MHT CET 2020 Chemistry Question Paper with Answer and Solution

(D) White phosphorus consists of discrete $P_4$ tetrahedral molecules.
In this structure,each phosphorus atom is linked to three other phosphorus atoms by single covalent bonds.
Due to the angular strain in the tetrahedral geometry,the $P-P-P$ bond angle is $60^{\circ}$.

(D) Let $y = \sin^{-1}\left(-\frac{1}{2}\right)$.
Then,$\sin(y) = -\frac{1}{2}$.
We know that the range of the principal value branch of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.
Thus,the principal value is $-\frac{\pi}{6}$.

(A) The given equation is $9x^2 - 12xy + 4y^2 = 0$.
Comparing this with the general homogeneous equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 9$,$2h = -12$ (so $h = -6$),and $b = 4$.
The condition for the lines to be coincident is $h^2 - ab = 0$.
Substituting the values: $(-6)^2 - (9)(4) = 36 - 36 = 0$.
Since the condition $h^2 - ab = 0$ is satisfied,the lines are coincident.

(D) The formula $C_4H_{11}N$ represents a saturated amine. Primary amines have the general structure $R-NH_2$.
For a four-carbon chain,the possible primary amine isomers are:
$1$. $CH_3-CH_2-CH_2-CH_2-NH_2$ ($n$-butylamine or $1$-aminobutane)
$2$. $CH_3-CH_2-CH(NH_2)-CH_3$ ($sec$-butylamine or $2$-aminobutane)
$3$. $CH_3-CH(CH_3)-CH_2-NH_2$ (isobutylamine or $2$-methyl-$1$-aminopropane)
$4$. $(CH_3)_3C-NH_2$ ($tert$-butylamine or $2$-methyl-$2$-aminopropane)
Thus,there are $4$ possible primary amines.

(A) At the mean position,the velocity of mass $M$ is maximum,given by $v_1 = A_1 \omega_1$,where $\omega_1 = \sqrt{\frac{K}{M}}$.
When mass $m$ is placed on $M$,the momentum is conserved during the collision (as the impulsive force acts vertically,not horizontally).
$M v_1 = (M + m) v_2$
Substituting $v_1 = A_1 \sqrt{\frac{K}{M}}$ and $v_2 = A_2 \omega_2 = A_2 \sqrt{\frac{K}{M+m}}$:
$M \left( A_1 \sqrt{\frac{K}{M}} \right) = (M + m) \left( A_2 \sqrt{\frac{K}{M + m}} \right)$
$A_1 \sqrt{MK} = A_2 \sqrt{(M + m)K}$
$\frac{A_1}{A_2} = \sqrt{\frac{M + m}{M}}$

(C) The chemical structure of hydroxyquinol ($1,2,4$-trihydroxybenzene) consists of a benzene ring with three hydroxy $(-OH)$ groups attached at the $1$,$2$,and $4$ positions.
By observing the structure,we can count the number of $-OH$ groups,which is $3$.
Therefore,the correct option is $C$.

(D) When $2-$methylpropan$-2-$ol (tert-butyl alcohol) is passed over heated alumina $(Al_2O_3)$ at $423 \ K$,it undergoes dehydration to form $2-$methylpropene (isobutylene).
The reaction is: $(CH_3)_3C-OH \xrightarrow{Al_2O_3, 423 \ K} CH_3-C(CH_3)=CH_2 + H_2O$.
$2-$methylpropene is commonly known as isobutylene.

(C) Sodium amide $(NaNH_{2})$ is prepared by passing dry ammonia gas over molten sodium at $300-400 \ ^{\circ}C$.
However,it is more commonly prepared by the reaction of sodium with liquid ammonia in the presence of a catalyst.
$Fe(NO_{3})_{3}$ (iron$(III)$ nitrate) acts as a catalyst for this reaction.
The reaction is represented as: $2Na + 2NH_{3} \xrightarrow{Fe(NO_{3})_{3}} 2NaNH_{2} + H_{2}$.

(D) The molecular formula of glucose is $C_6H_{12}O_6$. Let the oxidation number of carbon be $x$.
$6(x) + 12(+1) + 6(-2) = 0$
$6x + 12 - 12 = 0$
$6x = 0$
$x = 0$. Thus,the oxidation number of carbon in glucose is zero.

(D) The hydrolysis reaction of sucrose is given by:
$C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 \text{ (glucose)} + C_6H_{12}O_6 \text{ (fructose)}$
From the stoichiometry,$1 \ mol$ of sucrose $(342 \ g)$ produces $1 \ mol$ of glucose $(180 \ g)$.
Therefore,$68.4 \ g$ of sucrose will produce:
$\text{Mass of glucose} = \frac{180 \ g \times 68.4 \ g}{342 \ g} = 36.0 \ g$.

(D) The electronic configuration of $H_{2}$ is $(\sigma_{1s})^2$,giving a bond order of $\frac{2-0}{2} = 1$.
The electronic configuration of $Li_{2}$ is $(\sigma_{1s})^2, (\sigma_{1s}^*)^2, (\sigma_{2s})^2$,giving a bond order of $\frac{4-2}{2} = 1$.
However,in $Li_{2}$,the presence of inner $(\sigma_{1s})^2$ and $(\sigma_{1s}^*)^2$ electrons effectively shields the valence $\sigma_{2s}$ electrons from the nucleus,reducing the effective nuclear charge experienced by the bonding electrons and making the bond weaker compared to $H_{2}$.

(D) The octet rule states that atoms tend to form bonds such that they have $8$ electrons in their valence shell.
In the molecule $SF_6$,the central sulfur atom is bonded to $6$ fluorine atoms.
Each $S-F$ bond consists of $2$ electrons,so the sulfur atom has $6 \times 2 = 12$ electrons in its valence shell.
Since $12 > 8$,$SF_6$ is an example of an expanded octet and does not obey the octet rule.

(A) is the correct answer.
$1$. In $CH_{4}$ (Methane),the central carbon atom forms $4$ covalent bonds,sharing $8$ electrons in its valence shell,thus completing its octet.
$2$. In $SF_{6}$ (Sulphur hexafluoride),the central sulphur atom has $12$ electrons,which is an expanded octet.
$3$. In $AlCl_{3}$ (Aluminium chloride) and $BF_{3}$ (Boron trifluoride),the central atoms ($Al$ and $B$) have only $6$ electrons in their valence shells,representing an incomplete octet.

(B) The formula for formal charge $(F.C.)$ is:
$F.C. = V.E. - N.E. - \frac{1}{2} B.E.$
Where:
$V.E.$ = Valence electrons of the atom ($1$ for $H$)
$N.E.$ = Non-bonding electrons ($0$ for $H$)
$B.E.$ = Bonding electrons ($2$ for $H$)
Substituting the values for the hydrogen atom in $H_2O$:
$F.C. = 1 - 0 - \frac{1}{2}(2) = 1 - 1 = 0$
Therefore,the formal charge on the hydrogen atom is $0$.

(D) In $NH_{3}$ molecule,the nitrogen atom is $sp^{3}$ hybridized.
It has one lone pair and three bond pairs of electrons.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion.
Due to this,the bond angle decreases from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $107^{\circ} 18^{\prime}$.

(C) Tetrahedral geometry is associated with $sp^{3}$ hybridisation,where one $s$ orbital and three $p$ orbitals mix to form four equivalent $sp^{3}$ hybrid orbitals directed towards the corners of a regular tetrahedron with bond angles of $109.5^{\circ}$.

(B) In a water molecule $(H_2O)$,the central oxygen atom undergoes $sp^{3}$ hybridization.
Two of the $sp^{3}$ hybrid orbitals of oxygen overlap with the $1s$ orbitals of two hydrogen atoms to form two $O-H$ sigma bonds.
Therefore,the type of overlap involved is $sp^{3}-s$.

(A) The number of hybrid orbitals formed is always equal to the total number of atomic orbitals that participate in the hybridization process.
Here,the number of atomic orbitals involved is $1$ $(s)$ + $3$ $(p)$ + $1$ $(d)$ = $5$ orbitals.
Therefore,$5$ hybrid orbitals are formed.

(C) The central atom in $PCl_{5}$ is phosphorus $(P)$.
Phosphorus has an atomic number of $15$,and its ground state electronic configuration is $[Ne] 3s^{2} 3p^{3}$.
In the excited state,one electron from the $3s$ orbital is promoted to the $3d$ orbital,resulting in five unpaired electrons $(3s^{1} 3p^{3} 3d^{1})$.
These five orbitals ($one$ $s$,$three$ $p$,and $one$ $d$) undergo hybridization to form five equivalent $sp^{3}d$ hybrid orbitals.
These hybrid orbitals overlap with the $p$-orbitals of five chlorine atoms to form five $P-Cl$ sigma bonds,resulting in a trigonal bipyramidal geometry.

(B) The $PCl_{5}$ molecule has a trigonal bipyramidal geometry.
In this structure,three $Cl$ atoms are present in the equatorial plane,forming a trigonal arrangement around the central $P$ atom.
The bond angle between these equatorial $Cl-P-Cl$ bonds is $120^{\circ}$.
The other two $Cl$ atoms are in the axial positions,making a $90^{\circ}$ angle with the equatorial plane and a $180^{\circ}$ angle with each other.

(A) The $C-O-H$ bond angle in $CH_3OH$ (methanol) is $108.9^{\circ}$.
In methanol,the oxygen atom is $sp^3$ hybridized and has two lone pairs of electrons.
According to $VSEPR$ theory,the lone pair-lone pair repulsion is greater than the lone pair-bond pair repulsion,which in turn is greater than the bond pair-bond pair repulsion.
This repulsion causes the bond angle to deviate from the ideal tetrahedral angle of $109.5^{\circ}$ to $108.9^{\circ}$.

(A) According to the $VSEPR$ theory,the repulsive interaction between electron pairs follows the order: $lone \ pair-lone \ pair > lone \ pair-bond \ pair > bond \ pair-bond \ pair$.
This is because a $lone \ pair$ is under the influence of only one nucleus (the central atom),whereas a $bond \ pair$ is shared between two nuclei.
Consequently,the electron cloud of a $lone \ pair$ occupies more space around the central atom,leading to greater repulsion compared to $bond \ pair$ interactions.

(C) $1$. $ClF_{3}$ has $sp^{3}d$ hybridization with two lone pairs,resulting in a $T$-shaped geometry.
$2$. $XeF_{2}$ has $sp^{3}d$ hybridization with three lone pairs,resulting in a linear geometry.
$3$. $BeF_{2}$ has $sp$ hybridization with no lone pairs,resulting in a linear geometry.
$4$. $SO_{2}$ has $sp^{2}$ hybridization with one lone pair,resulting in a bent geometry.
$5$. Both $XeF_{2}$ and $BeF_{2}$ have linear geometry. However,in standard multiple-choice questions of this type,$BeF_{2}$ is the most fundamental example of linear geometry due to $sp$ hybridization.

(A) In an alkane, all carbon atoms are $sp^{3}$ hybridized.
The carbon-carbon $(C-C)$ bond is formed by the overlap of one $sp^{3}$ orbital from each carbon atom.
The bond length of a $C-C$ single bond is $1.54$ $\mathring{A}$, which is equivalent to $154$ $pm$.
All these bonds are sigma $(\sigma)$ bonds.

(C) The percentage of $s$-character in a hybrid orbital is calculated as:
$s$-character $= (\frac{1}{n}) \times 100$,where $n$ is the number of hybrid orbitals.
For $sp^{3}$ hybridization,there are $4$ hybrid orbitals,so $s$-character $= (\frac{1}{4}) \times 100 = 25 \%$.
In $CH_{4}$ (Methane),the carbon atom is $sp^{3}$-hybridized.
In $C_{2}H_{4}$ (Ethylene),carbon is $sp^{2}$-hybridized ($33.3 \% \,s$-character).
In $C_{2}H_{2}$ (Acetylene),carbon is $sp$-hybridized ($50 \% \,s$-character).
In $C_{6}H_{6}$ (Benzene),carbon is $sp^{2}$-hybridized ($33.3 \% \,s$-character).
Therefore,methane contains $25 \% \,s$-character.

(A) The ozone molecule $(O_3)$ exhibits resonance, where the two canonical forms contribute to the resonance hybrid.
In the resonance hybrid, the two $O-O$ bonds are equivalent due to the delocalization of $\pi$ electrons.
The experimental bond length for both $O-O$ bonds in the ozone molecule is $128 \ pm$, which is intermediate between the length of a single bond $(148 \ pm)$ and a double bond $(121 \ pm)$.

(B) According to Molecular Orbital Theory,the electronic configuration of the $O_2$ molecule ($16$ electrons) is:
$(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 = (\pi 2p_y)^2 (\pi^* 2p_x)^1 = (\pi^* 2p_y)^1$
The antibonding molecular orbitals (ABMOs) are $\sigma^* 1s$,$\sigma^* 2s$,$\pi^* 2p_x$,and $\pi^* 2p_y$.
The electrons in these antibonding orbitals are:
$\sigma^* 1s$: $2$ electrons
$\sigma^* 2s$: $2$ electrons
$\pi^* 2p_x$: $1$ electron
$\pi^* 2p_y$: $1$ electron
Total antibonding electrons = $2 + 2 + 1 + 1 = 6$ electrons.

(B) The electronic configuration of $B_{2}$ molecule ($10$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$.
Number of bonding electrons $(N_b)$ = $2+2+1+1 = 6$.
Number of antibonding electrons $(N_a)$ = $2+2 = 4$.
Bond order = $\frac{N_b - N_a}{2} = \frac{6 - 4}{2} = \frac{2}{2} = 1$.

(D) The electronic configuration of $N_{2}$ ($14$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.
Bond order is calculated as: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
Here,$N_b = 10$ and $N_a = 4$.
$\text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3$.
Thus,the bond order in $N_{2}$ is $3$.

(C) The electronic configuration of the $Be_{2}$ molecule ($8$ electrons) is: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2}$.
Here,the number of bonding electrons $(N_{b})$ is $4$ and the number of antibonding electrons $(N_{a})$ is $4$.
The bond order is calculated as: $\text{Bond Order} = \frac{N_{b} - N_{a}}{2} = \frac{4 - 4}{2} = 0$.
Since the bond order is $0$,the $Be_{2}$ molecule does not exist.

(C) When an electron is transferred from atom $A$ to atom $B$,$A$ loses one electron to form a cation $A^+$ and $B$ gains one electron to form an anion $B^-$.
Since the bond is formed by the electrostatic force of attraction between the oppositely charged ions ($A^+$ and $B^-$),it is an electrovalent or ionic bond.
Therefore,the compound $AB$ exhibits an electrovalent bond.

(D) An ionic compound is formed by the transfer of electrons between a metal and a non-metal.
$KI$ (Potassium Iodide) consists of a metal $(K^+)$ and a non-metal $(I^-)$,which are held together by strong electrostatic forces of attraction,making it an ionic compound.
$SO_{2}$,$ICl$,and $CHCl_{3}$ are covalent compounds formed by the sharing of electrons between non-metals.
Therefore,the correct option is $D$.

(B) The dipole moment $(\mu)$ is defined as the product of the magnitude of the charge $(Q)$ and the distance $(r)$ between the centers of positive and negative charges.
Mathematically,it is expressed as: $\mu = Q \times r$.
Therefore,the correct expression is $\mu = Q \times r$.

(D) In $CO_{2}$,the $C=O$ bond dipoles are equal in magnitude but oriented in opposite directions (at $180^{\circ}$).
Their vector sum is zero,so $CO_{2}$ has no net dipole moment.

(C) Benzene $(C_6H_6)$ is a non-polar molecule with no permanent dipole moment.
Ammonia $(NH_3)$ is a polar molecule with a permanent dipole moment.
When a polar molecule approaches a non-polar molecule,it induces a dipole in the non-polar molecule by distorting its electron cloud.
Therefore,the interaction between a permanent dipole $(NH_3)$ and an induced dipole (benzene) is known as a dipole - induced dipole interaction.

(B) The correct answer is $B$ (Ion-dipole interaction).
Magnesium chloride $(MgCl_2)$ dissociates in water into $Mg^{2+}$ and $Cl^-$ ions.
Water $(H_2O)$ is a polar molecule with a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atoms.
The interaction between the $Mg^{2+}$ ion and the partial negative charge on the oxygen atom of the water molecule is known as an ion-dipole interaction.
This is an attractive force between an ion and a polar molecule.

(B) The correct answer is $o-$nitrophenol.
In $o-$nitrophenol,there is intramolecular hydrogen bonding between the hydrogen atom of the $-OH$ group and the oxygen atom of the $-NO_2$ group,which are located on adjacent positions of the benzene ring. This type of hydrogen bonding occurs within the same molecule,resulting in a lower boiling point compared to its isomers.

(C) Soft soap is manufactured using $KOH$ (potassium hydroxide),also known as caustic potash.
Potassium soaps are more soluble in water compared to sodium soaps.
In their concentrated form,these potassium-based soaps are referred to as soft soaps.
Therefore,$KOH$ is the alkali used for this purpose.

(A) The element $La (Z=57)$ has the electronic configuration $[Xe] 5d^1 6s^2$. It belongs to Group-$3$ and Period-$6$.
$Ce (Z=58)$ is the first element of the lanthanoid series with the configuration $[Xe] 4f^1 5d^1 6s^2$.
In the long form of the periodic table,all lanthanoids (from $Z=57$ to $71$) are placed in Group-$3$ and Period-$6$ to avoid disrupting the table structure.
Therefore,both $La$ and $Ce$ belong to Group-$3$ and Period-$6$.

(C) According to the $IUPAC$ nomenclature for elements with atomic numbers greater than $100$,the temporary symbol is derived from the digits of the atomic number:
$1 = un$ $(u)$,$1 = un$ $(u)$,$6 = hex$ $(h)$.
Therefore,for atomic number $116$,the temporary name is $\text{Ununhexium}$ and the temporary symbol is $Uuh$.

(D) The elements of group-$2$ are Beryllium ($Be$,$Z=4$),Magnesium ($Mg$,$Z=12$),and Calcium ($Ca$,$Z=20$).
The third element of group-$2$ is Calcium $(Ca)$.
The atomic number of Calcium is $20$.
The electronic configuration of Calcium is $[Ar] 4s^2$.

(A) Electronegativity generally decreases down a group in the periodic table due to the increase in atomic size and shielding effect.
Group $15$ elements are $N, P, As, Sb, Bi$.
As we move from $N$ to $Bi$,the electronegativity decreases.
Therefore,$Bi$ (Bismuth) has the lowest electronegativity among the given elements.

(D) $BaCl_{2} \cdot 2H_{2}O$ contains interstitial water molecules.
In this salt,the water molecules are not coordinated to the metal ion but are trapped within the crystal lattice structure,specifically in the interstitial sites.
In contrast,in $CuSO_{4} \cdot 5H_{2}O$,four water molecules are coordinated to the $Cu^{2+}$ ion,and one is hydrogen-bonded.
Therefore,$BaCl_{2} \cdot 2H_{2}O$ is the correct example of a salt with interstitial water.

(A) $\delta$ bond is formed by the lateral overlap of two $d$-orbitals where all four lobes of one orbital overlap with the four lobes of another orbital.
This occurs when two $d_{xy}$ orbitals or two $d_{x^2-y^2}$ orbitals approach each other along the $z$-axis.
Specifically,the $d_{xy}$ orbitals (lying in the $xy$-plane) and $d_{x^2-y^2}$ orbitals (lying along the $x$ and $y$ axes) are capable of forming $\delta$ bonds due to their specific spatial orientation.
Therefore,the correct pair is $d_{xy}$ and $d_{x^2-y^2}$.

(B) $1$. Identify the longest carbon chain. The longest chain contains $8$ carbon atoms,so the parent alkane is octane.
$2$. Number the chain from the end that gives the substituents the lowest possible locants. Numbering from left to right gives substituents at positions $4, 3, 6, 6$. Numbering from right to left gives substituents at positions $3, 3, 6, 5$. Comparing the sets $(3, 3, 5, 6)$ and $(3, 4, 6, 6)$,the set $(3, 3, 5, 6)$ is lower.
$3$. The substituents are: bromo at position $5$,and methyl groups at positions $3, 3, 6$.
$4$. Combining these,the name is $5-$bromo$-3, 3, 6-$trimethyloctane.

(D) primary carbon atom is a carbon atom that is bonded to only one other carbon atom. In the structure of $3-$Ethyl$-2,4-$dimethylheptane,the terminal methyl $(-CH_3)$ groups are primary carbons.
Looking at the structure:
$1$. The $C-2$ position has one methyl group attached.
$2$. The $C-3$ position has an ethyl group attached,which ends in a methyl group.
$3$. The $C-4$ position has a methyl group attached.
$4$. The main chain heptane has two terminal methyl groups.
Counting these:
- Two methyls at $C-2$ (one is part of the chain,one is a substituent).
- One methyl at the end of the ethyl group at $C-3$.
- One methyl at $C-4$.
- Two terminal methyls of the heptane chain.
Total primary carbon atoms = $6$.

(B) The structure of isobutyl bromide is $(CH_3)_2CH-CH_2Br$.
To determine the $IUPAC$ name,we identify the longest carbon chain containing the functional group $(-Br)$.
The longest chain has $3$ carbon atoms,so the parent alkane is propane.
Numbering the chain starting from the carbon attached to the bromine atom gives the $1-$position to the bromine.
There is a methyl group at the $2-$position.
Thus,the $IUPAC$ name is $1-$bromo$-2-$methyl propane.

(A) The structure of acrolein is $CH_2=CH-CHO$.
It contains a carbon chain of $3$ carbon atoms with an aldehyde group $(-CHO)$ at position $1$ and a double bond starting at position $2$.
Therefore,the $IUPAC$ name is prop$-2-$enal.

(C) $1$. Identify the principal functional group: The $-OH$ group is the principal functional group,so the parent chain is a cyclopentanol. The carbon attached to the $-OH$ group is assigned position $1$.
$2$. Number the ring: To give the lowest possible locants to the substituents ($-Cl$ and $-CH_3$),we number the ring in the direction that assigns the lowest numbers to the substituents. Numbering clockwise gives the $-CH_3$ at position $2$ and the $-Cl$ at position $4$. Numbering counter-clockwise gives the $-Cl$ at position $3$ and the $-CH_3$ at position $5$.
$3$. Compare locant sets: The set $(2, 4)$ is lower than $(3, 5)$. Therefore,we number clockwise starting from the carbon with the $-OH$ group.
$4$. Assemble the name: The substituents are $4-$chloro and $2-$methyl. Alphabetical order dictates that chloro comes before methyl. Thus,the $IUPAC$ name is $4-$chloro$-2-$methylcyclopentanol.

(C) The structure of $3-$bromopropene is $CH_2=CH-CH_2Br$.
Counting the atoms in the structure:
There are $3$ carbon atoms $(C_3)$.
There are $5$ hydrogen atoms $(H_5)$.
There is $1$ bromine atom $(Br)$.
Therefore,the molecular formula is $C_3H_5Br$.

(A) Acid-catalysed hydration of alkenes follows Markovnikov's rule,where the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom.
$CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$ (Ethanol).
$CH_3CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH(OH)CH_3$ (Propan$-2-$ol).
$CH_3CH_2CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2CH(OH)CH_3$ (Butan$-2-$ol).
Methanol $(CH_3OH)$ cannot be prepared by this method as it requires at least two carbon atoms.
Propan$-1-$ol and Butan$-1-$ol are primary alcohols that cannot be formed by direct hydration of simple alkenes due to Markovnikov's rule.
Therefore,Ethanol is the correct answer.

(C) Acid catalyzed hydration of alkenes follows Markovnikov's rule,where the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom.
$Propan-1-ol$ $(CH_3CH_2CH_2OH)$ cannot be prepared by this method because the hydration of $propene$ $(CH_3CH=CH_2)$ yields $propan-2-ol$ as the major product according to Markovnikov's rule.
Therefore,$propan-1-ol$ is the correct answer.

(C) The acidity of alcohols depends on the stability of the alkoxide ion formed after the release of a proton $(H^+)$.
Alkyl groups are electron-donating ($+I$ effect) in nature.
As the number of alkyl groups attached to the carbon atom bearing the $-OH$ group increases,the electron density on the oxygen atom increases due to the $+I$ effect.
This destabilizes the alkoxide ion,thereby decreasing the acidic strength.
The order of acidity is: $CH_3OH > CH_3CH_2OH > (CH_3)_2CHOH > (CH_3)_3COH$.
Therefore,$CH_3OH$ is the most acidic.

(C) $1$. The reduction of acetonitrile $(CH_3CN)$ with sodium and alcohol (Mendius reduction) yields ethyl amine $(A)$:
$CH_3CN + 4[H] \xrightarrow{Na / \text{alcohol}} CH_3CH_2NH_2$ ($A$ = Ethyl amine).
$2$. The reaction of primary aliphatic amines with nitrous acid $(NaNO_2 / \text{dil. } HCl)$ produces unstable diazonium salts,which decompose to form alcohols,nitrogen gas,and water:
$CH_3CH_2NH_2 + HNO_2$ $\xrightarrow{NaNO_2 / \text{dil. } HCl} [CH_3CH_2N_2^+Cl^-]$ $\xrightarrow{H_2O} CH_3CH_2OH + N_2 \uparrow + HCl$.
Thus,compound $B$ is ethyl alcohol.

(A) dihydric phenol is a compound containing two hydroxyl $(-OH)$ groups attached directly to a benzene ring.
$1$. $\alpha$-naphthol is a monohydric phenol (contains one $-OH$ group).
$2$. Resorcinol ($1,3$-dihydroxybenzene),Hydroquinone ($1,4$-dihydroxybenzene),and Catechol ($1,2$-dihydroxybenzene) are all dihydric phenols.
Therefore,$\alpha$-naphthol is not a dihydric phenol.

(A) The solubility of alcohols in water depends on the balance between the hydrophilic hydroxyl $(-OH)$ group and the hydrophobic alkyl chain.
As the size of the hydrophobic alkyl chain increases,the solubility of the alcohol in water decreases.
All the given alcohols are isomers with the molecular formula $C_5H_{12}O$.
However,the solubility also depends on the branching of the alkyl chain.
Increased branching reduces the surface area of the hydrophobic part,which generally increases solubility.
Comparing the structures:
$1$. Pentan-$1$-ol (straight chain,$5$ carbons)
$2$. $2$-methyl butan-$2$-ol (branched,$5$ carbons)
$3$. Pentan-$2$-ol (branched,$5$ carbons)
$4$. $2,2$-Dimethyl propan-$1$-ol (highly branched,$5$ carbons)
Among these,Pentan-$1$-ol has the longest straight alkyl chain,which makes it the least soluble in water compared to the more branched isomers.

(C) The reaction of benzaldehyde $(C_{6}H_{5}CHO)$ with a Grignard reagent $(RMgX)$ followed by acid hydrolysis yields a secondary alcohol.
To obtain $1-$phenylethanol $(C_{6}H_{5}CH(OH)CH_{3})$,the Grignard reagent must provide the methyl group $(CH_{3})$.
Therefore,the reaction is:
$C_{6}H_{5}CHO + CH_{3}MgBr \rightarrow [C_{6}H_{5}CH(CH_{3})OMgBr]$
$[C_{6}H_{5}CH(CH_{3})OMgBr] + H_{2}O/H^{+} \rightarrow C_{6}H_{5}CH(OH)CH_{3} + Mg(OH)Br$
Thus,benzaldehyde reacts with $CH_{3}MgBr$ to produce $1-$phenylethanol.

(D) $LiAlH_4$ (Lithium aluminium hydride) is a selective reducing agent that reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_2OH)$ without affecting the carbon-carbon double bond $(C=C)$.
Therefore,$CH_3-CH=CH-CHO$ is converted to $CH_3-CH=CH-CH_2OH$ using $LiAlH_4$.

(C) The reaction of alcohols with Lucas reagent $(conc. \ HCl + ZnCl_2)$ follows the $S_N1$ mechanism,which depends on the stability of the carbocation intermediate.
Tertiary carbocations are the most stable,followed by secondary and then primary carbocations.
Therefore,tertiary alcohols react fastest with Lucas reagent at room temperature,resulting in immediate turbidity.
$2$-Methylpropan-$2$-ol is a tertiary alcohol,so it reacts immediately with Lucas reagent.

(A) The dehydration of secondary alcohols like $Butan-2-ol$ $(CH_3-CH(OH)-CH_2-CH_3)$ typically requires $60 \% \ H_2SO_4$ at a temperature of $373 \ K$.
This reaction follows the $E1$ mechanism,where the major product is $But-2-ene$ (Saytzeff product) and the minor product is $But-1-ene$ (Hofmann product).

(A) Step $1$: Ethanol $(CH_3CH_2OH)$ reacts with $NaBr$ in the presence of $H_2SO_4$ and heat to undergo nucleophilic substitution,forming ethyl bromide $(CH_3CH_2Br)$ as product '$A$'.
$CH_3CH_2OH + NaBr + H_2SO_4 \xrightarrow{\Delta} CH_3CH_2Br + NaHSO_4 + H_2O$
Step $2$: Ethyl bromide $(CH_3CH_2Br)$ reacts with magnesium metal in the presence of dry ether to form a Grignard reagent,ethyl magnesium bromide $(CH_3CH_2MgBr)$,which is product '$B$'.
$CH_3CH_2Br + Mg \xrightarrow{\text{Dry ether}} CH_3CH_2MgBr$

(B) Step $1$: $CH_{3}-CH_{2}-CH_{2}-OH + PCl_{3} \rightarrow CH_{3}-CH_{2}-CH_{2}-Cl (X) + H_{3}PO_{3}$.
Step $2$: $CH_{3}-CH_{2}-CH_{2}-Cl + alco.KOH \xrightarrow{\Delta} CH_{3}-CH=CH_{2} (Y) + KCl + H_{2}O$ (Dehydrohalogenation).
Step $3$: $CH_{3}-CH=CH_{2} + H_{2}O \xrightarrow{conc.H_{2}SO_{4}} CH_{3}-CH(OH)-CH_{3} (Z)$ (Acid-catalyzed hydration following Markovnikov's rule).
Thus,$Z$ is propan-$2$-ol,which is $(CH_{3})_{2}CH-OH$.

(A) The reaction of ethanol with $HCl$ to form ethyl chloride is known as the Grooves process.
In this reaction,anhydrous $ZnCl_{2}$ acts as a catalyst (Lewis acid) to facilitate the cleavage of the $C-O$ bond.
The reaction is: $C_{2}H_{5}OH + HCl \xrightarrow{\text{anhydrous } ZnCl_{2}} C_{2}H_{5}Cl + H_{2}O$.

(D) The correct answer is $D$ ($2-$Methylpropan$-2-$ol).
Primary and secondary alcohols can be oxidized to aldehydes/ketones using milder oxidizing agents like $PCC$ or $CrO_3$.
However,tertiary alcohols like $2-$methylpropan$-2-$ol are resistant to oxidation under normal conditions.
They require strong oxidizing agents such as acidic $KMnO_4$ and extreme conditions (high temperature) to undergo oxidation,which typically involves the cleavage of $C-C$ bonds to form smaller ketones or carboxylic acids.

(A) The final product is propanenitrile,which is $CH_3-CH_2-CN$.
This compound contains $3$ carbon atoms.
The reaction $B \xrightarrow{KCN, \Delta} CH_3-CH_2-CN$ indicates that $B$ must be an alkyl halide with $2$ carbon atoms,which is chloroethane $(CH_3-CH_2-Cl)$.
The reaction $A + SOCl_2 \xrightarrow{\text{pyridine}} B$ is the Darzens process for converting an alcohol into an alkyl chloride.
Therefore,$A$ must be ethanol $(CH_3-CH_2-OH)$.
Thus,$A$ is ethanol.

(C) The defensive froth emitted by the grasshopper $Romalea \ microptera$ contains several odorous compounds,including a chlorinated aromatic compound,$2, 5-$dichlorophenol.
This compound acts as a repellent to ants and is defensively useful to the grasshopper.
It is believed that this compound is derived from herbicides or herbicide derivatives ingested by the insect through its diet.

(A) dihydric phenol contains two hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. Resorcinol is benzene-$1,3$-diol,which is a dihydric phenol.
$2$. $m$-Cresol is a monohydric phenol ($3$-methylphenol).
$3$. Phloroglucinol is benzene-$1,3,5$-triol,which is a trihydric phenol.
$4$. Pyrogallol is benzene-$1,2,3$-triol,which is a trihydric phenol.
Therefore,the correct option is Resorcinol.

(A) In the Raschig process,chlorobenzene is hydrolyzed to phenol by reacting it with steam at a high temperature of $698 \ K$ in the presence of a catalyst.
The catalyst used in this reaction is calcium phosphate,represented by the chemical formula $Ca_3(PO_4)_2$.
The reaction is as follows:
$C_6H_5Cl + H_2O \xrightarrow{698 \ K, \ Ca_3(PO_4)_2} C_6H_5OH + HCl$
Therefore,the correct option is $A$.

(B) Carbolic acid is another name for phenol $(C_6H_5OH)$.
When phenol is heated with a mixture of concentrated nitric acid $(HNO_3)$ and concentrated sulphuric acid $(H_2SO_4)$,it undergoes electrophilic aromatic substitution (nitration) at the $2, 4,$ and $6$ positions.
This reaction results in the formation of $2,4,6$-trinitrophenol,which is commonly known as picric acid.
The chemical reaction is as follows:
$C_6H_5OH + 3HNO_3 \xrightarrow{conc. H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O$

(C) The reaction of toluene with $Cl_{2}$ in the presence of $hv$ (sunlight) is a free radical substitution reaction that occurs at the side chain,producing benzyl chloride $(A)$.
$C_{6}H_{5}CH_{3} + Cl_{2} \xrightarrow{hv} C_{6}H_{5}CH_{2}Cl + HCl$
Here,$A$ is benzyl chloride $(C_{6}H_{5}CH_{2}Cl)$.
When benzyl chloride $(A)$ is treated with $H_{2}O$ (hydrolysis),it undergoes nucleophilic substitution to form benzyl alcohol $(B)$.
$C_{6}H_{5}CH_{2}Cl + H_{2}O \xrightarrow{\Delta} C_{6}H_{5}CH_{2}OH + HCl$
Thus,$A$ is benzyl chloride and $B$ is benzyl alcohol.

(D) The reaction between benzene diazonium chloride and phenol in a mild alkaline medium is known as a coupling reaction.
In this reaction,the diazonium cation acts as an electrophile and attacks the electron-rich phenol ring at the para position.
Since an electrophile replaces a hydrogen atom on the aromatic ring,this process is classified as an electrophilic substitution reaction.

(A) When phenol is treated with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$,a formyl group $(-CHO)$ is introduced at the ortho position of the benzene ring.
This reaction is known as the Reimer-Tiemann reaction.
The reaction proceeds through the formation of a dichlorocarbene intermediate $(:CCl_2)$,which acts as an electrophile.
The final product obtained after acidification is salicylaldehyde (o-hydroxybenzaldehyde).

(D) When phenol reacts with dilute $HNO_3$ at low temperature,it undergoes electrophilic aromatic substitution to form a mixture of $o$-nitrophenol and $p$-nitrophenol.
$o$-nitrophenol is formed as the major product due to the presence of intra-molecular hydrogen bonding,which stabilizes the ortho isomer.

(A) The coupling reaction of benzene diazonium chloride with phenol involves the retention of the $-N=N-$ group,forming an azo dye ($p$-hydroxyazobenzene).
In other reactions like reduction with phosphinic acid $(H_3PO_2)$,hydrolysis with dilute $H_2SO_4$,or the Sandmeyer reaction with $HCl/Cu$,the diazonium group is replaced by other groups ($-H$,$-OH$,or $-Cl$ respectively),leading to the loss of the diazonium group.
Therefore,the correct option is $A$.

(A) When phenol reacts with bromine water $(Br_2/H_2O)$,it undergoes electrophilic aromatic substitution.
Due to the highly activating nature of the $-OH$ group,the reaction occurs at all available ortho and para positions simultaneously.
This results in the formation of a white precipitate of $2,4,6$-tribromophenol.

(A) The reaction is an example of the Etard reaction modification using acetic anhydride and chromic oxide.
Step $1$: $4-$Nitrotoluene reacts with acetic anhydride and $CrO_3$ at $273-278 \ K$ to form a gem-diacetate derivative $(X)$.
Step $2$: The gem-diacetate derivative $(X)$ undergoes acid-catalyzed hydrolysis $(H_3O^{+} / \Delta)$ to yield $4-$nitrobenzaldehyde $(Y)$.

(A) When phenol is heated with zinc dust,it undergoes reduction to form benzene and zinc oxide.
$C_{6}H_{5}OH + Zn \xrightarrow{\Delta} C_{6}H_{6} + ZnO$

(B) When an ether reacts with excess hot concentrated $HI$,the $C-O$ bonds are cleaved to form alkyl halides.
For methoxyethane $(CH_3OCH_2CH_3)$,the reaction with excess $HI$ proceeds as follows:
$CH_3OCH_2CH_3 + 2HI \xrightarrow{\Delta} CH_3I + CH_3CH_2I + H_2O$
The products formed are iodomethane $(CH_3I)$ and iodoethane $(CH_3CH_2I)$.

(B) $C_2H_5OH \xrightarrow[\text{Pyridine}]{SOCl_2} C_2H_5Cl$ $(A)$
$C_2H_5Cl + C_2H_5ONa \rightarrow C_2H_5-O-C_2H_5$ ($B$,Diethyl ether)
$C_2H_5-O-C_2H_5 + H_2O \xrightarrow[\Delta, \text{Pressure}]{\text{dil. } H_2SO_4} 2C_2H_5OH$ ($C$,Ethanol)
Thus,the product '$C$' is Ethanol.

(C) Sodium metal reacts with compounds containing an acidic hydrogen atom (like $-OH$ or $-COOH$ groups) to release hydrogen gas $(H_{2})$.
$CH_{3}CH_{2}OH$ (ethanol),$C_{6}H_{5}OH$ (phenol),and $CH_{3}COOH$ (acetic acid) all contain an acidic proton.
$CH_{3}OCH_{3}$ (dimethyl ether) does not contain any acidic hydrogen atom,therefore it does not react with sodium metal.

(B) The reaction of $\alpha$-chlorosodium propionate $(CH_3CHClCOONa)$ with aqueous sodium nitrite $(NaNO_2)$ involves the displacement of the chlorine atom by the nitrite group,followed by decarboxylation upon boiling.
The reaction proceeds as follows:
$CH_3CHClCOONa + NaNO_2 \rightarrow CH_3CH(NO_2)COONa + NaCl$
Upon boiling,the intermediate undergoes decarboxylation:
$CH_3CH(NO_2)COONa \xrightarrow{\Delta} CH_3CH_2NO_2 + CO_2 + NaCl$
Thus,the final product is nitroethane.

(B) The reaction of methoxybenzene $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.
In aryl alkyl ethers,the $O-CH_3$ bond is weaker than the $O-C_6H_5$ bond because the $O-C_6H_5$ bond has partial double bond character due to resonance.
Therefore,the $HI$ attacks the methyl group,resulting in the formation of phenol $(C_6H_5OH)$ and iodomethane $(CH_3I)$.
The reaction is: $C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$.

(B) The reaction of an ether with $HI$ involves the protonation of the ether oxygen followed by nucleophilic attack by the iodide ion $(I^-)$.
For an unsymmetrical ether like methoxy ethane $(CH_{3}-O-C_{2}H_{5})$,the reaction proceeds via an $S_{N}2$ mechanism.
The iodide ion attacks the less sterically hindered alkyl group.
Here,the methyl group $(CH_{3}-)$ is less hindered than the ethyl group $(C_{2}H_{5}-)$.
Therefore,the reaction is: $CH_{3}-O-C_{2}H_{5} + HI \rightarrow CH_{3}I + C_{2}H_{5}OH$.
The products obtained are iodomethane $(CH_{3}I)$ and ethanol $(C_{2}H_{5}OH)$.

(C) dihydric phenol contains two hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. $p$-Cresol is a monohydric phenol with a methyl group at the para position.
$2$. Phloroglucinol ($1,3,5$-trihydroxybenzene) is a trihydric phenol.
$3$. Catechol ($1,2$-dihydroxybenzene) is a dihydric phenol.
$4$. Pyrogallol ($1,2,3$-trihydroxybenzene) is a trihydric phenol.
Therefore,Catechol is the correct answer.

(D) dihydric phenol contains two hydroxyl $(-OH)$ groups attached to the benzene ring.
$1$. Catechol ($1,2$-dihydroxybenzene) is a dihydric phenol.
$2$. Resorcinol ($1,3$-dihydroxybenzene) is a dihydric phenol.
$3$. Quinol or Hydroquinone ($1,4$-dihydroxybenzene) is a dihydric phenol.
$4$. Hydroxyquinol ($1,2,4$-trihydroxybenzene) contains three hydroxyl groups,making it a trihydric phenol.
Therefore,Hydroxyquinol is not a dihydric phenol.

(B) The oxidation of an oxime with trifluoroperoxyacetic acid $(CF_3COOOH)$ results in the formation of a nitro compound.
For example,the oxidation of acetone oxime $(CH_3-C(=N-OH)-CH_3)$ yields $2-$nitropropane $(CH_3-CH(NO_2)-CH_3)$.
Generally,aldoximes yield $1^{\circ}$ nitroalkanes,while ketoximes yield $2^{\circ}$ nitroalkanes.

(D) The reaction of benzonitrile $(C_6H_5CN)$ with phenyl magnesium bromide $(C_6H_5MgBr)$ in dry ether proceeds via a nucleophilic addition to the nitrile group.
$1$. The phenyl group from the Grignard reagent attacks the electrophilic carbon of the nitrile group to form an imine magnesium bromide complex $(C_6H_5-C(C_6H_5)=NMgBr)$.
$2$. Upon subsequent acid hydrolysis $(H_3O^+)$,the imine complex is converted into a ketone.
$3$. The final product formed is benzophenone $(C_6H_5-CO-C_6H_5)$.

(A) Rosenmund reduction is the hydrogenation of an acid chloride $(R-COCl)$ to an aldehyde $(R-CHO)$ using a palladium catalyst poisoned with barium sulfate $(Pd/BaSO_4)$.
The reaction is:
$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$
Therefore,option $A$ represents the Rosenmund reduction.

(D) The reaction of benzoyl chloride with $H_{2}$ in the presence of $Pd$ supported on $BaSO_{4}$ is known as the Rosenmund reduction.
In this reaction,the acid chloride is reduced to the corresponding aldehyde.
$C_{6}H_{5}COCl + H_{2} \xrightarrow{Pd/BaSO_{4}} C_{6}H_{5}CHO + HCl$
Therefore,benzaldehyde is obtained.

(C) The reaction of a nitrile $(R-C \equiv N)$ with $DIBAl-H$ (Diisobutylaluminium hydride) followed by acidic hydrolysis $(H_{3}O^{+})$ is a standard method for the preparation of aldehydes.
$DIBAl-H$ acts as a selective reducing agent that reduces the nitrile group to an imine intermediate,which upon hydrolysis yields the corresponding aldehyde $(R-CHO)$.

(B) When calcium formate $(HCOO)_2Ca$ is subjected to dry distillation,it undergoes thermal decomposition to produce formaldehyde (methanal) and calcium carbonate $(CaCO_3)$.
The chemical reaction is as follows:
$(HCOO)_2Ca \xrightarrow{\Delta} HCHO + CaCO_3$
Thus,the correct product is methanal.

(C) When toluene is treated with a solution of chromyl chloride $(CrO_2Cl_2)$ in $CS_2$,a brown chromium complex is obtained,which on acid hydrolysis gives benzaldehyde. This reaction is known as the Etard reaction. The reaction is represented as follows:
$C_6H_5CH_3 + 2CrO_2Cl_2$ $\xrightarrow{CS_2} C_6H_5CH(OCrCl_2OH)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$

(C) The given reaction is the Rosenmund reduction,where an acid chloride is hydrogenated to an aldehyde using a poisoned catalyst,$Pd$ supported on $BaSO_{4}$.
In this reaction,benzoyl chloride $(C_{6}H_{5}COCl)$ reacts with $H_{2}$ in the presence of $Pd-BaSO_{4}$ to form benzaldehyde $(C_{6}H_{5}CHO)$ and $HCl$.
Therefore,'$A$' is benzoyl chloride.

(B) The hydrogenation of an acyl chloride $(RCOCl)$ to an aldehyde $(RCHO)$ using palladium $(Pd)$ supported on barium sulphate $(BaSO_4)$ is known as the $Rosenmund \ reduction$.
The $BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to a primary alcohol.

(D) The Stephen reaction involves the reduction of nitriles $(R-C \equiv N)$ to imines using stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,followed by hydrolysis to yield aldehydes $(R-CHO)$.
Option $(D)$ represents the correct chemical equation for the Stephen reaction:
$R-C \equiv N + 2[H]$ $\xrightarrow{SnCl_2/\text{dil. } HCl} R-CH=NH \cdot HCl$ $\xrightarrow{H_3O^+} R-CHO + NH_4Cl$

(A) The conversion of olefins (alkenes) into aldehydes is known as the hydroformylation or oxo process.
In this process,an alkene reacts with a mixture of $CO$ and $H_{2}$ in the presence of a metal catalyst,such as cobalt carbonyl $(Co_{2}(CO)_{8})$,to produce an aldehyde.
The reaction is represented as:
$CH_{3}-CH=CH_{2} \xrightarrow{CO/H_{2}/Co_{2}(CO)_{8}, \Delta} CH_{3}-CH_{2}-CH_{2}CHO$

(C) The Rosenmund reaction involves the hydrogenation of acid chlorides to aldehydes using a palladium catalyst supported on barium sulphate $(Pd/BaSO_4)$.
$BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.
The reaction is represented as:
$RCOCl + H_2 \xrightarrow{Pd/BaSO_4} RCHO + HCl$

(A) In the $Etard$ reaction,toluene is treated with chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ to form a brown chromium complex,which upon hydrolysis gives benzaldehyde. The reaction is:
$C_6H_5CH_3 + 2CrO_2Cl_2$ $\xrightarrow{CS_2} C_6H_5CH(OCrCl_2OH)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$.
Therefore,the correct option is $A$.

(B) The reaction of benzonitrile $(C_6H_5CN)$ with phenylmagnesium bromide $(C_6H_5MgBr)$ in the presence of dry ether,followed by acid hydrolysis,yields benzophenone $(C_6H_5COC_6H_5)$.
The chemical reaction is as follows:
$C_6H_5CN + C_6H_5MgBr$ $\xrightarrow{\text{dry ether}} C_6H_5C(NMgBr)C_6H_5$ $\xrightarrow{H_3O^+} C_6H_5COC_6H_5 + Mg(OH)Br + NH_3$

(A) The reaction of an acid chloride with a dialkylcadmium reagent is a standard method for the preparation of ketones. The general reaction is: $2RCOCl + R'_{2}Cd \rightarrow 2RCOR' + CdCl_{2}$. In the given reaction,the product is $CH_{3}-CO-CH_{2}-C_{6}H_{5}$ (benzyl methyl ketone). Comparing this with the general reaction,$R$ must be $CH_{3}$ and $R'$ must be $C_{6}H_{5}CH_{2}$. Thus,$A$ is $CH_{3}COCl$ (acetyl chloride).