Why do tertiary alkyl halides undergo S\textsubscript{N1 reaction at a faster rate?
Official Solution
Correct Option:
(1)
Step 1: The SN1 mechanism proceeds via formation of a carbocation intermediate. Step 2: Tertiary alkyl halides have a carbon atom bonded to three alkyl groups. These groups stabilize the positive charge on the carbocation through: - Hyperconjugation (delocalization of electrons from C–H bonds of adjacent alkyl groups), and - Inductive effect (electron-donating nature of alkyl groups). Step 3: This stability lowers the activation energy of the rate-determining step (carbocation formation), hence speeding up the reaction.
02
PYQ 2023
medium
chemistryID: cbse-cla
Why is chloroform stored in dark coloured air tight bottles?
Official Solution
Correct Option:
(1)
Step 1: Chloroform (\ch{CHCl3}) can oxidize in the presence of air and light to form phosgene (\ch{COCl2}), which is highly toxic. Step 2: To avoid this photochemical reaction, chloroform is stored in: - Dark coloured bottles to block light, - Air-tight containers to prevent oxygen exposure. Conclusion: This prevents degradation and ensures safe handling and storage of chloroform.
03
PYQ 2023
medium
chemistryID: cbse-cla
Write the major alkene that would be formed by dehydrohalogenation of 2-Bromopentane.
Official Solution
Correct Option:
(1)
Step 1: Dehydrohalogenation is an elimination reaction in which a hydrogen atom and a halogen atom are removed from adjacent carbon atoms, leading to the formation of an alkene. Step 2: The structure of 2-bromopentane is: Step 3: In the presence of alcoholic KOH (a strong base), elimination follows Zaitsev’s Rule, which favors the formation of the **more substituted alkene. Step 4: Removal of a -hydrogen from the third carbon (next to the carbon bearing bromine) leads to the formation of 2-pentene, which is more stable and thus the major product. Conclusion: The major alkene obtained from dehydrohalogenation of 2-bromopentane is 2-pentene.
04
PYQ 2023
medium
chemistryID: cbse-cla
Which would undergo SN2 reaction at a faster rate and why? Given compounds:
Compound A: CH3–CH2–Br
Compound B:
CH3
|
CH3–C–Br
|
CH3
Official Solution
Correct Option:
(1)
Step 1: The SN2 mechanism is a one-step reaction where the nucleophile attacks the carbon atom bonded to the leaving group from the backside. Step 2: The rate of SN2 reaction is highly sensitive to steric hindrance around the carbon bearing the leaving group. Step 3: - In Compound A (CH3–CH2–Br), the bromine is attached to a primary carbon, which is less hindered and easily accessible to the nucleophile. - In Compound B (CH3–C(CH3)2–Br), the bromine is attached to a tertiary carbon, which is highly hindered due to three bulky methyl groups, making backside attack very difficult. Conclusion: Compound A undergoes SN2 reaction at a much faster rate due to minimal steric hindrance.
05
PYQ 2023
medium
chemistryID: cbse-cla
Write the structure of 1-chloro-4-ethylcyclohexane.
Official Solution
Correct Option:
(1)
Step 1: Draw a cyclohexane ring with six carbon atoms. Step 2: Number the ring such that carbon-1 has a chlorine (Cl) atom. Step 3: Place an ethyl group ( ) on carbon-4 of the ring. Step 4: This gives the correct structure of 1-chloro-4-ethylcyclohexane.
06
PYQ 2023
medium
chemistryID: cbse-cla
What is an ambident nucleophile? Give one example.
Official Solution
Correct Option:
(1)
Step 1: Ambident nucleophiles have two nucleophilic sites, and can attack through either site. Step 2: For example, the cyanide ion ( ) can attack through carbon to form alkyl cyanides, or through nitrogen to form isocyanides. Step 3: The actual product depends on the conditions of the reaction and the nature of the substrate.
07
PYQ 2023
medium
chemistryID: cbse-cla
Which of the following reactions is a halogenated exchange reaction:
1
2
3
4
Bromination of toluene with and in dark
Official Solution
Correct Option:
(2)
Step 1: A halogen exchange reaction (also known as a halogen exchange or halogen replacement reaction) is a reaction where one halogen atom is replaced by another halogen atom. Step 2: In option (B), an alkyl halide reacts with sodium iodide ( ) in dry acetone. This results in the replacement of halogen with iodine, forming and sodium halide ( ). This is a classic halogen exchange reaction, commonly known as the Finkelstein reaction. Step 3: - Option (A) is an addition reaction (hydrohalogenation of alkene). - Option (C) is a conversion of an alcohol to an alkyl halide, not an exchange of halogens. - Option (D) is an electrophilic aromatic substitution reaction, not an exchange of halogens.
08
PYQ 2023
hard
chemistryID: cbse-cla
Arrange the following in increasing order of their boiling points: 1-chloropropane, 2-chloropropane, 1-chlorobutane (A) 2-chloropropane 1-chloropropane 1-chlorobutane
Official Solution
Correct Option:
(1)
Step 1: Boiling point depends on molecular weight, surface area, and branching. Step 2: Among the three compounds:
2-chloropropane is more branched, hence has lower surface area and lower boiling point.
1-chloropropane is linear and less branched, so it has a higher boiling point than 2-chloropropane.
1-chlorobutane has a larger molecular size and greater surface area than both propyl compounds, hence the highest boiling point.
09
PYQ 2023
easy
chemistryID: cbse-cla
Why is chlorobenzene less reactive towards nucleophilic substitution reaction?
Official Solution
Correct Option:
(1)
Step 1: In chlorobenzene, the chlorine atom is bonded to an sp2-hybridized carbon of the benzene ring. Step 2: The lone pair of electrons on chlorine can delocalize into the aromatic ring through resonance, creating structures where the C–Cl bond exhibits partial double bond character. This strengthens the bond and makes it difficult to break. Step 3: Additionally, the benzene ring is electron-rich, which repels nucleophiles (electron donors), making nucleophilic attack less favorable. Conclusion: Due to resonance and the resulting partial double bond character, as well as electronic repulsion from the aromatic ring, chlorobenzene is much less reactive towards nucleophilic substitution than alkyl halides.
10
PYQ 2024
medium
chemistryID: cbse-cla
Anisole reacts with HI to give:
1
2
3
4
Official Solution
Correct Option:
(1)
Anisole (C ) undergoes cleavage of the C-O bond when treated with hydroiodic acid (HI). The reaction proceeds via the SN1 or SN2 mechanism, depending on the conditions: Mechanism:
1. The iodide ion (I ) acts as a nucleophile and attacks the methyl group, breaking the ether bond.
2. This forms phenol (C OH) and methyl iodide (CH I).
3. Since the benzene ring is electron-rich, nucleophilic substitution does not occur at the ring. Thus, the final products are phenol and methyl iodide.
11
PYQ 2024
medium
chemistryID: cbse-cla
Which of the following alkyl halides will undergo S 1 reaction most readily?
1
2
3
4
Official Solution
Correct Option:
(1)
S 1 reactions favor alkyl halides that form stable carbocations. Tertiary alkyl halides undergo S 1 reactions faster due to increased carbocation stability. Iodine is the best leaving group among halogens, making tert-Butyl iodide ((CH ) I) the most reactive in an S 1 reaction.
12
PYQ 2024
easy
chemistryID: cbse-cla
In the following pair of compounds, which compound undergoes an 2 reaction faster and why?
Official Solution
Correct Option:
(1)
- The SN2 reaction rate depends on the leaving group. The better the leaving group, the faster the SN2 substitution. - Iodine (I ) is a better leaving group than bromine (Br ), due to the larger size and weaker bond of I with the carbon atom.
- Therefore, ethyl iodide (CH CH I) will undergo the SN2 reaction faster than ethyl bromide (CH CH Br).
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\end{center}
13
PYQ 2024
medium
chemistryID: cbse-cla
Which alkyl halide from the given options will undergo S 1 reaction faster?
1
(CH ) C–Br
2
(CH ) CH–Br
3
CH CH –Br
4
(CH ) C–CH Br
Official Solution
Correct Option:
(1)
The S 1 mechanism involves a two-step process:
1. The leaving group (in this case, Br ) departs, forming a carbocation.
2. The nucleophile (in this case, the solvent or other reactant) attacks the carbocation, resulting in the product. The rate-determining step in an S 1 reaction is the formation of the carbocation. The more stable the carbocation, the faster the reaction. Carbocation stability increases with the number of alkyl groups attached to the positively charged carbon, as these groups provide inductive stabilization and hyperconjugation. - Tertiary carbocations are the most stable due to the electron-donating effects of the three alkyl groups.
- Secondary carbocations are less stable.
- Primary carbocations are highly unstable and do not readily form. In this case, the alkyl halide (CH ) C–Br (option A) forms a tertiary carbocation, making it the most likely to undergo an S 1 reaction fastest. Step 1: The reaction mechanism for S 1 involves the formation of a carbocation. Step 2: Tertiary carbocations are the most stable and form the fastest, which is why (CH ) C–Br undergoes the reaction faster.
14
PYQ 2024
medium
chemistryID: cbse-cla
What happens when chlorobenzene and methyl chloride are treated with sodium metal in dry ether?
Official Solution
Correct Option:
(1)
When chlorobenzene and methyl chloride are treated with sodium metal in dry ether, a reaction known as the Wurtz-Fittig reaction occurs, leading to the formation of toluene. Sodium metal reduces the chlorine atom in chlorobenzene, and the methyl group from methyl chloride is added to the benzene ring. This results in the formation of toluene (C H CH ).
15
PYQ 2024
easy
chemistryID: cbse-cla
What type of nucleophilic substitution ( or ) occurs in the hydrolysis of 2-Bromobutane to form -Butan-2-ol? Give reason.
Official Solution
Correct Option:
(1)
Step 1: The reaction follows an mechanism. This is due to the formation of a planar carbocation, which can be attacked by the nucleophile from either side. This leads to racemization of the product.
The reaction is accompanied by racemization, and the intermediate carbocation is formed in an reaction.
16
PYQ 2024
medium
chemistryID: cbse-cla
Why does the presence of nitro groups at ortho- and para- positions in haloarenes increase their reactivity towards nucleophilic substitution reactions?
Official Solution
Correct Option:
(1)
The presence of nitro groups at the ortho- and para- positions in haloarenes increases their reactivity towards nucleophilic substitution reactions due to the following reasons: 1. Electron-Withdrawing Effect: - Nitro groups ( ) are strong electron-withdrawing groups due to their resonance and inductive effects. They pull electron density away from the aromatic ring, particularly at the ortho- and para- positions, making the carbon attached to the halogen more electrophilic. 2. Stabilization of the Transition State: - In nucleophilic substitution reactions, the intermediate or transition state is stabilized by electron-withdrawing groups. The nitro group at the ortho- or para- position stabilizes the negative charge on the transition state, making it easier for the nucleophile to attack the electrophilic carbon. 3. Increased Reactivity: - This increased electrophilicity of the carbon to which the halogen is attached makes the molecule more reactive towards nucleophilic substitution reactions, especially in the mechanism. Thus, nitro groups at ortho- and para- positions in haloarenes increase their reactivity towards nucleophilic substitution reactions by making the carbon more electrophilic and stabilizing the transition state.
17
PYQ 2024
medium
chemistryID: cbse-cla
Which halogen compound in the following pair will react faster in reaction and why?
Official Solution
Correct Option:
(1)
(a) The compound (ethyl chloride) will react faster in an reaction than (benzyl chloride). The main reasons are: 1. Steric Hindrance: - is a primary halide with less steric hindrance around the carbon bonded to the halogen. This allows the nucleophile to attack the electrophilic carbon more easily. - (benzyl chloride) is a benzyl halide. Although benzyl halides are generally reactive in , they still experience more steric hindrance than a simple alkyl halide like ethyl chloride. 2. Nucleophilic Substitution Mechanism: - reactions involve a backside attack of the nucleophile on the carbon bonded to the halogen, resulting in the inversion of configuration. Less steric hindrance on the carbon results in a faster reaction. Therefore, reacts faster in an reaction than .
18
PYQ 2024
medium
chemistryID: cbse-cla
What happens when ethyl chloride is treated with alcoholic potassium hydroxide?
Official Solution
Correct Option:
(1)
When ethyl chloride (C H Cl) is treated with alcoholic potassium hydroxide (KOH), it undergoes an elimination reaction. The result is the formation of ethene (C H ). The reaction follows the E2 mechanism, where the base abstracts a proton from a carbon atom adjacent to the one bonded to the halogen (chlorine), resulting in the elimination of HCl and the formation of a double bond.
Step 1: The KOH deprotonates the β-carbon (carbon adjacent to the one with the chlorine), leading to the formation of a double bond between the \alpha and \beta carbons.
Step 2: The product formed is ethene (C H ), and HCl is eliminated. Reaction:
19
PYQ 2024
medium
chemistryID: cbse-cla
The presence of –NO group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution reactions. Give reason to explain the above statement.
Official Solution
Correct Option:
(1)
The presence of the electron-withdrawing –NO group at the ortho or para positions of a halogenated benzene ring increases the reactivity of haloarenes towards nucleophilic substitution reactions. This is because:
- The –NO group is electron-withdrawing, meaning it pulls electron density away from the benzene ring through the inductive effect.
- This increases the partial positive charge on the carbon atom attached to the halogen (which is the site of nucleophilic attack).
- The increased positive charge on the carbon makes it more electrophilic, thus more susceptible to attack by nucleophiles, thereby increasing the rate of nucleophilic substitution reactions.
Step 1: The electron-withdrawing nature of the –NO group enhances the electrophilicity of the carbon bonded to the halogen.
Step 2: The nucleophile can more easily attack this carbon, especially at the ortho and para positions, where the electron-withdrawing effect of the NO group is strongest.
20
PYQ 2024
easy
chemistryID: cbse-cla
Which isomer of C4H9Cl has the lowest boiling point?
Official Solution
Correct Option:
(1)
The isomer of C4H9Cl with the lowest boiling point is tert-butyl chloride C4H9Cl. This is because branching reduces the surface area of the molecule, leading to weaker intermolecular forces (van der Waals forces) and consequently a lower boiling point compared to the straight-chain isomer.
21
PYQ 2024
medium
chemistryID: cbse-cla
Out of chlorobenzene and 2,4,6-trinitrochlorobenzene, which is more reactive towards nucleophilic substitution and why?
Official Solution
Correct Option:
(1)
2,4,6-Trinitrochlorobenzene is more reactive towards nucleophilic substitution because the nitro ( ) groups present on the benzene ring are electron-withdrawing in nature. These groups pull electron density away from the ring, making the carbon-chlorine bond weaker and easier to break, facilitating nucleophilic attack. In contrast, chlorobenzene does not have such electron-withdrawing groups, making it less reactive towards nucleophilic substitution.
22
PYQ 2024
easy
chemistryID: cbse-cla
Draw the structures of major monohalo products in each of the following reactions:
Official Solution
Correct Option:
(1)
(i) Reaction with HBr: The addition of HBr to styrene (C H CH=CH ) follows Markovnikov’s rule, leading to the formation of 1-bromoethylbenzene as the major product.
(ii) Bromination in UV light: In the presence of UV light, free radical substitution occurs, resulting in the formation of bromocyclohexane as the major product.
23
PYQ 2024
medium
chemistryID: cbse-cla
Which compound in the given pair will undergo S 2 reaction at a faster rate and why?
CH –CH –I and CH –CH –Br
Official Solution
Correct Option:
(1)
CH –CH –I will undergo the S 2 reaction at a faster rate than CH –CH –Br. This is because iodine is a better leaving group than bromine due to the larger size of the iodine atom, which makes it easier for the bond to break in the S 2 mechanism. Iodine forms a more stable anion, and its bond with carbon is weaker, allowing for a faster nucleophilic substitution.
Step 1: In S 2 reactions, the leaving group plays a significant role. Iodide (I ) is a better leaving group than bromide (Br ). Step 2: The larger the leaving group, the faster it will leave, resulting in a faster S 2 reaction.
24
PYQ 2025
hard
chemistryID: cbse-cla
Draw the structure of the major monohalo product for each of the following reactions:
Official Solution
Correct Option:
(1)
The reaction involves a bromination in the presence of heat, which typically leads to the formation of a benzylic bromination product. In this case, the hydrogen at the benzylic position (the carbon adjacent to the benzene ring) will be replaced by a bromine atom, forming 1-bromo-2-phenylethane as the major product. The structure of the product is:
25
PYQ 2025
easy
chemistryID: cbse-cla
Why are haloarenes less reactive towards nucleophilic substitution reactions? How does the presence of nitro (–NO₂) group at ortho- and para-positions in haloarenes increase the reactivity towards nucleophilic substitution reactions?
Official Solution
Correct Option:
(1)
Haloarenes are less reactive towards nucleophilic substitution reactions due to the partial double bond character between the carbon and halogen in the aromatic ring, which makes it harder for the halogen to leave. The presence of a nitro group ( ) at the ortho- or para-positions on the benzene ring activates the ring towards nucleophilic substitution. The nitro group is electron-withdrawing, which increases the positive character of the carbon attached to the halogen, making it more susceptible to attack by nucleophiles.
26
PYQ 2025
easy
chemistryID: cbse-cla
Assertion (A): n-Butyl chloride has higher boiling point than n-Butyl bromide. Reason (R): C—Cl bond is more polar than C—Br bond.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option:
(4)
The boiling point of n-Butyl chloride is actually lower than n-Butyl bromide due to the larger size and weaker Van der Waals forces in n-Butyl bromide, even though the C—Cl bond is more polar than the C—Br bond. Hence, the assertion is false, but the reason is true.
To solve this question, we need to understand the nature of nucleophilic bimolecular substitution reactions (SN2) involving alkyl halides.
1. Understanding the SN2 Mechanism: In the SN2 mechanism, the nucleophile attacks the carbon atom bonded to the leaving group from the opposite side of the leaving group. This results in the simultaneous displacement of the leaving group and the formation of a new bond with the nucleophile.
2. Inversion of Configuration: During the attack by the nucleophile, there is a backside attack on the carbon. This causes the configuration at the chiral center to invert, meaning that if the molecule was initially in the "R" configuration, it will now be in the "S" configuration, and vice versa.
3. Conclusion: Alkyl halides undergoing nucleophilic bimolecular substitution reactions (SN2) involve an inversion of configuration at the carbon center that is undergoing substitution.
Final Answer: The correct option is (C) inversion of configuration.
28
PYQ 2025
medium
chemistryID: cbse-cla
Assertion (A): The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. Reason (R): The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.
1
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
2
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option:
(2)
To solve this question, let's analyze the assertion and reason provided.
Assertion (A): The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.
Reason (R): The boiling points of alkyl chlorides, bromides, and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.
1. Boiling Points of Alkyl Halides: In alkyl halides, the boiling points depend on several factors such as molecular mass, polarity, and the type of halogen. The size of the halogen atom increases down the group in the periodic table, which leads to an increase in London dispersion forces. This results in a higher boiling point for alkyl iodides (RI) compared to alkyl bromides (RBr), alkyl chlorides (RCl), and alkyl fluorides (RF).
2. Explanation of Assertion (A): The order of boiling points is as follows: RI > RBr > RCl > RF. This is because iodine is the largest halogen and has the highest dispersion forces, leading to the highest boiling point. Fluorine, on the other hand, is the smallest halogen, and thus, RF has the lowest boiling point among alkyl halides.
3. Evaluation of Reason (R): It is true that the boiling points of alkyl halides (chlorides, bromides, and iodides) are higher than those of hydrocarbons with comparable molecular mass. This is due to the presence of dipole-dipole interactions and London dispersion forces between the halogen atoms and the alkyl group, which are absent in hydrocarbons.
4. Conclusion: Both the assertion and the reason are correct. Moreover, the reason explains the assertion because the boiling points of alkyl halides are higher than hydrocarbons due to intermolecular forces, and the trend in boiling points follows the pattern described in the assertion.
Final Answer: The correct option is that both the assertion and reason are true, and the reason explains the assertion.
29
PYQ 2025
medium
chemistryID: cbse-cla
Out of 2-Bromobutane, 1-Bromobutane, 2-Bromopropane and 1-Bromopropane, the molecule which is chiral in nature is:
1
2-Bromobutane
2
1-Bromobutane
3
2-Bromopropane
4
1-Bromopropane
Official Solution
Correct Option:
(1)
A chiral molecule is one that has a carbon atom bonded to four different groups, creating a non-superimposable mirror image (enantiomers). Among the options, 2-Bromobutane is chiral because the central carbon (C2) is bonded to four different groups: a methyl group, a hydrogen atom, a bromine atom, and an ethyl group. The other molecules do not have a chiral center.
30
PYQ 2025
medium
chemistryID: cbse-cla
Polyhalogen compounds have wide application in industries and agriculture. DDT is also a very important polyhalogen compound. It is a:
1
greenhouse gas
2
fertilizer
3
biodegradable insecticide
4
non-biodegradable insecticide
Official Solution
Correct Option:
(4)
DDT (Dichlorodiphenyltrichloroethane) is a well-known insecticide that is highly effective but is non-biodegradable. It accumulates in the environment and in living organisms, causing long-term ecological damage. Therefore, DDT is a non-biodegradable insecticide, making option (D) the correct choice.
31
PYQ 2025
medium
chemistryID: cbse-cla
Which of the following haloalkanes react with aqueous KOH most rapidly by reaction?
1
2-Chlorobutane
2
1-Bromobutane
3
2-Bromo-2-Methylpropane
4
2,2-Dimethyl-1-Chloropropane
Official Solution
Correct Option:
(3)
The mechanism proceeds through the formation of a carbocation. A more stable carbocation leads to a faster reaction. 2-Bromo-2-Methylpropane forms a more stable tertiary carbocation (due to hyperconjugation and inductive effects), making it react more rapidly by .
32
PYQ 2025
medium
chemistryID: cbse-cla
The synthesis of alkyl fluoride is best carried out by:
1
Finkelstein reaction
2
Swarts reaction
3
Sandmeyer reaction
4
Wurtz reaction
Official Solution
Correct Option:
(2)
Alkyl fluorides are best prepared by Swarts reaction, which involves halogen exchange using metal fluorides like SbF .
33
PYQ 2025
medium
chemistryID: cbse-cla
Which of the following haloalkanes is most reactive towards S 2 reaction ?
1
CH –CH –I
2
CH –CH –Br
3
CH –CH –Cl
4
CH –CH –F
Official Solution
Correct Option:
(1)
In S 2 reactions, the leaving group ability is key. Iodide ion (I ) is the best leaving group among halides. Thus, ethyl iodide is most reactive towards S 2.
34
PYQ 2025
easy
chemistryID: cbse-cla
A compound (A) with molecular formula , which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), , which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of (A), (B), (C) and (D). Write the chemical equation when compound (A) is reacted with alcoholic KOH.
Official Solution
Correct Option:
(1)
Compound (A): is butyl iodide (n-butyl iodide). Compound (B): The reaction of butyl iodide with alcoholic KOH gives butene (C H ). Compound (C): The reaction of butene with HI gives 2-iodobutane. Compound (D): When n-butyl iodide reacts with sodium in dry ether, butane is formed. The chemical reaction for the formation of (B) is:
35
PYQ 2025
medium
chemistryID: cbse-cla
Account for the following: (i) Allyl chloride is hydrolyzed more readily than n-propyl chloride. (ii) Isocyanides are formed when alkyl halides are treated with silver cyanide. (iii) Methyl chloride reacts faster with OH ion in reaction than t-butyl chloride.
Official Solution
Correct Option:
(1)
(i) Allyl chloride undergoes hydrolysis more readily than n-propyl chloride because the allyl group (C C) stabilizes the transition state in the mechanism through resonance, making the nucleophilic substitution faster. (ii) Isocyanides are formed when alkyl halides react with silver cyanide because silver cyanide is a strong nucleophile that replaces the halide ion, leading to the formation of isocyanides (RNC). (iii) Methyl chloride reacts faster with OH because the methyl group is small and does not hinder the nucleophile's approach, unlike t-butyl chloride, where the bulky tert-butyl group sterically hinders the attack.
36
PYQ 2025
medium
chemistryID: cbse-cla
A compound (A) with molecular formula which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.
Official Solution
Correct Option:
(1)
Upon treatment with alcoholic KOH, A undergoes elimination to form an alkene. The reaction follows an E2 mechanism, where the proton is abstracted and the leaving group is eliminated. Structures: (A) Alkyl halide (B) Alkene formed after elimination (C) Alcoholic KOH used (D) Possible side product from elimination
37
PYQ 2025
medium
chemistryID: cbse-cla
Why is chlorobenzene resistant to nucleophilic substitution reaction?
Official Solution
Correct Option:
(1)
Chlorobenzene is resistant to nucleophilic substitution due to the resonance stabilization of the C–Cl bond in the aromatic ring. The lone pair of electrons on the chlorine atom delocalizes into the benzene ring, making the carbon–chlorine bond stronger and less susceptible to attack by nucleophiles.
38
PYQ 2025
medium
chemistryID: cbse-cla
Arrange the following compounds in increasing order of their reactivity towards SN2 displacement:
Official Solution
Correct Option:
(1)
In SN2 reactions, the reactivity increases as steric hindrance decreases. Hence, the increasing order of reactivity is: This is because 1-Bromopentane, being a primary alkyl halide, has the least steric hindrance.
39
PYQ 2025
medium
chemistryID: cbse-cla
In the following pair of halogen compounds, which compound undergoes SN1 reaction faster and why?
Official Solution
Correct Option:
(1)
The compound undergoes the SN1 reaction faster. This is because the formation of a more stable carbocation is favored in the case of the tertiary carbocation formed by the second compound, whereas the first compound forms a primary carbocation, which is less stable.
40
PYQ 2025
medium
chemistryID: cbse-cla
Draw the structure of the major monohalo product for each of the following reactions:
Official Solution
Correct Option:
(1)
In this reaction, phenol (C6H5OH) reacts with hydrochloric acid (HCl) in the presence of heat to form chlorobenzene.
The hydroxyl group on the phenol is replaced by a chlorine atom through electrophilic aromatic substitution.
The structure of the product is:
41
PYQ 2025
easy
chemistryID: cbse-cla
Draw the structure of the major monohalo product for each of the following reactions:
Official Solution
Correct Option:
(1)
The reaction of toluene (methylbenzene) with HBr leads to the alkyl group (methyl) being substituted with a bromine atom at the benzylic position, forming 1-bromo-2-methylbenzene as the major product. The structure of the product is:
42
PYQ 2025
medium
chemistryID: cbse-cla
How do you convert: 2-Bromobutane to but-2-ene
Official Solution
Correct Option:
(1)
To solve the problem, we need to convert 2-Bromobutane to but-2-ene through an elimination reaction.
1. Understanding the Elimination Reaction: An elimination reaction involves the removal of a small molecule (like HBr) from a larger molecule to form a double bond.
2. Identifying the Reactant: The given compound is 2-bromobutane, with the structure: C₄H₉Br. In this structure, the bromine is attached to the second carbon atom.
3. Selecting the Elimination Reaction: The elimination of HBr from 2-bromobutane will lead to the formation of a double bond. This can be accomplished using a strong base, such as potassium hydroxide (KOH) or sodium ethoxide (NaOEt), in an alcohol solvent.
4. Reaction Mechanism: The base abstracts a proton (H⁺) from the β-carbon (the carbon adjacent to the carbon bearing the bromine). This leads to the formation of a double bond between the second and third carbon atoms, and the bromine (Br⁻) is eliminated as a leaving group.
5. Final Product: The product of this elimination is but-2-ene, which has the structure C₄H₈. The double bond can either be in the cis or trans configuration, but the trans isomer (trans-but-2-ene) is usually favored under these conditions.
Final Answer: The conversion of 2-bromobutane to but-2-ene is achieved through an elimination reaction, typically using a strong base like KOH or NaOEt.
43
PYQ 2025
medium
chemistryID: cbse-cla
How do you convert: Propene to 1-Iodopropane
Official Solution
Correct Option:
(1)
Conversion of Propene to 1-Iodopropane:
The conversion of propene (CH₃CH=CH₂) to 1-iodopropane (CH₃CH₂CH₂I) can be achieved by an electrophilic addition reaction with iodine (I₂) in the presence of a strong acid, such as hydroiodic acid (HI).
Reaction:
CH₃CH=CH₂ + HI → CH₃CH₂CH₂I
Mechanism: 1. Electrophilic Addition of HI: In this reaction, HI (hydroiodic acid) adds to the double bond of propene. The double bond in propene acts as a nucleophile, attacking the electrophilic hydrogen (H⁺) from HI. This leads to the formation of a carbocation intermediate. 2. Carbocation Formation: The intermediate carbocation is formed at the most stable position, which, in this case, is at the secondary carbon (CH₂) in propene, resulting in a secondary carbocation. 3. Nucleophilic Attack by I⁻: The iodide ion (I⁻) from HI attacks the carbocation, forming 1-iodopropane (CH₃CH₂CH₂I) as the final product.
Conditions: - The reaction occurs under acidic conditions (using HI), and heat may be applied to drive the reaction to completion.
Conclusion: Propene reacts with HI through an electrophilic addition mechanism to produce 1-iodopropane.
44
PYQ 2025
medium
chemistryID: cbse-cla
Why is chlorobenzene resistant to nucleophilic substitution reactions?
Official Solution
Correct Option:
(1)
Chlorobenzene is resistant to nucleophilic substitution due to the electron-withdrawing effect of the chlorine atom, which decreases the electron density on the benzene ring. This makes the carbon attached to the chlorine less electrophilic, which reduces its ability to undergo nucleophilic attack. Additionally, the resonance stabilization of the chlorine atom with the aromatic ring further makes the carbon-chlorine bond stronger, thus making nucleophilic substitution less favorable.
45
PYQ 2025
medium
chemistryID: cbse-cla
Arrange the following compounds in increasing order of their reactivity towards displacement: 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.
Official Solution
Correct Option:
(1)
Step 1: Understanding Mechanism. The mechanism involves a single step where the nucleophile attacks the electrophilic carbon, displacing the leaving group, all happening simultaneously. This mechanism is highly sensitive to steric hindrance; thus, less substituted carbons are more reactive due to easier accessibility for the nucleophile. 2-Bromo-2-methylbutane: This is a tertiary bromide and thus highly hindered, making reactions extremely slow. 2-Bromopentane: As a secondary bromide, it is less hindered than the tertiary but more so than the primary, placing it in the middle for reactivity. 1-Bromopentane: This is a primary bromide with the least steric hindrance, facilitating the fastest reaction. Therefore, the order of increasing reactivity towards displacement is:
46
PYQ 2025
easy
chemistryID: cbse-cla
In the following pair of halogen compounds, which compound undergoes reaction faster and why?
Official Solution
Correct Option:
(1)
Step 1: Understanding the Mechanism. The reaction proceeds via a two-step mechanism:
1. Formation of a carbocation (rate-determining step). 2. Nucleophilic attack on the carbocation. The rate of the reaction depends on the stability of the carbocation. Tertiary carbocations are more stable due to greater alkyl group substitution which provides better electron donation through hyperconjugation and inductive effects. This stability accelerates the formation of the carbocation, thus enhancing the reaction rate. Given two halogen compounds, the one with the tertiary halogenated carbon (i.e., more substituted carbon) will form a more stable carbocation and thus undergoes the reaction faster.
47
PYQ 2026
medium
chemistryID: cbse-cla
Which of the following is 'not' true about enantiomers?
1
They have the same chemical reactivity.
2
They have the same specific rotation.
3
They have the same melting or boiling point.
4
They have the same refractive index.
Official Solution
Correct Option:
(2)
Step 1: Understanding the Concept: Enantiomers are non-superimposable mirror images of each other. They share identical physical properties in an achiral environment. Step 2: Detailed Explanation: Enantiomers have identical melting points, boiling points, and refractive indices because their intermolecular forces are identical. They also show the same chemical reactivity towards achiral reagents. However, enantiomers differ in their interaction with plane-polarized light. They rotate the plane of polarized light by the same magnitude but in opposite directions (one is dextrorotatory, the other is levorotatory). Therefore, they do not have the "same" specific rotation; they have "equal and opposite" rotations. Step 3: Final Answer: Statement (B) is false.
48
PYQ 2026
medium
chemistryID: cbse-cla
Out of ‘X’ and ‘Y’ , which one will form product with inversion of configuration and why?
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Inversion of configuration (Walden Inversion) occurs in reactions. Step 2: Detailed Explanation: Compound 'Y' follows the mechanism as the rate depends on both reactants. In , the nucleophile ( ) attacks the carbon atom from the side opposite to the leaving group ( ). As the bond with the nucleophile forms and the bond with the leaving group breaks simultaneously, the configuration of the carbon atom flips like an umbrella in a strong wind. This results in a product with 100% inversion of configuration. Step 3: Final Answer: Y forms a product with inversion of configuration due to the backside attack of the nucleophile in the process.
49
PYQ 2026
easy
chemistryID: cbse-cla
Out of ‘X’ and ‘Y’, which one will undergo racemisation and why?
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Racemisation is a characteristic of the mechanism due to the formation of a planar carbocation intermediate. Step 2: Detailed Explanation: In an reaction (like that of compound X), the first step is the formation of a carbocation. A carbocation is hybridized and planar. The nucleophile can attack this planar intermediate from either side (front or back) with equal probability. This results in a 50:50 mixture of two enantiomers, leading to a racemic mixture. Although X (tert-butyl bromide) is achiral, if an optically active halide reacted similarly, it would undergo complete racemisation. Step 3: Final Answer: Compounds reacting via the mechanism (like X) undergo racemisation because of the planar carbocation intermediate.
50
PYQ 2026
medium
chemistryID: cbse-cla
Compound ‘X’ with molecular formula reacts with aqueous KOH to give an alcohol. The rate of this reaction depends only on the concentration of compound ‘X’. When an optically active isomer ‘Y’ of the compound ‘X’ was treated with aqueous KOH solution, the rate of reaction was found to be dependent on concentration of compound ‘Y’ and aqueous KOH both.
22(a). Write down the structural formula of both ‘X’ and ‘Y’.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Reactions where the rate depends only on the concentration of the alkyl halide follow the mechanism (Unimolecular). Reactions where the rate depends on both the alkyl halide and the nucleophile follow the mechanism (Bimolecular). Step 2: Detailed Explanation: 1. Compound 'X': The rate depends only on [X], so it is a alkyl halide. The only isomer of is 2-Bromo-2-methylpropane (tert-butyl bromide). Structural formula: . 2. Compound 'Y': The rate depends on both [Y] and [KOH], so it follows . It is stated to be optically active. Among isomers of , only 2-Bromobutane has a chiral center (C2 is attached to ). Structural formula: . Step 3: Final Answer: The structural formula of X is and Y is .
51
PYQ 2026
medium
chemistryID: cbse-cla
Why are haloarenes less reactive towards nucleophilic substitution reaction? Give two reasons.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Nucleophilic substitution involves the breaking of the C-X bond. In haloarenes, this bond is particularly strong and difficult to break compared to haloalkanes. Step 2: Detailed Explanation: 1. Resonance Effect: The lone pair of electrons on the halogen atom is in conjugation with the -electrons of the benzene ring. This results in resonance, which gives a partial double bond character to the C-X bond. Breaking a partial double bond is much harder than breaking a single bond. 2. Difference in Hybridization: In haloarenes, the halogen is attached to an hybridized carbon, whereas in haloalkanes it is attached to an carbon. An carbon is more electronegative (has more s-character), which holds the electron pair of the C-X bond more tightly, making the bond shorter and stronger. 3. Instability of Phenyl Cation: If the bond were to break, the resulting phenyl cation would be highly unstable because it cannot be stabilized by resonance. Step 3: Final Answer: Haloarenes are less reactive due to the partial double bond character from resonance and the stronger bond from hybridization.
52
PYQ 2026
medium
chemistryID: cbse-cla
p-Nitrotoluene with (heat)
Official Solution
Correct Option:
(1)
Bromination in presence of heat leads to side-chain bromination.
53
PYQ 2026
medium
chemistryID: cbse-cla
Write the major product in the following reactions:(1) Chlorobenzene + Na/dry ether
Official Solution
Correct Option:
(1)
This is Fittig reaction (coupling of aryl halides). Product: Biphenyl
54
PYQ 2026
medium
chemistryID: cbse-cla
Although chlorine shows strong −I effect, it is ortho–para directing in electrophilic aromatic substitution. Why?
Official Solution
Correct Option:
(1)
Chlorine has:
−I effect (electron withdrawing)
+R effect (lone pair donation)
Resonance donation increases electron density at ortho and para positions. Thus:
Deactivating overall
Ortho–para directing
55
PYQ 2026
medium
chemistryID: cbse-cla
Predict the alkene formed by dehydrohalogenation of 1-bromo-1-cyclohexylethane.
Which isomer of C H Br is most reactive towards reaction?
Official Solution
Correct Option:
(1)
reactions are favoured by least steric hindrance. Among isomers:
tert-butyl bromide → least reactive
isobutyl bromide → moderate
sec-butyl bromide → moderate
n-butyl bromide → least hindered
57
PYQ 2026
medium
chemistryID: cbse-cla
Racemic mixture
Official Solution
Correct Option:
(1)
A racemic mixture contains equal amounts of two enantiomers.
Optically inactive
Rotations cancel each other
58
PYQ 2026
medium
chemistryID: cbse-cla
Define the following terms: Ambident nucleophiles
Official Solution
Correct Option:
(1)
Ambident nucleophiles are nucleophiles that can attack through two different atoms. Examples:
CN (C or N attack)
NO
59
PYQ 2026
medium
chemistryID: cbse-cla
Why is chloroform stored in dark coloured bottles?
Official Solution
Correct Option:
(1)
Chloroform undergoes slow oxidation in presence of light and air to form poisonous phosgene gas ( ). Dark bottles prevent exposure to light and inhibit this reaction.
60
PYQ 2026
medium
chemistryID: cbse-cla
What type of halide is present in the compound:
Official Solution
Correct Option:
(1)
The chlorine atom is attached to a carbon involved in a double bond. Such halides are called vinylic halides.
61
PYQ 2026
medium
chemistryID: cbse-cla
Which of the following is more reactive towards reaction: 2-bromo-2-methylbutane or 1-bromopentane?
Official Solution
Correct Option:
(1)
reactions proceed via carbocation formation. Stability of carbocation determines reactivity.
2-bromo-2-methylbutane → forms tertiary carbocation
1-bromopentane → forms primary carbocation
Since tertiary carbocations are more stable:
62
PYQ 2026
medium
chemistryID: cbse-cla
Why is Ce in aqueous solution a good oxidising agent?
Official Solution
Correct Option:
(1)
Ce easily gets reduced to Ce .
Ce has stable configuration
High tendency to gain electron
Hence, Ce acts as a strong oxidising agent.
63
PYQ 2026
easy
chemistryID: cbse-cla
Consider the following reaction and identify A and B:
1
A = CH I, B = NaCl
2
A = CH OH, B = NaCl
3
A = CH CHO, B = NaCl
4
A = C H , B = CH I
Official Solution
Correct Option:
(1)
Concept:
This reaction represents the Finkelstein reaction, which involves halogen exchange in alkyl halides. It is an reaction where a chlorine atom is replaced by iodine using sodium iodide in dry acetone. General reaction:
The reaction proceeds forward because is insoluble in acetone and precipitates out, driving the reaction to completion. Step 1: Identify the reaction type. Presence of NaI in dry acetone strongly indicates Finkelstein reaction. Step 2: Apply halogen exchange. Step 3: Identify products.
A = CH I (methyl iodide)
B = NaCl (precipitate)
64
PYQ 2026
medium
chemistryID: cbse-cla
Complete the following equation:
Official Solution
Correct Option:
(1)
Concept:
This is an example of an azo coupling reaction. Aromatic diazonium salts react with activated aromatic compounds like phenol or aniline to form azo compounds ( ). Step 1: Reaction type. Benzenediazonium chloride reacts with phenol in alkaline medium. Step 2: Role of alkaline medium. In basic solution:
Phenol forms phenoxide ion.
Phenoxide is strongly activated toward electrophilic substitution.
Step 3: Position of coupling. Coupling occurs mainly at the para position relative to the –OH group due to less steric hindrance. Step 4: Product formed. The product is p-hydroxyazobenzene (an azo dye). Final Answer:
About Haloalkanes And Haloarenes - CBSE-CLASS-XII
Haloalkanes And Haloarenes is a vital chapter for CBSE-CLASS-XII aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
Frequently Asked Questions
Why focus on Haloalkanes And Haloarenes PYQs?
Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within Haloalkanes And Haloarenes carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
