Bromination of Alkene
Trending Questions
- CH3CHO+HCN→CH3CH(OH)CN
In a reaction of alkyl halide on chiral centres, there is _______.
Assertion : Addition of bromine to trans-2-butene yields meso-2, 3-dibromobutane
Reason : Bromine addition to an alkene is an electrophilic addition.
If both assertion and reason are true and the reason is the correct explanation of the assertion.
- If both assertion and reason are true but reason is not the correct explanation of the assertion.
- If assertion is true but reason is false.
- If the assertion and reason both are false.
If assertion is false but reason is true.
What is the main product of the reaction between 2-methyl propene with hydrogen bromide?
[RPMT 2002]
2-bromo-2 methyl propane
1-bromo butane
1-bromo-2 methyl propane
2-bromo butane
- (ii)>(iii)>(i)
- (iii)>(ii)>(i)
- (i)>(ii)>(iii)
- (ii)>(i)>(iii)
(a) β-Elimination reaction
(b) Follows Hoffmann rule
(c) Dehydrohalogenation reaction
(d) Dehydration reaction
- (a), (c) and (d) only
- (b) and (c) only
- (a), (b) and (d) only
- (a) and (c) only
- The methyl alcohol solvates the bromine
- The ion formed initially may react with Br− or CH3OH
- The reaction follows Markownikoff's rule
- This is a free-radical mechanism
- Polar aprotic solvent favour SN2 reaction
- Strong nucleophiles are preferred
- Nucleophile is involved in the RDS
- Rate depends on R-X only
- Nucleophile should be present within the molecule/internally.
- Generally, nucleophile and leaving group should be present anti to each other at 1, 2 position.
- Nucleophile and leaving group should be present syn to each other
- Concentration of nucleophile (external) should be less
CH3−CH=CH−CH3C6H5CO3H−−−−−−−→- CH3CH2ONa+C2H5Br→
- KCN+(CH3)3CBr→
- C2H5OH(Excess)+H2SO4140∘ C−−−−→
- Polar aprotic solvent favour SN1 reaction?
- Strong nucleophiles are preferred
- Nucleophile is involved in the RDS.
- Rate depends on R-X
Which of the following reactions will result in the formation of a chiral center in the product?
- Reaction I gives a substitution product whereas reaction II gives an elimination product.
- Reaction I gives an elimination product whereas reaction II gives a substitution product.
- Both reactions give elimination products.
- Both reactions give substitution products.
- X=H2SO4, Y=H3O+, Z=PCl5, heat
- X=HNO3, Y=Na2CO3, Z=H2SO4
- X=H2SO4/Hg2+, Y=PCl5/H2O, Z=K2Cr2O7/OH−
- X=H2SO4/Hg2+, Y=CH3MgBr/H2O, Z=K2Cr2O7/H+
The major product of this reaction is:
I) Benzene decolourises bromine water.
II) Benzene decolourises alkaline permanganate solution.
III) Acetylene decolourises alkaline permanganate solution.
IV) Acetylene decolourises bromine water.
- III and IV
- II and III
- I and IV
- I and II
- acetamide
- acetonitrile
- methyl acetate
- diethyl ether
- X=CH3CH2CH2CH2CH2Cl, Y=CH3CH2−C|ClH−CH2CH3
- X=CH3CH2−C|ClH−CH2CH3, Y=CH3CH2CH2−C|ClH−CH3
- X=CH3CH2−C|ClH−CH2CH3, Y=CH3−C|ClH−C|ClH=CH2CH3
- X=ClCH2−CH2−CH=CH−CH3, Y=CH3CH2−CH=CH−CH2Cl
What is the main product of the reaction between 2-methyl propene with hydrogen bromide?
[RPMT 2002]
1-bromo butane
1-bromo-2 methyl propane
2-bromo butane
2-bromo-2 methyl propane
H3C−CH2−C|CH3=CH2+HBr→A
A (predominantly) is
(JEE MAIN 2021)
- 2 identical mesomers
- 2 enantiomers and 2 mesomers
- 2 enantiomers
- 1 racemic and 2 enantiomers
- 1-Bromo-3-methylbutane
- 2-Bromo-3-methylbutane
- 2-Bromo-2-methylbutane
- 1-Bromo-2-methylbutane
- True
- False
- 2-Chloropropane
- Allyl chloride
- No reaction
- n-propyl chloride
- (CH3)3C−<CH3−CH2−<(CH3)2CH−
- CH3−CH2−<(CH3)2CH−<(CH3)3C−
- (CH3)3C−<(CH3)2CH−<CH3−CH2
- (CH3)2CH−<(CH3)3C−<CH3−CH2−