Problem 166
Question
Underlined carbon is sp \(^{3}\) hybridized in (a) \(\mathrm{CH}_{3} \mathrm{CH}=\underline{\mathrm{CH}}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\)
Step-by-Step Solution
Verified Answer
The sp³ hybridized carbon is in compound (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \underline{\mathrm{NH}}_{2}\).
1Step 1: Understand sp³ hybridization
sp³ hybridization occurs when a carbon atom forms four sigma (σ) bonds, resulting in a tetrahedral geometry with bond angles of about 109.5°. It involves mixing one s orbital and three p orbitals to form four equivalent hybrid orbitals.
2Step 2: Analyze each compound
Identify if the underlined carbon in each compound forms four sigma bonds, which indicates sp³ hybridization: - For (a) \(\mathrm{CH}_{3} \mathrm{CH}=\underline{\mathrm{CH}}_{2}\), the underlined carbon is part of a double bond, suggesting sp² hybridization, not sp³.- For (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \underline{\mathrm{NH}}_{2}\), the underlined carbon forms single bonds with neighboring atoms, suggesting sp³ hybridization.- For (c) \(\mathrm{CH}_{3} \underline{\mathrm{CO}}\mathrm{NH}_{2}\), the underlined carbon is part of a carbonyl group, suggesting sp² hybridization.- For (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \underline{\mathrm{C}}\equiv\mathrm{N}\), the underlined carbon forms a triple bond with nitrogen, suggesting sp hybridization.
3Step 3: Match the condition
The only compound where the underlined carbon is sp³ hybridized is (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \underline{\mathrm{NH}}_{2}\), as all bonds to the underlined carbon are single bonds, consistent with sp³ hybridization.
Key Concepts
sp³ HybridizationSigma BondsTetrahedral Geometry
sp³ Hybridization
In chemistry, understanding hybridization is crucial to predict the shape and bonding behavior of molecules. When we talk about sp³ hybridization, we specifically refer to the mixing of one s orbital and three p orbitals in a carbon atom. This combination creates four identical sp³ hybrid orbitals.
This concept is vital because it leads to the formation of four sigma (σ) bonds around the carbon atom, giving it a distinctive geometry. In sp³ hybridization, each bond is a result of the head-on overlap of orbitals, creating a strong sigma bond.
An sp³ hybridized carbon leads to a molecule with a tetrahedral shape. Consequently, all bond angles measure approximately 109.5°, creating a three-dimensional structure. This configuration is standard in many organic molecules, reflecting the versatility and stability of carbon.
This concept is vital because it leads to the formation of four sigma (σ) bonds around the carbon atom, giving it a distinctive geometry. In sp³ hybridization, each bond is a result of the head-on overlap of orbitals, creating a strong sigma bond.
An sp³ hybridized carbon leads to a molecule with a tetrahedral shape. Consequently, all bond angles measure approximately 109.5°, creating a three-dimensional structure. This configuration is standard in many organic molecules, reflecting the versatility and stability of carbon.
Sigma Bonds
Sigma bonds (σ bonds) are the strongest type of covalent chemical bond. They form when two orbitals overlap head-on. In the context of sp³ hybridization, these bonds are formed between hybridized orbitals of carbon and other atoms.
- They allow for the free rotation of bonded atoms around the bond axis.
- Each sp³ hybridized carbon forms four sigma bonds, contributing to the stability and geometry of the molecule.
Tetrahedral Geometry
When a carbon atom undergoes sp³ hybridization, it acquires a tetrahedral geometry. This unique shape occurs because the atom forms four identical sigma bonds that repel each other equally, leading to a symmetrical shape.
- Tetrahedral geometry involves bond angles of about 109.5°.
- It allows for the maximum spatial distribution of bonded valence electrons around the carbon atom.
Other exercises in this chapter
Problem 162
The number of geometrical isomers of the following alkene \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}\
View solution Problem 165
How many structural isomers are possible for monobromodichloro benzene?
View solution Problem 167
The incorrect IUPAC name is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOC}_{2} \mathrm{H}_{5}\) ethyl butanoate (b) CC(C)CC=O 3-methyl buta
View solution Problem 168
Racemic mixture is formed by mixing two (a) isomeric compounds (b) chiral compounds (c) meso compounds (d) enantiomers with chiral carbon
View solution