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Chapter 1: Problem 28

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 sp² 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⁺). 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 sp² 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 sp² 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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Most popular questions from this chapter

The IUPAC name of \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CHCl}-\mathrm{CH}_{2}-\mathrm{COOH}\) is (a) 3-Chloro-4-oxopentanoic acid (b) 4-Oxo-3-Chloropentanoic acid (c) 2-Oxo-3-Chloropentanoic acid (d) 1 -Carboxy-3-Chloro-4-ketobutane

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) \(\mathrm{CH}_{3}-\mathrm{CH}=\stackrel{+}{\mathrm{C}} \mathrm{H}\)

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.

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) \(4: 3\)

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) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}->\mathrm{CH}_{3}-\mathrm{CH}_{2}->\mathrm{CH}_{3}-\)
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