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Chapter 4: Problem 125
What is the product of addition of \(\mathrm{HCl}\) to \(\mathrm{CCl}_{3}-\mathrm{CH}=\mathrm{CH}_{2} ?\)
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The compound which gives yellow precipitate when heated with iodine and alkali is (a) propan - 1 -ol (b) methanol (c) tert-butyl alcohol (d) butan- \(2-\) ol
Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. The compound that gives a tertiary alcohol on reaction with \(\mathrm{CH}_{3} \mathrm{MgI}\) is
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{KI} \stackrel{\text { acetone }}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{I}+\mathrm{KBr} \quad \rightarrow(\mathrm{I})\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}+\mathrm{KI} \stackrel{\text { acetone }}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{I}+\mathrm{KBr} \quad \rightarrow(\mathrm{II})\) Reaction (II) is much slower than reaction (I) because (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) is sparingly soluble in acetone. (b) (I) follows \(\mathrm{S}_{\mathrm{N}} 1\) and (II) follows \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. (c) (I) follows \(\mathrm{S}_{\mathrm{N}} 2\) and (II) follows \(\mathrm{S}_{\mathrm{N}} 1\) mechanism. (d) steric hindrance in the case of reaction (II).
Match the elements of Column I to elements of Column II. There can be single or multiple matches. Column II (p) Nucleophilic substitution (q) Nitro alkane (r) Alkyl nitrite (s) El (t) E2
Preparation and reactions of alcohols Alcohols can by prepared from alkyl halides, epoxides, alkenes, etc. When treated with acid, \(\mathrm{C}-\mathrm{OH}\) bond is cleaved heterolytically to form a carbocation which could then undergo substitution, elimination or rearrangement depending on the conditions and stability of the carbocation. When \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) is refluxed with metallic sodium in dry ethers the major product obtained is (a) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\)
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