CUET UG Chemistry Booster Test -3 Classification of Halo Compounds
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QUESTION 1 OF 20
Arrange the following organohalogen compounds in decreasing order of their density (g/mL):
(A) n-C₃H₇I
(B) CCl₄
(C) CHCl₃
(D) n-C₃H₇Cl
QUESTION 2 OF 20
| Which statements are correct regarding the nature of the C | X bond? |
|---|---|
| (A) The C | F bond has the highest bond enthalpy among C—X bonds. |
| (B) Bond length increases progressively from C | F to C—I. |
| (C) The C | Cl bond has a slightly higher dipole moment (1.860 D) than the C—F bond (1.847 D). |
| (D) The dipole moment of the C | I bond is the highest among halogens. |
QUESTION 3 OF 20
Identify the stereochemical mechanism type: When an optically active monohalo compound like (–)-2-bromooctane reacts with sodium hydroxide to form (+)-octan-2-ol, the reaction predominantly proceeds via:
QUESTION 4 OF 20
What is the standard unit of dipole moment utilized in the text when detailing the properties of C—X bonds?
QUESTION 5 OF 20
Match List-I (Stereochemical terms) with List-II (Descriptions) relevant to halide reactions:
| List-I | List-II |
|---|---|
| 1. Retention of configuration | a. Preservation of spatial arrangement of bonds to an asymmetric centre |
| 2. Inversion of configuration | b. Characteristic of an SN2 mechanism in optically active halides |
| 3. Racemisation | c. Characteristic of an SN1 mechanism in optically active halides |
| 4. Chiral molecule | d. Non-superimposable on its mirror image |
QUESTION 6 OF 20
During the laboratory detection of a double bond, bromine is added in CCl₄ to an alkene. The resulting colourless compound formed is analytically classified as a:
QUESTION 7 OF 20
Properties of Chloroform (a trihalogen compound):
(A) It is slowly oxidized by air in the presence of light to the extremely poisonous phosgene gas.
(B) It is stored in closed dark coloured bottles completely filled to keep air out.
(C) It was heavily used as an antiseptic due to the liberation of free halogens.
(D) Chronic exposure may cause damage to the liver where it is metabolised to phosgene.
QUESTION 8 OF 20
Which of the following polyhalogen compounds is extremely stable, unreactive, non-toxic, easily liquefiable, and typically manufactured from tetrachloromethane by the Swarts reaction?
QUESTION 9 OF 20
QUESTION 10 OF 20
QUESTION 11 OF 20
What is the correct IUPAC name of the highly branched tertiary halide (CH₃)₃CBr?
QUESTION 12 OF 20
When a tertiary alkyl halide reacts with a bulkier nucleophile, which reaction route largely dominates due to severe steric reasons at the tetravalent carbon atom?
QUESTION 13 OF 20
Match List-I with List-II concerning reactivity orders of halides and related compounds:
| List-I | List-II |
|---|---|
| 1. Reactivity of alcohols with a given haloacid | a. Tertiary > Secondary > Primary |
| 2. SN2 reactivity of simple alkyl halides | b. Primary > Secondary > Tertiary |
| 3. SN1 reactivity of alkyl halides | c. 3° > 2° > 1° |
| 4. Bond enthalpies of the C—X bond | d. C—F > C—Cl > C—Br > C—I |
QUESTION 14 OF 20
Allylic halides show exceptionally high reactivity towards SN1 reactions primarily because:
QUESTION 15 OF 20
Arrange the following benzylic and related bromides in decreasing order of their SN1 reactivity:
(A) C₆H₅C(CH₃)(C₆H₅)Br
(B) C₆H₅CH(C₆H₅)Br
(C) C₆H₅CH(CH₃)Br
(D) C₆H₅CH₂Br
QUESTION 17 OF 20
What is the systematic IUPAC name for the vinylic halide CH₂=CHCl?
QUESTION 18 OF 20
Identify the reaction rule described by the following:
"In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms."
QUESTION 19 OF 20
Match List-I with List-II concerning the specific reactions of haloarenes:
| List-I | List-II |
|---|---|
| 1. Wurtz-Fittig reaction | a. Mixture of an alkyl halide and aryl halide treated with Na in dry ether |
| 2. Fittig reaction | b. Two aryl halides joined together using Na in dry ether |
| 3. Electrophilic substitution | c. Halogen acts as o, p-directing substituent due to stabilizing resonance |
| 4. Nucleophilic substitution | d. Haloarenes are extremely less reactive towards this category of reactions |
QUESTION 20 OF 20
Analytically, why are aryl halides exceptionally less reactive towards nucleophilic substitution compared to alkyl halides?
QUESTION 20 OF 20
Statements regarding nucleophiles attacking haloalkanes and benzylic halides:
(A) Ambident nucleophiles like cyanides can link through two different atoms.
(B) Reaction with AgCN mainly yields isocyanides due to its covalent nature.
(C) Reaction with KCN predominantly yields alkyl cyanides.
(D) The nitrite ion is incapable of acting as an ambident nucleophile.
Test Complete!
Answer Review
1 Arrange the following organohalogen compounds in decreasing order of their density (g/mL):
(A) n-C₃H₇I
(B) CCl₄
(C) CHCl₃
(D) n-C₃H₇Cl
Density generally increases with heavier halogen atoms. Iodo compounds have high density. n-C₃H₇Cl is the least dense among the given compounds.
Among the given organohalogen compounds, density is influenced by molecular mass and halogen content. Iodine-containing compounds generally have higher density because iodine is very heavy. Carbon tetrachloride and chloroform are also dense due to multiple chlorine atoms. The decreasing order is: n-C₃H₇I > CCl₄ > CHCl₃ > n-C₃H₇Cl So the correct sequence is: (A), (B), (C), (D)
- B → Places CCl₄ above n-C₃H₇I incorrectly.
- C → Places CHCl₃ above CCl₄ incorrectly.
- D → Places CHCl₃ above CCl₄ incorrectly.
Used
- Property Trend Analysis
Heavy halogen = higher density
2
| Which statements are correct regarding the nature of the C | X bond? |
|---|---|
| (A) The C | F bond has the highest bond enthalpy among C—X bonds. |
| (B) Bond length increases progressively from C | F to C—I. |
| (C) The C | Cl bond has a slightly higher dipole moment (1.860 D) than the C—F bond (1.847 D). |
| (D) The dipole moment of the C | I bond is the highest among halogens. |
C—F bond is strongest. Bond length increases down the halogen group. C—Cl has slightly higher dipole moment than C—F.
The C—F bond has the highest bond enthalpy because fluorine is small and forms a strong bond with carbon. Bond length increases from C—F to C—I because halogen atomic size increases down the group: F < Cl < Br < I So: C—F < C—Cl < C—Br < C—I Although fluorine is most electronegative, the C—Cl bond has slightly higher dipole moment than C—F because dipole moment depends on both charge separation and bond length. Statement D is incorrect because C—I does not have the highest dipole moment.
- B → Includes incorrect statement D.
- C → Includes incorrect statement D and omits A.
- D → Includes incorrect statement D and omits B.
Used
- Statement Verification
Bond length: F short, I long
3 Identify the stereochemical mechanism type: When an optically active monohalo compound like (–)-2-bromooctane reacts with sodium hydroxide to form (+)-octan-2-ol, the reaction predominantly proceeds via:
SN2 occurs by backside attack. Backside attack causes inversion. Optical activity changes due to stereochemical inversion.
In an SN2 reaction, the nucleophile attacks the carbon from the side opposite to the leaving group. This causes inversion of configuration, also called Walden inversion. Here, hydroxide ion replaces bromide in optically active 2-bromooctane, forming octan-2-ol with inverted configuration.
- A → SN1 usually gives racemisation, not clean inversion.
- C → Elimination gives alkene, not alcohol.
- D → Electrophilic substitution is not involved in haloalkane substitution.
Used
- Mechanism Identification
SN2 = Backside attack = Inversion
4 What is the standard unit of dipole moment utilized in the text when detailing the properties of C—X bonds?
Dipole moment measures bond polarity. C—X bonds are polar. Debye is the commonly used unit.
Dipole moment represents the extent of charge separation in a polar bond. In haloalkanes, the C—X bond is polar due to the higher electronegativity of halogens. In NCERT tables, dipole moment values are commonly expressed in Debye (D).
- B → SI unit, but not the standard unit used in the given text.
- C → Unit of bond enthalpy.
- D → Unit of bond length.
Used
- Unit Identification
Dipole = Debye
5 Match List-I (Stereochemical terms) with List-II (Descriptions) relevant to halide reactions:
| List-I | List-II |
|---|---|
| 1. Retention of configuration | a. Preservation of spatial arrangement of bonds to an asymmetric centre |
| 2. Inversion of configuration | b. Characteristic of an SN2 mechanism in optically active halides |
| 3. Racemisation | c. Characteristic of an SN1 mechanism in optically active halides |
| 4. Chiral molecule | d. Non-superimposable on its mirror image |
Retention means same configuration. SN2 gives inversion. SN1 gives racemisation. Chiral molecules are non-superimposable mirror images.
Retention of configuration means the spatial arrangement around the chiral carbon remains unchanged. Inversion of configuration is typical of SN2 reactions because of backside attack. Racemisation is commonly seen in SN1 reactions because the planar carbocation can be attacked from both sides. A chiral molecule is one that is non-superimposable on its mirror image. Thus: 1-a, 2-b, 3-c, 4-d
- B → Completely mismatches the stereochemical terms.
- C → Interchanges SN1 and SN2 features.
- D → Incorrectly assigns racemisation and chirality.
Used
- Concept Mapping
SN2 Inverts, SN1 Racemises
6 During the laboratory detection of a double bond, bromine is added in CCl₄ to an alkene. The resulting colourless compound formed is analytically classified as a:
Bromine adds across C=C. Two bromine atoms attach to adjacent carbons. Product is a vicinal dibromide.
Alkenes decolourise bromine in CCl₄ because bromine adds across the carbon-carbon double bond. Example: CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br The two bromine atoms are attached to adjacent carbon atoms. Such compounds are called vicinal dibromides.
- A → Gem-dibromide has both bromines on the same carbon.
- C → Product contains two halogens, not three.
- D → Product has two bromine atoms, not one.
Used
- Reaction Product Identification
Bromine across double bond = Vic-dibromide
7 Properties of Chloroform (a trihalogen compound):
(A) It is slowly oxidized by air in the presence of light to the extremely poisonous phosgene gas.
(B) It is stored in closed dark coloured bottles completely filled to keep air out.
(C) It was heavily used as an antiseptic due to the liberation of free halogens.
(D) Chronic exposure may cause damage to the liver where it is metabolised to phosgene.
Chloroform forms toxic phosgene in air and light. It is stored in dark filled bottles. Chronic exposure damages the liver.
Chloroform is slowly oxidized by air in the presence of light to form phosgene, which is highly poisonous. To prevent this oxidation, chloroform is stored in dark coloured bottles filled completely to keep air out. Chronic exposure to chloroform can cause liver damage because it may be metabolised to toxic phosgene. Statement C is incorrect because antiseptic action due to free iodine is associated with iodoform, not chloroform.
- A → Includes incorrect statement C.
- C → Includes incorrect statement C and omits A.
- D → Includes incorrect statement C and omits B.
Used
- Statement Elimination
Chloroform + Light + Air = Phosgene danger
8 Which of the following polyhalogen compounds is extremely stable, unreactive, non-toxic, easily liquefiable, and typically manufactured from tetrachloromethane by the Swarts reaction?
Freon-12 is stable and unreactive. It is non-corrosive and easily liquefiable. It is prepared from CCl₄ by Swarts reaction.
Freon-12, CCl₂F₂, is a chlorofluorocarbon. It is extremely stable, non-toxic, non-corrosive, and easily liquefiable. It is manufactured from carbon tetrachloride using hydrogen fluoride in the presence of antimony pentachloride, known as Swarts reaction.
- A → Phosgene is highly poisonous.
- C → DDT is an insecticide, not Freon.
- D → Iodoform is used as an antiseptic, not a refrigerant.
Used
- Property Identification
Freon = Stable refrigerant
9
Primary halides have low steric hindrance. SN2 favours less hindered substrates. Primary carbocations are unstable.
Primary alkyl halides have the halogen attached to a carbon bonded to only one alkyl group. This makes the carbon less crowded and more accessible to nucleophilic backside attack. Therefore, primary alkyl halides generally undergo SN2 reactions readily. They do not favour SN1 because primary carbocations are highly unstable.
- A → Primary halides do not exclusively undergo SN1.
- C → Primary carbocations are unstable.
- D → Haloalkanes have halogen attached to sp³ carbon, not sp² carbon.
Used
- Mechanism-Based Reasoning
Primary = Less crowded = SN2 faster
10
Secondary halides are intermediate in steric hindrance. They may undergo SN1 or SN2. Strong bases can favour elimination.
Secondary alkyl halides can react through different pathways depending on conditions. With suitable nucleophiles, they can undergo SN2 reactions. In polar protic solvents, they may form secondary carbocations and undergo SN1 reactions. With strong bases, β-elimination can occur to form alkenes. Therefore, secondary alkyl halides can undergo SN1, SN2, or elimination.
- A → Too restrictive; SN2 and elimination are also possible.
- B → Too restrictive; SN1 and elimination are also possible.
- D → Electrophilic aromatic substitution is not typical for alkyl halides.
Used
- Condition-Based Mechanism Analysis
2° halide = Flexible pathway
11 What is the correct IUPAC name of the highly branched tertiary halide (CH₃)₃CBr?
The parent chain is propane. Bromine is attached to carbon-2. A methyl group is also attached to carbon-2.
The structure (CH₃)₃CBr contains a central carbon attached to three methyl groups and one bromine atom. The longest chain contains three carbon atoms, so the parent hydrocarbon is propane. Both bromine and methyl substituents are present on carbon-2. Therefore, the correct IUPAC name is: 2-Bromo-2-methylpropane
- A → tert-Butyl bromide is the common name, not the IUPAC name.
- C → Represents neopentyl bromide, a primary halide.
- D → Represents a secondary halide, not the given structure.
Used
- Nomenclature Analysis
tert-Butyl bromide = 2-Bromo-2-methylpropane
12 When a tertiary alkyl halide reacts with a bulkier nucleophile, which reaction route largely dominates due to severe steric reasons at the tetravalent carbon atom?
Tertiary halides are highly crowded. Bulky nucleophiles cannot easily attack the carbon. They preferentially remove β-hydrogen, causing elimination.
Tertiary alkyl halides are sterically crowded around the carbon attached to the halogen. Bulky nucleophiles or bases find it difficult to perform backside attack required for SN2 substitution. Instead, they abstract a β-hydrogen, leading to dehydrohalogenation and formation of an alkene. Thus, elimination becomes the dominant pathway.
- A → SN2 is highly unfavourable in tertiary halides due to steric hindrance.
- C → Electrophilic substitution is not a typical reaction of tertiary alkyl halides.
- D → Wurtz-Fittig involves aryl halide and alkyl halide with sodium in dry ether.
Used
- Steric Hindrance Analysis
Bulky base + 3° halide = Elimination
13 Match List-I with List-II concerning reactivity orders of halides and related compounds:
| List-I | List-II |
|---|---|
| 1. Reactivity of alcohols with a given haloacid | a. Tertiary > Secondary > Primary |
| 2. SN2 reactivity of simple alkyl halides | b. Primary > Secondary > Tertiary |
| 3. SN1 reactivity of alkyl halides | c. 3° > 2° > 1° |
| 4. Bond enthalpies of the C—X bond | d. C—F > C—Cl > C—Br > C—I |
Alcohols react with haloacids in the order 3° > 2° > 1°. SN2 favours less hindered substrates. SN1 favours stable carbocations. C—F bond is strongest.
Reactivity of alcohols with haloacids follows: Tertiary > Secondary > Primary because tertiary alcohols form more stable carbocation-like intermediates. SN2 reactivity follows: Primary > Secondary > Tertiary because steric hindrance increases from primary to tertiary. SN1 reactivity follows: 3° > 2° > 1° because carbocation stability increases in the same order. Bond enthalpy order is: C—F > C—Cl > C—Br > C—I because bond strength decreases as halogen size increases. Therefore: 1-a, 2-b, 3-c, 4-d
- B → Interchanges alcohol and SN2 reactivity trends.
- C → Interchanges SN1 and SN2 trends.
- D → Incorrectly assigns bond enthalpy trend to alcohol reactivity.
Used
- Concept Mapping
SN2 hates crowding; SN1 loves carbocation stability
14 Allylic halides show exceptionally high reactivity towards SN1 reactions primarily because:
SN1 proceeds through carbocation formation. Allylic carbocations are resonance-stabilised. Resonance increases SN1 reactivity.
In SN1 reactions, the rate depends on the stability of the carbocation intermediate. Allylic halides form allylic carbocations after loss of halide ion. These carbocations are stabilised by resonance with the adjacent carbon-carbon double bond. Example: CH₂=CH–CH₂⁺ ↔ CH₂⁺–CH=CH₂ This resonance stabilization makes allylic halides highly reactive towards SN1 reactions.
- A → Molecular mass is not the reason.
- C → Describes vinylic halides, not allylic halides.
- D → Inversion is associated mainly with SN2 reactions.
Used
- Intermediate Stability Analysis
Allylic carbocation = Resonance stabilised
15 Arrange the following benzylic and related bromides in decreasing order of their SN1 reactivity:
(A) C₆H₅C(CH₃)(C₆H₅)Br
(B) C₆H₅CH(C₆H₅)Br
(C) C₆H₅CH(CH₃)Br
(D) C₆H₅CH₂Br
SN1 depends on carbocation stability. More phenyl/alkyl stabilization increases reactivity. The most substituted benzylic carbocation forms fastest. Cyanide is an ambident nucleophile. AgCN gives isocyanides. KCN gives cyanides. Nitrite is also ambident, so D is wrong.
Benzylic halides are reactive in SN1 because the carbocation formed is resonance stabilised by the aromatic ring. Carbocation stability order here is: C₆H₅C(CH₃)(C₆H₅)⁺ > C₆H₅CH(C₆H₅)⁺ > C₆H₅CH(CH₃)⁺ > C₆H₅CH₂⁺ The first carbocation is most stable because it is stabilised by two phenyl groups and one methyl group. The last is least stabilized among the given set. Therefore, decreasing SN1 reactivity is: (A), (B), (C), (D) Ambident nucleophiles can attack through two different atoms. Cyanide ion can attack through carbon or nitrogen. KCN is ionic and provides CN⁻ freely, so attack usually occurs through carbon, forming alkyl cyanides: R–X + KCN → R–CN AgCN is more covalent, so attack occurs mainly through nitrogen, producing isocyanides: R–X + AgCN → R–NC Statement D is incorrect because nitrite ion is also an ambident nucleophile and can attack through nitrogen or oxygen.
- B → Gives reverse order.
- C → Places C above B incorrectly.
- D → Places B above A incorrectly.
- B → Includes incorrect statement D.
- C → Includes incorrect statement D and omits B.
- D → Includes incorrect statement D and omits C.
Used
- Statement Elimination
KCN gives Cyanide, AgCN gives isocyanide
17 What is the systematic IUPAC name for the vinylic halide CH₂=CHCl?
Parent chain is ethene. Chlorine is attached to one double-bond carbon. IUPAC name is chloroethene.
CH₂=CHCl is a two-carbon alkene with chlorine attached to one carbon of the double bond. The parent compound is ethene, and the substituent is chloro. Therefore, the systematic IUPAC name is: Chloroethene
- A → Vinyl chloride is the common name.
- C → Allyl chloride is CH₂=CHCH₂Cl.
- D → 1-Chloroethane is a saturated haloalkane, CH₃CH₂Cl.
Used
- IUPAC Nomenclature
Vinyl chloride = Chloroethene
18 Identify the reaction rule described by the following:
"In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms."
Rule applies to elimination reactions. More substituted alkene is preferred. This is Zaitsev's rule.
During dehydrohalogenation of alkyl halides, a hydrogen atom and a halogen atom are removed from adjacent carbon atoms to form an alkene. According to Zaitsev's rule, the major product is the more substituted alkene, meaning the alkene with more alkyl groups attached to the double-bonded carbon atoms.
- A → Markovnikov's rule applies to addition reactions of unsymmetrical alkenes.
- C → Swarts reaction prepares alkyl fluorides.
- D → Finkelstein reaction prepares alkyl iodides.
Used
- Rule Identification
Zaitsev = More substituted alkene
19 Match List-I with List-II concerning the specific reactions of haloarenes:
| List-I | List-II |
|---|---|
| 1. Wurtz-Fittig reaction | a. Mixture of an alkyl halide and aryl halide treated with Na in dry ether |
| 2. Fittig reaction | b. Two aryl halides joined together using Na in dry ether |
| 3. Electrophilic substitution | c. Halogen acts as o, p-directing substituent due to stabilizing resonance |
| 4. Nucleophilic substitution | d. Haloarenes are extremely less reactive towards this category of reactions |
Wurtz-Fittig uses alkyl + aryl halide. Fittig couples two aryl halides. Haloarenes undergo o,p electrophilic substitution. Haloarenes resist nucleophilic substitution.
Wurtz-Fittig reaction involves treatment of a mixture of alkyl halide and aryl halide with sodium in dry ether to form alkylbenzene. Fittig reaction involves coupling of two aryl halide molecules using sodium in dry ether. Haloarenes undergo electrophilic substitution at ortho and para positions because halogens donate electron density through resonance. Haloarenes are much less reactive toward nucleophilic substitution because the C–X bond has partial double bond character due to resonance.
- B → Interchanges Wurtz-Fittig and Fittig reactions.
- C → Completely mismatches reaction descriptions.
- D → Incorrectly assigns electrophilic substitution and Fittig descriptions.
Used
- Concept Mapping
Wurtz-Fittig = Alkyl + Aryl; Fittig = Aryl + Aryl
20 Analytically, why are aryl halides exceptionally less reactive towards nucleophilic substitution compared to alkyl halides?
Halogen lone pair conjugates with benzene ring. C—Cl bond gains partial double bond character. Bond cleavage becomes difficult.
In aryl halides, the lone pair on halogen participates in resonance with the π-electrons of the benzene ring. This resonance gives the C—Cl bond partial double bond character. As a result, the bond becomes shorter and stronger than the C—Cl bond in alkyl halides. Because bond cleavage becomes difficult, aryl halides are less reactive towards nucleophilic substitution.
- B → sp² carbon has more s-character than sp³ carbon, not less.
- C → Phenyl cation is highly unstable, not stabilized.
- D → Nucleophiles are electron-rich, not electron-deficient.
Used
- Resonance-Based Reasoning
Haloarene C—X bond = partial double bond
20 Statements regarding nucleophiles attacking haloalkanes and benzylic halides:
(A) Ambident nucleophiles like cyanides can link through two different atoms.
(B) Reaction with AgCN mainly yields isocyanides due to its covalent nature.
(C) Reaction with KCN predominantly yields alkyl cyanides.
(D) The nitrite ion is incapable of acting as an ambident nucleophile.
Cyanide is an ambident nucleophile.
AgCN gives isocyanides.
KCN gives cyanides.
Nitrite is also ambident, so D is wrong.
Ambident nucleophiles can attack through two different atoms. Cyanide ion can attack through carbon or nitrogen.
KCN is ionic and provides CN⁻ freely, so attack usually occurs through carbon, forming alkyl cyanides:
R–X + KCN → R–CN
AgCN is more covalent, so attack occurs mainly through nitrogen, producing isocyanides:
R–X + AgCN → R–NC
Statement D is incorrect because nitrite ion is also an ambident nucleophile and can attack through nitrogen or oxygen.
B → Includes incorrect statement D.
C → Includes incorrect statement D and omits B.
D → Includes incorrect statement D and omits C.
Strategy Used
Statement Elimination
KCN gives Cyanide, AgCN gives isocyanide
