Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the given statements about the molecule \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2} \mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}=\mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) 1\. three carbons are sp \(^{3}\) hybridized 2, three carbons are sp \(^{2}\) hybridized 3\. two carbons are sp hybridized Of these statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) all are correct

Step-by-Step Solution

Answer

The answer is (d) all are correct.

Step 1: Identify hybridization of each carbon

Analyze the molecular structure: \[(CH_3)_2CH-CH=CH-C=C-CH=CH_2\]- The first two carbon atoms from the left are part of methyl groups (\(CH_3\)), which are connected by single bonds. Single-bonded carbon atoms are sp\(^3\) hybridized.- The carbon atom connected directly to the methyl groups has three hydrogens and one carbon surrounding it (4 single bonds), therefore is sp\(^3\) hybridized.- The next carbon atom is part of a double bond (\(-CH=CH-\)) and is sp\(^2\) hybridized.- Followed by another doubly-bonded carbon connected to a singly-bonded carbon and is also sp\(^2\) hybridized.- The fifth carbon atom then forms a triple bond (\(-C\equiv C-\)), this indicates sp hybridization.- Moving to the next carbon, also part of the triple bond, it's sp hybridized.- Finally, the double-bonded carbon has one hydrogen and is therefore sp\(^2\) hybridized.

Step 2: Count carbons for each hybridization

- **sp\(^3\) hybridized:** There are 3 carbons in total (two from the \(CH_3\) groups and one adjacent to the \(CH=CH\) segment).- **sp\(^2\) hybridized:** There are 3 carbons in total (two from the \(CH=CH\) segments and one from the \(CH=CH_2\) segment).- **sp hybridized:** There are 2 carbons in total from the \(C\equiv C\) segment.

Step 3: Compare with statements

Now that we know: - 3 carbons are sp\(^3\) hybridized,- 3 carbons are sp\(^2\) hybridized,- 2 carbons are sp hybridized,We can compare this with the given statements:1. Three carbons are sp\(^3\) hybridized: Correct.2. Three carbons are sp\(^2\) hybridized: Correct.3. Two carbons are sp hybridized: Correct.All three statements about the molecule are correct.

Key Concepts

sp3 Hybridization

In organic chemistry, understanding hybridization is crucial for comprehending molecular geometry and bonding properties. Let's start with **sp3 hybridization**. This happens when one 's' orbital combines with three 'p' orbitals from the same atom, creating four equivalent sp3 hybrid orbitals. These orbitals are oriented to form a tetrahedral geometry, which is common in carbon atoms that are single-bonded to other atoms or groups.

In the case of methane, for instance, the carbon atom is sp3 hybridized and bonds with four hydrogen atoms, forming a perfect tetrahedral shape. Each bond angle here is approximately 109.5°. Similarly, in the molecule given in the original exercise, the carbon atoms in the methyl groups \((CH_3)\) and the one connected to them with other single bonds demonstrate sp3 hybridization. Any carbon atom forming four sigma bonds with adjacent atoms is typically sp3 hybridized, ensuring a stable single-bond network. The 3D arrangement this hybridization provides contributes significantly to the molecule's properties and reactivity.

sp2 Hybridization

Next, let's explore **sp2 hybridization**, which is characteristic of carbon atoms involved in double bonds. In sp2 hybridization, one 's' orbital merges with two 'p' orbitals, forming three equivalent sp2 hybrid orbitals, and leaving one 'p' orbital unhybridized. This configuration typically leads to a planar geometry with bond angles of about 120° between the hybrid orbitals.

A classic example of sp2 hybridization is ethene (or ethylene), where each carbon atom forms two single bonds and one double bond with another carbon atom, utilizing its sp2 hybridized orbitals for the sigma bonds. Meanwhile, the unhybridized 'p' orbitals on the carbon atoms overlap side-to-side to form pi (π) bonds. Within the given molecular structure, carbon atoms in the \(-CH=CH-\) and \(=CH_2\) segments are sp2 hybridized. Their planar structure and ability to form rigidity in the molecular framework make these atoms essential for the molecule's stability.

sp Hybridization

Finally, we have the **sp hybridization**. This hybridization involves the combination of one 's' orbital with one 'p' orbital, resulting in two linearly oriented sp hybrid orbitals. This kind of hybridization is associated with carbon atoms that form triple bonds, as found in acetylene and certain parts of the given molecule structure.

The carbon-carbon triple bond in the molecule \((C\equiv C)\) is a classic example of sp hybridization. These carbon atoms form a linear geometry and have a bond angle of 180°. In a triple bond, one sigma bond is formed by the overlapping of sp orbitals, while two pi bonds are formed by the sideways overlap of the unhybridized 'p' orbitals. This configuration not only contributes to the linear shape but also gives the molecule unique properties such as significant rigidity and high bond energy. These features make sp hybridization a critical aspect of understanding reactivity in organic molecules.