Chapter 4: Problem 8
Starting with ethylene glycol, how is 1,4 -dioxane prepared?
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 4: Problem 8
Starting with ethylene glycol, how is 1,4 -dioxane prepared?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freen-butyl bromide can be prepared by treating \(n\) -butyl alcohol with \(\mathrm{KBr}\) and conc. \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) This process cannot be applied to prepare tertiary butyl bromide because (a) in tertiary alcohol, C - OH bond cleaves easily. (b) conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dehydrates tertiary alcohol to form an alkene. (c) tertiary alcohol does not react with \(\mathrm{KBr}\). (d) in primary alcohol, C - OH bond cleaves easily.
\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\mathrm{Mg} / \mathrm{ether}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{3}}{\longrightarrow} \mathrm{B}\) The product \(B\) in the above reaction is (a) 2 -phenyl- 3 -methyl butan- 2 -ol (b) 2 -phenylpentan- 2 -ol (c) 3 -phenyl butan-2-ol (d) propan-2-ol and ethyl benzene
How is \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\) prepared by a Grignard reaction with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\) and \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) as the starting materials?
The major product of dehydration of hex-4-en-2-ol would be (a) 1,4 -hexadiene (b) 2,4 -hexadiene (c) 1,5 -hexadiene (d) 2,3 -hexadiene
The most useful reagent to distinguish ethanol from propan-1-ol is (a) \(\mathrm{HCl}\) and \(\mathrm{ZnCl}_{2}\) (b) \(\mathrm{I}_{2}\) and \(\mathrm{NaOH}\) (c) Cu at \(570 \mathrm{~K}\) (d) \(\mathrm{KMnO}_{4}\)
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