Chapter 4: Problem 66
On boiling 2,2 -dichloropropane with aqueous alkali, the product obtained is (a) 2,2-propanediol (b) acetone (c) propanal (d) 2 -chloro-2-propanol
Chapter 4: Problem 66
On boiling 2,2 -dichloropropane with aqueous alkali, the product obtained is (a) 2,2-propanediol (b) acetone (c) propanal (d) 2 -chloro-2-propanol
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Get started for freeThe product \(\mathrm{P}\) in the following reaction is \(\mathrm{ClCH}_{2}-\mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{P}\)
Match the elements of Column I to elements of Column II. There can be single or multiple matches. Column I (a) \(\mathrm{S}_{\mathrm{N}} 1\) reaction (b) \(\mathrm{S}_{\mathrm{N}} 2\) reaction (c) El reaction (d) E2 reaction Column II (p) Carbocation (q) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{2}-\mathrm{CH}_{3}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} \stackrel{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}}{\longrightarrow}\) Major product (r) one step process (s) \(\mathrm{CH}_{3}-\mathrm{CHBr}-\mathrm{CH}_{3}+\mathrm{I}^{-} \stackrel{\text { acetoee }}{\longrightarrow}\) (t) Racemisation
Which of the following is the correct order of dipole moment of methyl halides? (a) \(\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{I}\) (c) \(\mathrm{CH}_{3} \mathrm{I}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{I}\) (d) \(\mathrm{CH}_{3} \mathrm{I}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{~F}\)
Treatment of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{MgBr}\) with \(\mathrm{I}_{2}\) gives \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2}\) I in good yield whereas, \(\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) on reaction with HBr gives mainly a rearranged product namely \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBrCH}_{2} \mathrm{CH}_{3}\) and not \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Br}\). Why?
If 4 -bromo-1-butanol is treated with sodium, a product of molecular formula \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) is formed. Arrive at the structure and propose a mechanism.
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