $(HCHO)$.
$3$. The reaction involves the initial formation of $o$- or $p$-hydroxymethylphenol derivatives, which then undergo cross-linking to form the polymer.
$4$. Therefore, the correct monomers are $Phenol$ and $Methanal$.

(B) Step $1$: Natural rubber is a linear polymer of isoprene and is soft and sticky.
Step $2$: To improve its physical properties, it is heated with sulphur. This process is known as vulcanization.
Step $3$: During vulcanization, sulphur forms cross-links at the reactive sites of double bonds, which makes the rubber harder, stronger, and more durable.
Step $4$: Therefore, vulcanization is the process that develops cross-links in elastomers.

(B) Step $1$: Identify the monomers of $Buna-S$. The name $Buna-S$ is derived from $Bu$ (for $1, 3$-butadiene), $na$ (for sodium, which acts as a catalyst), and $S$ (for styrene).
Step $2$: The polymerization reaction involves the copolymerization of $1, 3$-butadiene $(CH_2=CH-CH=CH_2)$ and styrene $(C_6H_5CH=CH_2)$.
Step $3$: Therefore, $Buna-S$ is prepared from $1, 3$-butadiene and styrene.

(B) Step $1$: Identify the properties required for nonstick cookware, such as high thermal stability and chemical inertness.
Step $2$: $Teflon$ (Polytetrafluoroethylene, $PTFE$) is a synthetic fluoropolymer of tetrafluoroethylene that has these properties.
Step $3$: Due to its low coefficient of friction and non-stick nature, it is widely used as a coating for nonstick pans and other cookware.
Step $4$: Therefore, the correct polymer is $Teflon$.

(D) $1$. Plastic bottles used for storing cooking oil and carbonated drinks are typically made from $Polyethylene \ terephthalate$ ($PET$ or $PETE$).
$2$. $PET$ is a thermoplastic polymer resin of the polyester family, known for its clarity, strength, and chemical resistance to oils and food products.
$3$. $PVC$ is used for pipes, $PS$ for disposable containers, and $LDPE$ for flexible films and bags.
$4$. Therefore, the correct option is $D$.

(C) $1$. Raincoats are made from water-resistant materials.
$2$. Polyvinyl chloride $(PVC)$, represented by the structure $- [CH_2 - CH(Cl)]_n -$, is a widely used thermoplastic polymer known for its water-repellent properties.
$3$. It is extensively used in the manufacturing of raincoats, handbags, water pipes, and floor coverings.
$4$. Therefore, the correct option is $C$.

(A) The preparation of viscose rayon involves several steps. The first stage is the steeping of cellulose (usually wood pulp) in a $17-20\%$ solution of $NaOH$ to form alkali cellulose.
Step $1$: Steeping (Formation of alkali cellulose).
Step $2$: Pressing.
Step $3$: Shredding.
Step $4$: Aging.
Step $5$: Xanthation.
Step $6$: Dissolving.
Step $7$: Ripening.
Therefore, the first stage is the treatment with $NaOH$.

(B) Step $1$: Disposable cups and plates are typically made from materials that are lightweight, inexpensive, and easy to mold.
Step $2$: $Polystyrene$ (specifically expanded polystyrene or $EPS$) is widely used for manufacturing disposable food service items like cups, plates, and containers due to its excellent insulating properties and low cost.
Step $3$: Therefore, the correct polymer is $Polystyrene$.

(B) $1$. $HDPE$ (High-Density Polyethylene) is a thermoplastic polymer known for its high strength-to-density ratio.
$2$. Due to its rigidity and chemical resistance, it is widely used in the manufacturing of blow-molded products such as shampoo bottles, detergent bottles, and milk jugs.
$3$. Shopping bags are typically made from $LDPE$ (Low-Density Polyethylene), while carpets are often made from fibers like nylon or polypropylene.
$4$. Therefore, the correct use of $HDPE$ among the given options is to prepare shampoo bottles.

(D) $LDP$ stands for Low Density Polyethene.
It is a tough, flexible, and chemically inert polymer.
It is primarily used in the manufacture of squeeze bottles, toys, and flexible pipes.
Buckets are typically made from $HDP$ (High Density Polyethene).
Pipes for irrigation are also made from $HDP$.
Non-stick cookware uses $PTFE$ (Teflon).
Therefore, the correct use of $LDP$ is for making squeeze bottles and flexible pipes.