Problem 33
Question
Give the major product that forms when (Z)-3-methyl2-pentene reacts with each of the following reagents: (a) \(\mathrm{HI} ;\) (b) \(\mathrm{H}_{2}\) in the presence of a platinum catalyst; (c) \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\).
Step-by-Step Solution
Verified Answer
The major products of the reactions are as follows: (a) 2-iodo-3-methylpentane, (b) 3-methylpentane, (c) 2-methyl-3-pentanol, and (d) (2R, 3R)-2,3-dibromo-3-methylpentane and its enantiomer (2S, 3S)-2,3-dibromo-3-methylpentane.
1Step 1: Reaction with Hydroiodic Acid (\(\mathrm{HI}\))
The reaction with \(HI\) will follow Markovnikov's rule, with the halogen attaching to the most substituted carbon of the alkene. The iodine will attach to the tertiary carbon, resulting in 2-iodo-3-methylpentane as the major product.
2Step 2: Reaction with Hydrogen (\(\mathrm{H}_{2}\)) in the Presence of a Platinum Catalyst
The reaction with hydrogen gas in the presence of a platinum catalyst will result in the hydrogenation of the alkene, leading to the addition of hydrogen across the double bond. This will yield 3-methylpentane.
3Step 3: Reaction with Water (\(\mathrm{H}_{2}\mathrm{O}\)) in \(\mathrm{H}_{2}\mathrm{SO}_{4}\)
The reaction with water in the presence of sulphuric acid will lead to the hydration of the alkene, following Markovnikov's rule. This will result in the formation of 2-methyl-3-pentanol.
4Step 4: Reaction with Bromine (\(\mathrm{Br}_{2}\)) in \(\mathrm{CCl}_{4}\)
The reaction with bromine in carbon tetrachloride will lead to the addition of bromine across the double bond in an anti fashion, resulting in the formation of (2R, 3R)-2,3-dibromo-3-methylpentane and its enantiomer (2S, 3S)-2,3-dibromo-3-methylpentane.
Key Concepts
Alkene ReactionsMarkovnikov's RuleHydrogenationHalogenation
Alkene Reactions
Alkenes are organic compounds with a carbon-carbon double bond. These double bonds are reactive centers, making alkenes prone to a variety of chemical reactions. Some common reactions include:
- Electrophilic Addition: This involves the addition of various electrophiles to the double bond, resulting in the formation of single bonds and subsequently, a more saturated molecule.
- Hydrogenation: The addition of hydrogen across the double bond, converting the alkene into an alkane.
- Halogenation: The addition of halogen atoms to the double bond.
Markovnikov's Rule
Markovnikov's Rule is a guiding principle for predicting the outcome of certain addition reactions involving alkenes. It states that in the addition of a protic acid to an alkene, the hydrogen atom connects to the less substituted carbon, while the more electronegative part of the reagent, such as a halogen, bonds with the more substituted carbon.
This happens because more substituted carbocations (intermediate positively charged ions) are more stable due to hyperconjugation and the dispersal of positive charge across adjacent carbon atoms. The stable formation of these carbocations leads to the major product being favored in reactions such as hydration and hydrohalogenation. Understanding this principle helps chemists predict how molecules will interact during a reaction and which product will be the major product formed.
Hydrogenation
Hydrogenation is the process of adding hydrogen (\(H_2\)) across an unsaturated molecule's double bond, turning it into a saturated molecule.
- Example Reaction: An alkene like 3-methylpentene undergoes hydrogenation to form 3-methylpentane when treated with hydrogen gas in the presence of a platinum catalyst.
- Catalysts Role: Catalysts such as platinum, palladium, or nickel are often used to expedite this reaction by providing a surface for the reaction to occur, lowering the activation energy required.
Halogenation
Halogenation involves the addition of halogens (such as bromine, chlorine) across an alkene's double bond. This reaction occurs in an anti fashion, meaning the halogen atoms add to opposite sides of the double bond, resulting in a trans product.In the presence of a solvent like carbon tetrachloride (\(CCl_4\)), halogenation results in the formation of dihalogenated products. For instance, when \(Br_2\) is added to a solution of (Z)-3-methyl-2-pentene, the product is a dibromo compound, specifically (2R, 3R)-2,3-dibromo-3-methylpentane along with its enantiomer.Halogenation is a valuable tool in organic chemistry as it allows for the creation of various significant compounds from simple alkenes, aiding in the exploration of stereochemistry and reactivity in synthetic pathways.
Other exercises in this chapter
Problem 29
Draw the structures of the products of each of the following reactions: (a) propene + hydrogen (Pt, heat) (b) 2 -butanol \(+\) heat (in the presence of sulfuric
View solution Problem 30
Predict the product(s) of the reaction of: (a) HCl with 2-chloro-1-propene (b) HCN with \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}\) (c) HCl with CH \(_{3} \
View solution Problem 34
Give the major product that forms when 1 -ethylcyclohexene reacts with each of the following reagents: (a) \(\mathrm{HI} ;\) (b) \(\mathrm{H}_{2}\) in the prese
View solution Problem 37
Alkylation of benzene can be accomplished by treating benzene with haloalkane (RX) in the presence of \(\mathrm{AlCl}_{3}\) The reaction is known as a Friedel-C
View solution