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 \(_{3} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\)

Short Answer

Expert verified
The products of the reactions are (a) \(\mathrm{CH}_{3}\mathrm{CHCl-CH}_{2}\mathrm{Cl},\) (b) \(\mathrm{CH}_{3}\mathrm{CH}-\mathrm{CH}_{2}\mathrm{CN},\) and (c) \(\mathrm{CH}_{3}\mathrm{CHCl-CH}(\mathrm{CH}_{3})_{2}\).

Step by step solution

01

Reaction of HCl with 2-chloro-1-propene

In this reaction, we have HCl reacting with 2-chloro-1-propene: \( \mathrm{HCl + CH}_{3}\mathrm{CH=CCl} \rightarrow ? \) Following Markovnikov's rule, the hydrogen atom from HCl will attach to the carbon atom in the double bond that is carrying more hydrogen atoms already. Here it's the middle carbon atom. The Cl from the HCl will attach to the other carbon atom in the double bond that originally carried the Cl in the reactant. The product is: \( \mathrm{CH}_{3}\mathrm{CHCl-CH}_{2}\mathrm{Cl} \)
02

Reaction of HCN with \(\mathrm{CH}_{3} \mathrm{CH=CH}_{2}\)

Here we have HCN reacting with propene: \( \mathrm{HCN} + \mathrm{CH}_{3}\mathrm{CH=CH}_{2} \rightarrow ? \) H-CN bond polarizes and C (carrying a partial positive charge) attacks the carbon atom of the double bond that has fewer hydrogen atoms bonded to it, due to electron-releasing effect of hydrogens, making it more rich in electrons. Thus C of CN attaches to the terminal carbon. The product is: \( \mathrm{CH}_{3}\mathrm{CH}-\mathrm{CH}_{2}\mathrm{CN} \)
03

Reaction of HCl with \(\mathrm{CH}_{3} \mathrm{CH=CH}(\mathrm{CH}_{3})_{2}\)

In this final reaction, we have HCl reacting with 4,4-dimethyl-2-pentene: \( \mathrm{HCl} + \mathrm{CH}_{3}\mathrm{CH=CH}(\mathrm{CH}_{3})_{2} \rightarrow ? \) Following Markovnikov's rule, the hydrogen from HCl will attach to the central carbon atom in the double bond that is carrying more hydrogen atoms. Hence, Cl attaches to the carbon atom next to the central carbon atom. The product is: \( \mathrm{CH}_{3}\mathrm{CHCl-CH}(\mathrm{CH}_{3})_{2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Markovnikov's Rule
Markovnikov's Rule is a principle that governs the outcome of addition reactions involving unsaturated compounds like alkenes and alkynes.

When an unsymmetrical alkene reacts with a hydrogen halide such as hydrochloric acid, Markovnikov's Rule helps predict which carbon atom of the double bond will connect to the hydrogen atom and which will connect to the halide (like chlorine).

According to Markovnikov's Rule, the hydrogen atom attaches to the carbon with the most hydrogen atoms already attached, leading to the more stable carbocation intermediate. The halide will then bond with the less substituted carbon atom. This rule is key in determining the major product in chemical reactions of alkenes like those presented in the exercise.
Chemical Reactions of Alkenes
Alkenes are hydrocarbons with at least one carbon-carbon double bond, which is the reactive site for many reactions. Chemical reactions of alkenes commonly involve addition reactions, where atoms or groups of atoms are added across the double bond.

Examples of these reactions include halogenation, hydrohalogenation, and hydration. In the presence of a halogen like chlorine or bromine, alkenes form dihalides. With hydrogen halides (HCl, HBr, etc.), alkenes form alkyl halides, following the Markovnikov's Rule. These reactions are vital in synthesizing a variety of organic compounds and are applied using the exercise's context.
Hydrochloric Acid Reactions
Hydrochloric acid (HCl) is known for its role in acid-base reactions, but it also participates in addition reactions with alkenes.

In these reactions, HCl provides a hydrogen atom and a chloride anion; following Markovnikov's Rule, the hydrogen atom bonds with the more substituted carbon of the alkene, while the chloride ion bonds with the other carbon atom at the double bond. This typically results in the formation of an alkyl chloride through hydrochlorination. The identification of products, as shown in the exercise, is foundational for students learning organic chemistry.
Hydrogen Cyanide Reactions
Hydrogen cyanide (HCN) is a polar molecule that can add across the double bonds in alkenes, leading to the formation of nitriles. In the reaction with alkenes, the carbon in the CN group attaches to the carbon with the most available electrons (usually the least substituted carbon), contrary to Markovnikov's Rule. This is because of the unique polarity of the H-CN bond.

The reaction between HCN and an alkene is an important synthetic process for generating carbon-nitrogen bonds, which are key structural features in many organic molecules. Understanding this reaction allows students to predict and explain the outcome of HCN additions to alkenes, as shown in the exercise.

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Most popular questions from this chapter

Draw a structure to represent the principal product of each of the following reactions: (a) 1 -pentanol \(+\) dichromate ion (acid solution) (b) butyric acid + ethanol (acid solution) (c) 2 -methyl-1-butene \(+\mathrm{HBr}\)

Write the structures of the isomers you would expect to obtain in the mononitration of \(m\) -methoxybenzaldehyde:

What is major product expected from the monobromination of 2,2,3 -trimethylpentane?

Write the formulas of the products formed from the reaction of 2 -butene with each of the following substances: (a) \(\mathrm{H}_{2} ;\) (b) \(\mathrm{Cl}_{2} ;\) (c) \(\mathrm{HCl} ;\) (d) \(\mathrm{H}_{2} \mathrm{O}\) (in acid).

The reduction of aldehydes and ketones with a suitable hydride-containing reducing agent is a good way of synthesizing alcohols. This approach would be even more effective if, instead of a hydride, we could use a source of nucleophilic carbon. Attack by a carbon atom on a carbonyl group would give an alcohol and simultaneously form a carbon-to-carbon bond. How can we make a C atom in an alkane nucleophilic? This was achieved by Victor Grignard, who created the organometallic reagent \(\mathrm{R}-\mathrm{MgBr},\) with the following reaction in diethyl ether: $$\mathrm{R}-\mathrm{Br}+\mathrm{Mg} \longrightarrow \mathrm{R}-\mathrm{MgBr}$$ The Grignard reagent is rarely isolated. It is formed in solution and used immediately in the desired reaction. The alkylmetal bond is highly polar, with the partial negative charge on the \(\mathrm{C}\) atom, which makes the C atom highly nucleophilic. The Grignard reagent \((\mathrm{R}-\mathrm{MgBr})\) can attack a carbonyl group in an aldehyde or ketone as follows: Addition of dilute aqueous acid solution to the metal alkoxide furnishes the alcohol. The important synthetic consequence of this procedure is that we have prepared a product with more carbon atoms than present in the starting material. A simple starting material can be transformed into a more complex molecule. (a) What is the product of the reaction between methanal and the Grignard reagent formed from 1-bromobutane after the addition of dilute acid? (b) By using a Grignard reagent, devise a synthesis for 2-hexanol. (c) By using a Grignard reagent, devise a synthesis for 2 -methyl- 2 -hexanol. (d) Grignard reagents can also be formed with aryl halides, such as chlorobenzene. What would be the product of the reaction between the Grignard reagent of chlorobenzene and propanone? Can you think of an alternative synthesis of this product, again using a Grignard reagent? (e) The basicity of the \(C\) atom bound to the magnesium in the Grignard reagent can be used to make Grignard reagents of terminal alkynes. Write the equation of the reaction between ethylmagnesium bromide and 1-hexyne. [Hint: Ethane is evolved.] (f) By using a Grignard reagent, suggest a synthesis for 2 -heptyn-1-ol.

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