(YELLOW) halogenoalkanes Flashcards

(YELLOW) halogenoalkanes

1
Q

describe the solubility of halogenoalkanes

A

However, halogenoalkane molecules are not polar enough to make even those molecules with the shortest chains soluble in water. There are no Hd+ atoms so hydrogen bonding with water cannot occur. They are immiscible in water. They will dissolve in organic solvents such as alcohols and ethers.

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2
Q

what kind of intermolecular forces are present

A

london forces

perminent dipole - perminent dipole

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3
Q

What is the trend in boiling point with increased chain length?

Explain your answer

A

increases

Increased chain length —> more electrons —> stronger London forces.

Require more heat energy to overcome them

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4
Q

What is the trend in boiling point as you go down the halogen group?

Explain your answer

A

increases

Increased number of electrons —> stronger London forces.

Require more heat energy to overcome them

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5
Q

What is the trend in boiling points 1o –> 2o –> 3o

Explain your answer.

A

decreases

Increased branching —> molecules further apart —>weaker London forces. (Also branched molecule —-> smaller dipole —> weaker London forces.)

So less heat energy required to overcome them and separate the molecules —-> vapour

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6
Q

how do you go from CH3CHOHCH3 to CH3CHClCH3

give reagents and conditions

This is not a very economical way to make large quantities as PCl5 is expensive, can you think of another reason?

A

Reagents phosphorus pentachloride Conditions room temp

CH3CHOHCH3 + PCl5 —> CH3CHClCH3 + POCl3(l) + HCl(g)

Hydrogen chloride gas produced is corrosive

could also use HCl and it would produce water instead

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7
Q

how do you get a bromoalkane from an alcohol

give reagents and conditions

also give the reaction need to make PBr3

A

1) From Bromine and moist red phosphorus (forming PBr3 in situ)Equation to make PBr3 :

P + 1½Br2 —> PBr3

3x(secondary alcoho)l + PBr —> 3 x(secondary halogenoalkane) + H3PO3

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8
Q

how do you get a iodoalkane from an alcohol

give the conditions and reagents

A

These can only be made from Iodine and red phosphorus (PI3 made ‘in situ’)

Equation: P + 1½I2 —-> PI3

conditions warmed + moist

3 x (secondary alcohol) + PI3 —> iodoalkane + H3PO3

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9
Q

what is the aparatus use to connect the tap funnel to the condencer

A

distillation head

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10
Q

what is the aparatus that is used to add the conc sulfuric acid in

A

tap funnel as its too corrosive to handle drop wise

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11
Q

Write an equation for the reaction of concentrated sulphuric acid with sodium bromide.

A

NaBr(s) + H2SO4(50% conc) —-> NaHSO4(aq) + HBr(g)

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12
Q

Why is it necessary to add the concentrated sulphuric acid slowly and cool the flask in cold water?

A

To avoid the reaction of 2HBr + H2SO4 —> SO2(g) + Br2(g) + 2H2O(l)

HBr acts as a reducing agent reducing H2SO4 to SO2 and being itself oxidised to Br2

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13
Q

Why is there a tube with an open-side arm between the condenser and the tap funnel?

A

So it is not a closed system where pressure of vapour could build up.

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14
Q

What changes did you see in the flask as you added the acid and then heated the mixture?

A

Steamy fumes (and orange fumes)

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15
Q

Can you use your observations to identify any of the products formed?

A

Steamy fumes of HBr, possibly SO2 and orange fumes of Br2

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16
Q

Explain what is meant by heating under reflux. Why is this often necessary when heating organic liquids?

A

Heated solution evaporates and vapour rises up vertical condenser.

Vapour condenses and drops back into the reaction vessel.

This avoids flammable vapours being released

It allows a longer reaction time with no loss of flask contents.

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17
Q

What is the reason for including a guard tube in the apparatus during the refluxing stage?

A

Contains soda-lime which reacts with any acidic gases (HBr) preventing their release

18
Q

Why is it necessary to heat the flask for such a long time?

A

To ensure complete reaction.

19
Q

can H2SO4 dissolve in water

A

yes

20
Q

info card read and recite

A

Purifying the product
The impure product is shaken with conc.HCl.Butan-1-ol dissolves in the conc. HCl.

1)Why are there two layers in the separating funnel when the product is shaken with aqueous reagents?

Bromobutane is immiscible in water/aqueous solutions as it only has permanent dipole forces and is unable to break up the stronger H-bonding between water molecules.

2) Suggest a reason why shaking the product with hydrochloric acid helps to remove unchanged butan-1-ol from the impure product.

Butan-1-ol is soluble in HCl because it does H-bonding

Conc. HCl protonates the butan-1-ol —> CH3CH2CH2CH2OH2+, which is ionic and therefore soluble

3) Which impurities are removed by shaking the product with a solution of sodium hydrogencarbonate? Acidic impurities e.g. original H2SO4, HBr from reaction, HCl added

Why is there a marked build up of pressure during the shaking? CO2 produced

Write an ionic equation for this reaction H+(aq) + HCO3-(aq) —> H2O(l) + CO2(g)

4) Suggest a reason why sodium hydrogencarbonate solution is used rather than sodium hydroxide solution. :OH- would act as a nucleophile substituting for Br —> butan-1-ol

5) How did the appearance of your product change when it was swirled with anhydrous sodium sulphate? Cloudy liquid (emulsion) became a clear colourless liquid.

How do you account for this?
Removal of water

6) Name another drying agent which could be used in place of the anhydrous sodium sulphate.

Anhydrous salt eg. CuSO4 , MgSO4, CaCl2, or silica gel crystals

21
Q

info card read and recite

A

All the impurities should now have been removed, except small amounts of organic impurities which will be dissolved in the bromobutane. These can be separated using distillation.

At first, you will see organic impurities coming out of solution, but nothing will condense in the receiver. Then the temperature of the vapour will rise.

The boiling point of 1-bromobutane is 102°C, we collect the fraction which distills over in the range 100 °C to 104°C.

Higher boiling point impurities will remain in the flask

22
Q

Explain the term ‘fraction’ to describe the sample of product collected during the final distillation.

A

Part of the whole sample having a narrow specified boiling point range

23
Q

How does the final stage allow you to check the identity and purity of your product?

A

Should all distil over at 102oC, any impurities will lower the B.Pt.

24
Q

Experience suggests that you will do well to get a yield over 50 %.

What reasons can you think of to account for a yield well under 100 %?

A

Reaction may not have gone to completion, side reactions produce other products, lengthy separation and purification process result in losses during transfer.

25
Q

Look carefully at the equation for the preparation of 1-bromobutane. What type of reaction is this?

CH3CH2CH2CH2OH(l) + HBr(g) —> CH3CH2CH2CH2Br(l) + H2O(l

A

Nucleophilic substitutio

26
Q

Most reactions of halogenoalkanes are nucleophilic substitution reactions.

A
27
Q

what is the definition of a substitution reaction

A

A substitution reaction Reaction where one atom or group of atoms is replaced by another atom or group of atoms.

28
Q

what is the definition of A nucleophile is

A

A nucleophile is An atom or molecule with a lone pair of electrons They will attack a d+C and form a new (dative) covalent bond

29
Q

how do you go from a halogenoalkane to an alcohol

give reagents
conditions and reaction type

A

Reagent: Dilute NaOH(aq) or KOH (aq) and the halogenoalkane e.g.chloromethane

Conditions: Dissolve halogenoalkane in ethanol and heat (60oC) with dilute aqueous NaOH/KOH

A nucleophile substitution

CH3Cl(ethanol) + NaOH(aq) —> CH3OH + NaCl

30
Q

info card

A

NUCLEOPHILC SUBSTITUTION MECHANISMS – SN1 AND SN2

The mechanism for the reaction of a halogenoalkane with a strong nucleophile such as OH- depends on the type of halogenoalkane. (primary, secondary or tertiary)

SN1 mechanism

Tertiary halogenoalkanes react with the SN1 mechanism which is a 2-step mechanism with

only 1 molecule involved in the slow step of the mechanism

e.g the reaction of 2-chloro-2-methylpropane with aqueous NaOH solution

step 1 (slow step)

step 2 (fast step)

What stabilises the intermediate carbocation formed after step 1? Methyl groups tend to donate electrons so stabilising the tertiary carbocation

SN2 mechanism

Primary halogenoalkanes react with aqueous NaOH by the SN2 mechanism which is a 1-step mechanism with 2 species involved in the slow step of the mechanism

e.g the reaction of chloroethane with aqueous NaOH solution

31
Q

Would you expect substitution reactions to happen more readily or less readily as you progress down the halogen group?

explain why

A

More readily.

Strength of the C-halogen bond decreases on descending the group. Larger halogen radius, more shielding of shared pair of electrons from the halogen nucleus, so shared pair of electrons held less strongly.

32
Q

Hydrolysis:-

· Put 1 cm3 of 20% potassium hydroxide in ethanol into a test tube add an equal volume of water.

· Add 3 drops of 2-bromo-2-methylpropane and shake the tube from side to side for a minute.

To test for bromide ions:-

· Acidify by adding an equal volume of 2 M nitric acid to neutralise the potassium hydroxide (test with indicator paper to ensure that your solution is acidic)

· Add a few drops of 0.02 M silver nitrate. If bromide ions are present a cream coloured precipitate of silver bromide will appear

What new organic compound has been formed?

Why was ethanol included in the reaction mixture?

Write an ionic equation for the formation of the precipitate:

A

Alcohol

To act as a solvent for the halogenoalkane

Ag+(aq) + Br-(aq) —> AgBr(s)

33
Q

add 3 halogeno alkanes cl , br and I in to test tubes , add silver nitrate and then put in a water bath

what are the observations
and what is the order that they appear in

Which is the most polar halogen–carbon bond?
Which is the biggest halogen–carbon bond energy?
Which factor appears to be the most important when considering the rates of reaction?

A

1-chlorobutane:
white ppt
last to appear

1-bromobutane 1-iodobutan:
cream coloured ppt
second to appear

1-iodobutan
yellow ppt
first to appear

Cd+-Cld-

C-Cl

Bond enthalpy is more important.

34
Q

Effect of nature of the carbon skeleton (primary, secondary or tertiary)

add halogenoalkane but whith diffrent degrees e.g primary secondary , tertiary , add silver nitrate and then put in a water bath

what are the observations ?

order in which ppt apears ?

A

1-bromobutane
cream ppt
last for ppt to apear

2-bromobutane
cream ppt
middle ppt to apear

2-bromo-2-methylpropane
cream/yellow ppt (so slghtly more yellow)
first to ppt

State the order of rate of reactivity of halogenoalkanes tertiary>secondary>primary

What is the nucleophile in these reactions? H2O

Is this nucleophile stronger or weaker than the OH- ion? weaker
Because the nucleophile is weak the mechanism here is SN1 .
Use this mechanism to explain the order of reactivity. C-X bond breaks leaving a carbocation, tertiary carbocations are more stable and exist long enough to react with H2O nucleophile

Why was it not necessary to acidify these solutions as you did in the reaction with NaOH?

Alkali was not added so there would not be a precipitate of silver oxide.

Explain why adding ethanol to the silver nitrate before adding the halogenoalkane would have been a good idea. To dissolve the halogenoalkane and improve reaction rate.

35
Q

what is a co-solvent

A

In chemistry, cosolvents are substances added to a primary solvent in small amounts to increase the solubility of a poorly-soluble compound.

36
Q

reaction of ammonia and halogenoalkane
what does it produce

its a nucliophilic substatutiuon

what are the conditions and reagents

give the general equation ?

A

Reagent: Excess NH3 and halogenoalkane

Conditions: alcoholic NH3 (i.e ammonia dissolved in ethanol) heated in a sealed tube

General equation

R-X + NH3 —> RNH2 + NH4X

37
Q

this is a nuecliophilic reaction with a halogenoalkane and water

give the reagents
conditions
and general equation

A

Reagent: H2O and halogenoalkane

Conditions: Dissolve halogenoalkane in ethanol, add H2O and warm.

General equation

R-X + H2O —> ROH + HX

HX will give a ppt if given AgNO3

38
Q

what is the definision of a eleimination reaction

A

Elimination reaction is the removal of one or more groups or atoms from adjacent C atoms forming an unsaturated molecule

39
Q

If a halogenoalkane is strongly heated with excess concentrated alkali dissolved in ethanol (not water) as a solvent then an elimination reaction occurs what is formed

what are the reagents
and concitions
give a general formula

A

an alkene

Reagent: Conc. KOH dissolved in ethanol

Conditions: Heat under reflux

CH3CHXCH3 + KOH —> CH2CH1CH3 + HBr

In this case the OH- ions act as a base removing H+.

(KOH is the reagent of choice as it is more soluble in ethanol.)

40
Q

info card read and recite

A

Put 2 cm3 of 20% potassium hydroxide in ethanol into a boiling tube

· Add 0.5 cm3 of 2-bromo-2-methylpropane and push a loose plug of ceramic fibres into the mixture.

· Arrange the apparatus for collection of gas.

· Heat gently and collect two to three test tubes of gas.

· Prevent suck-back by remove delivery tube from water before stopping heating.

· Test the gas for flammability; observation yellow flame at mouth of test tube

· Test it with bromine water, observation orange bromine water decolourised on shaking

· Test it with KMnO4 + H2SO4, observation purple manganate(VI) decolourised on shaking

So >C=C< functional group

What new gaseous compound has been formed? alkene

What type of reaction has taken place? Elimination of HBr from bromoalkane

An oily film may appear on the water surface, suggest what it is and how it got there.

unreacted halogenoalkane which has vapourised and passed down the delivery tube.

41
Q

this is a nucleophilic substatution
this equation involces increasing the chain length
how do you go from a primary alcohol to a nitrile

what are the reagents and conditions
give a general equation

A

Reagent: KCN and halogenoalkane

Conditions: ALCOHOLIC KCN (i.e potassium hydroxide dissolved in ethanol) heated under reflux

General equation

R-X + KCN —> R-CN + KX