Electrophilic addition reactions are characteristic of
A alkanes
B alkenes
C aromatics
D alcohols
π-electrons of alkenes attract electrophiles, leading to addition reactions.
In electrophilic addition, the first step is attack by
A nucleophile
B free radical
C electrophile
D base
The π-bond donates electrons to an electrophile.
Addition of HX to alkene proceeds via formation of
A carbanion
B free radical
C carbocation
D nitrene
Protonation of alkene forms a carbocation intermediate.
Stability of carbocation formed during addition depends mainly on
A steric hindrance
B inductive effect and hyperconjugation
C resonance only
D molecular mass
Electron donation stabilizes the positive charge.
Markovnikov’s rule applies to addition of
A Br₂
B H₂
C HX
D KMnO₄
It predicts orientation of hydrogen and halogen addition.
According to Markovnikov’s rule, the proton adds to the carbon
A having fewer hydrogens
B having more hydrogens
C having more alkyl groups
D attached to halogen
This leads to a more stable carbocation.
Anti-Markovnikov addition is observed in
A HCl addition
B HI addition
C HBr in presence of peroxide
D hydration of alkene
Free radical mechanism reverses orientation.
The peroxide effect is also called
A Kharasch effect
B Baeyer effect
C Saytzeff effect
D Hofmann effect
Discovered by Kharasch for HBr addition.
Why peroxide effect is not seen with HCl or HI
A bond energies are unsuitable
B carbocation is unstable
C peroxide decomposes
D halogen radicals are unstable
Chain propagation steps are energetically unfavourable.
Hydration of alkene in acidic medium follows
A free radical mechanism
B carbocation mechanism
C concerted mechanism
D SN2 mechanism
Protonation leads to carbocation intermediate.
Major product of acid-catalysed hydration follows
A anti-Markovnikov rule
B Markovnikov rule
C Hofmann rule
D no orientation
More stable carbocation determines product.
Oxymercuration–demercuration gives
A rearranged alcohol
B Markovnikov alcohol without rearrangement
C anti-Markovnikov alcohol
D ketone
Carbocation rearrangement is avoided.
Hydroboration–oxidation gives
A Markovnikov alcohol
B anti-Markovnikov alcohol
C ketone
D aldehyde
Boron adds to less substituted carbon.
Addition of Br₂ to alkene gives
A alkyl bromide
B vicinal dibromide
C geminal dibromide
D bromohydrin only
Bromine adds across double bond.
Bromine addition proceeds via
A carbocation
B bromonium ion
C free radical
D carbanion
Three-membered cyclic intermediate forms.
Bromohydrin formation requires
A Br₂ / CCl₄
B Br₂ / H₂O
C HBr / peroxide
D Br₂ / NaOH
OH replaces Br at more substituted carbon.
Bromohydrin formation follows
A Markovnikov orientation
B anti-Markovnikov orientation
C random orientation
D Saytzeff rule
OH attaches to more substituted carbon.
Hydrogenation of alkene requires
A UV light
B acid catalyst
C metal catalyst
D peroxide
Ni/Pd/Pt activates hydrogen.
Hydrogenation converts alkene into
A alcohol
B alkyne
C alkane
D ketone
Double bond is saturated.
Elimination reactions generally produce
A alkane
B alcohol
C alkene
D aldehyde
Removal of HX or H₂O creates π-bond.
Elimination of HX from alkyl halide using alcoholic KOH is
A SN1
B SN2
C E1
D E2
Strong base causes single-step elimination.
E2 reaction is
A two-step
B involves carbocation
C concerted
D radical
Bond breaking and formation occur simultaneously.
E1 reaction involves
A carbanion
B free radical
C carbocation
D bromonium ion
Similar to SN1, rearrangement possible.
Rearrangement is possible in
A E2 only
B E1 only
C both E1 and E2
D neither
Carbocation intermediate allows rearrangement.
Saytzeff’s rule predicts formation of
A least substituted alkene
B most substituted alkene
C terminal alkene only
D random alkene
More substituted alkene is more stable.
Hofmann elimination gives
A most substituted alkene
B least substituted alkene
C internal alkene only
D aromatic compound
Bulky base favours removal of least hindered proton.
Which base favours Hofmann elimination
A OH⁻
B OEt⁻
C tert-butoxide
D NH₃
Bulky base prefers less substituted β-hydrogen.
β-Elimination involves removal of
A H and X from same carbon
B H and X from adjacent carbons
C two hydrogens
D halogen only
Leads to formation of double bond.
Dehydration of alcohol using conc. H₂SO₄ is an example of
A substitution
B addition
C elimination
D rearrangement only
Water is eliminated to form alkene.
Dehydration of tertiary alcohol proceeds mainly via
A E2
B E1
C SN2
D free radical
Stable carbocation forms easily.
Order of ease of dehydration of alcohols is
A 1° > 2° > 3°
B 2° > 1° > 3°
C 3° > 2° > 1°
D all equal
Carbocation stability governs rate.
Which reaction gives trans-alkene preferentially
A E1
B E2
C SN1
D addition
Anti-periplanar elimination favours trans product.
Anti-periplanar geometry is required in
A E1
B E2
C SN1
D electrophilic addition
Proper orbital overlap needed.
Zaitsev rule is NOT followed when
A strong base is used
B bulky base is used
C high temperature
D polar solvent
Bulky base favours Hofmann product.
Elimination is favoured over substitution by
A aqueous KOH
B alcoholic KOH
C dilute acid
D cold conditions
Strong base and heat favour elimination.
Which alkyl halide favours elimination most
A primary
B secondary
C tertiary
D methyl
Stable carbocation and steric hindrance.
E1 reaction rate depends on
A base concentration
B nucleophile concentration
C substrate concentration
D solvent only
Rate-determining step is carbocation formation.
E2 reaction rate depends on
A substrate only
B base only
C both substrate and base
D temperature only
Bimolecular reaction.
Which solvent favours E1 reaction
A non-polar
B polar protic
C polar aprotic
D ether
Stabilises carbocation.
Which solvent favours E2 reaction
A polar protic only
B polar aprotic
C non-polar only
D acidic
Enhances base strength.
Which alkene is major in E1 elimination of tert-butyl bromide
A ethene
B propene
C 2-methylpropene
D but-1-ene
Most substituted alkene.
In elimination, β-hydrogen refers to hydrogen on
A same carbon as leaving group
B adjacent carbon
C terminal carbon only
D any carbon
Removal leads to double bond.
Dehydrohalogenation means
A removal of H₂
B removal of HX
C addition of HX
D addition of H₂
Typical elimination reaction.
Which reaction shows both substitution and elimination
A SN2
B SN1
C reaction with alcoholic KOH
D Wurtz reaction
Competing reactions occur.
Product distribution in elimination depends on
A base size
B substrate structure
C temperature
D all of these
All factors influence outcome.
Strong bulky base gives mainly
A Zaitsev product
B Hofmann product
C carbocation
D substitution
Least hindered alkene forms.
Which reaction has highest activation energy
A SN2
B E2
C E1
D electrophilic addition
Carbocation formation is energy demanding.
Elimination reactions are favoured at
A low temperature
B high temperature
C room temperature only
D freezing conditions
Entropy increase favours elimination.
Which statement is correct
A E1 is bimolecular
B E2 is unimolecular
C E1 involves carbocation
D E2 involves rearrangement
Key distinction between E1 and E2.
Correct statement is
A Markovnikov rule applies to elimination
B Hofmann rule predicts most substituted alkene
C Peroxide effect is radical based
D E2 allows rearrangement
HBr + peroxide proceeds via free radicals.