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

Question

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

Step-by-Step Solution

Answer

The final product \((Z)\) is propyne.

Step 1: Identify the first reaction

The first reaction is the conversion of \((\mathrm{CH}_{3})_{2}\mathrm{CHBr}\) using ethanolic KOH. Ethanolic KOH is a strong base that typically induces elimination reactions, resulting in the removal of HBr. The major product of this elimination is an alkene, specifically propene \((\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2)\).

Step 2: Identify the second reaction

The second reaction involves the treatment of propene \((\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2)\) with bromine \(\mathrm{Br}_2\). In the presence of \(\mathrm{Br}_2\), addition across the double bond occurs, producing 1,2-dibromopropane \((\mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{Br})\).

Step 3: Identify the third reaction

The third reaction utilizes \(\mathrm{NaNH}_2\) in liquid ammonia \(\mathrm{NH}_3\). This is a strong base that abstracts protons adjacent to bromine atoms, leading to the elimination of HBr. From 1,2-dibromopropane, elimination results in the formation of propyne \((\mathrm{CH}_3\mathrm{C}\equiv\mathrm{CH})\).

Step 4: Determine the final product

After applying the sequence of reactions: initial formation of propene from isopropyl bromide, conversion to 1,2-dibromopropane, and finally elimination to propyne, the final product \((Z)\) is identified as propyne.

Key Concepts

Elimination Reaction

An elimination reaction is a type of chemical reaction where two atoms or groups are removed from a molecule, often resulting in the formation of a double bond, or an alkene. In the case of a haloalkane like \((\mathrm{CH}_{3})_{2}\mathrm{CHBr}\), the reaction often involves the removal of a hydrogen atom and a halogen atom (bromine in this case). This process is facilitated by a strong base, such as ethanolic KOH.

• Strong bases like KOH donate electrons that help remove atoms from the molecule.
• The elimination results in the formation of water and the desired alkene product.
• This type of reaction decreases the saturation of the original molecule.

Elimination reactions are crucial in organic synthesis as they allow for the creation of more reactive molecules that can further react with other chemicals.

Alkene Formation

The formation of alkenes is a direct result of elimination reactions. When \((\mathrm{CH}_{3})_{2}\mathrm{CHBr}\) reacts with ethanolic KOH, the elimination of HBr occurs, leading to the creation of propene.

• Alkenes are hydrocarbons that contain at least one C=C double bond, making them unsaturated.
• The double bond is highly reactive and can undergo various types of addition reactions.
• In our example, propene, \((\mathrm{CH}_3\mathrm{CH} = \mathrm{CH}_2)\), is the alkene formed.

Alkenes are important intermediates in the synthesis of various organic compounds due to their reactive nature and ability to be transformed into other compounds.

Addition Reaction

An addition reaction is characterized by the addition of atoms or groups to a molecule, typically by breaking a double or triple bond. Following the formation of the alkene propene, the molecule undergoes an addition reaction with bromine (\(\mathrm{Br}_2\)).

• This leads to the addition of bromine atoms across the double bond in propene.
• The double bond in propene opens up to allow the formation of 1,2-dibromopropane \((\mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{Br})\).
• Addition reactions are often utilized to add functional groups to unsaturated molecules.

Addition reactions transform unsaturated molecules into saturated ones by adding atoms to make single bonds.

Halogenation

Halogenation involves the addition of halogens to organic compounds. This is a specific type of addition reaction where halogen molecules like \(\mathrm{Br}_2\) are added to molecules. In the case of propene, halogenation results in the formation of 1,2-dibromopropane.

• Halogenation increases the molecular weight of the compound by adding halogen atoms.
• The reaction with bromine proceeds readily with alkenes due to the rich electron cloud around the double bond.
• This reaction requires no special conditions, often occurring spontaneously when the reactants are mixed.

Halogenation is widely used in generating intermediates for polymer production and other chemical processes.

Propyne Formation

The final step in the sequence of reactions is the formation of propyne, a type of alkyne. This happens through an elimination reaction on the dibromo compound produced from halogenation. Using \(\mathrm{NaNH}_2\) in liquid ammonia as a strong base, \(\mathrm{CH}_3\mathrm{C} \equiv \mathrm{CH}\) forms by removing the bromine atoms and adjacent hydrogens.

• Alkynes like propyne are characterized by a carbon-carbon triple bond \(( \equiv )\), making them even less saturated than alkenes.
• Propyne, \((\mathrm{CH}_3\mathrm{C} \equiv \mathrm{CH})\), is considered more reactive due to the high electron density in the triple bond.
• This type of reaction is valuable in producing compounds used for further chemical transformations.

Through this reaction sequence, we see the transformation from a simple haloalkane to a versatile alkyne, highlighting the diverse nature of organic synthesis.