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Chapter 2: Problem 89

Choose the correct statement with regard to 3 -methyl but- 1 -ene (a) A tertiary alcohol is formed by hydration in the presence of an acid. (b) a primary alcohol is formed by oxymercuration reaction. (c) An optically active secondary alcohol is formed by hydroboration- oxidation. (d) The alkene is less reactive than ethene towards hydration under acid catalyzed conditions.

Short Answer

Expert verified
Answer: A tertiary alcohol is formed by hydration in the presence of an acid.

Step by step solution

01

Reaction 1: Hydration

3-Methyl but-1-ene can undergo hydration in the presence of an acid, which would result in an alcohol. The reaction proceeds with Markovnikov's rule, meaning that the alcohol will form at the more substituted carbon. In this case, 3-methyl but-1-ene would form a tertiary alcohol, which coincides with statement (a).
02

Reaction 2: Oxymercuration

Oxymercuration is a reaction that also involves the hydration of an alkene. This reaction proceeds without rearrangement and with Markovnikov's rule. For 3-methyl, but-1-ene, the primary alcohol will be formed. This matches statement (b).
03

Reaction 3: Hydroboration-oxidation

Hydroboration-oxidation is another reaction which leads to the formation of an alcohol from an alkene. This reaction proceeds with anti-Markovnikov's rule. In the case of 3-methyl but-1-ene, a secondary alcohol will be formed. However, this secondary alcohol will not have a chiral center and hence, it will not be optically active. Therefore, statement (c) is incorrect.
04

Reaction 4: Reactivity comparison with ethene

In a hydration reaction, under acid-catalyzed conditions, more substituted alkenes are more reactive than less substituted ones. This is because carbocations are formed during the reaction process, and more substituted carbocations are more stable than less substituted ones. Since 3-methyl but-1-ene is more substituted than ethene, it should be more reactive towards hydration under acid-catalyzed conditions. This contradicts statement (d). Considering all the reactions, the correct statement is: (a) A tertiary alcohol is formed by hydration in the presence of an acid.

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Most popular questions from this chapter

2-methylpropane on oxidation with \(\mathrm{KMnO}_{4}\) gives (a) acetone (b) t-butyl alcohol (c) isobutylene (d) t-butylperoxide

1,6 Dibromohexane can be converted to cyclohexane, when treated with \(\mathrm{Na}\) or \(\mathrm{Zn}\), in the presence of a catalyst. This reaction is known as (a) Wurtz coupling (b) Dieckmann cyclization (c) Freund's method (d) Diels Alder reaction

When \(\mathrm{n}\) -hexane is passed over heated alumina or chromic oxide catalyst under pressure, the product formed is (a) Iso hexane (b) cyclohexane (c) butane and ethane (d) Benzene

A hydrocarbon \(\mathrm{C}_{5} \mathrm{H}_{12}\) (A) on treatment with chlorine undergoes monochlorination by the replacement of the only \(1^{\circ}\) hydrogen to give(B) \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}\). When (B) is treated with Na metal in ether, the product obtained could be (a) \(\quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{CH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Cl}\)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) \(\stackrel{\text { alc } \mathrm{KOH} / \Delta}{\longrightarrow}(\mathrm{A})\). The product \((\mathrm{A})\) is (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH} \mathrm{CH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (d)
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