Problem 199 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following reactions will yield 2,2 dibromopropane? \(\quad\) (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHBr}+\mathrm{HBr} \longrightarrow\) (b) \(\mathrm{CH}=\mathrm{CH}+2 \mathrm{HBr} \longrightarrow\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow\) (d) \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH}+2 \mathrm{HBr}\)

Step-by-Step Solution

Answer

Reaction (d) will yield 2,2-dibromopropane.

Step 1: Analyze Product Formation

To form 2,2-dibromopropane, we need a reaction where a compound with three carbon atoms gains two bromine (Br) atoms at the second carbon atom. The structure of 2,2-dibromopropane is: CH3-C(Br)(Br)-CH3.

Step 2: Evaluate Each Reaction Option

Examine each option to determine if the reaction will result in the formation of 2,2-dibromopropane. (a) The reaction involves an alkene with a bromo group. Adding HBr once more will not result in the desired dibromo product. (b) The reactant has a terminal triple bond. The addition of 2 mol of HBr forms CH2Br-CH2Br, not 2,2-dibromopropane. (c) The reactant is an alkene and can undergo Markovnikov addition with HBr once to form 2-bromopropane, but not the dibromo derivative. (d) The reactant is a terminal alkyne. Exposing it to 2 mol of HBr will result in the desired structure, as both Br will add to the second carbon, forming CH3-C(Br)(Br)-CH3.

Step 3: Confirm the Correct Reaction

From the analysis, option (d) involves the addition of 2 mol of HBr to propyne (CH3-C≡CH), which will add both bromine atoms to the middle carbon atom forming 2,2-dibromopropane. The reaction proceeds via the mechanism of electrophilic addition typical for alkynes.

Step 4: Conclusion

The correct reaction is (d) CH3-C≡CH + 2 HBr → CH3-C(Br)(Br)-CH3, which yields 2,2-dibromopropane.

Key Concepts

Electrophilic Addition

Electrophilic addition is a fundamental reaction in organic chemistry that involves the addition of an electrophile to a compound with a double or triple bond. This process generally starts with an electron-rich site attacking an electron-poor electrophile, leading to the breaking of the pi bond and the formation of new sigma bonds.

This type of reaction is crucial when dealing with alkenes and alkynes, as it details how these compounds can interact with various reagents.

• An electrophile, often positively charged or with a significant dipole, seeks out the electron-rich regions of the molecule.
• The pi bond electrons from the alkene or alkyne are used to form a new bond with the electrophile, breaking the pi bond in the process.
• After the initial bond is formed, a second step typically involves another molecule or ion attacking what's left of the substrate, completing the addition process.

Understanding this concept is key to predicting and explaining the outcome of reactions like those in the original exercise, where propyne reacts with HBr.

Markovnikov Addition

Markovnikov's rule is an essential guideline used to determine the product distribution in addition reactions. When an alkene or alkyne reacts with a hydrogen halide, such as HBr, Markovnikov's rule predicts that the hydrogen atom will attach to the carbon with more hydrogen atoms, while the halide (like Br) will attach to the carbon with fewer hydrogen atoms.

This rule is important when predicting the products of electrophilic additions.

• In practice, it means adding the hydrogen portion of HBr to the less substituted carbon (having more hydrogens), and the bromine is added to the more substituted carbon (having fewer hydrogens), which helps stabilize the carbocation formed during the reaction.
• The rule is particularly useful in determining product stability since it guides the position of the new substituent, generally leading to a more stable carbocation intermediate.

In the context of the exercise, Markovnikov addition ensures bromine attaches to the strategically important carbon in forming 2,2-dibromopropane.

Alkynes

Alkynes are hydrocarbon compounds characterized by at least one carbon-carbon triple bond. This triple bond makes alkynes highly reactive and a common subject for electrophilic addition reactions.

Unlike alkenes, alkynes can undergo two successive additions of reagents like HBr because their triple bond contains two weak pi bonds available for reaction.

• The first addition of the reagent breaks one of the pi bonds, forming a vinyl halide.
• A second addition can then break the remaining pi bond, if conditions allow, resulting in the formation of a dibromo compound, much like the formation of 2,2-dibromopropane in the exercise.

Alkynes provide a versatile platform for creating a variety of key intermediates and products in organic synthesis.

Dibromo Compounds

Dibromo compounds are molecules that contain two bromine atoms, often resulting from the addition of bromine across an unsaturated bond. These compounds are significant in organic synthesis as intermediates or end products.

In the context of electrophilic addition, forming a dibromo compound typically involves reactions with sufficient bromine sources or through successive halogenation processes.

• This involves adding two bromine atoms originally associated with two bonds in the context of alkenes, or a sequential addition in the case of alkynes.
• For instance, in the exercise discussed, exposing propyne to two moles of HBr undergoes an addition reaction that results in the formation of 2,2-dibromopropane, a classic example of a dibromo compound.

Such transformations highlight the importance and utility of dibromo compounds in organic chemistry pathways.