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Chapter 1: Problem 105

which among the following compounds can exhibit geometrical isomerism? (a) 2 -phenyl but- 1 -ene (b) 1 -phenyl but- 2 -ene (c) 3 -phenyl but-1-ene (d) 1,1 -dichlorobut- 1 -ene

Short Answer

Expert verified
a) 2-phenyl but-1-ene b) 1-phenyl but-2-ene c) 3-phenyl but-1-ene d) 1,1-dichlorobut-1-ene Answer: Only compound (b) 1-phenyl but-2-ene can exhibit geometrical isomerism.

Step by step solution

01

Identify groups attached to the double bonds

For each of the given compounds, we need to examine the groups attached to the carbon atoms with the double bond and ensure that they are different from each other.
02

Analyze compound (a)

In 2 -phenyl but- 1 -ene, the double bond is between the first and second carbon atoms. The first carbon atom has H and CH2 attached, while the second carbon atom has H and phenyl group (C6H5) attached. As there are two hydrogen atoms attached to the double bond, it cannot exhibit geometrical isomerism.
03

Analyze compound (b)

In 1 -phenyl but-2-ene, the double bond is between the second and third carbon atoms. The second carbon has a phenyl group (C6H5) and H attached, and the third carbon has CH3 and H attached. Since different groups are attached to each carbon atom, this compound can exhibit geometrical isomerism.
04

Analyze compound (c)

In 3-phenyl but-1-ene, the double bond is between the first and second carbon atoms. The first carbon atom has H and CH2 attached, while the second carbon atom has H and phenyl group (C6H5) attached. As there are two hydrogen atoms attached to the double bond, it cannot exhibit geometrical isomerism.
05

Analyze compound (d)

In 1,1 -dichlorobut- 1 -ene, the double bond is between the first and second carbon atoms. The first carbon atom has two Cl atoms attached, while the second carbon atom has H and CH2CH2CH3 attached. As both the carbon atoms have Cl attached to them do not meet the criteria for geometrical isomerism, the compound cannot exhibit it.
06

Conclusion

Only compound (b) 1-phenyl but-2-ene can exhibit geometrical isomerism as it has different groups attached to the carbon atoms involved in the double bond and cannot be interconverted through rotation around the double bond.

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Most popular questions from this chapter

The boiling point of \(\mathrm{CH} 3 \mathrm{OH}\) is \(65^{\circ} \mathrm{C}\), while the boiling point of \(\mathrm{CH} 3 \mathrm{SH}\) is only \(6^{\circ} \mathrm{C}\). Explain this difference in boiling point.

A free radical is (a) generated by a heterolytic cleavage (b) paramagnetic (c) generally more stable than a carbanion (d) produced in an ionic reaction

The most stable free radical among the following is (a) C=CC(C)C (b) c1ccccc1 (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}\) (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\).

ClCCI ClC(Cl)(Cl)Cl The \(\mathrm{Cl}^{\text { }} \mathrm{Cl}\) bond angle in \(1,1,2,2\) -tetrachloroethene and tetrachloromethane respectively are about (a) \(90^{\circ}\) and \(109.5^{\circ}\) (b) \(109.5^{\circ}\) and \(120^{\circ}\) (c) \(120^{\circ}\) and \(109.5^{\circ}\) (d) \(120^{\circ}\) and \(120^{\circ}\)

The least stable cabocation among the following is (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}^{+}\) (b) \(\mathrm{F}_{3} \mathrm{C}-\mathrm{CH}_{2}\) (c) \(\mathrm{F}_{3} \mathrm{C}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\stackrel{+}{\mathrm{C}} \mathrm{H}\)
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