a) SN1
b) SN2
c) E1
d) E2
Answer: b) SN2
Explanation: In SN2 reactions, a nucleophile attacks the
electrophilic carbon, displacing the leaving group in a single step.
a) Alkyl group increases the rate.
b) Alkyl group decreases the rate.
c) Alkyl group has no effect.
d) The effect depends on the size of the alkyl group.
Answer: a) Alkyl group increases the rate.
Explanation: The rate of nucleophilic substitution increases
with the increase in the size of the alkyl group due to increased electron
density.
a) 1-Bromopropane
b) 2-Bromopropane
c) 1-Bromobutane
d) 2-Bromobutane
Answer: a) 1-Bromopropane
Explanation: The compound has a three-carbon chain with a
bromine atom attached to the first carbon, hence the name 1-Bromopropane.
a) NaOH
b) KCN
c) Zn
d) H₂O
Answer: a) NaOH
Explanation: Sodium hydroxide (NaOH) is used in nucleophilic
substitution reactions to convert haloalkanes into alcohols.
a) Methyl chloride
b) Ethyl bromide
c) Tertiary butyl chloride
d) Primary butyl iodide
Answer: c) Tertiary butyl chloride
Explanation: Tertiary haloalkanes undergo SN1 reactions faster
because of the stability of the tertiary carbocation.
a) CH₃Br
b) (CH₃)₃CCl
c) C₂H₅I
d) C₆H₅Cl
Answer: a) CH₃Br
Explanation: Methyl halides are most reactive in SN2 reactions
due to less steric hindrance around the carbon atom.
a) R-I > R-Br > R-Cl > R-F
b) R-Cl > R-Br > R-I > R-F
c) R-F > R-Cl > R-Br > R-I
d) R-Br > R-Cl > R-I > R-F
Answer: a) R-I > R-Br > R-Cl > R-F
Explanation: In SN1 reactions, the leaving ability of halides
follows the order I⁻ > Br⁻ > Cl⁻ > F⁻, which affects the reactivity.
a) Friedel–Crafts halogenation
b) Nucleophilic substitution reaction
c) Addition reaction
d) Reduction reaction
Answer: a) Friedel–Crafts halogenation
Explanation: Haloarenes can be prepared by Friedel–Crafts
halogenation, where a halogen reacts with an aromatic compound in the presence
of a Lewis acid.
a) CH₃COCH₃
b) CH₃CHO
c) CH₃CH₂OH
d) CH₃COOH
Answer: a) CH₃COCH₃
Explanation: Iodoform (CHI₃) is formed when methyl ketones
(such as acetone) are treated with iodine in the presence of a base.
a) CH₃CH₂Cl
b) (CH₃)₃CCl
c) C₆H₅Cl
d) CH₃Cl
Answer: a) CH₃CH₂Cl
Explanation: CH₃CH₂Cl is a primary alkyl halide because the
carbon attached to the halogen is attached to only one other carbon.
a) CH₃Cl
b) C₆H₅Cl
c) CH₃I
d) C₂H₅Cl
Answer: b) C₆H₅Cl
Explanation: Chlorobenzene (C₆H₅Cl) shows resonance structures
due to the delocalization of electrons in the aromatic ring.
a) Halogenation
b) Hydrogenation
c) Dehydrogenation
d) Nitration
Answer: a) Halogenation
Explanation: Haloalkanes are prepared by halogenation, where
halogens react with alkanes in the presence of UV light.
a) Butene
b) Butane
c) Butanol
d) 1-Butanol
Answer: c) Butanol
Explanation: When 1-bromobutane reacts with NaOH, it undergoes
nucleophilic substitution to form butanol.
a) SN1
b) SN2
c) E1
d) E2
Answer: b) SN2
Explanation: A primary alkyl halide reacts with a strong
nucleophile through the SN2 mechanism due to less steric hindrance.
a) It leads to the formation of alcohols.
b) It leads to the formation of alkynes.
c) It leads to the formation of alkenes.
d) It leads to the formation of haloarenes.
Answer: c) It leads to the formation of alkenes.
Explanation: Alcoholic KOH induces elimination (E2) reactions
to form alkenes from haloalkanes.
a) Phenol
b) Benzene
c) Chlorobenzene
d) Hydroquinone
Answer: a) Phenol
Explanation: Chlorobenzene reacts with NaOH under high
temperature and pressure to form phenol through nucleophilic substitution (the
Dow process).
a) CH₂=CHCl
b) CH₃CH₂Cl
c) C₆H₅Cl
d) (CH₃)₂CCl₂
Answer: a) CH₂=CHCl
Explanation: A vinylic halide is a halide attached to a
carbon-carbon double bond, like CH₂=CHCl.
a) CH₃Cl
b) C₂H₅Cl
c) (CH₃)₃CCl
d) C₆H₅Cl
Answer: c) (CH₃)₃CCl
Explanation: Tertiary alkyl halides undergo SN1 reactions due
to the stability of the tertiary carbocation.
a) Bromobenzyl
b) Benzylic bromide
c) Benzyl bromide
d) 1-Bromobenzene
Answer: c) Benzyl bromide
Explanation: The IUPAC name of C₆H₅CH₂Br is benzyl bromide, as
it consists of a benzene ring attached to a CH₂Br group.
a) C₂H₅Br with alcoholic KOH
b) C₆H₅Cl with aqueous NaOH
c) CH₃CH₂Cl with aqueous NaOH
d) CH₃I with alcoholic KOH
Answer: a) C₂H₅Br with alcoholic KOH
Explanation: Alcoholic KOH induces elimination (E2) reactions,
leading to the formation of an alkene from haloalkanes.
a) Electrophilic substitution
b) Nucleophilic substitution
c) Electrophilic addition
d) Nucleophilic addition
Answer: b) Nucleophilic substitution
Explanation: Chlorobenzene undergoes nucleophilic substitution
with NaOH under high pressure to form phenol.
a) It always follows an SN1 mechanism.
b) It always follows an SN2 mechanism.
c) The reaction mechanism depends on the structure of the alkyl group and the
leaving group.
d) It does not involve a transition state.
Answer: c) The reaction mechanism depends on the structure of
the alkyl group and the leaving group.
Explanation: The nucleophilic substitution mechanism (SN1 or
SN2) depends on factors such as the structure of the alkyl halide and the nature
of the leaving group.
a) CH₃Br
b) C₆H₅Cl
c) CH₂=CHCl
d) CH₃CH₂Cl
Answer: a) CH₃Br
Explanation: CH₃Br can react with sodium metal in a Wurtz
reaction to form an alkane.
a) Substitution occurs via SN1 mechanism.
b) Substitution occurs via SN2 mechanism.
c) Elimination reaction occurs.
d) No reaction occurs.
Answer: b) Substitution occurs via SN2 mechanism.
Explanation: Methyl iodide undergoes nucleophilic substitution
via the SN2 mechanism because methyl groups cause minimal steric hindrance.
a) CH₃Cl
b) C₂H₅Cl
c) C₆H₅Cl
d) (CH₃)₂CCl₂
Answer: a) CH₃Cl
Explanation: CH₃Cl is most reactive towards nucleophilic
substitution because the methyl group provides minimal steric hindrance.
a) 1-Propene
b) 2-Propene
c) 2-Butene
d) Propane
Answer: b) 2-Propene
Explanation: 2-Bromopropane undergoes elimination (E2) with
alcoholic KOH to form 2-propene.
a) CH₃Cl
b) CH₃I
c) (CH₃)₃CCl
d) C₂H₅Cl
Answer: c) (CH₃)₃CCl
Explanation: Tertiary alkyl halides undergo SN1 reactions
faster because of the stability of the tertiary carbocation.
a) SN1
b) SN2
c) E1
d) E2
Answer: b) SN2
Explanation: SN2 reactions do not involve the formation of a
carbocation intermediate; the nucleophile directly attacks the electrophilic
carbon.
a) Primary alkyl halides undergo substitution via SN1 mechanism.
b) Tertiary alkyl halides undergo substitution via SN2 mechanism.
c) Secondary alkyl halides undergo substitution via SN2 mechanism.
d) All alkyl halides can undergo nucleophilic substitution.
Answer: c) Secondary alkyl halides undergo substitution via SN2
mechanism.
Explanation: Secondary alkyl halides generally follow SN2
mechanisms, though they can also undergo SN1 depending on conditions.
a) Electrophilic substitution
b) Nucleophilic substitution
c) Addition-elimination mechanism
d) Hydrogenation
Answer: a) Electrophilic substitution
Explanation: Haloarenes are often prepared by electrophilic
substitution, such as in the Friedel-Crafts halogenation reaction.
a) CH₃Cl
b) C₂H₅Cl
c) (CH₃)₂CCl
d) C₆H₅Cl
Answer: a) CH₃Cl
Explanation: CH₃Cl undergoes nucleophilic substitution via the
SN2 mechanism the fastest due to the absence of steric hindrance around the
carbon atom.
a) Addition reaction
b) Substitution reaction
c) Elimination reaction
d) Polymerization reaction
Answer: b) Substitution reaction
Explanation: Haloalkanes predominantly undergo nucleophilic
substitution reactions, where the halogen atom is replaced by a nucleophile.
a) Substitution reaction
b) Elimination reaction
c) No reaction
d) Formation of an alkene
Answer: b) Elimination reaction
Explanation: When bromoethane reacts with alcoholic KOH, it
undergoes elimination (E2) to form ethene.
a) CH₃Cl
b) C₂H₅Cl
c) (CH₃)₃CCl
d) C₆H₅Cl
Answer: c) (CH₃)₃CCl
Explanation: Tertiary alkyl halides like (CH₃)₃CCl undergo
nucleophilic substitution via the SN1 mechanism due to the stability of the
tertiary carbocation formed during the reaction.
a) (CH₃)₃CCl
b) CH₃Cl
c) C₆H₅Cl
d) C₂H₅Cl
Answer: b) CH₃Cl
Explanation: Methyl halides like CH₃Cl undergo nucleophilic
substitution via the SN2 mechanism due to less steric hindrance around the
carbon atom.
a) Strong nucleophile
b) Polar protic solvent
c) Polar aprotic solvent
d) Weak nucleophile
Answer: b) Polar protic solvent
Explanation: SN1 reactions are favored by polar protic solvents
because they stabilize the carbocation intermediate and assist in the leaving
group departure.
a) It decreases the rate of the reaction
b) It does not affect the rate of the reaction
c) It increases the rate of the reaction
d) It stops the reaction completely
Answer: b) It does not affect the rate of the reaction
Explanation: In an SN1 reaction, the rate-determining step is
the formation of the carbocation, which is unaffected by the concentration of
the nucleophile.
a) 1-Iodopropane
b) 2-Iodopropane
c) Propene
d) Propane
Answer: a) 1-Iodopropane
Explanation: 1-Chloropropane undergoes nucleophilic
substitution with NaI in acetone, leading to the formation of 1-iodopropane.
a) Nucleophilic substitution
b) Electrophilic substitution
c) Elimination reaction
d) Addition reaction
Answer: c) Elimination reaction
Explanation: When a haloalkane reacts with alcoholic KOH at
high temperature, it undergoes an elimination (E2) reaction, leading to the
formation of an alkene.
a) Weak nucleophile and polar protic solvent
b) Strong nucleophile and polar aprotic solvent
c) Weak nucleophile and polar aprotic solvent
d) Strong nucleophile and non-polar solvent
Answer: b) Strong nucleophile and polar aprotic solvent
Explanation: SN2 reactions are favored by a strong nucleophile
and a polar aprotic solvent, which does not solvate the nucleophile too
strongly.
a) CH₃Cl
b) C₂H₅Cl
c) (CH₃)₃CCl
d) CH₃I
Answer: a) CH₃Cl
Explanation: CH₃Cl undergoes nucleophilic substitution via the
SN1 mechanism very slowly because it forms a very unstable methyl carbocation.
a) CH₃Cl
b) (CH₃)₂CHCl
c) CH₂=CHCl
d) C₆H₅Cl
Answer: b) (CH₃)₂CHCl
Explanation: (CH₃)₂CHCl, a secondary alkyl halide, is more
likely to undergo elimination (E2) with KOH, especially under high temperature
or in the presence of a strong base.
a) Butene
b) Butane
c) 1-Butene
d) 2-Butene
Answer: d) 2-Butene
Explanation: 2-Chlorobutane undergoes elimination (E2) with
NaOEt in ethanol to form 2-butene.
a) CH₃
b) OH
c) Cl
d) NH₂
Answer: c) Cl
Explanation: The chloride ion (Cl⁻) is the best leaving group
due to its ability to stabilize the negative charge when it departs during
nucleophilic substitution.
a) SN2
b) SN1
c) E2
d) E1
Answer: b) SN1
Explanation: Tertiary alkyl halides favor the SN1 mechanism due
to the stability of the tertiary carbocation intermediate and the solvent’s
ability to stabilize the intermediate.
a) Substitution reaction
b) Elimination reaction
c) Addition reaction
d) No reaction
Answer: b) Elimination reaction
Explanation: The reaction between an alkyl halide and a strong
base like KOH in ethanol usually leads to elimination, forming an alkene through
the E2 mechanism.
a) CH₃Br
b) CH₃Cl
c) C₆H₅Cl
d) (CH₃)₃CCl
Answer: a) CH₃Br
Explanation: CH₃Br undergoes nucleophilic substitution the
fastest because bromine is a better leaving group than chlorine and the methyl
group provides less steric hindrance.
a) SN1
b) SN2
c) E1
d) E2
Answer: a) SN1
Explanation: In an SN1 reaction, the halogen acts as an
electrophile after it leaves, forming a carbocation intermediate that can react
with the nucleophile.
a) CH₃Cl
b) C₂H₅Cl
c) (CH₃)₃CCl
d) CH₃I
Answer: c) (CH₃)₃CCl
Explanation: Tertiary alkyl halides like (CH₃)₃CCl are less
reactive towards nucleophilic substitution because the steric hindrance around
the carbon makes it difficult for the nucleophile to attack.
a) Nucleophilic substitution
b) Electrophilic addition
c) Elimination
d) Both a and c
Answer: d) Both a and c
Explanation: Alkyl halides can undergo both nucleophilic
substitution (SN1 and SN2) and elimination (E1 and E2) reactions.