MHT CET 2020 Chemistry Question Paper with Answer and Solution

(D) Successive members of a homologous series differ by a $-CH_2-$ group.
The molar mass of a carbon atom is $12 \ g \ mol^{-1}$ and that of two hydrogen atoms is $2 \times 1 \ g \ mol^{-1} = 2 \ g \ mol^{-1}$.
Therefore,the difference in molar mass between any two neighbouring alkanes is $12 + 2 = 14 \ g \ mol^{-1}$.

(D) The structure of isobutyl chloride is $(CH_3)_2CH-CH_2Cl$.
To determine the $IUPAC$ name,we identify the longest carbon chain containing the functional group.
The longest chain has $3$ carbon atoms (propane).
There is a methyl group at the $2^{nd}$ position and a chlorine atom at the $1^{st}$ position.
Therefore,the $IUPAC$ name is $1-$chloro$-2-$methylpropane.

(C) $Pentan-2-ol$ $(C_5H_{12}O)$ is an alcohol.
Functional isomers are compounds with the same molecular formula but different functional groups.
$Ethoxypropane$ $(C_2H_5-O-C_3H_7)$ has the molecular formula $C_5H_{12}O$ and belongs to the ether functional group.
Therefore,$ethoxypropane$ is a functional isomer of $pentan-2-ol$.
$Pentan-1-ol$ and $pentan-3-ol$ are positional isomers,while $pentan-2-one$ is a ketone with the formula $C_5H_{10}O$.

(C) The structure of isobutane is $(CH_3)_3CH$.
In this structure:
$1$. $A$ primary $(1^{\circ})$ carbon atom is attached to only one other carbon atom. There are three such $CH_3$ groups in isobutane.
$2$. $A$ secondary $(2^{\circ})$ carbon atom is attached to two other carbon atoms. There are no such carbon atoms in isobutane.
$3$. $A$ tertiary $(3^{\circ})$ carbon atom is attached to three other carbon atoms. The central carbon atom is attached to three $CH_3$ groups,so it is a tertiary carbon.
Thus,there are $3$ primary,$0$ secondary,and $1$ tertiary carbon atom in isobutane.

(C) The given molecule is $2$-methylpent-$1$-en-$4$-yne,which has the structure $CH_2=C(CH_3)-CH_2-C\equiv CH$.
Counting the bonds:
$1$. $\sigma$ bonds: There are $16$ $\sigma$ bonds in total ($1$ in $C=C$,$1$ in $C-C$,$1$ in $C-C$,$1$ in $C\equiv C$,and $12$ $C-H$ bonds).
$2$. $\pi$ bonds: There is $1$ $\pi$ bond in the double bond $(C=C)$ and $2$ $\pi$ bonds in the triple bond $(C\equiv C)$,totaling $3$ $\pi$ bonds.
Thus,the molecule contains $3 \pi$ and $16 \sigma$ bonds.

(C) The boiling point of alkanes depends on the surface area of the molecule.
As the branching in the alkane chain increases,the surface area decreases,leading to weaker van der Waals forces of attraction.
Therefore,the boiling point decreases as branching increases.
$n$-pentane is a straight-chain alkane with the largest surface area,followed by Isopentane (one branch),and Neopentane (two branches) has the smallest surface area.
Thus,the decreasing order of boiling point is: $n$-pentane $>$ Isopentane $>$ Neopentane.

(C) $I$. $CH_{3}-CH_{3}$ $(C_{2}H_{6})$: Contains only $\sigma$ bonds. (No $\pi$ bond)
$II$. $CH_{2}=CH_{2}$ $(C_{2}H_{4})$: Contains $1$ $\sigma$ bond and $1$ $\pi$ bond between carbons.
$III$. $CH \equiv CH$ $(C_{2}H_{2})$: Contains $1$ $\sigma$ bond and $2$ $\pi$ bonds between carbons.
$IV$. $CH_{3}-CH=CH_{2}$ $(C_{3}H_{6})$: Contains $1$ $\pi$ bond in the $C=C$ double bond.

(B) The percentage of $p$-character in a hybrid orbital is determined by its hybridization state.
In $sp$ hybridization,there is $50 \%$ $s$-character and $50 \%$ $p$-character.
In $sp^2$ hybridization,there is $33.33 \%$ $s$-character and $66.67 \%$ $p$-character.
In $sp^3$ hybridization,there is $25 \%$ $s$-character and $75 \%$ $p$-character.
Acetylene $(C_2H_2)$ has $sp$ hybridized carbon atoms,which corresponds to $50 \%$ $p$-character.

(B) The stability of free radicals is governed by the inductive effect and hyperconjugation.
The order of stability is $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$.
$R_3C^{\bullet}$ is a tertiary $(3^{\circ})$ radical,which is the most stable due to the maximum number of hyperconjugative structures and the electron-donating inductive effect of three $R$ groups.

(C) The reactivity of a carbocation is inversely proportional to its stability.
$CH_{3}^{+}$ is a methyl carbocation,which is the least stable among the given options because it lacks any electron-donating groups (such as alkyl groups) to stabilize the positive charge through $+I$ (inductive) or hyperconjugation effects.
Since it is the least stable,it is the most reactive.

(C) The methyl free radical is represented as $\cdot CH_3$.
In this structure,the carbon atom is bonded to three hydrogen atoms by three covalent bonds.
Each $C-H$ bond consists of $2$ electrons,contributing $3 \times 2 = 6$ electrons.
Additionally,there is one unpaired electron on the carbon atom.
Therefore,the total number of electrons around the carbon atom is $6 + 1 = 7$.

(C) The $(+R)$ effect (positive resonance effect) is shown by groups that donate electrons through resonance,typically having a lone pair of electrons on the atom directly attached to the conjugated system.
$-NH_2$,$-NHCOR$,and $-NR_2$ all possess a lone pair on the nitrogen atom and thus exhibit the $(+R)$ effect.
$-CN$ contains a multiple bond between carbon and nitrogen,which acts as an electron-withdrawing group via resonance,thus showing the $(-R)$ effect.

(D) The resonance effect ($R$ effect) is the polarity produced in a molecule by the interaction of two $\pi$-bonds or between a $\pi$-bond and a lone pair of electrons present on an adjacent atom.
Groups like $-CHO$,$-COOH$,and $-CN$ contain multiple bonds with electronegative atoms,which withdraw electron density from the conjugated system,thus showing a $(-R)$ effect.
Conversely,the $-OH$ group possesses lone pairs of electrons on the oxygen atom that can be donated into the conjugated system,thereby exhibiting a $(+R)$ effect.

(A) Resonance requires a conjugated system of $\pi$ bonds,such as alternating single and double bonds or the presence of lone pairs adjacent to a $\pi$ system.
$1$. $Aniline$,$nitrobenzene$,and $phenol$ all contain a benzene ring,which is a conjugated system of $\pi$ electrons,allowing for resonance.
$2$. $Cyclohexane$ $(C_6H_{12})$ is a saturated cyclic hydrocarbon where all carbon atoms are $sp^3$ hybridized and connected by single $\sigma$ bonds.
$3$. Since $cyclohexane$ lacks any $\pi$ bonds or a conjugated system,it does not exhibit resonance.

(A) The given compound is $2-$methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$.
There are $4$ distinct types of hydrogen atoms in $2-$methylbutane,which leads to the formation of $4$ different monobromo isomers upon bromination:
$1.$ $1-$bromo$-2-$methylbutane
$2.$ $2-$bromo$-2-$methylbutane
$3.$ $2-$bromo$-3-$methylbutane
$4.$ $1-$bromo$-3-$methylbutane
Therefore,the total number of isomers is $4$.

(A) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1$. $2-$Chloropentane: $CH_3-CHCl-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is a chiral center. Thus,it is optically active.
$2$. $3-$Chloropentane: $CH_3-CH_2-CHCl-CH_2-CH_3$. The $C-3$ atom is bonded to two identical ethyl groups $(-CH_2CH_3)$,so it is achiral.
$3$. $2-$Chloropropane: $CH_3-CHCl-CH_3$. The $C-2$ atom is bonded to two identical methyl groups $(-CH_3)$,so it is achiral.
$4$. $2-$Chloro$-2-$methylbutane: $CH_3-CCl(CH_3)-CH_2-CH_3$. The $C-2$ atom is bonded to two identical methyl groups $(-CH_3)$,so it is achiral.
Therefore,$2-$Chloropentane is the optically active compound.

(C) In the $Cahn-Ingold-Prelog$ $(CIP)$ priority rules,phantom atoms are used to represent multiple bonds by expanding them into single bonds.
- Acetaldehyde $(CH_3CHO)$ contains a $C=O$ double bond,which requires phantom atoms.
- Methyl cyanide $(CH_3CN)$ contains a $C \equiv N$ triple bond,which requires phantom atoms.
- Propionic acid $(CH_3CH_2COOH)$ contains a $C=O$ double bond,which requires phantom atoms.
- $n$-propyl alcohol $(CH_3CH_2CH_2OH)$ contains only single bonds between all atoms. Therefore,it does not require any phantom atoms.

(B) The number of optical isomers for a compound with $n$ asymmetric (chiral) carbon atoms is given by the formula $2^n$,provided there is no internal symmetry in the molecule.
Given $n = 3$,the number of optical isomers is $2^3 = 8$.

(A) According to the Cahn-Ingold-Prelog $(CIP)$ priority rules,phantom atoms are used to account for multiple bonds (double or triple bonds) by expanding them into single bonds.
In $CH_{3}COOH$ (acetic acid),the carbonyl group $(C=O)$ contains a double bond between carbon and oxygen.
To assign priorities,the $C=O$ bond is expanded by treating the carbon as bonded to two oxygens (one real,one phantom) and the oxygen as bonded to two carbons (one real,one phantom).
Since the other options ($CH_{3}CH_{2}Cl$,$CH_{3}CH_{2}NH_{2}$,and $CH_{3}CH_{2}CH_{2}OH$) contain only single bonds,they do not require the use of phantom atoms for priority assignment.
Therefore,$CH_{3}COOH$ is the correct answer.

(A) The number of optical isomers for a compound with $n$ asymmetric carbon atoms is given by the formula $2^{n}$.
Given that the compound has $n = 4$ asymmetric carbon atoms.
Therefore,the number of optical isomers is $2^{4} = 16$.

(A) compound is optically active if it lacks a plane of symmetry or center of symmetry (i.e.,it is chiral).
$3-$Chloropentane $(CH_3CH_2CHClCH_2CH_3)$ has a plane of symmetry passing through the $C-3$ atom,making it achiral and thus optically inactive.
$2-$Chloropentane,$2-$chloro$-3-$methylpentane,and $3-$chloro$-2-$methylpentane all contain at least one chiral carbon atom and lack internal symmetry,making them optically active.

(D) The general formula for calculating the number of optical isomers is $2^{n}$,where $n$ represents the number of asymmetric (chiral) carbon atoms.
An asymmetric carbon atom is a carbon atom bonded to $4$ different groups.
Therefore,for a molecule with $n$ chiral centers,the maximum number of optical isomers is $2^{n}$.

(C) The structure of $3,4-$dichloropentan$-2-$ol is $CH_3-CH(OH)-CH(Cl)-CH(Cl)-CH_3$.
This molecule has $3$ chiral centers (at $C2, C3,$ and $C4$).
Since the molecule is asymmetric (the groups attached to the chiral centers are different),the number of optical isomers is given by the formula $2^n$,where $n$ is the number of chiral centers.
Here,$n = 3$.
Therefore,the number of optical isomers = $2^3 = 8$.

(C) The separation of $Acetone$ and $Benzene$ from their mixture is effectively carried out by $Fractional \ distillation$ using a $fractionating \ column$.
This method is employed because the boiling points of $Acetone$ $(56 \ ^\circ C)$ and $Benzene$ $(80 \ ^\circ C)$ are relatively close,necessitating the use of a fractionating column to achieve efficient separation through repeated vaporization and condensation cycles.

(A) In gas chromatography,an inert carrier gas is required to transport the sample through the column. $Helium$ $(He)$ is the most commonly used carrier gas due to its inert nature,high thermal conductivity,and safety compared to $Hydrogen$. While $Argon$ is inert,$Helium$ is preferred in standard analytical applications.

(A) The correct method is $sublimation$.
$Ammonium \ chloride$ $(NH_4Cl)$ is a sublime substance,meaning it changes directly from a solid to a gas upon heating.
$Sodium \ chloride$ $(NaCl)$ does not sublime.
Therefore,when the mixture is heated,$NH_4Cl$ vaporizes and can be collected,leaving $NaCl$ behind.

(C) Magnesite is a mineral with the chemical formula $MgCO_3$.
It is a primary source of magnesium.
Limonite is an ore of iron,Cryolite is an ore of aluminium,and Magnetite is an ore of iron.

(D) The reaction of a mixture of two different alkyl halides with sodium metal in the presence of dry ether is known as the Wurtz reaction. $2CH_3I + 2C_2H_5I + 4Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + CH_3-CH_2-CH_2-CH_3 + CH_3-CH_2-CH_3 + 4NaI$. The products formed are a mixture of alkanes: ethane $(CH_3-CH_3)$,butane $(CH_3-CH_2-CH_2-CH_3)$,and propane $(CH_3-CH_2-CH_3)$.

(A) The reaction of isobutylene $(CH_3)_2C=CH_2$ with hydrogen bromide $(HBr)$ follows Markovnikov's rule.
According to this rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom having a lesser number of hydrogen atoms.
The reaction proceeds via the formation of a more stable $3^{\circ}$ carbocation intermediate.
$(CH_3)_2C=CH_2 + HBr \rightarrow (CH_3)_3C-Br$
The major product formed is $tert-$butyl bromide.

(C) The peroxide effect (Kharasch effect) is only applicable to the addition of $HBr$ to unsymmetrical alkenes.
It does not apply to $HCl$ or $HI$.
Therefore,the addition of $HCl$ to propene follows Markovnikov's rule $(M.R.)$ even in the presence of peroxide.
According to Markovnikov's rule,the negative part of the addendum $(Cl^-)$ attaches to the carbon atom with fewer hydrogen atoms.
Reaction: $CH_3-CH=CH_2 + HCl \xrightarrow{Na_2O_2} CH_3-CHCl-CH_3$ ($2-$Chloropropane).

(C) $(i)$ Propylene dibromide $(CH_3-CHBr-CH_2Br)$ $\xrightarrow{Zn, \Delta}$ Propene $(CH_3-CH=CH_2)$ $(A)$.
$(ii)$ Propene $(A)$ $\xrightarrow{HBr}$ $2-$Bromopropane $(CH_3-CHBr-CH_3)$ $(B)$ (following Markovnikov's rule).
$(iii)$ $2-$Bromopropane $(B)$ $\xrightarrow{Na, \text{ether}}$ $2,3-$Dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$ $(C)$ (Wurtz reaction).

(D) Diesel fuel consists mainly of paraffins,aromatics,and naphthenes.
Gasoline hydrocarbons typically contain $4-12$ carbon atoms with a boiling range between $30 \ ^\circ C$ and $210 \ ^\circ C$.
In contrast,diesel fuel contains hydrocarbons with approximately $15-18$ carbon atoms and a boiling range between $170 \ ^\circ C$ and $360 \ ^\circ C$.

(B) For an alkane with general formula $C_nH_{2n+2}$,the molar mass is $12n + 1(2n+2) = 14n + 2 = 240 \ g \ mol^{-1}$.
$14n = 238$,so $n = 17$.
The alkane is $C_{17}H_{36}$.
The mass percentage of hydrogen is $\frac{36}{240} \times 100 = 15 \ \%$.
$C_{17}H_{36}$ is a component found in diesel fuel.

(C) $Undecane$ is a volatile hydrocarbon secreted by cockroaches as a sex pheromone to attract the opposite gender of its species.

(A) The reaction sequence is as follows:
$1$. Compound $A$ is $n$-butane $(CH_{3}CH_{2}CH_{2}CH_{3})$.
$2$. Reaction with $Cl_{2}$ in the presence of $U.V.$ light (free radical substitution) gives $2$-chlorobutane $(B)$ as the major product: $CH_{3}CH_{2}CH_{2}CH_{3} + Cl_{2} \xrightarrow{h\nu} CH_{3}CH_{2}CHClCH_{3} (B) + HCl$.
$3$. $2$-chlorobutane $(B)$ reacts with $NaNO_{2}$ in $DMF$ (a polar aprotic solvent) via an $S_{N}2$ mechanism to form $2$-nitrobutane: $CH_{3}CH_{2}CHClCH_{3} + NaNO_{2} \xrightarrow{DMF} CH_{3}CH_{2}CH(NO_{2})CH_{3} + NaCl$.
Thus,compound $A$ is butane.

(C) The reaction of $n$-hexane $(CH_{3}-(CH_{2})_{4}-CH_{3})$ in the presence of $Cr_{2}O_{3}$ supported on alumina at $773 \ K$ and $10-20 \ atm$ pressure is known as aromatization or catalytic reforming.
In this process,$n$-hexane undergoes cyclization and dehydrogenation to form benzene $(C_{6}H_{6})$ and hydrogen gas $(H_{2})$.
The balanced chemical equation is:
$CH_{3}-(CH_{2})_{4}-CH_{3} \xrightarrow{Cr_{2}O_{3}, 773 \ K, 10-20 \ atm} C_{6}H_{6} + 4 \ H_{2}$
Thus,the product $X$ is $C_{6}H_{6} + 4 \ H_{2}$.

(C) Step $1$: $CH_{3}CH_{2}COOH + NaOH \rightarrow CH_{3}CH_{2}COONa (A) + H_{2}O$
Step $2$: $CH_{3}CH_{2}COONa + NaOH \xrightarrow{CaO, \Delta} CH_{3}CH_{3} (B) + Na_{2}CO_{3}$ (Decarboxylation)
Step $3$: $CH_{3}CH_{3} + HNO_{3} \xrightarrow{\Delta} CH_{3}CH_{2}NO_{2} (C) + H_{2}O$ (Nitration of alkane)
Thus,the product $C$ is nitroethane $(CH_{3}CH_{2}NO_{2})$.

(A) Step $1$: The reaction of ethane $(C_{2}H_{6})$ with bromine $(Br_{2})$ in the presence of $AlBr_{3}$ is a bromination reaction,which yields ethyl bromide $(A = C_{2}H_{5}Br)$.
Step $2$: The reaction of ethyl bromide $(C_{2}H_{5}Br)$ with silver acetate $(CH_{3}COOAg)$ is a nucleophilic substitution reaction (specifically,an esterification reaction).
Step $3$: The reaction proceeds as follows: $C_{2}H_{5}Br + CH_{3}COOAg \rightarrow CH_{3}COOC_{2}H_{5} + AgBr$.
Step $4$: The product $B$ is ethyl acetate,which is $CH_{3}COOC_{2}H_{5}$.

(C) The oxidation of $Cyclohexene$ with strong oxidizing agents like acidic $KMnO_4$ leads to the oxidative cleavage of the double bond.
This reaction breaks the ring and results in the formation of a dicarboxylic acid.
Specifically,$Cyclohexene$ reacts with $KMnO_4$ and dilute $H_2SO_4$ to produce $Hexanedioic$ acid,which is commonly known as adipic acid.
The reaction is represented as: $Cyclohexene + 4[O] \xrightarrow{KMnO_4 / dil. H_2SO_4, \Delta} HOOC-(CH_2)_4-COOH$ (Adipic acid).

(A) Hydrogen resembles halogens in several properties:
$1$. Both exist as diatomic molecules $(H_2, F_2, Cl_2, Br_2, I_2)$.
$2$. Both have high ionisation enthalpy values.
$3$. Both form uninegative ions $(H^-, F^-, Cl^-, Br^-, I^-)$ by gaining one electron.
$4$. Both form covalent compounds with non-metals.
Among the given options,the ionisation enthalpy of hydrogen is comparable to that of halogens,making it the most appropriate similarity in terms of periodic trends.

(D) In the Bosch process,water gas $(CO + H_{2})$ is reacted with steam to produce $CO_{2}$ and $H_{2}$.
$Fe_{2}O_{3}$ acts as a catalyst,while $Cr_{2}O_{3}$ is used as a promoter to increase the efficiency of the catalyst.

(C) The hydride $NiH$ is an example of a metallic or non-stoichiometric hydride.
These hydrides are often electron-deficient,meaning they do not have a sufficient number of electrons to form normal covalent bonds,which is why they are also referred to as non-stoichiometric hydrides.

(C) $HF$ is a molecular (covalent) hydride because it is formed by the sharing of electrons between hydrogen and a non-metal (fluorine).
$KH$,$NaH$,and $LiH$ are examples of saline or ionic hydrides,which are formed by the transfer of electrons from an alkali metal to hydrogen.

(B) The formula of hydrolith is $CaH_{2}$.
Calcium hydride $(CaH_{2})$ is a salt-like binary hydride that is commonly used as a reducing agent and as a convenient source of hydrogen gas.

(B) In the gas phase,the $H-O-O-H$ dihedral angle in $H_{2}O_{2}$ is $111.5^{\circ}$,while the $O-O-H$ bond angle is $94.8^{\circ}$.
In the solid phase,the $H-O-O-H$ dihedral angle is $90.2^{\circ}$ and the $O-O-H$ bond angle is $101.9^{\circ}$.
The question asks for the $H-O-O-H$ bond angle (often referred to as the dihedral angle in this context) in the gas phase,which is $111.5^{\circ}$. However,based on the provided image and options,the question is specifically asking for the $O-O-H$ bond angle in the gas phase,which is $94.8^{\circ}$.

(D) The standard state of a substance is its most stable physical form at a specified temperature (usually $298 \text{ K}$ or $25^{\circ} C$) and $1 \text{ bar}$ (or $1 \text{ atm}$) pressure.
At $25^{\circ} C$ and $1 \text{ atm}$ pressure,water exists as a liquid $(H_{2}O_{(l)})$,not as a gas $(H_{2}O_{(g)})$.
Therefore,$H_{2}O_{(g)}$ is not in its standard state under these conditions.

(B) In the crystalline structure of ice,each oxygen atom is surrounded by four hydrogen atoms.
Two of these hydrogen atoms are covalently bonded to the oxygen atom,while the other two are linked through hydrogen bonding.
This arrangement creates a tetrahedral geometry around each oxygen atom.

(A) The boiling point of heavy water $(D_2O)$ is $101.4^{\circ} C$ $(374.4 \ K)$.
This is higher than the boiling point of ordinary water $(H_2O)$,which is $100^{\circ} C$ $(373 \ K)$,due to stronger hydrogen bonding in $D_2O$.

(B) The correct answer is $B$ (Petrol).
Petrol is a mixture of hydrocarbons containing carbon atoms typically in the range of $C_{6}$ to $C_{8}$.
It consists of various alkanes (linear or branched) and cycloalkanes.
Specifically,petrol is composed of alkanes ranging from pentane $(C_{5}H_{12})$ to octane $(C_{8}H_{18})$,making it the correct choice for the given range.

(A) Paraffin wax is a white or colourless soft solid that is derived from petroleum,coal,or oil shale.
It consists of a mixture of hydrocarbon molecules,specifically alkanes.
The range of carbon atoms in these alkanes typically falls between $C_{20}$ and $C_{40}$,but the most common range cited for commercial paraffin wax is $C_{21}$ to $C_{30}$.
It is widely used in applications such as candles,wax paper,polishes,cosmetics,and electrical insulation.

(D) The reaction of propanenitrile $(CH_3CH_2CN)$ with ethylmagnesium iodide $(C_2H_5MgI)$ in the presence of dry ether proceeds as follows:
$1$. Nucleophilic attack of the ethyl group from the Grignard reagent on the carbon atom of the nitrile group leads to the formation of an imine magnesium complex: $CH_3CH_2CN + C_2H_5MgI \rightarrow CH_3CH_2C(C_2H_5)=NMgI$.
$2$. Acid hydrolysis $(H_3O^+)$ of this imine complex results in the formation of a ketone and ammonia: $CH_3CH_2C(C_2H_5)=NMgI + H_2O/H^+ \rightarrow CH_3CH_2COCH_2CH_3 + NH_3 + Mg(OH)I$.
$3$. The product formed is $CH_3CH_2COCH_2CH_3$,which is $3-$pentanone (pentan-$3-$one).
Therefore,the correct option is $D$.

(A) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in the presence of dry ether forms an intermediate imine complex $(A)$.
This imine complex is then hydrolyzed by $H_3O^+$ to yield benzophenone $(C_6H_5COC_6H_5)$ as the final product '$B$'.
The reaction is: $C_6H_5CN + C_6H_5MgBr$ $\rightarrow C_6H_5C(NMgBr)C_6H_5$ $\xrightarrow{H_3O^+} C_6H_5COC_6H_5 + NH_3 + Mg(Br)(OH)$.

(B) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the carbonyl compound.
$1$. Acetone $(CH_3COCH_3)$ has $6$ $\alpha$-hydrogens.
$2$. Benzophenone $(C_6H_5COC_6H_5)$ has no $\alpha$-hydrogen atom attached to the carbonyl carbon.
$3$. Acetaldehyde $(CH_3CHO)$ has $3$ $\alpha$-hydrogens.
$4$. Acetophenone $(C_6H_5COCH_3)$ has $3$ $\alpha$-hydrogens.
Therefore,Benzophenone does not undergo aldol condensation.

(D) The reaction of a Grignard reagent $(CH_3MgBr)$ with a carbonyl compound followed by acid hydrolysis $(H_3O^{+})$ produces an alcohol.
In the given reaction,the product is tert-butyl alcohol,which is $(CH_3)_3C-OH$.
Since the Grignard reagent provides one methyl group $(CH_3)$,the carbonyl compound $A$ must provide the remaining three-carbon skeleton with a carbonyl group.
Acetone $(CH_3COCH_3)$ reacts with $CH_3MgBr$ to form an intermediate complex,which upon hydrolysis yields tert-butyl alcohol.
The reaction is: $CH_3COCH_3 + CH_3MgBr$ $\rightarrow (CH_3)_3C-OMgBr$ $\xrightarrow{H_3O^{+}} (CH_3)_3C-OH + Mg(OH)Br$.
Therefore,$A$ is Acetone.

(B) The reactivity of carbonyl compounds towards nucleophilic addition reactions like $HCN$ addition depends on two factors: steric hindrance and electronic effects.
$1$. Steric hindrance: Smaller groups around the carbonyl carbon facilitate the attack of the nucleophile $(CN^-)$.
$2$. Electronic effects: Electron-withdrawing groups increase the electrophilicity of the carbonyl carbon.
Among the given options,$HCHO$ (formaldehyde) has the smallest hydrogen atoms attached to the carbonyl carbon,resulting in the least steric hindrance.
Therefore,$HCHO$ is the most reactive towards $HCN$.

(A) The Clemmensen reduction is a chemical reaction where aldehydes or ketones are reduced to alkanes using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
Option $A$ represents the general reaction for the Clemmensen reduction: $>C=O + 4[H] \xrightarrow{Zn-Hg / \text{conc. } HCl / \Delta} >CH_2 + H_2O$.
Option $B$ is the Rosenmund reduction.
Option $C$ is the Wolff-Kishner reduction.
Option $D$ is the Stephen reduction.

(D) Formaldehyde $(HCHO)$ undergoes polymerization to form different products depending on the conditions.
$1$. Paraformaldehyde is a linear polymer of formaldehyde with the formula $HO(CH_2O)_n H$.
$2$. Trioxane $(C_3H_6O_3)$ is the cyclic trimer of formaldehyde,which is a solid at room temperature.
$3$. Paraldehyde is the cyclic trimer of acetaldehyde,not formaldehyde.
Therefore,the correct answer is Trioxane.

(B) Methanal $(HCHO)$ reacts with ammonia $(NH_3)$ to form urotropine (hexamethylenetetramine).
The chemical reaction is as follows:
$6HCHO + 4NH_3 \rightarrow (CH_2)_6N_4 + 6H_2O$

(B) The haloform reaction is given by compounds containing the $CH_{3}CO-$ group or the $CH_{3}CH(OH)-$ group.
Acetaldehyde $(CH_{3}CHO)$,acetone $(CH_{3}COCH_{3})$,and acetophenone $(C_{6}H_{5}COCH_{3})$ all contain the $CH_{3}CO-$ group and thus give a yellow precipitate of iodoform $(CHI_{3})$ with $I_{2}$ and $NaOH$.
Benzaldehyde $(C_{6}H_{5}CHO)$ does not contain the $CH_{3}CO-$ group,so it does not undergo the haloform reaction.

(D) The reaction $R-CN + H_2O \xrightarrow{H^+} R-COOH$ is the hydrolysis of a nitrile,which yields a carboxylic acid,not an amine.
- Option $A$: Ammonolysis of alkyl halides yields amines.
- Option $B$: Reduction of nitro compounds yields amines.
- Option $C$: Reduction of oximes yields amines.
- Option $D$: Hydrolysis of nitriles yields carboxylic acids.

(B) The reduction of nitro compounds $(R-NO_2)$ with $Zn$ and $NH_4Cl$ in a neutral medium is a selective reduction process.
This reaction yields $N$-alkylhydroxylamine $(R-NHOH)$ as the major product '$A$'.
The reaction is represented as:
$R-NO_2 + 4[H] \xrightarrow{Zn / NH_4Cl} R-NHOH + H_2O$

(C) In the Gattermann reaction,chlorine or bromine can be introduced into the arene ring by treating the arene diazonium salt solution with the corresponding halogen acid in the presence of copper powder.
Option $C$ represents the Gattermann reaction where $ArN_{2}^{+}X^{-}$ reacts with $Cu/HBr$ to form $ArBr$.

(D) The boiling point of amines depends on the extent of intermolecular hydrogen bonding.
Primary amines $(R-NH_2)$ have two hydrogen atoms available for hydrogen bonding,secondary amines $(R_2NH)$ have one,and tertiary amines $(R_3N)$ have none.
Therefore,the general order of boiling points is $1^{\circ} > 2^{\circ} > 3^{\circ}$ for isomeric amines.
$n-Butylamine$ $(1^{\circ})$ has the highest boiling point,followed by $Diethylamine$ $(2^{\circ})$,and $Ethyl$ dimethylamine $(3^{\circ})$ has the lowest.
The correct order is $n-Butylamine > Diethylamine > Ethyl$ dimethylamine.

(D) In the aqueous phase,the basic strength of amines is determined by a combination of three factors: inductive effect,solvation effect (solvation of the conjugate acid),and steric hindrance.
For methyl-substituted amines,the order is $(CH_{3})_{2}NH > CH_{3}NH_{2} > (CH_{3})_{3}N > NH_{3}$.
This is because the secondary amine $(CH_{3})_{2}NH$ has the optimal balance of inductive effect and solvation,while the tertiary amine $(CH_{3})_{3}N$ experiences significant steric hindrance,reducing its basicity compared to the secondary and primary amines.

(A) $1$. The reaction of acetanilide with a nitrating mixture $(conc. HNO_3 + conc. H_2SO_4)$ leads to electrophilic aromatic substitution,primarily yielding $p$-nitroacetanilide as product $A$ due to the steric hindrance at the ortho position.
$2$. The subsequent step involves the hydrolysis of the amide group $(-NHCOCH_3)$ in $p$-nitroacetanilide using acid $(H^+)$ or base $(OH^-)$.
$3$. This hydrolysis converts the $-NHCOCH_3$ group back into an amino group $(-NH_2)$,resulting in the formation of $p$-nitroaniline as product $B$.

(D) The reaction sequence is: $CH_{3}CH_{2}NO_{2}$ $\xrightarrow[\Delta]{HNO_{2}} A$ $\xrightarrow{LiAlH_{4}} B$.
Nitroalkanes like nitroethane $(CH_{3}CH_{2}NO_{2})$ react with nitrous acid $(HNO_{2})$ to form nitrolic acids or pseudonitroles. However,in the context of standard organic synthesis problems,the reduction of a nitro group $(NO_{2})$ using $LiAlH_{4}$ yields a primary amine.
$A$ is the nitro compound $CH_{3}CH_{2}NO_{2}$.
Reduction of $CH_{3}CH_{2}NO_{2}$ with $LiAlH_{4}$ gives $CH_{3}CH_{2}NH_{2}$.
Therefore,$B$ is $CH_{3}CH_{2}NH_{2}$ (Ethanamine).

(B) The reaction of benzene diazonium chloride with aniline is a coupling reaction,not a replacement reaction. In this reaction,the diazonium group $(-N_2^+Cl^-)$ is retained in the product,$p$-aminoazobenzene,where the nitrogen atoms form an azo linkage $(-N=N-)$ between the two benzene rings.
In contrast,the other options involve the replacement of the diazonium group:
$1$. Reaction with $KI$ replaces the $-N_2^+Cl^-$ group with $-I$.
$2$. Reaction with $CuCl/HCl$ (Sandmeyer reaction) replaces the $-N_2^+Cl^-$ group with $-Cl$.
$3$. Reaction with $H_3PO_2$ (hypophosphorus acid) replaces the $-N_2^+Cl^-$ group with $-H$ (reduction to benzene).

(B) Primary nitroalkanes $(R-CH_2-NO_2)$ undergo hydrolysis when boiled with concentrated mineral acids like $HCl$ or $H_2SO_4$ to form the corresponding carboxylic acid $(R-COOH)$ and hydroxylamine $(NH_2OH)$.
The reaction is represented as: $R-CH_2-NO_2 + H_2O \xrightarrow{HCl/\Delta} R-COOH + NH_2OH$.

(A) The reaction of benzene diazonium chloride with aniline in a mild alkaline medium is a coupling reaction.
This reaction leads to the formation of $p$-aminoazobenzene,which is a yellow dye.
The reaction is: $C_6H_5N_2Cl + C_6H_5NH_2 \xrightarrow{pH \approx 4-5} C_6H_5-N=N-C_6H_4-NH_2 + HCl$.

(C) Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines.
It involves the nucleophilic substitution of an alkyl halide with the phthalimide anion.
Aromatic primary amines,such as $Aniline$,cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion under these conditions.

(D) The carbylamine test is a characteristic reaction given only by primary $(1^{\circ})$ amines,whether aliphatic or aromatic.
In this reaction,the primary amine reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine),which has a foul smell.
$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$.
Among the given options:
$A$. Ethylamine $(CH_3CH_2NH_2)$ is a primary amine.
$B$. Sec. butylamine $(CH_3CH_2CH(NH_2)CH_3)$ is a primary amine.
$C$. Isopropylamine $((CH_3)_2CHNH_2)$ is a primary amine.
$D$. Dimethylamine $((CH_3)_2NH)$ is a secondary $(2^{\circ})$ amine.
Therefore,Dimethylamine does not give the carbylamine test.

(A) The reduction of ketoximes $(R_2C=NOH)$ using sodium $(Na)$ in ethanol $(C_2H_5OH)$ is a standard method for the preparation of primary amines ($1^{\circ}$ amines).
For example,acetoxime $(CH_3-C(=NOH)-CH_3)$ upon reduction with $Na/C_2H_5OH$ yields isopropylamine $(CH_3-CH(NH_2)-CH_3)$,which is a $1^{\circ}$ amine.

(C) The Hoffmann elimination reaction involves the conversion of a quaternary ammonium halide into an alkene.
First,the quaternary ammonium halide is treated with moist silver oxide $(Ag_{2}O / H_{2}O)$ to form a quaternary ammonium hydroxide.
This hydroxide is then subjected to strong heating $(\Delta)$,which causes $\beta$-elimination to yield an alkene as the major product.
Therefore,the reagent used is $Ag_{2}O / H_{2}O, \Delta$.

(A) The reaction of aniline with acetic anhydride in the presence of pyridine is an acetylation reaction.
Aniline reacts with acetic anhydride to form $N$-phenylacetamide,which is commonly known as Acetanilide.
The reaction is as follows:
$C_6H_5NH_2 + (CH_3CO)_2O \xrightarrow{\text{pyridine}} C_6H_5NHCOCH_3 + CH_3COOH$
Thus,the product $A$ is Acetanilide.

(B)
$-OCH_3$ is an electron-donating group $(EDG)$ due to the $+M$ effect,which increases the electron density on the nitrogen atom of the aniline ring,thereby increasing the basic strength of substituted aniline.

(B) The correct answer is $B$.
When benzene diazonium chloride reacts with ethanol,it undergoes a reduction reaction where the diazonium group is replaced by a hydrogen atom.
The ethanol is oxidized to acetaldehyde.
The reaction is as follows:
$C_6H_5N_2Cl + CH_3CH_2OH \xrightarrow{\Delta} C_6H_6 + CH_3CHO + HCl + N_2$
Thus,benzene is formed.

(B) The reaction sequence is as follows:
$1$. $C_2H_5NH_2 \xrightarrow{HNO_2} C_2H_5OH$ (Product $A$ is Ethanol).
$2$. $C_2H_5OH \xrightarrow{PCl_5} C_2H_5Cl$ (Product $B$ is Chloroethane).
$3$. $C_2H_5Cl \xrightarrow{NH_3 (alc.)} C_2H_5NH_2$ (Product $C$ is Ethanamine).
Therefore,the end product $C$ is Ethanamine.

(C) $(C)$
Sandmeyer's reaction is the process where the diazonium group $(-N_{2}^{+}X^{-})$ is replaced by a halogen (like $-Cl$ or $-Br$) using the corresponding cuprous salt ($CuCl$ or $CuBr$) in the presence of its acid ($HCl$ or $HBr$).
The chemical equation is: $Ar-N_{2}^{+}X^{-} \xrightarrow{CuCl / HCl} Ar-Cl + N_{2} \uparrow$

(D) The basicity of amines in the aqueous phase depends on the combined effect of inductive effect,solvation effect,and steric hindrance.
For methyl-substituted amines,the order of basicity is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
Thus,dimethylamine is the most basic among the given options.

(B) In an alkaline medium,nitroalkanes react with bromine only if they possess at least one $\alpha$-hydrogen atom.
The reaction involves the formation of a carbanion at the $\alpha$-carbon,which then reacts with bromine.
Option $B$,which is $2$-methyl-$2$-nitropropane $((CH_3)_3CNO_2)$,is a tertiary nitroalkane and does not have any $\alpha$-hydrogen atom.
Therefore,it cannot form the required carbanion and does not react with bromine in an alkaline medium.

(A) The reaction in which a diazonium group $(-N_2^+)$ is replaced by a halogen or a cyano group using copper$(I)$ salts ($CuCl/HCl$,$CuBr/HBr$,or $CuCN/KCN$) is known as the Sandmeyer reaction.
In this reaction,the diazonium salt is treated with the corresponding cuprous halide in the presence of its respective acid to yield aryl halides or aryl nitriles.

(D) The Mendius reduction involves the reduction of nitriles (alkyl cyanides) to primary amines using sodium $(Na)$ in ethanol $(C_2H_5OH)$.
The reaction is represented as:
$R-C \equiv N + 4[H] \xrightarrow{Na / C_2H_5OH} R-CH_2-NH_2$ (primary amine).

(B) The reaction of phthalimide with an alkyl halide in the presence of a base followed by alkaline hydrolysis is known as the Gabriel phthalimide synthesis.
This method is specifically used for the preparation of primary aliphatic amines $(R-NH_2)$.
It cannot be used for the preparation of aromatic amines $(Ar-NH_2)$ because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

(D) The reaction with benzene sulphonyl chloride (Hinsberg's reagent) is used to distinguish between primary,secondary,and tertiary amines.
Primary amines $(R-NH_{2})$ react with benzene sulphonyl chloride to form $N$-alkylbenzene sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen.
This acidic hydrogen makes the product soluble in excess of potassium hydroxide $(KOH)$,resulting in a clear solution.
Secondary amines $(R_{2}NH)$ form $N,N$-dialkylbenzene sulphonamides,which lack an acidic hydrogen and are insoluble in $KOH$.
Tertiary amines $(R_{3}N)$ do not react with benzene sulphonyl chloride.
Among the given options,$CH_{3}NH_{2}$ is a primary amine,so it forms a clear solution.

(A) The reaction of benzamide $(C_{6}H_{5}CONH_{2})$ with water in the presence of an acid $(HCl)$ and heat is an acid-catalyzed hydrolysis reaction.
The amide group $(-CONH_{2})$ is hydrolyzed to a carboxylic acid group $(-COOH)$ and ammonium chloride $(NH_{4}Cl)$ is formed as a byproduct.
The chemical equation is:
$C_{6}H_{5}CONH_{2} + H_{2}O + HCl \xrightarrow{\Delta} C_{6}H_{5}COOH + NH_{4}Cl$
Thus,the product $A$ is benzoic acid $(C_{6}H_{5}COOH)$.

(D) The reaction of benzamide with bromine and aqueous sodium hydroxide is known as the Hofmann bromamide degradation reaction. In this reaction,an amide is converted into a primary amine with one fewer carbon atom. The reaction is as follows: $C_6H_5CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$. The final product obtained is aniline $(C_6H_5NH_2)$.

(C)
Primary amines can be obtained by the reduction of alkyl cyanide with sodium and ethanol. The chemical reaction is:
$R-C \equiv N + 4[H] \xrightarrow{Na/ethanol} R-CH_2-NH_2$
This reaction is known as Mendius reduction.

(C) When glucose is treated with bromine water ($Br_2$ water),it undergoes mild oxidation.
In this reaction,the aldehydic group $(-CHO)$ of glucose is selectively oxidized to a carboxylic acid group $(-COOH)$,resulting in the formation of gluconic acid.
The reaction is:
$CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2/H_2O} COOH(CHOH)_4CH_2OH$

(C) Glucose reacts with hydroxylamine $(NH_2OH)$ to form an oxime,which is known as glucoxime.
The reaction is:
$CHO(CHOH)_4CH_2OH + NH_2OH \rightarrow CH=NOH(CHOH)_4CH_2OH + H_2O$
Therefore,the reagent used is hydroxylamine.

(A) The chemical formula of saccharic acid (also known as $D$-glucaric acid) is $C_6H_{10}O_8$.
Based on the structure provided:
- There are two $COOH$ groups,each containing $2$ oxygen atoms ($2 \times 2 = 4$ oxygen atoms).
- There are four $OH$ groups attached to the carbon chain,each containing $1$ oxygen atom ($4 \times 1 = 4$ oxygen atoms).
- Total number of oxygen atoms $= 4 + 4 = 8$.

(A) Glucose contains $5$ hydroxyl $(-OH)$ groups ($4$ secondary and $1$ primary).
When glucose reacts with excess acetic anhydride,all $5$ hydroxyl groups are acetylated to form glucose pentaacetate.
The reaction is as follows:
$CHO(CHOH)_4CH_2OH + 5(CH_3CO)_2O \xrightarrow{Pyridine} CHO(CHOCOCH_3)_4CH_2OCOCH_3 + 5CH_3COOH$
As shown in the reaction,$5$ moles of acetic acid $(CH_3COOH)$ are produced for every mole of glucose reacted.

(B) The reaction of glucose with hydrogen cyanide $(HCN)$ is a nucleophilic addition reaction. The carbonyl group $(CHO)$ of glucose reacts with $HCN$ to form a cyanohydrin derivative known as glucose cyanohydrin. The reaction is represented as follows:
$CHO(CHOH)_4CH_2OH + HCN \rightarrow CN-CH(OH)(CHOH)_4CH_2OH$

(A) When glucose is treated with bromine water ($Br_{2}$ water),it undergoes mild oxidation.
The aldehydic group $(-CHO)$ of glucose is oxidized to a carboxylic acid group $(-COOH)$,while the secondary alcoholic groups remain unaffected.
This reaction produces gluconic acid.
The reaction is:
$CHO(CHOH)_{4}CH_{2}OH + [O] \xrightarrow{Br_{2}/H_{2}O} COOH(CHOH)_{4}CH_{2}OH$
(Glucose) $\rightarrow$ (Gluconic acid)

(C)
When glucose is treated with bromine water ($Br_2$ water),it undergoes mild oxidation to form gluconic acid.
In this reaction,the aldehydic group $(-CHO)$ is selectively oxidized to a carboxylic acid group $(-COOH)$,while the secondary alcoholic groups remain unaffected.
This reaction confirms that the carbonyl group in glucose is an aldehydic group.
The reaction is:
$CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2/H_2O} COOH(CHOH)_4CH_2OH$
(Glucose $\rightarrow$ Gluconic acid)

(B) $1$. The hydrolysis of maltose yields two molecules of glucose $(X = \text{Glucose})$.
$2$. When glucose is treated with excess dilute nitric acid $(\text{dil. HNO}_3)$,it acts as a strong oxidizing agent and oxidizes both the terminal primary alcohol group $(-CH_2OH)$ and the aldehydic group $(-CHO)$ to carboxylic acid groups $(-COOH)$.
$3$. This oxidation results in the formation of saccharic acid (also known as glucaric acid).

(A) Lactose is a disaccharide with the formula $C_{12}H_{22}O_{11}$.
On hydrolysis in the presence of an acid or the enzyme lactase,it undergoes cleavage to produce one molecule of $D-(+)$-glucose and one molecule of $D-(+)$-galactose.
The chemical reaction is:
$C_{12}H_{22}O_{11} + H_2O$ $\xrightarrow{H_3O^+ \text{ or lactase}} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Galactose})$
Therefore,one mole of lactose yields $1.0 \ mol$ of glucose and $1.0 \ mol$ of galactose.

(A) Raffinose is a trisaccharide with the molecular formula $C_{18}H_{32}O_{16}$.
Upon complete acid-catalyzed hydrolysis,one mole of raffinose reacts with two moles of water to yield one mole each of glucose,fructose,and galactose.
The chemical equation is: $C_{18}H_{32}O_{16} + 2H_2O$ $\xrightarrow{H^+} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose}) + C_6H_{12}O_6 (\text{Galactose})$.

(B) Cellulose is a linear polysaccharide consisting of $D$-glucose units joined by $\beta$-glycosidic linkages between $C-1$ of one glucose unit and $C-4$ of the adjacent glucose unit.
Therefore,cellulose contains $1 \rightarrow 4$ $\beta$-glycosidic linkages.
The correct option is $(B)$.

(C) Dexon is a biodegradable polymer prepared from the monomers lactic acid and glycolic acid.
It is a copolymer formed by the condensation polymerization of these two hydroxy acids,resulting in the formation of ester linkages.
The reaction is as follows:
$nCH_3-CH(OH)-COOH + nHO-CH_2-COOH \rightarrow [-O-CH(CH_3)-CO-O-CH_2-CO-]_n + nH_2O$

(D) Sucrose is a disaccharide formed by the condensation of one molecule of $\alpha-D$-glucopyranose and one molecule of $\beta-D$-fructofuranose.
In this structure,the glycosidic linkage is formed between the $C-1$ carbon of $\alpha-D$-glucopyranose and the $C-2$ carbon of $\beta-D$-fructofuranose.
Therefore,the correct option is $D$.