Chapter 26: Problem 101
1-chlorobutane on reaction with alcoholic potash gives (a) 1 -butene (b) 1 -butanol (c) 2 -butene (d) 2-butanol
Short Answer
Expert verified
The product is (a) 1-butene.
Step by step solution
01
Identify the Reaction Type
The reaction of 1-chlorobutane with alcoholic potash typically involves the elimination reaction known as dehydrohalogenation, where a hydrogen halide (HCl in this case) is eliminated.
02
Reaction Prediction
In dehydrohalogenation, the chlorine atom and a hydrogen atom from the adjacent carbon are removed, resulting in the formation of a double bond between the carbons.
03
Determine the Product Formation
From 1-chlorobutane, when the chlorine atom and a hydrogen atom from the adjacent carbon atom (beta-carbon) are removed, it results in the formation of 1-butene. This is due to the formation of the least substituted alkene, following Zaitsev's rule.
04
Chemical Equation
The reaction of 1-chlorobutane with alcoholic KOH can be represented as: \[ CH_3CH_2CH_2CH_2Cl + KOH_{(alc)}
ightarrow CH_2=CHCH_2CH_3 + KCl + H_2O \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Elimination Reaction
In organic chemistry, an elimination reaction is a type of reaction where a molecule loses atoms or groups to form a double or triple bond. These reactions are crucial for creating unsaturated compounds from saturated ones. One common form of elimination is dehydrohalogenation. Here, a hydrogen atom and a halogen atom are removed, typically resulting in the formation of an alkene.
Elimination reactions can follow two common mechanisms: E1 and E2. In an E1 mechanism, the reaction occurs in two steps with the formation of a carbocation intermediate, while E2 involves a single concerted step where the base simultaneously removes a hydrogen atom, and the leaving group departs. This creates a double bond in one swift motion.
Elimination reactions can follow two common mechanisms: E1 and E2. In an E1 mechanism, the reaction occurs in two steps with the formation of a carbocation intermediate, while E2 involves a single concerted step where the base simultaneously removes a hydrogen atom, and the leaving group departs. This creates a double bond in one swift motion.
- In our case, the reaction of 1-chlorobutane with alcoholic potash is an E2 elimination reaction.
- This type of reaction is favored in the presence of a strong base like potash (KOH).
Zaitsev's Rule
Zaitsev's rule is a guiding principle for predicting the products of elimination reactions, especially dehydrohalogenation. According to this rule, the more substituted alkene will generally be the major product. This means that the double bond is more likely to form between the carbon atoms with fewer hydrogen atoms.
But why does this happen? Alkenes where the carbon atoms involved in the double bond are attached to more alkyl groups (carbon chains) are more stable. This stability is attributed to hyperconjugation and the dispersal of electrons, which lower the overall energy of the molecule.
But why does this happen? Alkenes where the carbon atoms involved in the double bond are attached to more alkyl groups (carbon chains) are more stable. This stability is attributed to hyperconjugation and the dispersal of electrons, which lower the overall energy of the molecule.
- However, in the reaction of 1-chlorobutane with alcoholic potash, Zaitsev's rule helps indicate that 1-butene will form, despite it being slightly less substituted than the alternative.
- This outcome is typically due to the sterics in specific reactions, where forming the less substituted alkene might be the easier or faster pathway.
1-Chlorobutane Reaction
1-Chlorobutane is an alkyl halide, and when it reacts with alcoholic potash (KOH), an elimination reaction occurs. In this context, the chlorine atom and a hydrogen atom from the adjacent carbon (known as beta-carbon) are removed. This results in a dehydrohalogenation process, where the atoms lost form a molecule of hydrochloric acid (HCl), and most commonly, a double bond forms between the two carbons, creating an alkene.
The outcome of these reactions tends to align with Zaitsev's rule. However, in the case of 1-chlorobutane, 1-butene forms. This is because the system favors the less hindered transition state allowing for the creation of a product even if less substituted.
The outcome of these reactions tends to align with Zaitsev's rule. However, in the case of 1-chlorobutane, 1-butene forms. This is because the system favors the less hindered transition state allowing for the creation of a product even if less substituted.
- When 1-chlorobutane reacts with KOH in an alcoholic medium, 1-butene is often the final product.
- This is a perfect illustration of how elimination reactions can produce alkenes from alkyl halides.