ChemistryQ351–400 of 806 questions
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351
ChemistryDifficultMCQMHT CET · 2026
The rate law for the reaction $A + B \rightarrow P$ is given by $\text{rate} = k[A]^2[B]$. The rate constant of the reaction at $300 \text{ K}$ is $6.0 \text{ M}^{-2} \text{s}^{-1}$. Calculate the rate of the reaction when $[A] = 1 \text{ M}$ and $[B] = 0.2 \text{ M}$. (in $\text{ M s}^{-1}$)
A
$0.6$
B
$2$
C
$8$
D
$4$
Solution
(A) Step $1$: Identify the given values.
Rate constant $k = 6.0 \text{ M}^{-2} \text{s}^{-1}$.
Concentration $[A] = 1 \text{ M}$.
Concentration $[B] = 0.2 \text{ M}$.
Step $2$: Use the rate law expression.
$\text{Rate} = k[A]^2[B]$
Step $3$: Substitute the values into the equation.
$\text{Rate} = 6.0 \times (1)^2 \times (0.2)$
$\text{Rate} = 6.0 \times 1 \times 0.2$
$\text{Rate} = 1.2 \text{ M s}^{-1}$.
Note: The calculated value is $1.2 \text{ M s}^{-1}$. Since this is not among the options, the question options are incorrect.
Rate constant $k = 6.0 \text{ M}^{-2} \text{s}^{-1}$.
Concentration $[A] = 1 \text{ M}$.
Concentration $[B] = 0.2 \text{ M}$.
Step $2$: Use the rate law expression.
$\text{Rate} = k[A]^2[B]$
Step $3$: Substitute the values into the equation.
$\text{Rate} = 6.0 \times (1)^2 \times (0.2)$
$\text{Rate} = 6.0 \times 1 \times 0.2$
$\text{Rate} = 1.2 \text{ M s}^{-1}$.
Note: The calculated value is $1.2 \text{ M s}^{-1}$. Since this is not among the options, the question options are incorrect.
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352
ChemistryMediumMCQMHT CET · 2026
Which of the following is an example of a second-order reaction?
A
$2 H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)$
B
$CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$
C
$2 NO_2(g) + F_2(g) \rightarrow 2NO_2F(g)$
D
$2 NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(l)$
Solution
(C) Step $1$: Analyze the rate laws for the given reactions.
Step $2$: The reaction $2 NO_2(g) + F_2(g) \rightarrow 2NO_2F(g)$ follows the rate law $Rate = k[NO_2][F_2]$. This is a second-order reaction (first order with respect to $NO_2$ and first order with respect to $F_2$, total order $1+1=2$).
Step $3$: $2 H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)$ is a first-order reaction.
Step $4$: $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ is a fractional order reaction (order $1.5$).
Step $5$: $2 NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(l)$ is a third-order reaction.
Step $6$: Therefore, option $C$ is the correct answer.
Step $2$: The reaction $2 NO_2(g) + F_2(g) \rightarrow 2NO_2F(g)$ follows the rate law $Rate = k[NO_2][F_2]$. This is a second-order reaction (first order with respect to $NO_2$ and first order with respect to $F_2$, total order $1+1=2$).
Step $3$: $2 H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)$ is a first-order reaction.
Step $4$: $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ is a fractional order reaction (order $1.5$).
Step $5$: $2 NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(l)$ is a third-order reaction.
Step $6$: Therefore, option $C$ is the correct answer.
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353
ChemistryDifficultMCQMHT CET · 2026
For the reaction $2NOBr(g) \rightarrow 2NO(g) + Br_2(g)$, the rate law is $r = k[NOBr]^2$. If the rate constant $k = 1.62 \times 10^{-2} \text{ L mol}^{-1} \text{ s}^{-1}$ and the concentration of $[NOBr] = 2 \times 10^{-3} \text{ mol L}^{-1}$, what is the rate of reaction?
A
$6.48 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}$
B
$3.24 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}$
C
$1.62 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}$
D
$8.10 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}$
Solution
(A) Step $1$: Identify the given values: $k = 1.62 \times 10^{-2} \text{ L mol}^{-1} \text{ s}^{-1}$ and $[NOBr] = 2 \times 10^{-3} \text{ mol L}^{-1}$.
Step $2$: Use the rate law expression $r = k[NOBr]^2$.
Step $3$: Substitute the values into the equation: $r = (1.62 \times 10^{-2}) \times (2 \times 10^{-3})^2$.
Step $4$: Calculate the square of the concentration: $(2 \times 10^{-3})^2 = 4 \times 10^{-6} \text{ mol}^2 \text{ L}^{-2}$.
Step $5$: Multiply by the rate constant: $r = 1.62 \times 10^{-2} \times 4 \times 10^{-6} = 6.48 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}$.
Step $2$: Use the rate law expression $r = k[NOBr]^2$.
Step $3$: Substitute the values into the equation: $r = (1.62 \times 10^{-2}) \times (2 \times 10^{-3})^2$.
Step $4$: Calculate the square of the concentration: $(2 \times 10^{-3})^2 = 4 \times 10^{-6} \text{ mol}^2 \text{ L}^{-2}$.
Step $5$: Multiply by the rate constant: $r = 1.62 \times 10^{-2} \times 4 \times 10^{-6} = 6.48 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}$.
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354
ChemistryMediumMCQMHT CET · 2026
Which of the following reactions has an overall order of $1.5$?
A
$2H_2O_2(g) \rightarrow 2H_2O(l) + O_2(g)$
B
$H_2(g) + I_2(g) \rightarrow 2HI(g)$
C
$CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$
D
$2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(l)$
Solution
(C) Step $1$: Analyze the rate laws for the given reactions.
Step $2$: The decomposition of acetaldehyde, $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$, follows the rate law $Rate = k[CH_3CHO]^{1.5}$.
Step $3$: The overall order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
Step $4$: For the decomposition of acetaldehyde, the order is $1.5$.
Step $5$: Other reactions listed are typically first or second order. Thus, option $C$ is correct.
Step $2$: The decomposition of acetaldehyde, $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$, follows the rate law $Rate = k[CH_3CHO]^{1.5}$.
Step $3$: The overall order of a reaction is the sum of the powers of the concentration terms in the rate law expression.
Step $4$: For the decomposition of acetaldehyde, the order is $1.5$.
Step $5$: Other reactions listed are typically first or second order. Thus, option $C$ is correct.
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355
ChemistryDifficultMCQMHT CET · 2026
For the reaction $2NOBr(g) \rightarrow 2NO(g) + Br_2(g)$, the rate law is $r = k[NOBr]^2$. If the rate constant $k = 1.62 \text{ M}^{-1} \text{ s}^{-1}$ and the concentration of $NOBr$ is $5 \times 10^{-3} \text{ M}$, what is the rate of the reaction?
A
$4.05 \times 10^{-5} \text{ M s}^{-1}$
B
$4.05 \times 10^{-6} \text{ M s}^{-1}$
C
$1.62 \times 10^{-5} \text{ M s}^{-1}$
D
$8.10 \times 10^{-6} \text{ M s}^{-1}$
Solution
(A) Given:
Rate law: $r = k[NOBr]^2$
Rate constant $k = 1.62 \text{ M}^{-1} \text{ s}^{-1}$
Concentration $[NOBr] = 5 \times 10^{-3} \text{ M}$
Substitute the values into the rate law:
$r = (1.62 \text{ M}^{-1} \text{ s}^{-1}) \times (5 \times 10^{-3} \text{ M})^2$
$r = 1.62 \times 25 \times 10^{-6} \text{ M s}^{-1}$
$r = 40.5 \times 10^{-6} \text{ M s}^{-1}$
$r = 4.05 \times 10^{-5} \text{ M s}^{-1}$
Rate law: $r = k[NOBr]^2$
Rate constant $k = 1.62 \text{ M}^{-1} \text{ s}^{-1}$
Concentration $[NOBr] = 5 \times 10^{-3} \text{ M}$
Substitute the values into the rate law:
$r = (1.62 \text{ M}^{-1} \text{ s}^{-1}) \times (5 \times 10^{-3} \text{ M})^2$
$r = 1.62 \times 25 \times 10^{-6} \text{ M s}^{-1}$
$r = 40.5 \times 10^{-6} \text{ M s}^{-1}$
$r = 4.05 \times 10^{-5} \text{ M s}^{-1}$
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356
ChemistryMediumMCQMHT CET · 2026
Which of the following is an example of a fractional order reaction?
A
$2H_2O_2(g) \rightarrow 2H_2O(l) + O_2(g)$
B
$H_2(g) + I_2(g) \rightarrow 2HI(g)$
C
$CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$
D
$2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(l)$
Solution
(C) Step $1$: Analyze the order of the given reactions.
Step $2$: The decomposition of acetaldehyde $(CH_3CHO)$ follows a chain mechanism where the rate law is given by $\text{Rate} = k[CH_3CHO]^{3/2}$.
Step $3$: Since the exponent of the concentration term is $3/2$ (which is $1.5$), this reaction is of fractional order.
Step $4$: The other reactions listed are typically elementary or follow integer order kinetics under standard conditions.
Step $5$: Therefore, option $C$ is the correct example of a fractional order reaction.
Step $2$: The decomposition of acetaldehyde $(CH_3CHO)$ follows a chain mechanism where the rate law is given by $\text{Rate} = k[CH_3CHO]^{3/2}$.
Step $3$: Since the exponent of the concentration term is $3/2$ (which is $1.5$), this reaction is of fractional order.
Step $4$: The other reactions listed are typically elementary or follow integer order kinetics under standard conditions.
Step $5$: Therefore, option $C$ is the correct example of a fractional order reaction.
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357
ChemistryMediumMCQMHT CET · 2026
Which of the following is an elementary reaction?
A
$O_3(g) \rightarrow O_2(g) + O(g)$
B
$2NO_2Cl(g) \rightarrow 2NO_2(g) + Cl_2(g)$
C
$2NO_2(g) + F_2(g) \rightarrow 2NO_2F(g)$
D
$2NO(g) + Cl_2(g) \rightarrow 2NOCl(g)$
Solution
(A) An elementary reaction is a single-step reaction that occurs in one go without any intermediate.
Option $A$: The decomposition of ozone $O_3(g) \rightarrow O_2(g) + O(g)$ is a single-step unimolecular elementary reaction.
Option $B$, $C$, and $D$ are complex reactions involving multiple steps and intermediates.
Therefore, the correct answer is $A$.
Option $A$: The decomposition of ozone $O_3(g) \rightarrow O_2(g) + O(g)$ is a single-step unimolecular elementary reaction.
Option $B$, $C$, and $D$ are complex reactions involving multiple steps and intermediates.
Therefore, the correct answer is $A$.
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358
ChemistryMediumMCQMHT CET · 2026
What is the order and molecularity of the following elementary reaction? $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$, rate $= k[NO_2]^2$
A
The reaction is second order and bimolecular.
B
The reaction is first order and bimolecular.
C
The reaction is second order and unimolecular.
D
The reaction is zero order and bimolecular.
Solution
(A) $1$. The order of a reaction is the sum of the powers of the concentration terms in the rate law expression. Here, the rate $= k[NO_2]^2$, so the order is $2$.
$2$. The molecularity of an elementary reaction is the number of reacting species (atoms, ions, or molecules) taking part in the reaction. In the reaction $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$, two molecules of $NO_2$ are involved, so the molecularity is $2$.
$3$. $A$ reaction with molecularity $2$ is called bimolecular.
$4$. Therefore, the reaction is second order and bimolecular.
$2$. The molecularity of an elementary reaction is the number of reacting species (atoms, ions, or molecules) taking part in the reaction. In the reaction $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$, two molecules of $NO_2$ are involved, so the molecularity is $2$.
$3$. $A$ reaction with molecularity $2$ is called bimolecular.
$4$. Therefore, the reaction is second order and bimolecular.
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359
ChemistryDifficultMCQMHT CET · 2026
Calculate the rate constant of a first-order reaction $A \rightarrow B$ having a rate of $5.4 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1}$ and concentration $[A] = 0.3 \text{ M}$.
A
$1.8 \times 10^{-5} \text{ s}^{-1}$
B
$1.8 \times 10^{-4} \text{ s}^{-1}$
C
$2.5 \times 10^{-5} \text{ s}^{-1}$
D
$3.2 \times 10^{-4} \text{ s}^{-1}$
Solution
(A) For a first-order reaction, the rate law is given by: $\text{Rate} = k[A]$.
Given: $\text{Rate} = 5.4 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1}$ and $[A] = 0.3 \text{ M} = 0.3 \text{ mol dm}^{-3}$.
Rearranging the formula for the rate constant $k$: $k = \frac{\text{Rate}}{[A]}$.
Substituting the values: $k = \frac{5.4 \times 10^{-6}}{0.3} \text{ s}^{-1}$.
$k = 18 \times 10^{-6} \text{ s}^{-1} = 1.8 \times 10^{-5} \text{ s}^{-1}$.
Given: $\text{Rate} = 5.4 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1}$ and $[A] = 0.3 \text{ M} = 0.3 \text{ mol dm}^{-3}$.
Rearranging the formula for the rate constant $k$: $k = \frac{\text{Rate}}{[A]}$.
Substituting the values: $k = \frac{5.4 \times 10^{-6}}{0.3} \text{ s}^{-1}$.
$k = 18 \times 10^{-6} \text{ s}^{-1} = 1.8 \times 10^{-5} \text{ s}^{-1}$.
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360
ChemistryMediumMCQMHT CET · 2026
Which of the following factors increases the rate of a unimolecular nucleophilic substitution $(S_N1)$ reaction?
A
Aprotic solvent
B
Stronger nucleophile
C
Weaker nucleophile
D
Polar protic solvent
Solution
(D) $1$. The rate-determining step of an $S_N1$ reaction is the formation of a carbocation intermediate.
$2$. Polar protic solvents (like $H_2O$ or $ROH$) stabilize the carbocation intermediate and the leaving group through solvation (hydrogen bonding), thereby lowering the activation energy.
$3$. $A$ stronger nucleophile does not affect the rate of an $S_N1$ reaction because the nucleophile is involved only in the second, fast step.
$4$. Therefore, a polar protic solvent increases the rate of an $S_N1$ reaction.
$2$. Polar protic solvents (like $H_2O$ or $ROH$) stabilize the carbocation intermediate and the leaving group through solvation (hydrogen bonding), thereby lowering the activation energy.
$3$. $A$ stronger nucleophile does not affect the rate of an $S_N1$ reaction because the nucleophile is involved only in the second, fast step.
$4$. Therefore, a polar protic solvent increases the rate of an $S_N1$ reaction.
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361
ChemistryMediumMCQMHT CET · 2026
Identify the major product of the dehydrohalogenation of $CH_3-CH_2-CH(Br)-CH_3$ using an alcoholic base.
A
$CH_3-CH_2-CH=CH_2$
B
$CH_3-CH=CH-CH_3$
C
$CH_3-(CH_2)_2CH_3$
D
$CH_3-(CH_2)_2CHO$
Solution
(B) $1$. Dehydrohalogenation follows $Saytzeff's$ rule, which states that the more substituted alkene is the major product.
$2$. The reactant is $2-bromobutane$ $(CH_3-CH_2-CH(Br)-CH_3)$.
$3$. Elimination of $HBr$ can occur from $C_1$ or $C_3$ to form $but-1-ene$ or $but-2-ene$ respectively.
$4$. $But-2-ene$ $(CH_3-CH=CH-CH_3)$ is a disubstituted alkene, making it more stable than the monosubstituted $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$.
$5$. Therefore, $CH_3-CH=CH-CH_3$ is the major product.
$2$. The reactant is $2-bromobutane$ $(CH_3-CH_2-CH(Br)-CH_3)$.
$3$. Elimination of $HBr$ can occur from $C_1$ or $C_3$ to form $but-1-ene$ or $but-2-ene$ respectively.
$4$. $But-2-ene$ $(CH_3-CH=CH-CH_3)$ is a disubstituted alkene, making it more stable than the monosubstituted $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$.
$5$. Therefore, $CH_3-CH=CH-CH_3$ is the major product.
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362
ChemistryEasyMCQMHT CET · 2026
Which of the following compounds is formed when chloroform $(CHCl_3)$ is exposed to air and light?
A
Phosgene
B
Phosphine
C
Carbon dioxide
D
Carbon tetrachloride
Solution
(A) When chloroform $(CHCl_3)$ is exposed to air and light, it undergoes slow oxidation to form an extremely poisonous gas called phosgene ($COCl_2$, carbonyl chloride).
The chemical equation is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$
The chemical equation is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$
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363
ChemistryMediumMCQMHT CET · 2026
Which of the following compounds is obtained when $Bromoethane$ is reacted with silver acetate?
A
propanal
B
acetic acid
C
ethyl acetate
D
methyl acetate
Solution
(C) The reaction between $Bromoethane$ $(C_2H_5Br)$ and silver acetate $(CH_3COOAg)$ is a nucleophilic substitution reaction.
$C_2H_5Br + CH_3COOAg \rightarrow CH_3COOC_2H_5 + AgBr$
In this reaction, the acetate ion $(CH_3COO^-)$ acts as a nucleophile and replaces the bromide ion $(Br^-)$ to form ethyl acetate $(CH_3COOC_2H_5)$.
$C_2H_5Br + CH_3COOAg \rightarrow CH_3COOC_2H_5 + AgBr$
In this reaction, the acetate ion $(CH_3COO^-)$ acts as a nucleophile and replaces the bromide ion $(Br^-)$ to form ethyl acetate $(CH_3COOC_2H_5)$.
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364
ChemistryMediumMCQMHT CET · 2026
Which of the following factors favors an $S_N1$ reaction?
A
Strong nucleophile
B
Non-polar solvent
C
Bulky alkyl groups on the carbon atom attached to the halogen atom
D
Small alkyl groups on the carbon atom attached to the halogen atom
Solution
(C) $1$. The $S_N1$ reaction mechanism proceeds via the formation of a carbocation intermediate.
$2$. The rate-determining step is the formation of this carbocation.
$3$. Bulky alkyl groups around the carbocation center stabilize the positive charge through the inductive effect and hyperconjugation, making the formation of the carbocation easier.
$4$. Therefore, tertiary alkyl halides (which have bulky groups) favor $S_N1$ reactions over primary or secondary alkyl halides.
$5$. Strong nucleophiles favor $S_N2$ reactions, and polar protic solvents are required for $S_N1$ reactions.
$2$. The rate-determining step is the formation of this carbocation.
$3$. Bulky alkyl groups around the carbocation center stabilize the positive charge through the inductive effect and hyperconjugation, making the formation of the carbocation easier.
$4$. Therefore, tertiary alkyl halides (which have bulky groups) favor $S_N1$ reactions over primary or secondary alkyl halides.
$5$. Strong nucleophiles favor $S_N2$ reactions, and polar protic solvents are required for $S_N1$ reactions.
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365
ChemistryEasyMCQMHT CET · 2026
What is the general molecular formula of alkyl halides?
A
$C_nH_{2n+2}X$
B
$C_nH_{2n+1}X$
C
$C_nH_{2n}X$
D
$C_nH_{2n-2}X$
Solution
(B) Alkyl halides are derived from alkanes by replacing one hydrogen atom with a halogen atom $(X)$.
The general formula for an alkane is $C_nH_{2n+2}$.
Replacing one $H$ atom with $X$ gives: $C_nH_{2n+2-1}X = C_nH_{2n+1}X$.
Therefore, the correct option is $B$.
The general formula for an alkane is $C_nH_{2n+2}$.
Replacing one $H$ atom with $X$ gives: $C_nH_{2n+2-1}X = C_nH_{2n+1}X$.
Therefore, the correct option is $B$.
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366
ChemistryEasyMCQMHT CET · 2026
Which of the following halogen derivatives forms poisonous phosgene gas when it comes in contact with air and light?
A
$CHCl_3$
B
$PCl_3$
C
$C_6H_5Cl$
D
$CH_2Cl_2$
Solution
(A) Chloroform $(CHCl_3)$ undergoes slow oxidation in the presence of air and light to form a highly poisonous gas known as phosgene $(COCl_2)$.
The chemical equation is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$
The chemical equation is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$
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367
ChemistryEasyMCQMHT CET · 2026
The method of preparation of alkyl fluoride given by the reaction $2R-Br + Hg_2F_2 \rightarrow 2R-F + Hg_2Br_2$ is known as:
A
Finkelstein reaction
B
Swartz reaction
C
Sandmeyer reaction
D
Wurtz reaction
Solution
(B) Step $1$: The reaction involves the exchange of a bromine atom in an alkyl bromide with a fluorine atom using a metallic fluoride like $Hg_2F_2$.
Step $2$: This specific method for the synthesis of alkyl fluorides by heating alkyl chlorides or bromides in the presence of metallic fluorides (such as $AgF$, $Hg_2F_2$, $CoF_2$, or $SbF_3$) is called the Swartz reaction.
Step $2$: This specific method for the synthesis of alkyl fluorides by heating alkyl chlorides or bromides in the presence of metallic fluorides (such as $AgF$, $Hg_2F_2$, $CoF_2$, or $SbF_3$) is called the Swartz reaction.
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368
ChemistryEasyMCQMHT CET · 2026
In benzylic halides, the halogen atom is bonded to
A
$sp$ hybridized carbon atom.
B
$sp^2$ hybridized carbon atom.
C
$sp^3$ hybridized carbon atom.
D
$dsp^2$ hybridized carbon atom.
Solution
(C) $1$. $A$ benzylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom, which is further attached to an aromatic ring.
$2$. The general structure is $C_6H_5-CH_2-X$, where $X$ is the halogen atom.
$3$. In the group $-CH_2-X$, the carbon atom is bonded to two hydrogen atoms, one halogen atom, and one phenyl group, making it $sp^3$ hybridized.
$2$. The general structure is $C_6H_5-CH_2-X$, where $X$ is the halogen atom.
$3$. In the group $-CH_2-X$, the carbon atom is bonded to two hydrogen atoms, one halogen atom, and one phenyl group, making it $sp^3$ hybridized.
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369
ChemistryMediumMCQMHT CET · 2026
Which of the following is a correct feature of the $SN^2$ mechanism?
A
The $SN^2$ mechanism proceeds mainly with racemization.
B
The $SN^2$ mechanism includes a transition state with penta-coordinated carbon.
C
The $SN^2$ mechanism involves the formation of a planar carbocation intermediate in the first step.
D
The $SN^2$ mechanism involves the formation of a carbocation with $sp^2$ hybridized carbon.
Solution
(B) $1$. The $SN^2$ mechanism is a concerted, single-step reaction.
$2$. It involves a backside attack by the nucleophile, leading to an inversion of configuration (Walden inversion), not racemization.
$3$. During the reaction, the nucleophile and the leaving group are both partially bonded to the central carbon atom in a transition state, where the carbon atom is penta-coordinated (bonded to five groups).
$4$. $SN^2$ reactions do not involve the formation of carbocation intermediates.
$5$. Therefore, the correct feature is that it includes a transition state with a penta-coordinated carbon.
$2$. It involves a backside attack by the nucleophile, leading to an inversion of configuration (Walden inversion), not racemization.
$3$. During the reaction, the nucleophile and the leaving group are both partially bonded to the central carbon atom in a transition state, where the carbon atom is penta-coordinated (bonded to five groups).
$4$. $SN^2$ reactions do not involve the formation of carbocation intermediates.
$5$. Therefore, the correct feature is that it includes a transition state with a penta-coordinated carbon.
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370
ChemistryMediumMCQMHT CET · 2026
Which of the following alkyl halides undergoes $SN^1$ reaction most readily?
A
$(CH_3)_3C-F$
B
$(CH_3)_3C-Cl$
C
$(CH_3)_3C-Br$
D
$(CH_3)_3C-I$
Solution
(D) $1$. The rate-determining step in an $SN^1$ reaction is the formation of a carbocation by the departure of the leaving group.
$2$. The ease of departure of the leaving group depends on the strength of the $C-X$ bond.
$3$. The bond dissociation energy decreases as the size of the halogen atom increases $(I > Br > Cl > F)$.
$4$. Therefore, the $C-I$ bond is the weakest and breaks most easily, making the iodide ion the best leaving group.
$5$. Thus, $(CH_3)_3C-I$ undergoes $SN^1$ reaction most readily.
$2$. The ease of departure of the leaving group depends on the strength of the $C-X$ bond.
$3$. The bond dissociation energy decreases as the size of the halogen atom increases $(I > Br > Cl > F)$.
$4$. Therefore, the $C-I$ bond is the weakest and breaks most easily, making the iodide ion the best leaving group.
$5$. Thus, $(CH_3)_3C-I$ undergoes $SN^1$ reaction most readily.
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371
ChemistryMediumMCQMHT CET · 2026
Which of the following alkyl halides is hydrolyzed most rapidly by the $S_N2$ mechanism?
A
$CH_3-Br$
B
$(CH_3)_3C-Br$
C
$CH_3-CH_2-Br$
D
$(CH_3)_2CH-Br$
Solution
(A) The rate of $S_N2$ reaction depends on the steric hindrance around the electrophilic carbon atom.
$1$. The order of reactivity for $S_N2$ reactions is: $Methyl \ halide > Primary \ (1^\circ) > Secondary \ (2^\circ) > Tertiary \ (3^\circ)$.
$2$. $CH_3-Br$ is a methyl halide, which has the least steric hindrance.
$3$. $(CH_3)_3C-Br$ is a tertiary halide, $(CH_3)_2CH-Br$ is a secondary halide, and $CH_3-CH_2-Br$ is a primary halide.
$4$. Therefore, $CH_3-Br$ reacts most rapidly.
$1$. The order of reactivity for $S_N2$ reactions is: $Methyl \ halide > Primary \ (1^\circ) > Secondary \ (2^\circ) > Tertiary \ (3^\circ)$.
$2$. $CH_3-Br$ is a methyl halide, which has the least steric hindrance.
$3$. $(CH_3)_3C-Br$ is a tertiary halide, $(CH_3)_2CH-Br$ is a secondary halide, and $CH_3-CH_2-Br$ is a primary halide.
$4$. Therefore, $CH_3-Br$ reacts most rapidly.
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372
ChemistryMediumMCQMHT CET · 2026
What is the product $P$ obtained in the following reaction? $CH_3CH_2Br + AgCN_{(alc)} \rightarrow P$
A
$CH_3CH_2NC$
B
$CH_3CH_2CN$
C
$CH_3CH_2CONH_2$
D
$CH_3CH_2NH_2$
Solution
(A) $1$. $AgCN$ is a covalent compound.
$2$. In $AgCN$, the nitrogen atom has a lone pair of electrons available for nucleophilic attack, while the carbon atom is not free to act as a nucleophile.
$3$. Therefore, the reaction follows an $S_N2$ mechanism where the nitrogen atom attacks the alkyl halide.
$4$. The reaction is: $CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr$.
$5$. The product $P$ is ethyl isocyanide $(CH_3CH_2NC)$.
$2$. In $AgCN$, the nitrogen atom has a lone pair of electrons available for nucleophilic attack, while the carbon atom is not free to act as a nucleophile.
$3$. Therefore, the reaction follows an $S_N2$ mechanism where the nitrogen atom attacks the alkyl halide.
$4$. The reaction is: $CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr$.
$5$. The product $P$ is ethyl isocyanide $(CH_3CH_2NC)$.
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373
ChemistryMediumMCQMHT CET · 2026
Which among the following is an allylic halide?
A
$CH_3 - CH_2 - CH_2 - X$
B
$C_6H_5 - CH_2 - X$
C
$CH_2 = CH - CH_2 - X$
D
$CH_3 - CH = CH - X$
Solution
(C) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C)$.
In the structure $CH_2 = CH - CH_2 - X$, the halogen atom $X$ is attached to a carbon atom that is adjacent to the $C=C$ double bond.
Therefore, $CH_2 = CH - CH_2 - X$ is an allylic halide.
In the structure $CH_2 = CH - CH_2 - X$, the halogen atom $X$ is attached to a carbon atom that is adjacent to the $C=C$ double bond.
Therefore, $CH_2 = CH - CH_2 - X$ is an allylic halide.
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374
ChemistryEasyMCQMHT CET · 2026
Which of the following reactions is used for the preparation of alkyl fluorides from alkyl chlorides?
A
Sandmeyer reaction
B
Fittig reaction
C
Swarts reaction
D
Finkelstein reaction
Solution
(C) The preparation of alkyl fluorides is best achieved by heating alkyl chlorides or alkyl bromides in the presence of a metallic fluoride such as $AgF$, $Hg_2F_2$, $CoF_2$, or $SbF_3$. This specific reaction is known as the $Swarts$ reaction.
Step $1$: The general reaction is $R-X + AgF \rightarrow R-F + AgX$ (where $X = Cl, Br$).
Step $2$: Since $Finkelstein$ reaction is used for alkyl iodides and $Sandmeyer$ or $Fittig$ reactions are for aryl halides, the correct choice is the $Swarts$ reaction.
Step $1$: The general reaction is $R-X + AgF \rightarrow R-F + AgX$ (where $X = Cl, Br$).
Step $2$: Since $Finkelstein$ reaction is used for alkyl iodides and $Sandmeyer$ or $Fittig$ reactions are for aryl halides, the correct choice is the $Swarts$ reaction.
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375
ChemistryDifficultMCQMHT CET · 2026
Identify the product obtained when $2-\text{chloro}-2-\text{methylbutane}$ is reacted with sodium metal in the presence of dry ether.
A
$2,2,3,3-\text{tetramethyloctane}$
B
$3,3,4,4-\text{tetramethyloctane}$
C
$2,2,3,3-\text{tetramethylhexane}$
D
$3,3,4,4-\text{tetramethylhexane}$
Solution
(D) The reaction of an alkyl halide with sodium metal in the presence of dry ether is known as the Wurtz reaction.
$2-\text{chloro}-2-\text{methylbutane}$ is a tertiary alkyl halide with the structure $CH_3-CH_2-C(CH_3)(Cl)-CH_3$.
In the Wurtz reaction, two molecules of the alkyl halide couple to form a higher alkane.
Two radicals of $2-\text{methylbutan-2-yl}$ $(CH_3-CH_2-C(CH_3)_2^{\bullet})$ combine at the tertiary carbon atoms.
The resulting structure is $CH_3-CH_2-C(CH_3)_2-C(CH_3)_2-CH_2-CH_3$.
Naming the longest chain: the longest chain has $6$ carbons (hexane), with methyl groups at positions $3$ and $4$.
The $IUPAC$ name is $3,3,4,4-\text{tetramethylhexane}$.
$2-\text{chloro}-2-\text{methylbutane}$ is a tertiary alkyl halide with the structure $CH_3-CH_2-C(CH_3)(Cl)-CH_3$.
In the Wurtz reaction, two molecules of the alkyl halide couple to form a higher alkane.
Two radicals of $2-\text{methylbutan-2-yl}$ $(CH_3-CH_2-C(CH_3)_2^{\bullet})$ combine at the tertiary carbon atoms.
The resulting structure is $CH_3-CH_2-C(CH_3)_2-C(CH_3)_2-CH_2-CH_3$.
Naming the longest chain: the longest chain has $6$ carbons (hexane), with methyl groups at positions $3$ and $4$.
The $IUPAC$ name is $3,3,4,4-\text{tetramethylhexane}$.
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376
ChemistryMediumMCQMHT CET · 2026
Identify the product obtained when $2$-chlorobutane is heated with an aqueous solution of potassium hydroxide.
A
$But-1-ene$
B
$But-2-ene$
C
$Butan-1-ol$
D
$Butan-2-ol$
Solution
(D) Step $1$: The reaction of an alkyl halide with an aqueous solution of $KOH$ is a nucleophilic substitution reaction.
Step $2$: In this reaction, the hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the chloride ion $(Cl^-)$ from the $2$-chlorobutane.
Step $3$: The reaction is: $CH_3CH_2CHClCH_3 + KOH_{(aq)} \rightarrow CH_3CH_2CH(OH)CH_3 + KCl$.
Step $4$: The product formed is $butan-2-ol$.
Step $2$: In this reaction, the hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the chloride ion $(Cl^-)$ from the $2$-chlorobutane.
Step $3$: The reaction is: $CH_3CH_2CHClCH_3 + KOH_{(aq)} \rightarrow CH_3CH_2CH(OH)CH_3 + KCl$.
Step $4$: The product formed is $butan-2-ol$.
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377
ChemistryMediumMCQMHT CET · 2026
Identify reagent $B$ in the following reaction: $R - X \xrightarrow{B} \text{Nitroalkane}$
A
$AgNO_2$
B
$NaOR$
C
$KCN \text{ (alc.)}$
D
$KNO_2$
Solution
(A) $1$. The reaction of alkyl halides $(R-X)$ with silver nitrite $(AgNO_2)$ yields nitroalkanes $(R-NO_2)$ as the major product.
$2$. This occurs because the $Ag-O$ bond is covalent, making the nitrogen atom the nucleophilic center.
$3$. In contrast, reaction with potassium nitrite $(KNO_2)$ is ionic and yields alkyl nitrites $(R-ONO)$ as the major product.
$4$. Therefore, reagent $B$ is $AgNO_2$.
$2$. This occurs because the $Ag-O$ bond is covalent, making the nitrogen atom the nucleophilic center.
$3$. In contrast, reaction with potassium nitrite $(KNO_2)$ is ionic and yields alkyl nitrites $(R-ONO)$ as the major product.
$4$. Therefore, reagent $B$ is $AgNO_2$.
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378
ChemistryDifficultMCQMHT CET · 2026
Identify the product $B$ in the following reaction: $CH_3Br \xrightarrow{KCN} A \xrightarrow{Na / C_2H_5OH} B$
A
Ethane
B
Ethanamine
C
Ethanol
D
Methanol
Solution
(B) Step $1$: Reaction of $CH_3Br$ with $KCN$ is a nucleophilic substitution reaction. $CH_3Br + KCN \rightarrow CH_3CN + KBr$. Thus, $A$ is $CH_3CN$ (ethanenitrile).
Step $2$: Reduction of $CH_3CN$ with $Na / C_2H_5OH$ (Mendius reduction) yields a primary amine. $CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3CH_2NH_2$. Thus, $B$ is $CH_3CH_2NH_2$ (ethanamine).
Step $2$: Reduction of $CH_3CN$ with $Na / C_2H_5OH$ (Mendius reduction) yields a primary amine. $CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3CH_2NH_2$. Thus, $B$ is $CH_3CH_2NH_2$ (ethanamine).
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379
ChemistryMediumMCQMHT CET · 2026
Which reagent is used to convert an $Alkyl \ halide$ $(R-X)$ into an $Alkyl \ nitrite$ $(R-O-N=O)$?
A
$Silver \ nitrite$ $(AgNO_2)$
B
$Silver \ cyanide$ $(AgCN)$
C
$Potassium \ nitrite$ $(KNO_2)$
D
$Silver \ carboxylate$ $(RCOOAg)$
Solution
(A) $1$. $Alkyl \ halides$ react with $Silver \ nitrite$ $(AgNO_2)$ to form $Alkyl \ nitrites$ $(R-O-N=O)$ as the major product.
$2$. This occurs because $AgNO_2$ is a covalent compound, and the oxygen atom is more nucleophilic than the nitrogen atom.
$3$. In contrast, $Potassium \ nitrite$ $(KNO_2)$ is an ionic compound, which provides the $NO_2^-$ ion, leading to the formation of $Nitroalkanes$ $(R-NO_2)$ as the major product.
$2$. This occurs because $AgNO_2$ is a covalent compound, and the oxygen atom is more nucleophilic than the nitrogen atom.
$3$. In contrast, $Potassium \ nitrite$ $(KNO_2)$ is an ionic compound, which provides the $NO_2^-$ ion, leading to the formation of $Nitroalkanes$ $(R-NO_2)$ as the major product.
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380
ChemistryMediumMCQMHT CET · 2026
Which among the following is $NOT$ a feature of $S_N2$ mechanism?
A
Single step mechanism
B
Backside attack of nucleophile
C
Formation of planar carbocation intermediate
D
Involves simultaneous bond breaking and bond forming
Solution
(C) $1$. The $S_N2$ mechanism is a concerted, single-step reaction.
$2$. In this mechanism, the nucleophile attacks from the backside of the carbon atom, leading to an inversion of configuration.
$3$. Bond breaking and bond formation occur simultaneously through a pentacoordinate transition state.
$4$. $S_N2$ reactions do not involve the formation of a carbocation intermediate; carbocation formation is a characteristic of $S_N1$ mechanisms.
$5$. Therefore, the formation of a planar carbocation intermediate is $NOT$ a feature of the $S_N2$ mechanism.
$2$. In this mechanism, the nucleophile attacks from the backside of the carbon atom, leading to an inversion of configuration.
$3$. Bond breaking and bond formation occur simultaneously through a pentacoordinate transition state.
$4$. $S_N2$ reactions do not involve the formation of a carbocation intermediate; carbocation formation is a characteristic of $S_N1$ mechanisms.
$5$. Therefore, the formation of a planar carbocation intermediate is $NOT$ a feature of the $S_N2$ mechanism.
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381
ChemistryEasyMCQMHT CET · 2026
In the Finkelstein reaction, which one of the following reagents is used?
A
$HI$ in dry acetone
B
$NaI$ in dry acetone
C
$NaOI$ in dry acetone
D
$NaI$ in dry ether
Solution
(B) The Finkelstein reaction is a type of nucleophilic substitution reaction $(S_N2)$ used to prepare alkyl iodides from alkyl chlorides or alkyl bromides.
In this reaction, the alkyl halide is treated with sodium iodide $(NaI)$ in the presence of dry acetone.
Acetone acts as a solvent in which $NaI$ is soluble, while the byproduct $NaCl$ or $NaBr$ is insoluble and precipitates out, driving the reaction forward according to Le Chatelier's principle.
Therefore, the correct reagent is $NaI$ in dry acetone.
In this reaction, the alkyl halide is treated with sodium iodide $(NaI)$ in the presence of dry acetone.
Acetone acts as a solvent in which $NaI$ is soluble, while the byproduct $NaCl$ or $NaBr$ is insoluble and precipitates out, driving the reaction forward according to Le Chatelier's principle.
Therefore, the correct reagent is $NaI$ in dry acetone.
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382
ChemistryMediumMCQMHT CET · 2026
What is the major product formed when $2$-bromopentane reacts with alcoholic $KOH$?
A
$Pent-1-ene$
B
$Pent-2-ene$
C
$Pentane$
D
$Pentan-2-ol$
Solution
(B) Step $1$: The reaction of $2$-bromopentane with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
Step $2$: According to $Saytzeff's$ rule, in elimination reactions, the more substituted alkene is the major product.
Step $3$: The reaction is: $CH_3-CH(Br)-CH_2-CH_2-CH_3 + KOH (alc.) \rightarrow CH_3-CH=CH-CH_2-CH_3 + KBr + H_2O$.
Step $4$: $Pent-2-ene$ is more substituted than $Pent-1-ene$, hence it is the major product.
Step $2$: According to $Saytzeff's$ rule, in elimination reactions, the more substituted alkene is the major product.
Step $3$: The reaction is: $CH_3-CH(Br)-CH_2-CH_2-CH_3 + KOH (alc.) \rightarrow CH_3-CH=CH-CH_2-CH_3 + KBr + H_2O$.
Step $4$: $Pent-2-ene$ is more substituted than $Pent-1-ene$, hence it is the major product.
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383
ChemistryEasyMCQMHT CET · 2026
Which of the following is a dihydric alcohol?
A
Glycerol
B
Ethylene glycol
C
Sorbitol
D
Methanol
Solution
(B) $1$. $A$ dihydric alcohol is an alcohol that contains two hydroxyl $(-OH)$ groups in its molecule.
$2$. Glycerol $(CH_2OH-CHOH-CH_2OH)$ is a trihydric alcohol.
$3$. Ethylene glycol $(HOCH_2-CH_2OH)$ contains two $-OH$ groups, making it a dihydric alcohol.
$4$. Sorbitol $(C_6H_{14}O_6)$ is a hexahydric alcohol.
$5$. Methanol $(CH_3OH)$ is a monohydric alcohol.
$6$. Therefore, the correct option is $B$.
$2$. Glycerol $(CH_2OH-CHOH-CH_2OH)$ is a trihydric alcohol.
$3$. Ethylene glycol $(HOCH_2-CH_2OH)$ contains two $-OH$ groups, making it a dihydric alcohol.
$4$. Sorbitol $(C_6H_{14}O_6)$ is a hexahydric alcohol.
$5$. Methanol $(CH_3OH)$ is a monohydric alcohol.
$6$. Therefore, the correct option is $B$.
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384
ChemistryMediumMCQMHT CET · 2026
Which of the following ethers, on hydrolysis, gives two different products that are successive members of the homologous series?
A
$CH_3OCH_3$
B
$CH_3OCH_2CH_3$
C
$CH_3CH_2OCH_2CH_2CH_3$
D
$CH_3OCH_2CH_2CH_3$
Solution
(B) $1$. Hydrolysis of an ether $(R-O-R')$ yields two alcohols: $R-OH$ and $R'-OH$.
$2$. For the products to be successive members of a homologous series, they must differ by a $-CH_2-$ group.
$3$. In $CH_3OCH_2CH_3$ (Methoxyethane), hydrolysis gives $CH_3OH$ (Methanol) and $CH_3CH_2OH$ (Ethanol).
$4$. Methanol $(CH_3OH)$ and Ethanol $(CH_3CH_2OH)$ are successive members of the alcohol homologous series, differing by one $-CH_2-$ group.
$2$. For the products to be successive members of a homologous series, they must differ by a $-CH_2-$ group.
$3$. In $CH_3OCH_2CH_3$ (Methoxyethane), hydrolysis gives $CH_3OH$ (Methanol) and $CH_3CH_2OH$ (Ethanol).
$4$. Methanol $(CH_3OH)$ and Ethanol $(CH_3CH_2OH)$ are successive members of the alcohol homologous series, differing by one $-CH_2-$ group.
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385
ChemistryEasyMCQMHT CET · 2026
Identify the trihydric phenol from the following.
A
Hydroquinone
B
Catechol
C
$o$-Cresol
D
Pyrogallol
Solution
(D) $1$. $A$ trihydric phenol is a benzene ring substituted with three hydroxyl $(-OH)$ groups.
$2$. Hydroquinone is a dihydric phenol ($1,4$-dihydroxybenzene).
$3$. Catechol is a dihydric phenol ($1,2$-dihydroxybenzene).
$4$. $o$-Cresol is a monohydric phenol ($2$-methylphenol).
$5$. Pyrogallol is a trihydric phenol ($1,2,3$-trihydroxybenzene).
$6$. Therefore, the correct option is $D$.
$2$. Hydroquinone is a dihydric phenol ($1,4$-dihydroxybenzene).
$3$. Catechol is a dihydric phenol ($1,2$-dihydroxybenzene).
$4$. $o$-Cresol is a monohydric phenol ($2$-methylphenol).
$5$. Pyrogallol is a trihydric phenol ($1,2,3$-trihydroxybenzene).
$6$. Therefore, the correct option is $D$.
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386
ChemistryMediumMCQMHT CET · 2026
Which among the following compounds has the highest solubility in water?
A
Phenol
B
p-Cresol
C
o-Nitrophenol
D
p-Nitrophenol
Solution
(D) $1$. Solubility in water depends on the ability of the compound to form hydrogen bonds with water molecules.
$2$. $o-Nitrophenol$ exhibits intramolecular hydrogen bonding, which reduces its ability to form intermolecular hydrogen bonds with water.
$3$. $p-Nitrophenol$ exhibits intermolecular hydrogen bonding with water molecules due to the presence of the polar $-NO_2$ group and the $-OH$ group, making it more soluble than the others.
$4$. Phenol and $p-Cresol$ have hydrophobic hydrocarbon parts that limit their solubility compared to $p-Nitrophenol$.
$2$. $o-Nitrophenol$ exhibits intramolecular hydrogen bonding, which reduces its ability to form intermolecular hydrogen bonds with water.
$3$. $p-Nitrophenol$ exhibits intermolecular hydrogen bonding with water molecules due to the presence of the polar $-NO_2$ group and the $-OH$ group, making it more soluble than the others.
$4$. Phenol and $p-Cresol$ have hydrophobic hydrocarbon parts that limit their solubility compared to $p-Nitrophenol$.
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387
ChemistryMediumMCQMHT CET · 2026
Which among the following has the highest melting point?
A
Phenol
B
o-Nitrophenol
C
p-Nitrophenol
D
p-Cresol
Solution
(C) $1$. $o-Nitrophenol$ exhibits intramolecular hydrogen bonding, which reduces the intermolecular forces of attraction.
$2$. $p-Nitrophenol$ exhibits strong intermolecular hydrogen bonding, leading to the association of molecules.
$3$. Due to this strong intermolecular association, $p-Nitrophenol$ requires more energy to break the lattice structure, resulting in a significantly higher melting point compared to $Phenol$, $o-Nitrophenol$, and $p-Cresol$.
$2$. $p-Nitrophenol$ exhibits strong intermolecular hydrogen bonding, leading to the association of molecules.
$3$. Due to this strong intermolecular association, $p-Nitrophenol$ requires more energy to break the lattice structure, resulting in a significantly higher melting point compared to $Phenol$, $o-Nitrophenol$, and $p-Cresol$.
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388
ChemistryMediumMCQMHT CET · 2026
Which of the following is a suitable method for the preparation of ether via the $Williamson$ synthesis?
A
Reacting $sodium \ methoxide$ and $tertiary \ butyl \ bromide$
B
Reacting $sodium \ tertiary \ butoxide$ and $methyl \ bromide$
C
Reacting $methyl \ alcohol$ and $ethene$
D
Reacting $methyl \ alcohol$ and $methyl \ bromide$
Solution
(B) $1$. The $Williamson$ synthesis involves the reaction of an alkoxide ion with a primary alkyl halide to form an ether.
$2$. In option $B$, $sodium \ tertiary \ butoxide$ ($CH_3)_3CONa$ reacts with $methyl \ bromide$ $(CH_3Br)$. Since $methyl \ bromide$ is a primary alkyl halide, $S_N2$ reaction occurs efficiently to form $tert-butyl \ methyl \ ether$.
$3$. In option $A$, the use of a tertiary alkyl halide $(tertiary \ butyl \ bromide)$ with an alkoxide leads to an elimination reaction (forming an alkene) rather than substitution, making it unsuitable for ether synthesis.
$2$. In option $B$, $sodium \ tertiary \ butoxide$ ($CH_3)_3CONa$ reacts with $methyl \ bromide$ $(CH_3Br)$. Since $methyl \ bromide$ is a primary alkyl halide, $S_N2$ reaction occurs efficiently to form $tert-butyl \ methyl \ ether$.
$3$. In option $A$, the use of a tertiary alkyl halide $(tertiary \ butyl \ bromide)$ with an alkoxide leads to an elimination reaction (forming an alkene) rather than substitution, making it unsuitable for ether synthesis.
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389
ChemistryMediumMCQMHT CET · 2026
What type of mechanism is followed by the dehydration of alcohol to form ether at $413 \text{ K}$?
A
Nucleophilic substitution unimolecular reaction $(S_N1)$
B
Nucleophilic substitution bimolecular reaction $(S_N2)$
C
Electrophilic substitution reaction
D
Electrophilic addition reaction
Solution
(B) Step $1$: The dehydration of primary alcohols to ethers at $413 \text{ K}$ involves the protonation of the alcohol molecule by an acid catalyst.
Step $2$: $A$ second molecule of alcohol acts as a nucleophile and attacks the protonated alcohol.
Step $3$: This attack occurs via an $S_N2$ mechanism, where the nucleophile displaces a water molecule from the protonated alcohol.
Step $4$: Finally, deprotonation yields the ether product. Thus, the mechanism is a nucleophilic substitution bimolecular reaction $(S_N2)$.
Step $2$: $A$ second molecule of alcohol acts as a nucleophile and attacks the protonated alcohol.
Step $3$: This attack occurs via an $S_N2$ mechanism, where the nucleophile displaces a water molecule from the protonated alcohol.
Step $4$: Finally, deprotonation yields the ether product. Thus, the mechanism is a nucleophilic substitution bimolecular reaction $(S_N2)$.
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390
ChemistryMediumMCQMHT CET · 2026
Which of the following compounds is obtained by Williamson synthesis?
A
only simple ethers
B
only mixed ethers
C
simple and mixed ethers
D
esters
Solution
(C) Williamson synthesis is a reaction of an alkoxide ion $(R-O^-)$ with an alkyl halide $(R'-X)$ to form an ether $(R-O-R')$.
Step $1$: If $R$ and $R'$ are the same alkyl groups, a simple (symmetrical) ether is formed.
Step $2$: If $R$ and $R'$ are different alkyl groups, a mixed (unsymmetrical) ether is formed.
Step $3$: Therefore, Williamson synthesis is used to prepare both simple and mixed ethers.
Step $1$: If $R$ and $R'$ are the same alkyl groups, a simple (symmetrical) ether is formed.
Step $2$: If $R$ and $R'$ are different alkyl groups, a mixed (unsymmetrical) ether is formed.
Step $3$: Therefore, Williamson synthesis is used to prepare both simple and mixed ethers.
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391
ChemistryMediumMCQMHT CET · 2026
What is the product of the reaction between $CH_3ONa$ and $CH_3CH_2Br$?
$CH_3ONa + CH_3CH_2Br \rightarrow$ ?
$CH_3ONa + CH_3CH_2Br \rightarrow$ ?
A
Formation of $CH_3OCH_2CH_3 + NaBr$
B
Formation of $CH_3OH + NaBr$
C
Formation of $CH_3CH_2OH$
D
Formation of $(CH_3)_2O$
Solution
(A) $1$. The reaction between sodium methoxide $(CH_3ONa)$ and bromoethane $(CH_3CH_2Br)$ is a Williamson ether synthesis.
$2$. $CH_3ONa$ acts as a strong nucleophile $(CH_3O^-)$.
$3$. The nucleophile attacks the electrophilic carbon of the ethyl bromide $(CH_3CH_2Br)$ via an $S_N2$ mechanism.
$4$. The bromide ion $(Br^-)$ acts as a leaving group, resulting in the formation of ethyl methyl ether $(CH_3OCH_2CH_3)$ and sodium bromide $(NaBr)$.
$5$. The balanced chemical equation is: $CH_3ONa + CH_3CH_2Br \rightarrow CH_3OCH_2CH_3 + NaBr$.
$2$. $CH_3ONa$ acts as a strong nucleophile $(CH_3O^-)$.
$3$. The nucleophile attacks the electrophilic carbon of the ethyl bromide $(CH_3CH_2Br)$ via an $S_N2$ mechanism.
$4$. The bromide ion $(Br^-)$ acts as a leaving group, resulting in the formation of ethyl methyl ether $(CH_3OCH_2CH_3)$ and sodium bromide $(NaBr)$.
$5$. The balanced chemical equation is: $CH_3ONa + CH_3CH_2Br \rightarrow CH_3OCH_2CH_3 + NaBr$.
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392
ChemistryMediumMCQMHT CET · 2026
Which among the following does $NOT$ undergo Williamson's synthesis?
A
$C_2H_5Br$
B
$CH_3-CH(CH_3)-CH_2Br$
C
$C_6H_5Br$
D
$C_6H_5-CH_2Br$
Solution
(C) Williamson's synthesis involves an $S_N2$ reaction between an alkyl halide and an alkoxide ion.
For the reaction to proceed, the alkyl halide must be primary $(1^\circ)$ or secondary $(2^\circ)$ to minimize steric hindrance.
$C_6H_5Br$ (bromobenzene) does not undergo $S_N2$ reactions because the $C-Br$ bond has partial double-bond character due to resonance, and the $sp^2$ hybridized carbon atom is sterically hindered.
Therefore, $C_6H_5Br$ does not undergo Williamson's synthesis.
For the reaction to proceed, the alkyl halide must be primary $(1^\circ)$ or secondary $(2^\circ)$ to minimize steric hindrance.
$C_6H_5Br$ (bromobenzene) does not undergo $S_N2$ reactions because the $C-Br$ bond has partial double-bond character due to resonance, and the $sp^2$ hybridized carbon atom is sterically hindered.
Therefore, $C_6H_5Br$ does not undergo Williamson's synthesis.
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393
ChemistryEasyMCQMHT CET · 2026
What happens when $phenol$ reacts with bromine water?
A
Brown coloured liquid is obtained
B
Colourless gas evolves
C
White precipitate formed
D
Pink colored solution is obtained
Solution
(C) When $phenol$ reacts with bromine water, it undergoes electrophilic substitution at all ortho and para positions simultaneously.
This results in the formation of $2,4,6-tribromophenol$, which appears as a white precipitate.
The chemical equation is: $C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH + 3HBr$.
This results in the formation of $2,4,6-tribromophenol$, which appears as a white precipitate.
The chemical equation is: $C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH + 3HBr$.
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394
ChemistryMediumMCQMHT CET · 2026
An organic compound with the molecular formula $C_6H_6O$ dissolves in $NaOH$, gives a characteristic colour with neutral $FeCl_3$ and on treatment with bromine water gives a tri-bromo derivative. Which is that compound among the following?
A
Alcohol
B
Ether
C
Ketone
D
Phenol
Solution
(D) $1$. The molecular formula $C_6H_6O$ corresponds to the degree of unsaturation $U = 6 - \frac{6}{2} + 1 = 4$, which suggests an aromatic ring.
$2$. The compound dissolves in $NaOH$, indicating it is acidic in nature.
$3$. It gives a characteristic violet colour with neutral $FeCl_3$, which is a standard test for phenolic groups.
$4$. It reacts with bromine water to form a tri-bromo derivative $(2,4,6-\text{tribromophenol})$, which is characteristic of phenol due to the strong activating effect of the $-OH$ group.
$5$. Therefore, the compound is phenol.
$2$. The compound dissolves in $NaOH$, indicating it is acidic in nature.
$3$. It gives a characteristic violet colour with neutral $FeCl_3$, which is a standard test for phenolic groups.
$4$. It reacts with bromine water to form a tri-bromo derivative $(2,4,6-\text{tribromophenol})$, which is characteristic of phenol due to the strong activating effect of the $-OH$ group.
$5$. Therefore, the compound is phenol.
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395
ChemistryDifficultMCQMHT CET · 2026
In the following sequence of the reaction, the final product "$C$" is:
$Phenol \xrightarrow[NaOH]{CHCl_3} A \xrightarrow[H_3O^+]{} B \xrightarrow[Oxidation]{} C$
$Phenol \xrightarrow[NaOH]{CHCl_3} A \xrightarrow[H_3O^+]{} B \xrightarrow[Oxidation]{} C$
A
B
C
D
Solution
(B) The given reaction sequence is the Reimer-Tiemann reaction followed by oxidation:
$1$. $Phenol$ reacts with $CHCl_3$ and $NaOH$ to form $o-hydroxybenzaldehyde$ (Salicylaldehyde) as the major product $(A)$.
$2$. The intermediate $A$ is already $Salicylaldehyde$. The step $\xrightarrow[H_3O^+]{} B$ typically refers to the acidification of the phenoxide intermediate to obtain the final aldehyde product $B$.
$3$. Oxidation of $Salicylaldehyde$ $(B)$ using an oxidizing agent converts the $-CHO$ group into a $-COOH$ group, resulting in $Salicylic \ acid$ $(C)$.
Therefore, the final product $C$ is $Salicylic \ acid$.
$1$. $Phenol$ reacts with $CHCl_3$ and $NaOH$ to form $o-hydroxybenzaldehyde$ (Salicylaldehyde) as the major product $(A)$.
$2$. The intermediate $A$ is already $Salicylaldehyde$. The step $\xrightarrow[H_3O^+]{} B$ typically refers to the acidification of the phenoxide intermediate to obtain the final aldehyde product $B$.
$3$. Oxidation of $Salicylaldehyde$ $(B)$ using an oxidizing agent converts the $-CHO$ group into a $-COOH$ group, resulting in $Salicylic \ acid$ $(C)$.
Therefore, the final product $C$ is $Salicylic \ acid$.
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396
ChemistryEasyMCQMHT CET · 2026
Identify the name of the reaction when phenol reacts with chloroform in the presence of aqueous $NaOH$.
A
Kolbe's reaction
B
Reimer-Tiemann reaction
C
Williamson synthesis
D
Friedel-Crafts acylation
Solution
(B) Step $1$: The reaction of phenol with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ at $340 \text{ K}$ followed by hydrolysis leads to the formation of salicylaldehyde (o-hydroxybenzaldehyde).
Step $2$: This specific chemical reaction is known as the Reimer-Tiemann reaction.
Step $3$: The electrophile involved in this reaction is dichlorocarbene $(:CCl_2)$.
Step $2$: This specific chemical reaction is known as the Reimer-Tiemann reaction.
Step $3$: The electrophile involved in this reaction is dichlorocarbene $(:CCl_2)$.
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397
ChemistryMediumMCQMHT CET · 2026
Identify '$Z$' in the following reaction: $Ar-OH + Cl-C(=O)-R \xrightarrow{\text{Pyridine}} Z + HCl$
A
$Ar-O-R$
B
$Ar-O-C(=O)-R$
C
$R-C(=O)-O-Ar$
D
$Ar-C(=O)-R$
Solution
(B) Step $1$: The reaction between a phenol $(Ar-OH)$ and an acid chloride $(R-COCl)$ in the presence of a base like pyridine is known as the Schotten-Baumann reaction.
Step $2$: The lone pair on the oxygen atom of the phenol attacks the electrophilic carbonyl carbon of the acid chloride.
Step $3$: The chloride ion $(Cl^-)$ acts as a leaving group, and the pyridine neutralizes the $HCl$ produced to drive the reaction forward.
Step $4$: The product formed is an ester, specifically an aryl ester, with the structure $Ar-O-C(=O)-R$.
Step $2$: The lone pair on the oxygen atom of the phenol attacks the electrophilic carbonyl carbon of the acid chloride.
Step $3$: The chloride ion $(Cl^-)$ acts as a leaving group, and the pyridine neutralizes the $HCl$ produced to drive the reaction forward.
Step $4$: The product formed is an ester, specifically an aryl ester, with the structure $Ar-O-C(=O)-R$.
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398
ChemistryMediumMCQMHT CET · 2026
Identify the product $P$ obtained in the following reaction: $C_6H_5OH + HCHO \xrightarrow{\text{Acid}} P$
A
B
C
D
Solution
(D) The reaction between phenol $(C_6H_5OH)$ and formaldehyde $(HCHO)$ in the presence of an acid catalyst is an electrophilic aromatic substitution reaction.
Phenol is an ortho/para-directing group due to the electron-donating $-OH$ group.
Formaldehyde acts as an electrophile in the presence of acid.
The reaction leads to the formation of hydroxybenzyl alcohol (also known as saligenin) as the major product, specifically the ortho-isomer ($o$-hydroxybenzyl alcohol).
Thus, the product $P$ is $o$-hydroxybenzyl alcohol.
Phenol is an ortho/para-directing group due to the electron-donating $-OH$ group.
Formaldehyde acts as an electrophile in the presence of acid.
The reaction leads to the formation of hydroxybenzyl alcohol (also known as saligenin) as the major product, specifically the ortho-isomer ($o$-hydroxybenzyl alcohol).
Thus, the product $P$ is $o$-hydroxybenzyl alcohol.
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399
ChemistryMediumMCQMHT CET · 2026
In the following reaction, condition '$A$' is -
A
Reduction with Zinc dust
B
Dehydration with $H_2SO_4 / 443 \text{ K}$
C
Treatment with $CS_2 / \text{low temperature}$
D
Oxidation with $CrO_3$
Solution
(D) The reaction shows the oxidation of phenol to $p$-benzoquinone.
Phenol undergoes oxidation in the presence of chromic acid ($CrO_3$ in acidic medium) to form a conjugated diketone known as $p$-benzoquinone.
Therefore, the correct condition '$A$' is oxidation with $CrO_3$.
Phenol undergoes oxidation in the presence of chromic acid ($CrO_3$ in acidic medium) to form a conjugated diketone known as $p$-benzoquinone.
Therefore, the correct condition '$A$' is oxidation with $CrO_3$.
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400
ChemistryDifficultMCQMHT CET · 2026
How many moles of hydrogen gas are liberated when $6 \text{ moles}$ of ethyl alcohol reacts with sodium metal?
A
$2$
B
$3$
C
$4$
D
$6$
Solution
(B) The chemical reaction between ethyl alcohol $(C_2H_5OH)$ and sodium metal $(Na)$ is:
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \uparrow$
From the balanced chemical equation, $2 \text{ moles}$ of ethyl alcohol produce $1 \text{ mole}$ of hydrogen gas $(H_2)$.
Therefore, $6 \text{ moles}$ of ethyl alcohol will produce:
$\frac{1}{2} \times 6 = 3 \text{ moles}$ of $H_2$ gas.
Thus, the correct option is $B$.
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \uparrow$
From the balanced chemical equation, $2 \text{ moles}$ of ethyl alcohol produce $1 \text{ mole}$ of hydrogen gas $(H_2)$.
Therefore, $6 \text{ moles}$ of ethyl alcohol will produce:
$\frac{1}{2} \times 6 = 3 \text{ moles}$ of $H_2$ gas.
Thus, the correct option is $B$.
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