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Chemistry > M4 Alkenes > Flashcards

M4 Alkenes Flashcards

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1
Q

What are alkenes?

A

Unsaturated hydrocarbons containing a C=C double bond, comprising of pie-bond and a sigma bond, with restricted rotation of the pie bond.

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2
Q

What is a pi bond?

A

Sideways overlap of adjacent p-orbitals above and below the bonding carbon atoms.

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3
Q

What is a sigma bond?

A

Overlap of orbitals directly between the bonding atoms.

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4
Q

Describe the double bond

A
  • Each atom has four electrons in its outer shell and can use these electrons to form bonds.
  • For each carbon atom of the double bond, three of the four electrons are used in three sigma bonds, one to the other carbon atom, of the double bond, and the other two electrons to two other atoms.
  • this leaves one electron on each carbon atom of the double bond not involved in a sigma bond. This electron is in a P-orbital. A pie-bond is formed by the sideways overlap of two p-orbitals, One from each carbon atom of the double bond. Each carbon atom contributes one electrons to the electron pair in the pie bond. The pi electron density is concentrated above and below the line joining the nuclei of the bonding atoms.
  • The pi bond locks the two carbon atoms in position and prevents them from rotating around the double pond. This makes the geometry of the alkenes different from alkanes, where rotation is possible around every atom.
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5
Q

Describe the shape around the double bond

A

The shape around each of the double bond is trigonal planar because:
- there are three regions of electron density around each of the carbon atoms
- three regions repel each other as far as possible, so the bond angle around each carbon atom is 120°
- all of the atoms are in the same plane

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6
Q

Define stereoisomers

A
  • Stereoisomers have the same structural formula but a different arrangement of the atoms in space
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7
Q

Define E/Z isomerism

A
  • Stereoisomerism around the double bond arises because rotation about the double bond is restricted and the groups attached to each carbon atom are their full fixed relative to each other. The reason for the rigidity is the position of the pie bond’s electron density above and below the plane of the sigma bond.
  • If a molecule has a C=C double bond and different groups attached to each carbon atom of the double bond, it will have E/Z isomerism.
    (E) = diagonal
    (Z) = next to each other
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8
Q

What are the Cahn-Ingold-Prelog priority rules to identify the E and the Z stereoisomers?

A
  • If the groups of higher priority are on the same side of the double bond, the compound is the Z isomer
  • If the groups of higher priority are digitally placed across the double bond, the compound is the E isomer
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9
Q

How to assign priority?

A
  • The higher the atomic number, the higher the priority
  • If two atoms attached to a carbon atom in the double bond are the same, you will need to find the first point of difference. The group which has the higher atomic number at the first point of difference is given the higher priority.
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10
Q

Describe the reactivity of alkenes

A
  • Alkenes are much more reactive than alkanes because of the presence of the pi-bond. The C=C double bond is made up of a sigma-bond and a pi-bond, and the pi-bond electron density is concentrated above and below the plane of the sigma-bond.
  • Being on the outside of the double bond the pi-electrons are more exposed than the electrons in the sigma bond. A pi-bond readily breaks and alkanes undergo addition reactions relatively easily.
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11
Q

Describe hydrogenation of alkenes

A
  • When an alkene, such as propane is mixed, with hydrogen and passed over a nickel catalyst at 423K, an addition reaction takes place to form an alkane.
  • This addition reaction in which hydrogen is added across a double bond, is known as hydrogenation.
    • propane + hydrogen = propane
  • All C=C bonds react with hydrogen in this way, the double bonds are both hydrogenated, requiring two molecules of hydrogen per molecule of buta-1,3-diene
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12
Q

How to test for alkenes

A
  • The reaction of alkenes with bromine can be used to identify if there is a C=C bond present.
  • When bromine water (orange) is added to an alkene, bromine adds across the double bond and the orange colour disappears, indicating the presence of a C=C bond.
  • If there is no colour change, there is no addition reaction as it is a saturated compound.
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13
Q

Describe the reaction of alkenes with hydrogen halides

A
  • Alkenes react with gaseous hydrogen halides at room temperature to form haloalkanes.
  • Alkenes also react with concentrated hydrochloric/hydrobromic acid with are solutions of hydrogen halides in water.
    propane + hydrochloric acid —> 1-chloropropane or 2-chloropropane
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14
Q

Describe the hydration reactions of alkenes

A
  • Alcohols are formed when Alkenes react with steam (H2O (g) ) in the presence of a phosphoric acid catalyst H2PO4.
  • Steam adds across the double bond. This addition reaction is used widely in industry to produce ethanol from ethene. As with the addition with hydrogen halides, there are two possible products.

propane + H2O (g) —> propan-1-ol or propan-2-ol

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15
Q

What are electrophilic addition reactions?

A

Alkenes usually take part in addition reactions to form saturated compounds. The mechanism is called electrophilic addition.
- the double bond in an alkene with represents a region of high electron density because of the presence of the pi-electrons
- the high electron density of the pi-electrons attracts electrophiles
- an electrophile is an atom or group of atoms that is attracted to an electron-rich centre and accepts an electron pair. An electrophile is usually a positive ion or molecule containing an atom with a partial positive charge

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16
Q

Define an electrophile

A

An electron pair acceptor

17
Q

Describe the mechanism for electrophilic addition of but-2-ene and hydrogen bromide

A
  1. Bromine is more electronegative than hydrogen so hydrogen bromide is polar and contains the dipole H+—Br-
  2. The electron pair in the pi-bond is attracted to the partially positive hydrogen atom, causing the double bond to break
  3. A bond forms between the hydrogen atom of the H-Br molecule and a carbon atom that was part of the double bond
  4. The H—Br bond breaks by heterolytic fission, with the electron pair going to the bromine atom.
  5. A bromide ion (Br-) and a carbocation are formed. A carbocation contains a positively charged carbon atom.
  6. The final step in the Br- ion reacts with the carbocation to form the addition product.
18
Q

What is Markownikoff’s rule in which of the two possible isomers is the major product?

A
  • The major product from an addition of a hydrogen halide to an unsymmetrical alkene is the one where hydrogen adds to the carbon with the most hydrogen already attached.
19
Q

Describe the stability of carbocations

A
  • Carbocations are classified by the number of alkyl groups attached to the positively charged carbon atom.
  • Tertiary carbocations (with 3 R groups) are the most stable, and primary cations are the least stable.
  • Each alkyl group donates and pushes electrons towards the positive charge of the carbocation. The positive charge is spread over the alkyl groups. The more alkyl groups attached to the positively charged carbon atom, the more the charge is spread out, making the ion more stable.
20
Q

Describe addition polymerisation

A
  • Unsaturated alkene molecules undergo addition polymerisation to produce long saturated chains containing no double bonds. Many different polymers can be formed, each with its own specific properties depending on the monomer. Industrial polymerisation is carried out at high temperature and high pressure using catalysts:
  • addition polymers have high molecular masses
  • synthetic polymers are usually named after the monomer that reacts to form their giant molecules
21
Q

What is a repeat unit?

A
  • The specific arrangement of atoms in the polymer molecule that repeats over again
22
Q

What are the benefits of processing waste polymers by combustion for energy consumption?

A
  • Polymers are difficult to recycle as they have a high stored energy value. Waste polymers can be incinerated to produce heat, generating steam to drive a turbine generating electricity.
23
Q

What are the benefits of feedstock recycling?

A
  • Feedstock recycling describes the chemical and thermal processes that can reclaim monomers, gases or oil from waste polymers.
  • The products can reclaim monomers, gases or oil from waste polymers. The products from feedstock recycling resemble those produced from crude oil in refineries.
  • These materials can be used as raw materials for the production of new polymers. A major advantage of feedstock recycling is that it is able to handle unsorted and unwashed polymers.
24
Q

What are the benefits of PVC recycling?

A
  • The disposal and recycling of PVC is hazardous due to high chlorine content and the range of additives present in the polymer.
  • Dumping PVC in landfill is not sustainable, and when burnt, PVC releases hydrogen chloride and other pollutants.
  • Previously recycling involved grinding PVC and reusing it to manufacture new products. New technology uses solvents to dissolve the polymer. High-grade PVC is then recovered by precipitation from the solvent, and the solvent is used again.
25
Q

What are the benefits of biodegradable polymers?

A
  • Biodegradable polymers are broken down by microorganisms into water, carbon dioxide and biological compounds. These polymers are usually made from starch or cellulose, or contain additives that alter the structure of traditional polymers so that microorganisms can break them down.
  • Compostable polymers degrade and leave no visible or toxic residues. Compostable polymers based on poly(lactic acid) are becoming more common as an alternative to alkene-based products.
  • Supermarket bags made from plant starch can be used as bin liners for food waste so the waste and bag can be composed together, compostable plates, cups and food trays made from sugar cane fibres are replacing expanded polystyrene.
26
Q

What are the benefits of using photodegradable polymers?

A
  • Photodegradable oil-based polymers contain bonds that are weakened by absorbing light to start degradation, where the use of plant-based polymers is not possible.

Chemistry (15 decks)

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