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Topic 6.3 (Halogenoalkanes) > Halogenoalkane reactions and mechanisms > Flashcards

Halogenoalkane reactions and mechanisms Flashcards

1
Q

Reaction 1 is the reaction between a halogenoalkane and water in warm conditions, which produces

A

ROH, an alcohol

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2
Q

halogenoalkane

A

ROH, an alcohol

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3
Q

Reaction 3 is the reaction between a halogenoalkane and potassium cyanide (KCN) dissolved in ethanol in heat under reflux, producing

A

RCN, a nitrile

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4
Q

Reaction 4 is the reaction between a halogenoalkane and ammonia solution, forming

A

RNH2, a primary amine

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5
Q

In Reaction 2, (KOH), the attacking nucleophile is

A

the OH- ion

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6
Q

nitriles are

A

organic compounds containing the C-CN group

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7
Q

primary amines are

A

compounds containing the C-NH2 group

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8
Q

nucleophilic substitution reactions are

A

reactions in which an attacking nucleophile replaces an existing atom or group in a molecule

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9
Q

an ethanoic solution is

A

one in which ethanol is the solvent

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10
Q

an elimination reaction is

A

one in which a molecule loses atoms attached to adjacent carbon atoms, forming a double C=C

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11
Q

reactions in which nitriles are formed are used to

A

extend the carbon chain as the original halogenoalkane gains a CN group

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12
Q

the equation for the conversion of bromoethane into propanenitrile is

A

C2H5Br + KCN = C2H5CN + Br

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13
Q

in Reaction 4, heating the halogenoalkane with ammonia, a …………….. tube is used because……

A

sealed tube is used because the NH3 gas would escape otherwise

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14
Q

The attacking nucleophile in Reaction 4 (halogenoalkane and ammonia) is

A

NH3 molecule

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15
Q

the equation for the reaction between 1-iodobutane and ammonia gas to give butylamine is

A

CH3CH2CH2CH2I + NH3 = CH3CH2CH2CH2NH2 + HI

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16
Q

Ammonia, NH3, is

A

a base

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17
Q

the organic product, CH3CH2CH2CH2NH2 is

A

a base

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18
Q

the inorganic product, HI is

A

an acid

19
Q

In Reaction 4, the products react because

A

the organic product is a base and the inorganic product is an acid

20
Q

in Reaction 4, the products react to form

A

a salt

21
Q

The first substep of Reaction 4, which shows the formation of a salt, is

A

C4H9I + NH3 = C4H9NH3+ + I-

22
Q

The product of the first substep, C4H9NH3, is ………… and not …………

A

is a salt and not an amine, so is not the final step

23
Q

To produce a high yield of amine,

A

ammonia is used in excess

24
Q

to produce a high yield of amine, ammonia is used

A

in excess

25
Q

if used in excess, the ammonia will react in a second substep in Reaction 4 to to produce

A

the amine

26
Q

the equation which shows the second substep in Reaction 4, where excess ammonia reacts with the salt to produce butylamine is

A

C4H9NH3+I- + NH3 = C4H9NH2 + NH4+I-

27
Q

the two substeps of Reaction 4 can be combined in the equation

A

C4H9I + 2NH3 = C4H9NH2 + NH4+I-

28
Q

Reaction 2, between a halogenoalkane and potassium hydroxide, starts when

A

the oxygen in the hydroxide ion donates a lone pair of electrons to the electron-deficient carbon atom and the formation of a C-O bond

29
Q

while the oxygen in the OH- ion donates a lone pair of electrons to the carbon atom to form a C-O bond in Reaction 2, the electrons in the C-X bond

A

move to the X atom, breaking the C-X bond

30
Q

‘Hetero’ means

A

different

31
Q

heterolytic fission means

A

the shared pair of electrons in a bond both go to one atom, meaning the other does not gain any electrons

32
Q

the bond breaking of the C-Br bond in Reaction 2 is

A

heterolytic fission

33
Q

The first step for the mechanism of Reaction 4 (halogenoalkane and ammonia) is

A

the donation of a lone pair of electrons from the nitrogen of an ammonia molecule to the electron deficient carbon atom and the formation of a C-X bond

34
Q

while the nitrogen in the ammonia molecule donates a lone pair of electrons to the electron-deficient carbon to form a C-N bond, the electrons in the C-X bond

A

move to the X atom, resulting in the breaking of the C-X bond in heterolytic fission

35
Q

the second step for the mechanism of Reaction 4 involves the excess ammonia, which acts as

A

a base, removing a hydrogen ion from the ion formed in the first step of the mechanism, forming the primary amine and an NH4+ ion

36
Q

Reaction 2, between a halogenoalkane and aqueous potassium hydroxide can also be performed with the potassium hydroxide in not water but

A

ethanol, as ethanoic potassium hydroxide

37
Q

when a halogenoalkane is heated with ethanoic potassium hydroxide, the OH- ion acts as a

A

base and not a nucleophile

38
Q

when a halogenoalkane is heated with ethanoic potassium hydroxide, OH- ion acts as a base and not

A

a nucleophile

39
Q

when ethanoic KOH is heated with a halogenoalkane, then the OH- ion acts as a base, reacting with a

A

H+ ion

40
Q

when ethanoic KOH is heated with a halogenoalkane, the OH- ion reacts with a H+ ion which is attached to

A

a carbon atom next to the C-X bond

41
Q

predict where the double bond will form when 2-bromopropane reacts with ethanoic potassium hydroxide: (CH3-CHBr-CH3 + KOH)

A

the OH- ion acts as a base, so reacts with an H+ ion, and as this is a halogenoalkane reaction, the H+ ion which reacts is bonded to the carbon atom next to the C-Br bond, so the double bond will occur here: CH2=CH-CH3

42
Q

when halogenoalkanes react with ethanoic KOH, which hydrogen ion (H+) reacts with the OH- ion in the reaction mixture?

A

the H+ ion which reacts is the one attached to a carbon atom next to the C-X bond in the halogenoalkane molecule

43
Q

the equation for the reaction between 2-bromopropane and ethanoic potassium hydroxide is:

A

CH3-CHBr-CH3 + KOH(alc) = CH2=CH-CH3 + KBr + H2O

44
Q

the reaction between a halogenoalkane and ethanoic potassium hydroxide when heated is an elimination reaction. explain why.

A

the OH- ion in the reaction mixture (provided by the ethanoic KOH) acts as a base, and so reacts with a H+ ion while the K+ ion reacts with the Br- ion in the halogenoalkane. The atoms removed (H+ and Br-) are not replaced, as a double bond forms between the two carbons which lose an atom each and are adjacent (next to each other), so the atoms lost are ‘eliminated’ from the halogenoalkane molecule.

Topic 6.3 (Halogenoalkanes) (3 decks)

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