Identify a statement which is not correct among the following (a) \(\mathrm{C}_{6} \mathrm{H}_{5}\) group in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{M}_{8} \mathrm{Cl}\) acts as an electrophile. (b) The two nonbonded electrons of a singlet carbene occupy the same orbital. (c) The stability of the alkyl carbocation is influenced by resonance. (d) Free radicals are formed in a homolytic bond fission.

Short Answer

Expert verified
(a) \(\mathrm{C}_{6} \mathrm{H}_{5}\) group in \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{M}_{8} \mathrm{Cl}\) acts as an electrophile. (b) Singlet carbene is less reactive than triplet carbene. (c) Resonance stabilization is a factor that determines the stability of alkyl carbocation. (d) Free radicals are formed through homolytic bond fission of covalent bonds. Answer: (a) \(\mathrm{C}_{6} \mathrm{H}_{5}\) group in \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{M}_{8} \mathrm{Cl}\) acts as an electrophile.

Step by step solution

01

Option (a) Analysis

In option (a), we have a compound \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{M}_{8} \mathrm{Cl}\), where \(\mathrm{C}_{6} \mathrm{H}_{5}\) is a phenyl group. Electrophiles are electron-deficient species that seek electrons to complete their valence shell. However, the phenyl group is not electron-deficient and does not act as an electrophile. Therefore, this statement is incorrect.
02

Option (b) Analysis

In option (b), we deal with singlet carbene. Carbene is a molecule that contains a neutral carbon atom with a valence of two and two unshared valence electrons. In a singlet carbene, the two non-bonded electrons are in the same orbital, making the carbene less reactive than triplet carbene. This statement is correct.
03

Option (c) Analysis

Option (c) deals with the stability of alkyl carbocation, which is influenced by resonance. Resonance stabilization occurs when a bonding scheme can be represented using two or more electron-pushing diagrams (resonance structures), providing delocalization of electrons; thus, the carbocation is better stabilized. This statement is correct.
04

Option (d) Analysis

Option (d) refers to the formation of free radicals through homolytic bond fission. Homolytic bond fission is the process in which the electrons in a covalent bond are divided equally between the two bonded atoms, forming free radicals (highly reactive species with an unpaired electron). This statement is also correct. In conclusion, the incorrect statement among the given options is: (a) \(\mathrm{C}_{6} \mathrm{H}_{5}\) group in \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{M}_{8} \mathrm{Cl}\) acts as an electrophile.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The specific rotation of an isomer of an alkyl halide X containing one chiral arbon atom is \(+40^{\circ} .\) During a reaction in solution, it undergoes partial racemization. If the specific rotation of the equilibrium mixture is \(+10^{\circ}\), the ratio of moles of \((+) X\) and \((-) X\) at equilibrium is (a) \(2: 3\) (b) \(5: 3\) (c) \(3: 2\) (d) \(4: 3\)

Four pairs of compounds (A) - (D) are given below. Indicate, in each pair, which compound can be more effectively hydrogen bonded. (a) \(\mathrm{CH}_{3} \underset{(\mathrm{I})}{\mathrm{CH}_{2}} \mathrm{OH}\) and \(\mathrm{CH}_{3}-\underset{(\mathrm{IH})}{\mathrm{O}-\mathrm{CH}_{3}}\) (b) \(\underset{\text { (I) }}{\left(\mathrm{CH}_{3}\right)}_{3} \mathrm{~N}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (c) \(\mathrm{HOCH}_{2}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CH}_{2} \mathrm{OH}\) and \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}_{\text {(II) }}{\mathrm{C}} \mathrm{HOHCH}_{3}\) (d) \(\mathrm{CH}_{3} \underset{\text { (I) }}{\mathrm{CH}_{2}} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}\) (a) \((\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{I})\) (b) (A) - (I), (B) - (II), (C) - (I), (D) - (II) (c) \((\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{II})\) (d) (A) - (I), (B) - (II), (C) - (I), (D) - (I)

Which of the following are correct statements? (a) \(\mathrm{BF}_{3}\) and are: \(\mathrm{CCl}_{2}\) electrophiles. (b) Dimethylsulphoxide and dimethyl formamide (DMF) are aprotic solvents. (c) In butadiene \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) all the three \(\mathrm{C}-\mathrm{C}\) bond lengths are equal. (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}\) has low reactivity because chlorine is attached to an \(\mathrm{sp}^{2}\) carbon.

IUPAC name of the compound (a) Trans-3,4-dimethylhex-3-ene (b) Cis- 3,4 -dimethylhex-3-ene (c) Cis- 3,4 -dimethylhex-4-ene (d) Cis-2,3-diethylbut-2-ene

\(\mathrm{Cl} \mathrm{CH}_{2}-\mathrm{CHCl}-\mathrm{CH}=\mathrm{CH}_{2}\) is known by IUPAC nomenclature as (a) \(1,2-\) dichloro but \(-3\) -ene (b) \(3,4-\) dichloro but-2-ene (c) \(3,4-\) dichloro but-1-ene (d) 1,2 -dichloro butene-3

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free