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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 incorrect resonance structure of diazomethane is (a) \(\mathrm{H}_{2} \stackrel{\ominus}{\cdots}-\mathrm{N} \equiv \cdots\) (b) \(\mathrm{CH}_{3}-\mathrm{N}=\cdots \mathrm{N}\) (c) \(\mathrm{H}_{2} \mathrm{C}=\stackrel{\oplus}{\mathrm{N}}=\cdots \mathrm{N}^{\cdot}\) (d) \(\mathrm{H}_{2} \mathrm{C} \rightarrow \ddot{\mathrm{N}}=\cdots^{\ominus}\)

A chemical reaction is one in which old bonds are broken and new bonds are made. During the course of these changes a variety of intermediates are formed before a starting material is converted to final products. Formation of these intermediates depend on several factors like bond energies, kinetics of the reactions etc. Identify the correct statement among the following. (a) An electrophile has always a positive charge. (b) \(\mathrm{CN}^{-}\) can act as an ambident nucleophile. (c) The carbon species A formed is a methyl carbanion. (d) In \(\mathrm{CH}_{3} \mathrm{MgBr}\), the methyl group carries slight positive charge.

Two pairs of compounds are given below, which one in each is more acidic and why? (i) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{H}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (ii) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) and CCC(C)CCO

The hyperconjugative effect of the group \(\mathrm{R}\) in the compound \(\mathrm{R}-\mathrm{CH}=\mathrm{CH}_{2}\) follows the order. (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}>\mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CH}_{2}-\) (b) \(\mathrm{CH}_{3}->\mathrm{CH}_{3} \mathrm{CH}_{2}->\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}->\mathrm{CH}_{3}-\mathrm{CH}_{2}->\mathrm{CH}_{3}-\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}->\mathrm{CH}_{3}-\mathrm{CH}_{2}->\mathrm{CH}_{3}-\)

Configuration of a chiral molecule can be changed only if (a) the compound is rotated out of the plane of paper. (b) the confirmation is changed by rotation. (c) the Fischer projection is changed to the Newman projection. (d) the bond to the chiral carbon is broken.
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