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When aniline is treated with chloroform in the presence of alcoholic KOH, the product forms is
Options
(a) aniline
(b) benzene
(c) phenyl isocyanide
(d) phenol
Correct Answer:
phenyl isocyanide
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Which one of the following is not an aldose
- Using the Gibb’s energy change,ΔG⁰ = +63.3 kJ, for the following reaction
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Topics: Haloalkenes and Haloarenes
(78)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which one of the following is not an aldose
- Using the Gibb’s energy change,ΔG⁰ = +63.3 kJ, for the following reaction
- H₂O₂ reduces MnO₄⁻ ion to
- Carbon atom in C₂(CN)₄ are
- Which is weakest acid
Topics: Haloalkenes and Haloarenes (78)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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