4.1.2: Alkanes Flashcards

1
Q

Intermolecular forces in alkanes:

A

London forces

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2
Q

London forces

A

Temporary instantaneous mutually induced dipole-dipole attractions.

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3
Q

Why are alkanes non-polar?

A

• All bonds are σ bonds
• σ bonds are formed by head on overall of atomic or hybrid orbitals
• Carbon and hydrogen have fairly similar electronegativities
∴ No permanent dipole

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4
Q

General formula for alkanes

A

C𝒏H₂𝒏+₂

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5
Q

Trends in MBP

A
  • Longer chain = higher MBP

* More branching = lower MBP due to intermolecular forces

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6
Q

Why are alkanes unreactive with many reagents?

A
  • High bond enthalpy

* Very low polarity of the σ bonds

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7
Q

Free radical

A

A species with a single unpaired electron.

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8
Q

Why do you not get 100% yield in a reaction?

A
  • Reaction may be reversible
  • Unfavourable conditions
  • Slow reaction
  • Byproducts may be formed
  • Competing reactions may lead to alternate products
  • Products may react with each other to form something else
  • Mechanical losses
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9
Q

What is a sigma bond?

A

Overlap of orbitals directly between the bonding atoms.

Enables free rotation of the σ bond.

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10
Q

Equation to balance complete combustion of any alkane?

A

CxHʏ +(x+ ʏ/4)O₂ ⟶ xCO₂ + ʏ/2H₂O

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11
Q

Each successive alkane needs how much more oxygen for complete combustion?

A

1 ½ extra O₂ molecules required per extra CH₂

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12
Q

Why is incomplete combustion bad?

A

It produces CO which is toxic.

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13
Q

By what mechanism does bromine react with alkanes?

A

Free radical substitution

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14
Q

Limitations of radical substitution in synthesis

A

Forms a mixture of organic products
Products can substitute further
Substitution can occur at different points along the carbon chain –> difficult to control

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15
Q

Bromination of alkanes - initiation step

A

Br₂ ⟶ Br• + •Br

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16
Q

Bromination of alkanes - propagation steps

A

1) CH₄ + Br• ⟶ •CH₃ + HBr

2) •CH₃ + Br₂ ⟶ CH₃Br + Br•

17
Q

Bromination of alkanes - termination step

A

Several options
Br• + •Br ⟶ Br₂
•CH₃ + •CH₃ ⟶ C₂H₆
•CH₃ + •Br ⟶ CH₃Br

Further substitution can replace all H atoms with Br to give CBr₄