Question

In the following reactions: (a) ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_3\stackrel{\mathrm{NCS}}{⟶}$ (b) ${\mathrm{CH}}_2^{14}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{OH}\stackrel{{\mathrm{soCl}}_2}{⟶}$ If ${\mathrm{CH}}_2^{14}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{Cl}$ is $\mathrm{P}$ and ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2\mathrm{Cl}$ is $\mathrm{Q}$, then major product obtained is (A) Pin both reaction (B) P in reaction (a) and $\mathrm{Q}$ in reaction (b) (C) P in reaction (b) and Q in reaction (a) (D) $\mathrm{P},\mathrm{Q}$ in reaction (a) and $\mathrm{Q}$ in reaction (b)

Short answer

The short answer is: The major product obtained is (C) P in reaction (b) and Q in reaction (a).

Step-by-step solution

Reaction (a):

For the first reaction, we have: ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_3$ with $\mathrm{NCS}$ as the reagent. This is an electrophilic addition reaction of a nucleophile (NCS) to an alkene, where the $NC{S}^{-}$ ion will add across the double bond. The major product in this reaction will be the one following Markovnikov's rule, where the nucleophile will attack the less substituted carbon atom in the alkene. Thus, the major product should be: ${\mathrm{CH}}_2(\mathrm{NCS})-\mathrm{CH}-{\mathrm{CH}}_3$ Now, let's compare this product with P and Q: P: ${\mathrm{CH}}_2^{14}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{Cl}$ Q: ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2\mathrm{Cl}$ Neither P nor Q matches the major product obtained in reaction (a).

Reaction (b):

Now, let's analyze the second reaction, which is: ${\mathrm{CH}}_2^{14}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{OH}$ with ${\mathrm{SOCl}}_2$ as the reagent. This reaction involves the conversion of an alcohol ($\mathrm{OH}$ group) to an alkyl chloride ($\mathrm{Cl}$ group) using thionyl chloride (${\mathrm{SOCl}}_2$) as the reagent. The reaction is straightforward; the $\mathrm{OH}$ group will be replaced by a $\mathrm{Cl}$ group: ${\mathrm{CH}}_2^{14}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{Cl}$ Comparing this product with P and Q: P: ${\mathrm{CH}}_2^{14}=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{Cl}$ Q: ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2\mathrm{Cl}$ The major product obtained in reaction (b) matches with P. Hence, the correct option is (C) P in reaction (b) and Q in reaction (a).

Key concepts

Markovnikov's Rule

Markovnikov's rule plays a critical role in understanding how reagents add across the double bonds in alkenes during chemical reactions. Simply put, it predicts that when a protic acid, like hydrogen bromide (HBr), reacts with an asymmetric alkene, the acid's hydrogen (H) will attach itself to the carbon with the most hydrogen atoms already present. Meanwhile, the halide (such as Br) will bond to the carbon with fewer hydrogen atoms.

This preference occurs because the most stable carbocation intermediate is formed on the more substituted carbon, leading to more stable products. In a classroom setting, this might be visualized using molecular models or illustrated by drawing potential carbocation intermediate structures. Knowing Markovnikov's rule is essential when predicting the major product of an electrophilic addition reaction which is the basis of many JEE Organic Chemistry problems.

Electrophilic Addition Reaction

When it comes to organic chemistry, electrophilic addition reactions are a staple, especially in the synthesis of new molecules from alkenes. These reactions involve an electrophile, which, by definition, is a species with a strong affinity for electrons. The alkene, with its electron-rich double bond, acts like a magnet for electrophiles, allowing the reaction to proceed.

Typically, the double bond opens up, and two new bonds form: one with the electrophile and another with a nucleophile, which could be a part of the electrophilic reagent or another species in the reaction mixture. This makes alkenes highly versatile for creating a variety of functional groups, including alcohols, halides, and others. A classic example that might be discussed in JEE prep sessions is the addition of hydrohalic acids (like HBr) to alkenes, which proceeds through a carbocation intermediate.

Alcohol to Alkyl Chloride Conversion

In converting an alcohol to an alkyl chloride, several reagents can be employed, but one of the most efficient methods is using thionyl chloride (SOCl2). This chemical offers a clean and straightforward transformation where the OH group in the alcohol is replaced by a chlorine atom.

Noteworthy is the mechanism of this reaction, known as nucleophilic substitution, which typically follows an SN2 pathway for primary alcohols and an SNi pathway for secondary or tertiary ones. It's an excellent illustration of the adaptability of nucleophilic substitution reactions in different chemical environments. The ease of this transformation makes it a useful tool for students tackling synthetic challenges in their JEE Organic Chemistry syllabus, providing a straightforward way to convert between functional groups and diversify molecular structures.