Question

What products would you expect from the following reactions? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{Cl}_{2} \longrightarrow ?\) (b) C#CC1CCCC1 \(\mathrm{HBr} \longrightarrow ?\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CCH}_{3}+1 \mathrm{HBr} \longrightarrow ?\)

Short answer

(a) \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CCl}_2\mathrm{CHCl}_2 \); (b) \( \mathrm{CH}_2 \mathrm{CBr} \mathrm{C}1\mathrm{CCCC}1 \); (c) \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CBr} \equiv \mathrm{CH}_2 \).

Step-by-step solution

Determine Reaction Type (a)

The reaction involves an alkyne, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH} \), reacting with chlorine, \( \mathrm{Cl}_{2} \). Given the conditions of excess chlorine, expect two-step chlorination: first to the dichloroalkene, and then to the tetrachloride. The alkyne will be both di-substituted on the double bond, then further fully substituted as single bonds are formed.

Predict Product for Reaction (a)

When adding excess \( \mathrm{Cl}_{2} \) to an alkyne, it reacts across the triple bond forming 1,2-dichloroalkene first, followed by further chlorination to add an additional two chlorine atoms across the formerly present double bond to form a tetra-chloride. The final product is thus \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CCl}_2\mathrm{CHCl}_2 \).

Determine Reaction Type (b)

The reaction involves an alkyne \( \mathrm{C} \equiv \mathrm{C} \mathrm{C}1\mathrm{CCCC}1 \) reacting with \( \mathrm{HBr} \). This is a typical electrophilic addition reaction, where the alkyne will react with hydrogen bromide to form a haloalkene.

Predict Product for Reaction (b) using Markovnikov's Rule

According to Markovnikov's rule, in the addition of \( \mathrm{HBr} \) to unsymmetrically substituted alkynes like \( \mathrm{C} \equiv \mathrm{C} \), the hydrogen atom attaches to the carbon with more hydrogen atoms, while the bromine attaches to the more substituted carbon. The resulting product will be \( \mathrm{CH}_2 \mathrm{CBr} \mathrm{C}1\mathrm{CCCC}1 \).

Determine Reaction Type (c)

This reaction involves adding \( \mathrm{HBr} \) to an alkyne. It will also go through electrophilic addition. The terminal alkyne \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CCH}_{3} \) will react with \( \mathrm{HBr} \).

Predict Product for Reaction (c) using Markovnikov's Rule

According to Markovnikov's rule, after adding \( \mathrm{HBr} \) to the terminal alkyne, the bromine atom will end up on the more substituted carbon due to the formation of a more stable carbocation intermediate. Therefore, the product of this reaction will be 2-bromo-hexene: \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CBr} \equiv \mathrm{CH}_2 \).

Key concepts

Alkyne Reactions

Alkynes, which are hydrocarbons containing carbon-carbon triple bonds, engage in a variety of reactions that are pivotal in organic chemistry. These reactions often involve breaking the triple bond and creating different products. One common type of reaction that alkynes undergo is halogenation. For example, when an alkyne reacts with diatomic chlorine, the multiple bonds can be replaced with chlorine atoms. In the exercise, an alkyne reacts with an excess of chlorine, leading to tetrachlorination. This results first in a vicinal dichloroalkene and ultimately in a saturated tetrachloroalkane. Such transformations are significant as they allow the synthesis of diverse compounds with varying degrees of halogenation, which serve as valuable intermediates in chemical synthesis.

Electrophilic Addition

Electrophilic addition is a fundamental reaction mechanism involving the addition of an electrophile to a compound with electrons available for binding, such as an alkene or an alkyne. In the context of alkynes, as seen in the exercise, candidates like hydrogen halides (HBr) are common electrophiles. The reaction starts with the electrophile attacking the electron-rich region of the triple bond, leading to the formation of a carbocation intermediate. This intermediate is then rapidly attacked by the nucleophile, forming a new single bond at what was previously the site of the triple bond.

• This type of reaction is commonplace in converting alkynes to haloalkenes.
• It is particularly useful because it leads to selective bond formation.

Understanding electrophilic addition helps predict the outcomes of various alkyne reactions, such as the halogenation and hydrohalogenation processes commonly used in organic synthesis.

Markovnikov's Rule

Markovnikov's Rule is an essential principle in organic chemistry which predicts the outcome of addition reactions involving unsymmetrical reagents and substrates. According to this rule, when an alkene or an alkyne, such as in our exercise, undergoes electrophilic addition with hydrogen halides like HBr, the hydrogen atom from the reagent tends to bond to the carbon with the greater number of hydrogen atoms initially attached. This provides stability through the formation of a more stable carbocation intermediate, typically on the more substituted carbon, leading to the product with the more stable configuration.

• This rule is essential for understanding regioselectivity in addition reactions.
• It serves as a guide to anticipating how molecules transform through reactions that form new carbon-hydrogen or carbon-halogen bonds.

Grasping Markovnikov's Rule allows chemists and students to predict the major product of a reaction efficiently, exemplified by the formation of 2-bromo products in the given exercise, which arises from the addition of HBr to alkynes.