Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following sequence of reactions \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\text { Ethanolic KOH }}{\longrightarrow}(\mathrm{X}) \stackrel{\mathrm{Br}_{2}}{\longrightarrow}(\mathrm{Y})\) \(\frac{\mathrm{NaNH}_{2}}{\mathrm{liq} \cdot \mathrm{NH}_{3}}\) The end product \((Z)\) is (a) propane (b) propyne (c) propene (d) propan-2-al

Step-by-Step Solution

Answer

The end product (Z) is propyne (b).

Step 1: Identify the First Reaction

The starting compound is 2-bromo-propane, \ ((\mathrm{CH}_{3})_{2}\mathrm{CHBr}). It is treated with ethanolic KOH. This is a dehydrohalogenation reaction, where the bromine is removed along with a hydrogen atom from an adjacent carbon, forming a double bond. The product of this reaction is propene: \[(\mathrm{CH}_{3})_{2}\mathrm{C} = \mathrm{CH}_{2}\] This compound is indicated as (X).

Step 2: Bromination of Propene

Propene (X) reacts with \(\mathrm{Br}_{2}\) to form a dibromo compound by an addition reaction. This involves the addition of bromine across the double bond. The product is 1,2-dibromopropane:\[\mathrm{BrCH}_{2}\mathrm{CHBrCH}_{3}\]This compound is referred to as (Y).

Step 3: Dehydrohalogenation Using Sodium Amide

The compound (Y) is then treated with sodium amide \((\mathrm{NaNH}_{2})\) in liquid ammonia. \(\mathrm{NaNH}_{2}\) is a strong base that causes dehydrohalogenation, removing the two bromine atoms as hydrogen bromide \((\mathrm{HBr})\). This results in the formation of a triple bond, yielding propyne:\[\mathrm{HC} \equiv \mathrm{CCH}_{3}\]This alkyne is the final product, indicated as (Z).

Key Concepts

Dehydrohalogenation

Dehydrohalogenation is a chemical reaction used to remove a hydrogen halide (such as HBr) from organic molecules. The reaction often results in the formation of a carbon-carbon double bond, turning an alkyl halide into an alkene. In the exercise, 2-bromo-propane undergoes dehydrohalogenation through the use of ethanolic KOH, where KOH serves as a base to facilitate the removal of hydrogen and bromine from adjacent carbons. This process is critical because it leads to the formation of propene, a simpler alkene.
This type of reaction is important in synthetic organic chemistry as it allows for the transformation of saturated hydrocarbons into their unsaturated counterparts, enabling further chemical modifications. Dehydrohalogenation is characterized by:

• The consumption of a base, such as KOH or NaNH₂, to remove hydrogen halides.
• The resultant formation of a double bond in the product molecule.
• The reaction often involves heat or reflux to shift the equilibrium towards alkene formation.

Understanding dehydrohalogenation helps students grasp how minor changes in molecular structure can significantly alter the chemical properties and reactivity of organic compounds.

Bromination

Bromination involves the addition of bromine (\( Br_2 \)) to an organic compound, often across a double bond. For propene (\((\mathrm{CH}_{3})_{2}C = \mathrm{CH}_{2}\)), bromination is an addition reaction where both bromine atoms add across the double bond, converting the alkene into a dibromoalkane. This specific reaction is useful in extending the framework of the molecule by increasing the atoms directly attached to the carbon backbone.
The propene reacts with \( Br_2 \) in the following manner, transforming into 1,2-dibromopropane:

• The \( Br_2 \) molecule approaches the double bond in propene, causing the bond to break and a temporary cyclic bromonium ion to form.
• This ring opens as the second bromine atom attacks the intermediate, leading to the final dibromo compound.

Bromination is an example of electrophilic addition, where the electrophile \( Br_2 \) seeks electron-rich areas like double bonds to react. It’s a common step in preparing compounds for further transformations such as elimination reactions.

Alkyne Formation

The formation of an alkyne from a dibromoalkane involves a second dehydrohalogenation reaction. In our reaction sequence, sodium amide (\( \mathrm{NaNH}_2 \)) is used in liquid ammonia (\( \mathrm{NH}_{3} \)) to perform a strong elimination. Sodium amide is a potent base and can remove hydrogen bromide twice from the dibromo compound, resulting in the formation of an alkyne with a carbon-carbon triple bond.
This final transformation from 1,2-dibromopropane to propyne (\( \mathrm{HC} \equiv \mathrm{CCH}_3 \)) follows several principles of organic chemistry:

• The reaction conditions are deliberately harsh to disrupt stable bonds and allow for the formation of the triple bond.
• This process shifts the molecule from a saturated state (with single bonds only) to a highly unsaturated one, increasing reactivity for future chemical manipulations.
• The triple bond gives the alkyne distinct physical and chemical properties, such as higher acidity of terminal hydrogen atoms, compared to alkanes and alkenes.

Alkyne formation is crucial in synthetic pathways because alkynes serve as key intermediates in creating complex organic molecules.