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Chapter 2: Problem 101

An alkene which is least reactive towards electrophilic addition, among the following is (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{Cl}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\) (d) \(\mathrm{ClCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\)

Short Answer

Expert verified
(a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{Cl}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\) (d) \(\mathrm{ClCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) Answer: (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{Cl}\)

Step by step solution

01

Analyze the given alkene structures

Let's analyze each alkene structure to identify the factor that can affect the electron density around the double bond. (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{Cl}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\) (d) \(\mathrm{ClCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\)
02

Identify electron-donating and electron-withdrawing groups

Alkyl groups (like \(\mathrm{CH}_{3}\)) have electron-donating properties, hence they can enhance the electron density around the alkene double bonds. On the other hand, halogen atoms (like \(\mathrm{-Cl}\)) have electron-withdrawing properties, which can decrease the electron density around the alkene double bonds.
03

Compare the electron densities

Let's compare the electron densities around the double bonds for each alkene: (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{Cl}\): There is an electron-withdrawing chlorine attached to the double bond. (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\): There are two electron-donating alkyl groups attached to the double bond. (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\): There are four electron-donating alkyl groups attached to the double bond. (d) \(\mathrm{ClCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\): There is an electron-withdrawing chlorine attached to an atom connected to the double bond, but not directly connected to the double bond.
04

Determine the alkene with least reactivity

From the analysis above, we can see that option (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{Cl}\) has an electron-withdrawing group (chlorine) directly attached to the double bond, which leads to its decreased electron density, making it the least reactive alkene towards electrophilic addition among the given options. Therefore, the correct answer is (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{Cl}\).

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Most popular questions from this chapter

Alkenes are a highly reactive group of compounds as they undergo a wide variety of reactions like hydrogenation, ozonolysis, halogenation, epoxidation and halohydrin formation. Many of the reactions are regioselective or stereoselective reactions. When bromine is added to a solution of 1 -hexene in methanol, the product formed is (a) \(\mathrm{BrCH}_{2}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) a mixture of \(\mathrm{BrCH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) and \(\mathrm{BrCH}_{2}-\mathrm{CH}\left(\mathrm{OCH}_{3}\right)-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{COCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) \(\stackrel{\text { alc } \mathrm{KOH} / \Delta}{\longrightarrow}(\mathrm{A})\). The product \((\mathrm{A})\) is (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH} \mathrm{CH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (d)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3} \stackrel{\mathrm{Br}_{2} / \mathrm{hv}}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{Mg} / \mathrm{Ether}}{\longrightarrow}(\mathrm{B}) \stackrel{\mathrm{D}_{2} \mathrm{O}}{\longrightarrow}(\mathrm{C})\) The major product \(C\) formed is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{D}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CHD} \mathrm{CH}_{3}\) (d)

Which of the following on hydration will give an aldehyde? (a) \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (c) \(\mathrm{HC} \equiv \mathrm{CH}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{CH}\)

Which of the following reactions give an alkyne? (a) sodium fumarate \(\frac{\text { Kolbe's }}{\text { electrolysis }}\) (b) \(\mathrm{CH}_{3} \mathrm{CBr}_{2} \mathrm{CHBr}_{2} \longrightarrow \underset{\mathrm{Zn} \text { /alcohol } / \Delta}{\longrightarrow}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBr}_{2} \stackrel{\mathrm{NaNH}_{2} / \Delta}{\longrightarrow}\) (d) \(\mathrm{CH}_{3} \mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{Br} \stackrel{\mathrm{NaNH}_{2}}{\longrightarrow}\)
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