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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.

Short Answer

Expert verified
a) BF3 and CCl2 are electrophiles. b) Dimethylsulphoxide (DMSO) and dimethylformamide (DMF) are examples of aprotic solvents. c) In butadiene, all three carbon-carbon bond lengths are equal. d) The double bond in the molecule CH2=CH-Cl has low reactivity because a chlorine atom is attached to an sp虏 carbon. Answer: a) True b) True c) False d) True

Step by step solution

01

Statement (a) - Electrophiles

BF3 (boron trifluoride) and CCl2 (carbon dichloride) are electrophiles because they are electron-deficient species. Both molecules have central atoms with incomplete octets, which makes them highly reactive and eager to accept a pair of electrons from another species to complete their octets. Therefore, statement (a) is correct.
02

Statement (b) - Aprotic Solvents

Dimethylsulphoxide (DMSO) and dimethylformamide (DMF) are aprotic solvents. Aprotic solvents are those that do not have an acidic hydrogen atom; in other words, they cannot donate a proton (H鈦). Both DMSO and DMF fit this description, as they lack an acidic hydrogen atom. Thus, statement (b) is correct.
03

Statement (c) - Butadiene Bond Lengths

In butadiene (H2C=CH-CH=CH2), there are three C-C bonds: two carbon-carbon double bonds and one carbon-carbon single bond. The statement claims that all three bond lengths are equal, which is incorrect. Double bonds are shorter than single bonds due to the presence of a pi bond in addition to a sigma bond, which pulls the atoms closer together. So, statement (c) is incorrect.
04

Statement (d) - Chlorine Reactivity

In the molecule CH2=CH-Cl, chlorine is attached to an sp虏 carbon. Chlorine is more electronegative than carbon, so it withdraws electron density from the double bond, making the double bond less reactive. The statement claims that this molecule has low reactivity because chlorine is attached to an sp虏 carbon, which is true. Therefore, statement (d) is correct. In summary, statements (a), (b), and (d) are correct, while statement (c) is incorrect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrophiles
Electrophiles are essential actors in organic chemistry reactions. They act like sponges, thirsty for pairs of electrons to satisfy their electron-deficient nature. Molecules such as BF鈧 (boron trifluoride) and CCl鈧 (dichlorocarbene) are perfect examples of electrophiles because they have central atoms that do not have a complete octet. Without a full octet, these compounds become highly reactive and readily engage in chemical reactions by accepting electrons from other species like nucleophiles.

To understand electrophiles better:
  • They are typically positively charged or neutral with an incomplete octet.
  • They can be atoms or ions willing to accept electron pairs.
  • They play a critical role in reactions like electrophilic substitution or addition reactions.
Electrophiles are often contrasted with nucleophiles, which are electron-rich species looking to donate electrons. Together, these two groups form the backbone of many organic reactions.
Aprotic Solvents
Aprotic solvents are unique because they do not donate hydrogen ions, making them different from their protic counterparts. Dimethylsulphoxide (DMSO) and dimethylformamide (DMF) fall under this category because they do not contain any acidic hydrogen atoms capable of breaking free as protons (H鈦).

Characteristics of aprotic solvents include:
  • The absence of a hydrogen atom bonded to an electronegative atom, like oxygen or nitrogen, prevents them from donating protons.
  • They have high dielectric constants, allowing them to dissolve a wide range of ionic compounds.
  • They are often used in reactions where protic solvents might interfere by participating in or slowing down the reaction.
Aprotic solvents stabilize ionic intermediates by their strong solubilizing power, thereby enhancing the nucleophilicity of anions in solution and making them ideal in many substitution and elimination reactions.
Bond Lengths
Bond lengths in organic molecules provide crucial insights into their structure and stability. In butadiene (H鈧侰=CH-CH=CH鈧), this concept becomes apparent. Although the statement that all carbon-carbon (C-C) bonds in butadiene are the same length is incorrect, it brings up important ideas in bond chemistry.

In a molecule like butadiene:
  • C=C double bonds consist of one sigma bond and one pi bond, making them shorter than single C-C sigma bonds.
  • The alternating single and double bonds create a conjugated system, affecting the overall bond lengths and strengths.
  • This uneven bond length distribution can influence the molecule's reactivity and properties.
Generally, bond lengths are determined by the types of bonds and atoms involved. Shorter bonds, like double or triple bonds, have more shared electrons pulling the nuclei closer together, compared to longer single bonds.
Reactivity
Reactivity in organic chemistry essentially deals with how eager a molecule is to undergo a chemical reaction. For example, in the case of CH鈧=CH-Cl, the presence of chlorine significantly influences its reactivity. Here, chlorine's attachment to an sp虏 hybridized carbon atom makes for interesting chemistry.

Influencing factors of reactivity involve:
  • Electronegativity differences: With chlorine being more electronegative, it pulls electron density away from the carbon, affecting the molecule's reactivity.
  • Bond type: The presence of a pi-bond in an sp虏 hybridized carbon contributes additional reaction pathways like addition reactions.
  • Steric and electronic factors: These can either hinder or promote reactivity based on the atoms around the reactive site.
Understanding reactivity allows chemists to predict and control the outcomes of chemical reactions, selecting and modifying conditions to favor desired products.

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Most popular questions from this chapter

Directions: Each question contains Statement- 1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 Stereoisomers which do not have an object and mirror image relationship can be termed as diastereomers. and Statement 2 An organic compound having only one chiral centre cannot have diastereoisomers.

The correct increasing order of electron donating effect of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-,\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\) and \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\) groups when attached to benzene or unsaturated system is (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-<\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-<\mathrm{CH}_{3}-\mathrm{CH}_{2}-\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-<\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-<\mathrm{CH}_{3}-\mathrm{CH}_{2}-\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}<\mathrm{CH}_{3}-\mathrm{CH}_{2}-<\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\)

Which of the following carbocations is most stable? (a) Trityl (b) benzyl (c) Diphenyl (d) \(\mathrm{t}\) -butyl

The correct decreasing order of acidity is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}>\mathrm{HC}=\mathrm{CH}>\mathrm{H}_{2} \mathrm{O}>\mathrm{NH}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}>\mathrm{NH}_{3}>\mathrm{HC}=\mathrm{CH}\) (c) \(\mathrm{HC}=\mathrm{CH}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}>\mathrm{H}_{2} \mathrm{O}>\mathrm{NH}_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}>\mathrm{H}_{2} \mathrm{O}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}>\mathrm{HC}=\mathrm{CH}\)

Which of the following statements are correct? (a) Singlet carbene is paramagnetic. (b) In the two species \(\mathrm{CN}^{-}\) and \(\mathrm{CO}\), the common feature is that bond order is three and they are isoelectronic. (c) Between \(\mathrm{CH}_{3} \mathrm{Cl}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\), the former has a higher dipole moment. (d) In the species \(\mathrm{CH}_{3}-\mathrm{C}^{2} \mathrm{H}+\mathrm{HCN} \stackrel{\mathrm{H}_{3}}{\longrightarrow}\) the order of acidity with respect to the positions indicated is \((2)>(1)>(3)\)

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