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Chapter 4: Problem 41

The compound with zero dipole moment is (a) 1,2 -dichlorobenzene (b) 1,3 -dichlorobenzene (c) 1,4 -dichlorobenzene (d) 1,4 -dihydroxybenzene

Short Answer

Expert verified
The compound with a zero dipole moment is 1,4-dichlorobenzene.

Step by step solution

01

Understand the structure of benzene

Benzene is a cyclic compound with 6 carbon atoms, each bonded to another carbon atom and a hydrogen atom. Each carbon forms 3 covalent bonds, with one double bond between two carbon atoms. The double bonds have resonance across the ring, making the compound symmetric and planar.
02

Locate the position of Chlorine atoms in each dichlorobenzene isomer

We have three isomers of dichlorobenzene: (a) 1,2 -dichlorobenzene: Chlorine atoms are located on adjacent carbon atoms. (b) 1,3 -dichlorobenzene: Chlorine atoms are located on carbon atoms with one carbon atom between them. (c) 1,4 -dichlorobenzene: Chlorine atoms are located on carbon atoms with two carbon atoms between them.
03

Analyze the dipole moment in 1,2-dichlorobenzene

In 1,2-dichlorobenzene, the Chlorine atoms are on adjacent carbon atoms. The two carbon-chlorine bonds create dipole moments that do not cancel each other out due to a lack of symmetry. This compound has a nonzero dipole moment.
04

Analyze the dipole moment in 1,3-dichlorobenzene

In 1,3-dichlorobenzene, the Chlorine atoms are located on carbon atoms with one carbon atom between them. This configuration is also not symmetrical, and the individual dipole moments from the carbon-chlorine bonds do not cancel each other out. This compound also has a nonzero dipole moment.
05

Analyze the dipole moment in 1,4-dichlorobenzene

In 1,4-dichlorobenzene, the Chlorine atoms are located on opposite sides of the benzene ring, with two carbon atoms between them. This configuration is symmetrical, and the individual dipole moments from the carbon-chlorine bonds cancel each other out since they are oppositely directed. This compound has a zero dipole moment.
06

Analyze the dipole moment in 1,4-dihydroxybenzene

In 1,4-dihydroxybenzene, the Hydroxyl (OH) groups are located on opposite sides of the benzene ring, with two carbon atoms between them. This configuration is symmetrical, and the individual dipole moments from the carbon-oxygen bonds cancel each other out since they are oppositely directed. This compound has a zero dipole moment.
07

Determine the final answer

From our analysis, the compound with zero dipole moment is 1,4-dichlorobenzene (c).

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Most popular questions from this chapter

Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Which of the following is the best method to convert 3-methylbut-1-ene to 3 -methyl butan-2-ol? (a) Acid catalyzed hydration. (b) Addition of con. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) followed by hydrolysis. (c) Hydroboration-oxidation. (d) Oxymercuration-demercuration.

The major product formed when 2-Fluoropentane is treated with \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is (a) pent-1-ene (b) cis pent- 2 -ene (c) trans pent-2-ene

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\mathrm{Mg} / \mathrm{ether}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{3}}{\longrightarrow} \mathrm{B}\) The product \(B\) in the above reaction is (a) 2 -phenyl- 3 -methyl butan- 2 -ol (b) 2 -phenylpentan- 2 -ol (c) 3 -phenyl butan-2-ol (d) propan-2-ol and ethyl benzene

The correct nucleophilicity order among the following is (a) \(\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}>\mathrm{CH}_{3}^{-}\) (b) \(\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}\) (c) \(\mathrm{OH}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{CH}_{3}^{-}>\mathrm{F}^{-}\) (d) \(\mathrm{CH}_{3}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}>\mathrm{NH}_{2}^{-}\)

It is feasible to prepare tertiary butyl ethyl ether by treating ethyl bromide with sodium tertiary butoxide, rather than treating tertiary butyl bromide with sodium ethoxide because (a) sodium ethoxide will not react with t-butyl bromide. (b) the first reaction is faster and gives more yield. (c) t-butyl bromide undergoes dehydrohalogenation to give alkene in presence of sodium ethoxide. (d) t-butyl bromide undergoes rearrangement.
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