Organic-Alkenes Flashcards
How can alkenes be produced from halogenoalkanes?
By elimination-
With- ethanolic NaOH and heat.
Removes a hydrogen halide (HX) to form an alkene.
HX in the sense- H from the alkane and then x from the alkane are remove
not electrophilic addition! elimination
What happens to the eliminated H⁺ and X⁻ in the elimination reaction?
H⁺ reacts with ethanolic OH⁻ to form H₂O, and X⁻ reacts with Na⁺ to form NaX.
How does NaOH concentration affect the reaction of halogenoalkanes?
Ethanolic NaOH: Causes elimination to form an alkene.
Aqueous NaOH: Causes nucleophilic substitution to form an alcohol.
How can alkenes be produced from alcohols?
By dehydration, which involves the elimination of water.
-Use a hot catalyst (e.g., Al₂O₃)
OR
-Use a concentrated acid (e.g., H₂SO₄).
example of dehydration of alcohols?
Ethanol is dehydrated to form ethene.
How are smaller alkenes collected in the dehydration of alcohols?
Smaller alkenes (e.g., ethene) are gases at room temperature and can be collected over water.
What are the conditions for the reaction of alkenes with steam?
Reaction with H₂O (g) in the presence of a H₃PO₄ catalyst under heat.
What is the product when ethene reacts with steam?
Ethanol
What happens when alkenes react with hydrogen halides at room temperature?
State the state of HX
Alkenes form haloalkanes via electrophilic addition.
HX is a gas
Explain the mechanism for the reaction of HBr with ethene.
- The H atom in HBr acts as an electrophile, attacking the electron-rich C=C bond.
- A carbocation intermediate forms.
- The Br⁻ ion attacks the carbocation to form bromoethane.
Describe the reaction of alkenes with halogens like Br₂.
ELECTROPHILIC ADDIITON
1. Br₂ is non-polar but becomes polarized near the electron-rich C=C bond.
- The double bond attacks Br⁻, breaking the Br–Br bond heterolytically.
- A bromonium ion intermediate forms, followed by attack by Br⁻.
- The product is a dihaloalkane (e.g., ethene + Br₂ → 1,2-dibromoethane).
How can aqueous bromine be used to test for alkenes?
Alkenes decolorize orange bromine water, indicating the presence of a C=C bond.
What is the product of alkene oxidation with cold, dilute acidified KMnO₄?
A diol
(e.g., ethene + KMnO₄ → ethane-1,2-diol).
What happens when alkenes are oxidized with hot, concentrated acidified KMnO₄?
How is this reaction used?
The C=C bond ruptures, forming carboxylic acids, ketones, or CO₂, depending on the alkene’s structure.
Example: Propene oxidizes to ethanoic acid and CO₂.
USED- To determine the position of C=C bonds in larger molecules.
What happens during addition polymerization?
Give exmaples of polymers of ethene and propene
C=C bonds broken and join together to form long-chain polymers
Ethene:
* poly(ethene)
* (polyethylene)
Propene:
* poly(propene)
* (polypropylene)
Example:
What determines the stability of carbocations?
The inductive effect of alkyl groups, which donate electron density to stabilize the positive charge.
Stability order: Tertiary > Secondary > Primary.
How does the stability of carbocations affect reaction outcomes?
The major product forms from the most stable carbocation intermediate (Markovnikov addition).
Halogen example:
The hydrogen atom from HX adds to the carbon with more hydrogen atoms, forming the more stable carbocation. The halogen bonds to the more substituted carbon.
What happens when an alkene with two hydrogen atoms on the same carbon is oxidized with hot, concentrated acidified KMnO₄?
The carbon is fully oxidized to form carbon dioxide (CO₂).
What happens when an alkene with one hydrogen and one alkyl group on the same carbon is oxidized with hot, concentrated acidified KMnO₄?
The carbon is oxidized to form a carboxylic acid.
Example:Ethene (CH₂=CH₂) produces ethanoic acid (CH₃COOH).
Aldehyde->Carboxylic acid
What happens when an alkene with two alkyl groups on the same carbon is oxidized with hot, concentrated acidified KMnO₄?
The carbon is oxidized to form a ketone.
Example:
But-2-ene (CH₃CH=CHCH₃) produces propanone (CH₃COCH₃).
What is the overall result of oxidative cleavage of alkenes by hot, concentrated acidified KMnO₄?
- 2 H atoms → CO₂
- 1 H atom + 1 alkyl group → Carboxylic acid
- 2 alkyl groups → Ketone
What are the products when propene is oxidized by hot, concentrated acidified KMnO₄?
The CH₂ group (with 2 H atoms) is oxidized to CO₂.
The CH group (with 1 H atom and 1 alkyl group) is oxidized to ethanoic acid (CH₃COOH).
What are the products when 2-methylpropene is oxidized by hot, concentrated acidified KMnO₄?
The CH2 group (with 2 H atoms ) forms CO2.
The C atom (with 2 alkyl groups) forms acetic acid (CH₃COOH).
