Question

Which of the following organic compounds will exhibit Geometrical Isomerism ?

• A. Butanol-2
• B. Butene-1
• C. Butene-2
• D. Butyne-1

Hint:

For geometrical (cis-trans) isomerism in alkenes (compounds with \(C=C\)): 1. Find the \(C=C\) double bond. 2. Look at each carbon of the double bond separately. 3. Each carbon of the double bond MUST have {two different groups} attached to it. Example: For \(R_1R_2C=CR_3R_4\), you need \(R_1 \neq R_2\) AND \(R_3 \neq R_4\). % Option (c) 1-Butene (\(CH_2=CH-CH_2CH_3\)): First carbon has two H's (same) \(\rightarrow\) No. % Option (d) 2-Butene (\(CH_3-CH=CH-CH_3\)): Each double-bonded carbon has one H and one \(CH_3\) (different) \(\rightarrow\) Yes.

Answer 1

The Correct Option is C

Concept: Geometrical isomerism (also known as cis-trans isomerism) is a type of stereoisomerism that can occur in molecules with restricted rotation around a bond, typically a carbon-carbon double bond (\(C=C\)) or in cyclic compounds. For geometrical isomerism around a \(C=C\) double bond, two conditions must be met: \begin{enumerate} % Option (W) There must be restricted rotation around the \(C=C\) bond (which is inherent to double bonds). % Option (X) Each carbon atom of the double bond must be attached to two different atoms or groups. Step 1: Analyze the structures of the given compounds % Option (Y) (1) Butanol-2 (2-Butanol): \(CH_3-CH(OH)-CH_2-CH_3\). This is an alcohol and does not have a \(C=C\) double bond relevant for geometrical isomerism in its main chain. (It has a chiral center and can exhibit optical isomerism, but not geometrical isomerism in the typical alkene sense). % Option (Z) (2) Butene-1 (1-Butene): \(CH_2=CH-CH_2-CH_3\). Let's look at the carbons of the double bond (\(C_1=C_2\)): % Option ([) \(C_1\) is attached to two Hydrogen atoms (H, H). Since these two groups are identical, geometrical isomerism is not possible around this bond. % Option (\) \(C_2\) is attached to a Hydrogen atom (H) and an Ethyl group (\(-CH_2CH_3\)). Since \(C_1\) has two identical groups (H, H), 1-Butene does not exhibit geometrical isomerism. % Option (]) (3) Butene-2 (2-Butene): \(CH_3-CH=CH-CH_3\). Let's look at the carbons of the double bond (\(C_2=C_3\)): % Option (^) \(C_2\) is attached to a Hydrogen atom (H) and a Methyl group (\(-CH_3\)). These are different. % Option (_) \(C_3\) is attached to a Hydrogen atom (H) and a Methyl group (\(-CH_3\)). These are different. Since both \(C_2\) and \(C_3\) are each attached to two different groups, 2-Butene can exhibit geometrical isomerism (cis-2-butene and trans-2-butene). % Option (`) {cis}-2-Butene: Methyl groups are on the same side of the double bond. % Option (a) {trans}-2-Butene: Methyl groups are on opposite sides of the double bond. % Option (b) (4) Butyne-1 (1-Butyne): \(HC \equiv C-CH_2-CH_3\). This molecule has a carbon-carbon triple bond (\(C \equiv C\)). Triple bonds are linear around the alkyne carbons and do not exhibit geometrical isomerism of the cis-trans type. Step 2: Identifying the compound exhibiting geometrical isomerism Based on the analysis, Butene-2 (2-Butene) meets the criteria for geometrical isomerism.