Question

Which of the following compounds do not have \(sp^3\) carbon atom(s)?
I. Acetone II. Acetic
acid III. Buta-1,3-diene IV. Propyne V. Naphthalene

• A. I, II only
• B. II, III only
• C. IV, V only
• D. III, V only

Hint:

Hybridization of carbon can be identified from the number of \(\sigma\)-bonds: \[ sp^3 \rightarrow 4 \sigma \text{ bonds} \] \[ sp^2 \rightarrow 3 \sigma \text{ bonds} \] \[ sp \rightarrow 2 \sigma \text{ bonds} \] Aromatic ring carbons and alkene carbons are generally \(sp^2\) hybridized.

Answer 1

The Correct Option is D

Step 1: Examine Acetone \((CH_3COCH_3)\).
The structure of acetone is \[ CH_3-CO-CH_3 \] The carbonyl carbon \((C=O)\) is \[ sp^2 \] hybridized, whereas both methyl carbon atoms are \[ sp^3 \] hybridized.
Hence, acetone contains \(sp^3\) carbon atoms.

Step 2: Examine Acetic acid \((CH_3COOH)\).
The structure is \[ CH_3-COOH \] The carboxyl carbon is \[ sp^2 \] hybridized, while the methyl carbon is \[ sp^3 \] hybridized.
Hence, acetic acid contains \(sp^3\) carbon atoms.

Step 3: Examine Buta-1,3-diene \((CH_2=CH-CH=CH_2)\).
In buta-1,3-diene, every carbon atom participates in a double bond.
Therefore, all four carbon atoms are \[ sp^2 \] hybridized.
Hence, there is no \(sp^3\) carbon atom present.

Step 4: Examine Propyne \((CH_3-C\equiv CH)\).
The methyl carbon atom is \[ sp^3 \] hybridized.
The two carbon atoms involved in the triple bond are \[ sp \] hybridized.
Therefore, propyne contains one \(sp^3\) carbon atom.

Step 5: Examine Naphthalene \((C_{10}H_8)\).
Naphthalene consists of two fused benzene rings.
Every carbon atom in the aromatic ring system is \[ sp^2 \] hybridized.
Hence, naphthalene contains no \(sp^3\) carbon atom.

Step 6: Identify the correct compounds.
Compounds having no \(sp^3\) carbon atoms are III. Buta-1,3-diene and V. Naphthalene

Step 7: Final conclusion.
Therefore, \[ \boxed{\text{III, V only}} \] Hence, the correct option is \[ \boxed{(4)} \]