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Chapter 4: Problem 88

The major product formed when 2-Fluoropentane is treated with \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is (a) pent-1-ene (b) cis pent- 2 -ene (c) trans pent-2-ene

Short Answer

Expert verified
Answer: (c) trans pent-2-ene

Step by step solution

01

Write down the starting compound and the reagent

In this case, the starting compound is 2-Fluoropentane (\(\mathrm{CH}_{3} \mathrm{CHF} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)), and the reagent for the elimination reaction is ethoxide ion (C2H5O-) in ethanol solvent (C2H5OH).
02

Identify the α- and β-carbons

First, label the carbon that has the fluorine attached as the α-carbon. In this case, it's the second carbon in the chain. Then, the carbon(s) directly connected to the α-carbon are the β-carbons. Here, we have two β-carbons: one to the left and one to the right of the α-carbon.
03

Apply the Zaitsev's rule to analyze the possible products

According to Zaitsev's rule, when there are multiple β-carbons to choose from, the hydrogen will be removed from the β-carbon that, when forming the double bond, generates the more substituted alkene (the alkene with more carbon atoms attached to the double bond). In this case, both β-carbons have the same degree of substitution - they are each bound to two other carbon atoms.
04

Determine the major product formed as an alkene

Since both β-carbons are equivalent in terms of substitution, we will form a double bond between the α-carbon and one of the β-carbons. In this case, we will end up with pent-2-ene (\(\mathrm{CH}_{3} \mathrm{CH} = \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{3}\)).
05

Analyze the possibility of cis or trans configuration

The alkene we formed in step 4 has two hydrogen atoms on the α-carbon and two methyl groups on the β-carbon. The configuration can be either cis (both methyl groups on the same side of the double bond) or trans (methyl groups on opposite sides of the double bond). Since the trans isomer tends to have lower steric hindrance than the cis isomer, it is generally more stable and is thus the major product formed.
06

Match the major product with the answer choices

Comparing the major product, trans pent-2-ene, with the answer choices, we find that the answer is (c) trans pent-2-ene.

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Most popular questions from this chapter

Reaction of an alkyl halide (A) with alcoholic KOH forms a product which on ozonolysis gives a mixture of propanone and methanal. The structure of the alkyl halide (A) is (a) 2 -Bromobutane (b) 1,4 -Dibromobutane (c) 2 -Bromo-2-methyl propane (d) 2,3 -Dibromobutane

Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Which of the following is the best method to convert 3-methylbut-1-ene to 3 -methyl butan-2-ol? (a) Acid catalyzed hydration. (b) Addition of con. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) followed by hydrolysis. (c) Hydroboration-oxidation. (d) Oxymercuration-demercuration.

Benzyl chloride and o-chlorotoluene can be distinguished using (a) alcoholic \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{NaNO}_{2}\) and \(\mathrm{HCl}\) (c) hot alkaline \(\mathrm{KMnO}_{4}\) (d) both (a) and (c)

How is \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\) prepared by a Grignard reaction with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\) and \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) as the starting materials?

\(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{3}+\mathrm{Cl}_{2} \stackrel{800 \mathrm{~K}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{HCl}\). This reaction takes place mainly by (a) \(\mathrm{S}_{\mathrm{N}} 1\) mechanism (b) \(\mathrm{S}_{\mathrm{N}} 2\) mechanism (c) free radical mechanism (d) E1 mechanism
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