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Chapter 2: Problem 39

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) \(\stackrel{\text { alc } \mathrm{KOH} / \Delta}{\longrightarrow}(\mathrm{A})\). 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)

Short Answer

Expert verified
Short Answer: The structure of product (A) obtained from the given reaction using alcoholic KOH and heat is (a) \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3}\). This elimination reaction (E2) forms a double bond between the second and third carbons in the molecule.

Step by step solution

01

Identify the reaction

In this case, the reaction involves alcoholic KOH and heat, which is indicative of an elimination reaction. Specifically, it is an E2 (bimolecular elimination) reaction, in which both the substrate and base are involved in the rate-determining step, and a β-hydrogen and a halogen atom (iodine in this case) are eliminated, resulting in the formation of a double bond.
02

Locate the elimination site

In an E2 reaction, a β-hydrogen and a halogen atom are eliminated. In the given compound \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\), we can locate the iodine atom on the third carbon. The β-hydrogens are present on the second and fourth carbons. Since we already have a double bond present in the molecule, it's more likely that the elimination occurs at the site with a single bond adjacent to the halogen atom - which is the second carbon.
03

Perform the elimination

Now that the reaction site has been identified, perform the elimination reaction by removing the iodine atom from the third carbon and a hydrogen atom from the adjacent second carbon. This creates a double bond between the second and third carbons: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \stackrel{\text { alc } \mathrm{KOH}/ \Delta}{\longrightarrow}\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3}\)
04

Identify the product (A)

Based on the reaction performed, the structure of product (A) can be determined to be \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3}\). Comparing the structure with the given options, the correct answer is: (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH} \mathrm{CH}_{3}\)

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Most popular questions from this chapter

An organic compound (A) \(\mathrm{C}_{5} \mathrm{H}_{8}\) does not react with Tollens reagent but reacts with \(\mathrm{H}_{2}\) in the presence of Lindlar catalyst to give (B). Aqueous chlorine water reacts with (B) to give (C) which when reacted with hot alcoholic KOH gives (D). The compound (D) is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\) (c) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (d)

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{ICl} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CClCH}_{2} \mathrm{I}\) is an example of (a) anti Markovnikov addition to the double bond (b) a regioselective reaction (c) an electrophilic substitution reaction (d) a stereospecific reaction

The compound which gives methanal, ethanal and glyoxal on ozonolysis is (a) Hexa-1,3-diene (b) Penta- 1,3 -diene (c) Penta- 1,4 -diene (d) Penta-1,2-diene

An organic compound ( \(3.0 \mathrm{~g}\) ) gave on complete combustion, \(3.476 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(1.422 \mathrm{~g}\) of water. The molecular weight of the compound is 228 . Give molecular formula of the compound.

HI does not exhibit peroxide effect because (a) HI is a reducing agent. (b) the iodine free radicals formed recombine readily to form the iodine molecule. (c) the bond energy of HI is very high that it does not undergo homolytic fission to give iodine radicals. (d) Free radical reactions work well when both the propagation steps are endothermic.
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