Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify \(Z\) in the following series: \(\mathrm{CH}_{2}=\mathrm{CH}_{2} \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{X} \stackrel{\text { Hydrolysis }}{\rightarrow} \mathrm{Y} \stackrel{\mathrm{I}_{2} \text { Excess }}{\longrightarrow} \mathrm{Z}\) (a) \(\mathrm{CHI}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Step-by-Step Solution

Answer

The compound Z is \(\mathrm{CHI}_{3}\).

Step 1: React Ethylene with HBr

The reaction between ethylene, \( \mathrm{CH}_{2}=\mathrm{CH}_{2} \), and HBr follows Markovnikov's rule, where the bromine atom attaches to the more substituted carbon. This converts ethylene into bromoethane (\( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Br} \)). Let this intermediate be compound X.

Step 2: Hydrolysis of Bromoethane

Bromoethane undergoes hydrolysis, where the bromine atom is replaced by an -OH group, converting bromoethane into ethanol (\( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \)). This intermediate is compound Y.

Step 3: React Ethanol with Excess Iodine

The reaction of ethanol with excess iodine, in the presence of a base, leads to the formation of iodoform. However, ethanol (\( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \)) doesn't form iodoform directly. Instead, ethanol is oxidized first to acetaldehyde (\( \mathrm{CH}_{3}\mathrm{CHO} \)), which then reacts with iodine to form iodoform (\( \mathrm{CHI}_{3} \)). Thus, the final product Z is iodoform.

Key Concepts

Markovnikov's rule

Markovnikov's rule is an important guideline used in organic chemistry to predict the outcome of addition reactions of alkenes. In the case of ethylene reacting with hydrogen bromide (HBr), Markovnikov's rule helps us determine where the bromine atom will attach. According to this rule, when a protic acid (like HBr) is added to an unsymmetrical alkene, the acid's hydrogen prefers to bind to the carbon with fewer alkyl substituents. This results in the halogen (bromine) attaching to the carbon with more substituents.
For ethylene (\[\mathrm{CH}_{2}=\mathrm{CH}_{2}\]), both carbons have the same number of substituents initially. However, as the reaction proceeds, the intermediate formed will stabilize to form bromoethane (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Br}\]). Markovnikov's rule helps explain why the bromine attaches to the end carbon, forming this specific product. This is a fundamental concept in predicting the outcomes of many addition reactions in organic chemistry.

Hydrolysis

Hydrolysis is a chemical process where a compound reacts with water, leading to the breaking of bonds and often the formation of new substances. This reaction is common in organic transformations, such as when you convert bromoethane into ethanol.
In the hydrolysis of bromoethane (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Br}\]), the bond between bromine and the ethyl group is broken. Water (\[\mathrm{H}_{2}\mathrm{O}\]) supplies an -OH group to the ethyl group, resulting in the formation of ethanol (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\]). As a result, the bromine atom is typically released as hydrobromic acid (\[\mathrm{HBr}\]), though in practice it might be neutralized or captured by other ions in solution. Hydrolysis is fundamental in organic synthesis for transforming one functional group into another.

Iodoform Reaction

The iodoform reaction is a notable test for the presence of certain alcohols and ketones containing a methyl group next to a carbonyl group or a secondary alcohol group. It's particularly used to identify ethanol and secondary alcohols, which can oxidize to corresponding ketones or aldehydes.
In the context of ethanol oxidation, the reaction begins with ethanol (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\]), which is first oxidized to acetaldehyde (\[\mathrm{CH}_{3}\mathrm{CHO}\]). This intermediate is crucial, as acetaldehyde can proceed with iodoform reaction when treated with iodine and a base. The final step converts acetaldehyde to yellow crystalline iodoform (\[\mathrm{CHI}_{3}\]). The reaction is not only a functional group transformation but also serves as a qualitative test in identification procedures.

Bromoethane

Bromoethane is a simple alkyl halide that plays a critical role in organic chemistry, particularly as an intermediate in various reactions. It is produced through the addition of HBr to ethylene, following Markovnikov's rule.
This compound, formally known as ethyl bromide (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Br}\]), acts as an excellent starting material for further synthesis. In organic reactions, bromoethane can participate in nucleophilic substitution reactions, where the bromine atom can be readily exchanged with a nucleophile like hydroxide ions (\[\mathrm{OH}^{-}\]) in hydrolysis. This versatility makes bromoethane a staple compound in both laboratory synthesis and various industrial applications.

Ethanol

Ethanol is a prominent alcohol in organic chemistry, often recognized for its versatility both as a solvent and as a reactant. It naturally arises during the hydrolysis of bromoethane, where the -Br group is substituted by -OH.
This simple alcohol (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\]) is crucial in many reactions. Being a primary alcohol, ethanol's oxidation pathway leads to the production of acetaldehyde, which is a precursor in various industrial applications. Moreover, ethanol's participation in the iodoform reaction underscores its reactivity, particularly in reactions involving excess iodine and bases. Alongside its uses in synthesis, ethanol's everyday roles include use as fuel, an antiseptic, and a recreational substance due to its psychoactive properties. Its ubiquitous presence in both chemical and non-chemical contexts makes understanding ethanol essential for any study of organic chemistry.