HALOGENOALKANES Flashcards

1
Q

State and explain the trend in polarity of C-X (halogen) bonds going down the group

A

from C-F to C-I bond polarity decreases as you go down the group. This is because Electronegativity decreases as you go down the group, so the difference in electronegativity between carbon and the halogen decreases, therefore the polarity of the bond is less.

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2
Q

Name the two forces that interact between Halogenoalkane molecules

A

Van der Waals forces
Permanent dipole-dipole interactions

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3
Q

What is the trend in boiling point as the length of the haloalkane chain increases?

A

VdWs increase as the relative molecular mass/ number of electrons increases. Therefore as chain length increases, boiling point increases.

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4
Q

What is the pattern of boiling points in different halogenoalkanes

A

VdWs increase due to the greater molecular mass of halogens further down the group, the boiling point increases. Even though the permanent dipole-dipole interactions are less, because the C-X bond becomes less polar.

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5
Q

Are halogenoalkanes soluble in water?

A

Not soluble or only slightly soluble.

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6
Q

What is a nucleophile?

A

A species with a lone pair of electrons

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7
Q

Which 3 nucleophiles do halogenoalkanes react with?

A

Cyanide ions
Hydroxide ions
Ammonia

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8
Q

How does nucleophilic substitution work?

A

The C-X bond is polar leaving the carbon slightly positive. The lone pair on the species is very negative so it is attracted to the positive carbon which forms a new bond, kicking out the halogen

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9
Q

What does nucleophilic substitution of a halogenoalkane by OH produce

A

an Alcohol
(+ an ionic substance halide e.g Sodium Bromide from Sodium Hydroxide)

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10
Q

Iodoethane + Potassium Hydroxide&raquo_space;>

A

Ethanol + Potassium iodide

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11
Q

What does nucleophilic substitution of a halogenoalkane by CN produce

A

a Nitrile
(+ ionic halide compound e.g Sodium Bromide from Sodium Hydroxide)

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12
Q

Bromoethane + Potassium Cyanide&raquo_space;> (hint add 1 C)

A

Propanenitrile + Potassium Bromide

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13
Q

What does nucleophilic substitution of a halogenoalkane by NH3 produce

A

an Amine
(+ ammonium halide)

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14
Q

Bromoethane + Ammonia&raquo_space;>

A

Ethylamine + Ammonium Bromide

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15
Q

When drawing a mechanism, how many arrows are there in total for OH and CN

A

2

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16
Q

When drawing a mechanism, how many arrows are there in total for NH3

A

3

17
Q

Where do the 2 arrows go in OH mechanism

A

Lone pair on the O to the slightly positive Carbon

C-X bond to the halogen

18
Q

Where do the 2 arrows go in CN mechanism

A

From lone pair on the C of the CN to the slightly positive C in the chain

From the C-X bond to the halogen

19
Q

Where do the arrows go in the NH3 mechanism

A

From lone pair on the N to the slighty postive C

From the C-X bond to the halogen

From one of the N-H bonds to the H

20
Q

What are the conditions of an elimination reaction?

A

Hot ethanolic

HYDROXIDE ION ONLY

21
Q

Does the OH act as a base or nucleophile in elimination

A

Base

22
Q

What does elimination of a halogenoalkane form?

A

Alkene
Halide ion
Water

23
Q

How does elimination work?

A
  • OH attacks a hydrogen that is attached to the carbon that is attached to the halogen
    X-C-C-H
  • OH (accepts a proton) takes H to form water
  • The electrons from the C-X bond go to form a C=C double bond

-The halogen C now has 5 bonds so kicks out the halogen

24
Q

What is the chemical formula of Ozone?

A

O3

25
Q

Is ozone very reactive?

A

Yes

26
Q

What are CFCs

A

Chlorofluorocarbons

27
Q

Are CFCs reactive?

A

Not particularly

28
Q

What is the Ozone layer and what does it do?

A

1) The ozone layer is found in the stratosphere and is important for preventing harmful UV rays from entering our atmosphere

29
Q

What were CFCs used for?

A

Refrigerants and aerosols

30
Q

How do CFCs damage the ozone layer?

A
  • CFCs were broken down by UV light to form UV radicals
  • The radicals react with ozone to produce o2, this damages and depletes the ozone layer
  • The process is through free radical substitution
  • One chlorine radical can cause many more ozone molecules to be broken down. This is because at the end of the propagation steps, more chlorine radicals are produced.
31
Q

To what extent has the ozone layer recovered?

A

The ozone layer has still not fully recovered

Even though there are no more CFCs in use, chlorine radicals are still being produced which break down ozone.

The ozone layer has recovered to an extent since CFCs were banned worldwide in 1987

32
Q

What has replaced CFCs and why are they better?

A

Hydrogenfluorocarbons HFCs
HFCs break down more easily in the atmosphere (lower) instead of becoming a radical