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Chapter 6: Problem 19

Reaction of 2-methyl-2-pentene with each reagent is regioselective. Draw a structural formula for the product of each reaction, and account for the observed regioselectivity. (a) HI (b) \(\mathrm{HBr}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) in the presence of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{Br}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{Hg}(\mathrm{OAc})_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
(a) HI (b) HBr (c) H2O in the presence of H2SO4 (d) Br2 in H2O (e) Hg(OAc)2 in H2O Answer: (a) The product formed is 2-iodo-2-methylpentane, and the regioselectivity is Markovnikov. (b) The product formed is 2-bromo-2-methylpentane, and the regioselectivity is Markovnikov. (c) The product formed is 2-methyl-2-pentanol, and the regioselectivity is Markovnikov. (d) The product formed is (1S,2R)-2-bromo-2-methylpentan-1-ol, and the regioselectivity is anti-Markovnikov. (e) The product formed is 2-methyl-2-pentanol, and the regioselectivity is Markovnikov.

Step by step solution

01

Draw the structural formula of 2-methyl-2-pentene

First, draw the structural formula of the given alkene, 2-methyl-2-pentene. This molecule has a double bond between the second and third carbons and a methyl group at the second carbon.
02

Reaction with HI (Hydroiodic Acid)

The reaction between 2-methyl-2-pentene and HI undergoes a Markovnikov addition, where the hydrogen attaches to the least-substituted carbon of the double bond while the iodine attaches to the most-substituted carbon. This is because the carbocation intermediate formed in the reaction is more stable when the positive charge is on the more substituted carbon. The product formed is 2-iodo-2-methylpentane.
03

Reaction with HBr (Hydrobromic Acid)

The reaction between 2-methyl-2-pentene and HBr also follows the Markovnikov addition rule, similar to the reaction with HI. The hydrogen from HBr attaches to the least-substituted carbon of the double bond, and the bromine attaches to the most-substituted carbon, forming 2-bromo-2-methylpentane as the product.
04

Reaction with \(\mathrm{H}_{2} \mathrm{O}\) in the presence of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (Acid-Catalyzed Hydration)

In the presence of a strong acid like \(\mathrm{H}_{2} \mathrm{SO}_{4}\), the reaction between 2-methyl-2-pentene and water undergoes acid-catalyzed hydration. This reaction also follows the Markovnikov rule, with the hydrogen from \(\mathrm{H}_{2} \mathrm{O}\) attaching to the least-substituted carbon of the double bond, and the oxygen attaching to the most-substituted carbon. The product formed is 2-methyl-2-pentanol.
05

Reaction with \(\mathrm{Br}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) (Bromohydrin Formation)

The reaction between 2-methyl-2-pentene and \(\mathrm{Br}_{2}\) in the presence of \(\mathrm{H}_{2} \mathrm{O}\) forms a bromohydrin. In this reaction, the bromine gets added to the most-substituted carbon of the double bond, and the \(\mathrm{OH}\) group gets added to the least-substituted carbon. The stereochemistry of this reaction is anti, meaning that the bromine and \(\mathrm{OH}\) group will be placed on opposite sides of the double bond. The product formed is (1S,2R)-2-bromo-2-methylpentan-1-ol.
06

Reaction with \(\mathrm{Hg}(\mathrm{OAc})_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) (Oxymercuration-Demercuration)

The reaction of 2-methyl-2-pentene with \(\mathrm{Hg}(\mathrm{OAc})_{2}\) in the presence of water undergoes an oxymercuration-demercuration process, resulting in the Markovnikov addition of water across the double bond. The hydrogen attaches to the least-substituted carbon, and the \(\mathrm{OH}\) group attaches to the most-substituted carbon of the double bond. The product formed is 2-methyl-2-pentanol, which is the same as the one formed in step 4.

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Most popular questions from this chapter

When 2-pentene is treated with \(\mathrm{Cl}_{2}\) in methanol, three products are formed. Account for the formation of each product (you need not explain their relative percentages).

The acid-catalyzed hydration of 3,3-dimethyl-1-butene gives 2,3 -dimethyl-2-butanol as the major product. Propose a mechanism for the formation of this alcohol.

Consider the following reaction. Cyclohexane-1, 4-dicarbaldehyde (a) Draw a structural formula for the compound with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{12-}\) (b) Do you predict the product to be the cis isomer, the trans isomer, or a mixture of cis and trans isomers? Explain. (c) Draw a suitable stereorepresentation for the more stable chair conformation of the dicarbaldehyde formed in this oxidation.

Draw a structural formula for the cycloalkene with the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{10}\) that reacts with \(\mathrm{Cl}_{2}\) to give each compound.

In this chapter, we studied the mechanism of the acid-catalyzed hydration of an alkene. The reverse of this reaction is the acid-catalyzed dehydration of an alcohol. Propose a mechanism for the acid-catalyzed dehydration of 2-propanol to propene.
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