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Match the following $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) } \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^{\oplus} & \text { (p) } \begin{array}{l} \mathrm{sp}^{3} \text { (carbon carry- } \\ \text { ing the charge) } \end{array} \\ \begin{array}{ll} \text { (b) } \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C} \mathrm{H}_{2} & \text { (q) } \mathrm{sp}^{2} \text { (carbon carry- } \\ \text { ing the charge) } \end{array} \\ \text { (c) } \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2} & \text { (r) Nucleophile } \\ \begin{array}{ll} \text { (d) } \mathrm{CH}_{3}-\mathrm{C}^{+}=\mathrm{O} & \text { (s) Electrophile } \\ & \text { (t) Planar } \\ \hline \end{array} \end{array} $$

Short Answer

Expert verified
(a) - (q), (b) - (p), (c) - (r), (d) - (s, t)

Step by step solution

01

Analyze the Structure of (a)

The ion \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^{\oplus} \) involves a carbon with a positive charge (carbocation) at the end of the chain. A carbon atom carrying a positive charge (a carbocation) typically has an \( \mathrm{sp}^{2} \) hybridization because one of the \( \mathrm{sp}^{3} \) orbitals becomes vacant. Match (a) with Column-II (q) \( \mathrm{sp}^{2} \).
02

Determine the Hybridization of (b)

In \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C} \mathrm{H}_{2} \), there is no formal charge on any carbon. The molecule is neutral and the carbon atoms are \( \mathrm{sp}^{3} \) hybridized because they form four sigma bonds. Therefore, (b) corresponds to Column-II (p) \( \mathrm{sp}^{3} \).
03

Identify the Nature of (c)

The group \( \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2} \) indicates a benzyl group, which can participate in reactions as a nucleophile due to the presence of the electron-rich benzene ring. Therefore, (c) matches with Column-II (r) Nucleophile.
04

Determine the Properties of (d)

For the ion \( \mathrm{CH}_{3}-\mathrm{C}^{+}=\mathrm{O} \), the carbon with the positive charge (carbocation) is connected to an oxygen atom through a double bond, likely giving it \( \mathrm{sp}^{2} \) hybridization as well, suggesting that it is planar. The presence of the positive charge also makes it an electrophile. Therefore, (d) corresponds with Column-II (s) Electrophile and (t) Planar.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbocation Hybridization
Carbocations are carbon atoms that carry a positive charge. When a carbon becomes a carbocation, its electron configuration changes, which alters its hybridization. Typically, a neutral carbon atom forms four sigma bonds, making it \(\text{sp}^3\) hybridized. However, when it carries a positive charge, as seen in carbocations, one of these hybrid orbitals becomes vacant.This leads to \(\text{sp}^2\) hybridization, where the carbon forms three sigma bonds and one p-orbital remains empty. With \(\text{sp}^2\) hybridization:
  • The carbon atom forms a planar structure.
  • It features a 120-degree bond angle.
  • It has three \(\text{sp}^2\) hybrid orbitals and one unoccupied p-orbital.
Understanding this hybridization is crucial, as it impacts the stability and reactivity of carbocations. Factors such as hyperconjugation, resonance, and inductive effects can all influence the stability of these species.
Nucleophilic Reaction
Nucleophiles are species that donate a pair of electrons to an electrophile. This characteristic makes them rich in electron density, allowing them to form new chemical bonds in reactions. Often, nucleophiles have lone pairs or \(\pi\)-electron systems, like the benzyl group.In nucleophilic reactions:
  • The nucleophile attacks an electron-deficient atom or group, often a positively charged or neutral atom with an empty orbital.
  • This leads to the formation of a new covalent bond.
  • Such reactions are crucial in substitution reactions, like \(\text{S}_\text{N}2\) and \(\text{S}_\text{N}1\).
For example, the benzyl group \(\text{C}_6\text{H}_5\text{-CH}_2\) can act as a nucleophile, using its electron-rich aromatic ring to form new bonds. Nucleophiles play key roles in organic synthesis, helping to create more complex molecular architectures.
Electrophilic Reaction
Electrophiles are eager to accept electrons. They are electron-deficient and seek out nucleophiles to achieve stable electronic configurations, making them essential players in many chemical reactions.In electrophilic reactions:
  • An electrophile accepts a pair of electrons from a nucleophile to form a new covalent bond.
  • Reactions often involve carbocations, as they are excellent electrophiles due to their positive charge.
  • This kind of reaction is commonly seen in electrophilic substitution and addition mechanisms.
Carbocations like \(\text{CH}_3\text{-C}^+=\text{O}\) can serve as electrophiles, using their positive charge to attract nucleophiles. Understanding electrophilic reactions is crucial for manipulating reaction pathways, enabling chemists to synthesize desired compounds efficiently.

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Most popular questions from this chapter

In the following benzyl/allyl system or [R]C=C [R]c1ccccc1 ( \(\mathrm{R}\) is alkyl group) decreasing order of inductive effect is (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}->\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}->\mathrm{CH}_{3} \mathrm{CH}_{2}-\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}->\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}->\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\) (c) \(\left.\mathrm{CH}_{3}\right)_{2} \mathrm{CH}->\mathrm{CH}_{3} \mathrm{CH}_{2} \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}->\mathrm{CH}_{3} \mathrm{CH}_{2}->\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\)

Which of the following statements are correct. (1) \(\overline{\mathrm{N}} \mathrm{H}_{2}\) is better nucleophile than \(\mathrm{NH}_{3}\) but latter \(\left(\mathrm{NH}_{3}\right)\) is better nucleophile than \(\mathrm{NH}_{4}^{+}\) (2) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}^{-}\)is better nucleophile than CC(=O)O (3) \(\mathrm{OH}^{-}\)is better nucleophile than \(\mathrm{SH}^{-}\)and \(\mathrm{H}_{2} \mathrm{O}\), but \(\mathrm{H}_{2} \mathrm{O}\) is better nucleophile than \(\mathrm{H}_{3} \mathrm{O}^{+}\) (4) \(\mathrm{ClO}^{-1}\) is weaker nucleophile than \(\mathrm{ClO}_{4}^{-}\) (a) 1,2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) \(1,2,3\) and 4

Match the following: \(\begin{array}{ll}\text { List I } & \text { List II }\end{array}\) (Type of reaction) (Phenomenon) 1\. \(\mathrm{SN}_{1}^{2}\) (i) Walden inversion 2\. SN (ii) Carbanion intermediate 3\. \(\mathrm{E}_{2}\) (iii) Anti-periplanar configuration 4\. \(\mathrm{E}_{1 \mathrm{cc}}\) (iv) Carbocation intermediate The correct matching is: \(1 \quad 2 \quad 3 \quad 3\) (a) (i) (iv) (iii) (ii) (b) (i) (ii) (iii) (iv) (c) (iii) (ii) (i) (iv) (d) (iv) (iii) (ii) (i)

The correct order of increasing acid strength of the compound: (1) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (2) \(\mathrm{MeOCH}_{2} \mathrm{CO}_{2} \mathrm{H}\) (3) \(\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}\) (4) \((\mathrm{Me})_{2}-\overline{\mathrm{CO}_{2}} \mathrm{H}\) (a) \(2<4<1<3\) (b) \(4<1<3<2\) (c) \(4<1<2<3\) (d) \(1<4<3<2\)

The correct order of basicity of the following compound is (1) CC(=N)N (2) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (3) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (4) CC(N)=O (5) \(\mathrm{CH}_{3} \mathrm{CN}\) (a) \(2>1>3>4>5\) (b) \(1>3>2>4>5\) (c) \(3>1>2>5>4\) (d) \(1>3>2>5>4\).

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