Reagents such as HCI, HBr, and HOH (H_O) can add across carbon-carbon double and triple bonds, with H forming a bond to one of the carbon atoms in the multiple bond and \(\mathrm{Cl}, \mathrm{Br},\) or OH forming a bond to the other carbon atom in the multiple bond. In some cases, two products are possible. For the major organic product, the addition occurs so that the hydrogen atom in the reagent attaches to the carbon atom in the multiple bond that already has the greater number of hydrogen atoms bonded to it. With this rule in mind, draw the structure of the major product in each of the following reactions. a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{2}}{\longrightarrow}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{HBr} \longrightarrow\)

Short Answer

Expert verified
For the major organic products in each reaction, considering Markovnikov's rule, we have: a. CH3CH2CHOHCH3 b. CH3CH2CHBrCH3 c. CH3CH2CHBrCH2Br

Step by step solution

01

a. Reaction of CH3CH2CH=CH2 + H2O

In this reaction, the double bond is between the second and third carbon atoms. Following Markovnikov's rule, the H in H2O will attach to the second carbon atom, and the OH will attach to the third carbon atom. The structure of the major product will be: CH3CH2CHOHCH3.
02

b. Reaction of CH3CH2CH=CH2 + HBr

In this reaction, the double bond is also between the second and third carbon atoms. Following Markovnikov's rule, the H in HBr will attach to the second carbon atom, and the Br will attach to the third carbon atom. The structure of the major product will be: CH3CH2CHBrCH3.
03

c. Reaction of CH3CH2C≡CH + 2 HBr

In this reaction, the triple bond is between the third and fourth carbon atoms. Since there are two moles of HBr, the reaction will proceed in two steps. In the first step, following Markovnikov's rule, the H in HBr will attach to the third carbon atom, and the Br will attach to the fourth carbon atom. This will result in the formation of a double bond between the third and fourth carbon atoms: CH3CH2CH=CHBr. In the second step, the double bond will undergo another addition reaction, with H attaching to the third carbon atom and Br attaching to the fourth carbon atom. The final structure of the major product will be: CH3CH2CHBrCH2Br.

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Key Concepts

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

Markovnikov's rule
Understanding Markovnikov's rule is essential when predicting the outcome of addition reactions involving alkenes and alkynes. The rule states that during an addition reaction, the hydrogen atom from the adding molecule (like HCl or HBr) will bond to the carbon atom that already has more hydrogen atoms attached. Essentially, the 'rich get richer,' meaning that the more substituted carbon gets the hydrogen, leading to the formation of the most stable carbocation intermediate. This rule helps in identifying the major product when there are two possible outcomes.

For example, in the reaction of propene with HCl, hydrogen will attach to the carbon with more hydrogens (the first carbon), and chlorine will bond to the less substituted carbon (the second carbon). This produces 2-Chloropropane as the major product, which is consistent with Markovnikov's rule.
Alkene reactions
Alkenes are hydrocarbons with double bonds, characterized by the general formula CnH2n. Reactions involving alkenes usually include the addition of atoms or groups to the carbons of the double bond, turning the double bond into a single bond. These addition reactions include hydrohalogenation, hydration, hydrogenation, and halogenation.

Alkenes are reactive due to the presence of the electron-rich double bond which acts as a site for electrophilic attack. For instance, during hydration of alkenes, water adds across the double bond in the presence of an acid catalyst, leading to the formation of alcohols. The place where the constituents add will follow Markovnikov's rule, as seen in the hydration of 1-butene to form 2-butanol.
Alkyne reactions
Alkynes, on the other hand, are hydrocarbons with a carbon-carbon triple bond and have the general formula CnH2n-2. Similar to alkenes, alkynes undergo addition reactions; however, they have two pi bonds that can react, which allows them to add two equivalents of a reagent. The reactions can stop at the alkene stage after the addition of one equivalent of a reactant or proceed to an alkane with two additions.

For instance, an alkyne like 1-butyne can react with two equivalents of HBr to yield 1,2-dibromobutane as the final product. The addition will proceed stepwise, adhering to Markovnikov's rule, with the first equivalent adding across the triple bond and the second reacting with the resulting double bond.
Hydrohalogenation
Hydrohalogenation is an addition reaction where a hydrogen halide (HCl, HBr, HI) adds across the double or triple bond of alkenes or alkynes. The resulting product is a haloalkane. This reaction is regioselective and follows Markovnikov's rule, which dictates that the hydrogen atom from the hydrogen halide will attach to the carbon with more hydrogen atoms.

For example, the addition of HBr to 1-butene will result in 2-bromobutane, as the hydrogen will preferentially add to the first carbon, and bromine to the second carbon. It's key to remember that in the presence of peroxides, hydrohalogenation of alkenes can proceed via a 'peroxide effect' or 'anti-Markovnikov addition', but this is an exception rather than the rule.
Hydration of alkenes
The hydration of alkenes involves adding water (H2O) across the double bond to form an alcohol. This reaction requires an acid catalyst, such as sulfuric acid, to proceed. The addition of water follows Markovnikov's rule, where the hydrogen attaches to the carbon with more hydrogens, and the hydroxyl group (OH) attaches to the carbon with fewer hydrogen atoms.

Considering the alkene but-1-ene reacting with water, the major product is 2-butanol, with the -OH group attaching to the second carbon. This transformation turns an unsaturated hydrocarbon into a functionalized, oxygen-containing molecule, expanding the compound's chemical reactivity and applicability.

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Most popular questions from this chapter

Mimosine is a natural product found in large quantities in the seeds and foliage of some legume plants and has been shown to cause inhibition of hair growth and hair loss in mice. a. What functional groups are present in mimosine? b. Give the hybridization of the eight carbon atoms in mimosine. c. How many \(\sigma\) and \(\pi\) bonds are found in mimosine?

ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right),\) butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right),\) and styrene \(\left(\mathbf{C}_{8} \mathbf{H}_{8}\right)\) a. Draw two repeating units of ABS plastic assuming that the three monomer units react in a 1: 1: 1 mole ratio and react in the same order as the monomers listed above. b. A sample of ABS plastic contains \(8.80 \%\) N by mass. It took 0.605 g Br_ to react completely with a 1.20-g sample of ABS plastic. What is the percent by mass of acrylonitrile, butadiene, and styrene in this polymer sample? c. ABS plastic does not react in a 1: 1: 1 mole ratio among the three monomer units. Using the results from part b, determine the relative numbers of the monomer units in this sample of ABS plastic.

Cis-trans isomerism is also possible in molecules with rings. Draw the cis and trans isomers of 1,2 -dimethylcyclohexane. In Exercise \(41,\) you drew all of the noncyclic structural and geometric isomers of \(\mathbf{C}_{4} \mathrm{H}_{7} \mathrm{F}\). Now draw the cyclic structural and geometric isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{F}\).

Draw the five structural isomers of hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right).\)

Draw a structural formula for each of the following. a. 3 -methylpentanoic acid b. ethyl methanoate c. methyl benzoate d. 3 -chloro- 2,4 -dimethylhexanoic acid

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