Halogenoalkane reactions and mechanisms Flashcards
Reaction 1 is the reaction between a halogenoalkane and water in warm conditions, which produces
ROH, an alcohol
halogenoalkane
ROH, an alcohol
Reaction 3 is the reaction between a halogenoalkane and potassium cyanide (KCN) dissolved in ethanol in heat under reflux, producing
RCN, a nitrile
Reaction 4 is the reaction between a halogenoalkane and ammonia solution, forming
RNH2, a primary amine
In Reaction 2, (KOH), the attacking nucleophile is
the OH- ion
nitriles are
organic compounds containing the C-CN group
primary amines are
compounds containing the C-NH2 group
nucleophilic substitution reactions are
reactions in which an attacking nucleophile replaces an existing atom or group in a molecule
an ethanoic solution is
one in which ethanol is the solvent
an elimination reaction is
one in which a molecule loses atoms attached to adjacent carbon atoms, forming a double C=C
reactions in which nitriles are formed are used to
extend the carbon chain as the original halogenoalkane gains a CN group
the equation for the conversion of bromoethane into propanenitrile is
C2H5Br + KCN = C2H5CN + Br
in Reaction 4, heating the halogenoalkane with ammonia, a …………….. tube is used because……
sealed tube is used because the NH3 gas would escape otherwise
The attacking nucleophile in Reaction 4 (halogenoalkane and ammonia) is
NH3 molecule
the equation for the reaction between 1-iodobutane and ammonia gas to give butylamine is
CH3CH2CH2CH2I + NH3 = CH3CH2CH2CH2NH2 + HI
Ammonia, NH3, is
a base
the organic product, CH3CH2CH2CH2NH2 is
a base
the inorganic product, HI is
an acid
In Reaction 4, the products react because
the organic product is a base and the inorganic product is an acid
in Reaction 4, the products react to form
a salt
The first substep of Reaction 4, which shows the formation of a salt, is
C4H9I + NH3 = C4H9NH3+ + I-
The product of the first substep, C4H9NH3, is ………… and not …………
is a salt and not an amine, so is not the final step
To produce a high yield of amine,
ammonia is used in excess
to produce a high yield of amine, ammonia is used
in excess
if used in excess, the ammonia will react in a second substep in Reaction 4 to to produce
the amine
the equation which shows the second substep in Reaction 4, where excess ammonia reacts with the salt to produce butylamine is
C4H9NH3+I- + NH3 = C4H9NH2 + NH4+I-
the two substeps of Reaction 4 can be combined in the equation
C4H9I + 2NH3 = C4H9NH2 + NH4+I-
Reaction 2, between a halogenoalkane and potassium hydroxide, starts when
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
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
move to the X atom, breaking the C-X bond
‘Hetero’ means
different
heterolytic fission means
the shared pair of electrons in a bond both go to one atom, meaning the other does not gain any electrons
the bond breaking of the C-Br bond in Reaction 2 is
heterolytic fission
The first step for the mechanism of Reaction 4 (halogenoalkane and ammonia) is
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
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
move to the X atom, resulting in the breaking of the C-X bond in heterolytic fission
the second step for the mechanism of Reaction 4 involves the excess ammonia, which acts as
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
Reaction 2, between a halogenoalkane and aqueous potassium hydroxide can also be performed with the potassium hydroxide in not water but
ethanol, as ethanoic potassium hydroxide
when a halogenoalkane is heated with ethanoic potassium hydroxide, the OH- ion acts as a
base and not a nucleophile
when a halogenoalkane is heated with ethanoic potassium hydroxide, OH- ion acts as a base and not
a nucleophile
when ethanoic KOH is heated with a halogenoalkane, then the OH- ion acts as a base, reacting with a
H+ ion
when ethanoic KOH is heated with a halogenoalkane, the OH- ion reacts with a H+ ion which is attached to
a carbon atom next to the C-X bond
predict where the double bond will form when 2-bromopropane reacts with ethanoic potassium hydroxide: (CH3-CHBr-CH3 + KOH)
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
when halogenoalkanes react with ethanoic KOH, which hydrogen ion (H+) reacts with the OH- ion in the reaction mixture?
the H+ ion which reacts is the one attached to a carbon atom next to the C-X bond in the halogenoalkane molecule
the equation for the reaction between 2-bromopropane and ethanoic potassium hydroxide is:
CH3-CHBr-CH3 + KOH(alc) = CH2=CH-CH3 + KBr + H2O
the reaction between a halogenoalkane and ethanoic potassium hydroxide when heated is an elimination reaction. explain why.
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.
