In the nitration of benzene with HNO₃/H₂SO₄, the rate-determining step involves attack by:
ANO₂⁺ (nitronium ion)
BNO₂• (nitrogen dioxide radical)
CHNO₃ (nitric acid molecule)
DNO₃⁻ (nitrate ion)
Correct Answer:
A. NO₂⁺ (nitronium ion)
EXPLANATION
The HNO₃/H₂SO₄ mixture generates NO₂⁺ (nitronium ion), which is the electrophile attacking the benzene ring in the rate-determining step. This is a classic electrophilic aromatic substitution.
The addition of HBr to propene in the presence of peroxides follows:
AMarkovnikov's rule to give 2-bromopropane
BAnti-Markovnikov's rule to give 1-bromopropane
CFree radical mechanism to give equal amounts of both isomers
DIonic mechanism to give 2-bromopropane
Correct Answer:
B. Anti-Markovnikov's rule to give 1-bromopropane
EXPLANATION
Peroxides initiate free radical mechanism (Kharasch effect). In free radical addition, HBr adds anti-Markovnikov to propene, giving 1-bromopropane as the major product. The Br radical adds first to the terminal carbon.
In the dehydration of 2-methylbutan-2-ol, the major product is:
A2-methylbut-1-ene
B2-methylbut-2-ene
CBut-1-ene
DPent-2-ene
Correct Answer:
B. 2-methylbut-2-ene
EXPLANATION
Dehydration of 2-methylbutan-2-ol follows Zaitsev's rule, producing the most stable (most substituted) alkene. 2-methylbut-2-ene is a trisubstituted alkene and is the major product.
The reaction of phenol with excess bromine in water produces:
A2,4,6-tribromophenol
B4-bromophenol
C2,4-dibromophenol
D2,6-dibromophenol
Correct Answer:
A. 2,4,6-tribromophenol
EXPLANATION
Phenol is highly activated towards electrophilic aromatic substitution due to the electron-donating -OH group. Bromine can add at all three ortho and para positions with excess bromine, giving 2,4,6-tribromophenol.