Halogenoalkane reactions and mechanisms

Question 1

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

Answer 1

ROH, an alcohol

Question 2

halogenoalkane

Answer 2

ROH, an alcohol

Question 3

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

Answer 3

RCN, a nitrile

Question 4

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

Answer 4

RNH2, a primary amine

Question 5

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

Answer 5

the OH- ion

Question 6

nitriles are

Answer 6

organic compounds containing the C-CN group

Question 7

primary amines are

Answer 7

compounds containing the C-NH2 group

Question 8

nucleophilic substitution reactions are

Answer 8

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

Question 9

an ethanoic solution is

Answer 9

one in which ethanol is the solvent

Question 10

an elimination reaction is

Answer 10

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

Question 11

reactions in which nitriles are formed are used to

Answer 11

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

Question 12

the equation for the conversion of bromoethane into propanenitrile is

Answer 12

C2H5Br + KCN = C2H5CN + Br

Question 13

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

Answer 13

sealed tube is used because the NH3 gas would escape otherwise

Question 14

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

Answer 14

NH3 molecule

Question 15

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

Answer 15

CH3CH2CH2CH2I + NH3 = CH3CH2CH2CH2NH2 + HI

Question 16

Ammonia, NH3, is

Answer 16

a base

Question 17

the organic product, CH3CH2CH2CH2NH2 is

Answer 17

a base

Question 18

the inorganic product, HI is

Answer 18

an acid

Question 19

In Reaction 4, the products react because

Answer 19

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

Question 20

in Reaction 4, the products react to form

Answer 20

a salt

Question 21

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

Answer 21

C4H9I + NH3 = C4H9NH3+ + I-

Question 22

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

Answer 22

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

Question 23

To produce a high yield of amine,

Answer 23

ammonia is used in excess

Question 24

to produce a high yield of amine, ammonia is used

Answer 24

in excess

Question 25

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

Answer 25

the amine

Question 26

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

Answer 26

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

Question 27

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

Answer 27

C4H9I + 2NH3 = C4H9NH2 + NH4+I-

Question 28

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

Answer 28

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

Question 29

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

Answer 29

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

Question 30

'Hetero' means

Answer 30

different

Question 31

heterolytic fission means

Answer 31

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

Question 32

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

Answer 32

heterolytic fission

Question 33

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

Answer 33

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

Question 34

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

Answer 34

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

Question 35

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

Answer 35

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

Question 36

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

Answer 36

ethanol, as ethanoic potassium hydroxide

Question 37

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

Answer 37

base and not a nucleophile

Question 38

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

Answer 38

a nucleophile

Question 39

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

Answer 39

H+ ion

Question 40

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

Answer 40

a carbon atom next to the C-X bond

Question 41

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

Answer 41

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

Question 42

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

Answer 42

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

Question 43

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

Answer 43

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

Question 44

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

Answer 44

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.