Problem 111 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction sequence, \(\mathrm{Z}\) will be Glycerol \(\stackrel{\mathrm{KHSO}_{4} / \Delta}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{Zn}-\mathrm{Hg} / \mathrm{conc} \mathrm{HCl} / \Delta}{\longrightarrow}\) \(\mathrm{Y} \longrightarrow{\mathrm{NBS} / \mathrm{CCl}_{4}}{\longrightarrow} \mathrm{Z}\) (a) 1,2 -dibromopropane (b) 1-bromopropane (c) 2-bromopropane (d) 3-bromopropene

Step-by-Step Solution

Answer

The final product \( \mathrm{Z} \) is 1-bromopropane.

Step 1: Dehydration of Glycerol

Glycerol undergoes dehydration in the presence of potassium hydrogen sulfate (\( \mathrm{KHSO}_{4} \)) and heat (\( \Delta \)) to form acrolein (\( \mathrm{CH}_{2}=\mathrm{CHCHO} \)). In this step, glycerol loses a molecule of water to form an unsaturated aldehyde known as acrolein.

Step 2: Clemmensen Reduction

The acrolein (\( \mathrm{CH}_{2}=\mathrm{CHCHO} \)) is then reduced using zinc amalgam (\( \mathrm{Zn}-\mathrm{Hg} \)) and concentrated hydrochloric acid (\( \mathrm{conc.} \: \mathrm{HCl} \)) under heat to yield propanal (\( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CHO} \)). The \( \mathrm{C=O} \) group is converted to \( \mathrm{CH_{2}} \), forming an aldehyde with one fewer oxygen atom.

Step 3: Formation of Alkyl Bromide

Propanal (\( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CHO} \)) is subjected to an allylic bromination using N-bromosuccinimide (NBS) and carbon tetrachloride (\( \mathrm{CCl}_{4} \)). In this reaction, a bromine is added to the position next to the terminal carbon (allylic position) to produce 1-bromopropane (\( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Br} \)), making \( \mathrm{Z} \) the desired product.

Key Concepts

Dehydration of Glycerol

The dehydration of glycerol is a key chemical transformation that involves the removal of water molecules from glycerol to form a new compound. In this process, glycerol, which is a trihydroxy alcohol (\( \mathrm{C_3H_8O_3} \)), is treated with potassium hydrogen sulfate (\( \mathrm{KHSO_4} \)) and heat. This leads to the breakdown of glycerol into acrolein (\( \mathrm{CH_2=CHCHO} \)), a simple unsaturated aldehyde.

The mechanism begins with the protonation of one of the hydroxyl groups in glycerol. This is followed by a series of eliminations where water is removed sequentially from glycerol. This dehydration is facilitated by the acidic environment produced by \( \mathrm{KHSO_4} \). The final product, acrolein, retains a double bond and an aldehyde group, which are important for subsequent reactions. Understanding this conversion is crucial for students studying organic mechanisms, as it highlights how alcohols can be transformed into unsaturated aldehydes through dehydration.

• Glycerol: \( \mathrm{C_3H_8O_3} \)
• Acrolein (product): \( \mathrm{CH_2=CHCHO} \)
• Key reagent: \( \mathrm{KHSO_4} \)
• Temperature requirement: heat (\( \Delta \))

Clemmensen Reduction

Clemmensen reduction is an important reaction in organic chemistry used to transform carbonyl groups into hydrocarbons. In the context of this exercise, acrolein (\( \mathrm{CH_2=CHCHO} \)) undergoes Clemmensen reduction to form propanal.\( \mathrm{Zn-Hg} \) amalgam and concentrated hydrochloric acid (\( \mathrm{conc. HCl} \)) under heat act as the reducing agents in this transformation. The reduction converts the carbonyl group (\( \mathrm{C=O} \)) of acrolein into a methylene group (\( \mathrm{CH_2} \)), thereby forming propanal (\( \mathrm{CH_3CH_2CHO} \)).

This reduction is highly effective for acid-sensitive compounds since it uses an acidic medium. It's a vital reaction for students to understand because it highlights how ketones and aldehydes can be converted to alkanes, paving the way for various synthetic routes in organic chemistry. Clemmensen reduction is particularly useful when other reduction methods, such as catalytic hydrogenation, could affect other sensitive functional groups present in the molecule.

• Acrolein: \( \mathrm{CH_2=CHCHO} \)
• Propanal (product): \( \mathrm{CH_3CH_2CHO} \)
• Reduction agents: \( \mathrm{Zn-Hg} \) and \( \mathrm{conc. HCl} \)

Allylic Bromination

The final transformation in this sequence is allylic bromination, a method to selectively introduce bromine to an allylic carbon. This reaction employs N-bromosuccinimide (NBS) in carbon tetrachloride (\( \mathrm{CCl_4} \)). The allylic position is the carbon atom adjacent to a double bond, which makes it particularly reactive due to the potential for resonance stabilization.

In propanal (\( \mathrm{CH_3CH_2CHO} \)), the hydrogen atoms adjacent to the terminal alkene can be replaced by a bromine atom to form 1-bromopropane (\( \mathrm{CH_3CH_2CH_2Br} \)). NBS is preferred for these reactions because it provides a controlled, steady release of bromine, which helps in selectively attacking the allylic position without affecting other parts of the molecule.

Understanding allylic bromination is important for students because it introduces them to selective bromination reactions and highlights the utility of using radical mechanisms to achieve specificity in organic transformations.

• Propanal: \( \mathrm{CH_3CH_2CHO} \)
• Product: 1-bromopropane (\( \mathrm{CH_3CH_2CH_2Br} \))
• Key reagents: NBS and \( \mathrm{CCl_4} \)