Assignment 14D
Introduction:
In this assignment I will be explaining the chemical reactions carried out practically,
assessing the importance of the conditions chosen for the reactions carried out
practically and evaluating the importance of the reaction conditions for the reactions.
Researching methods for organic compounds for the following 3 types of organic
compounds:
● 1. Preparation of a compound without a carbonyl functional group of how to
prepare propan-1-ol from 1-chloropropane.
● 2. Preparation of a compound with a carbonyl functional group of how to prepare
ethanal from ethanol.
● 3. Preparation of an aromatic compound of how to prepare nitrobenzene from
benzene.
P5
Reaction 1 (Preparation of a compound without a carbonyl functional group of how to
prepare propan-1-ol from 1-chloropropane)
Balanced chemical equation:
CH₃CH₂CH₂Cl + NaOH → CH₃CH₂CH₂OH + NaCl
The functional groups involved are Haloalkane (-Cl) and Primary alcohol (-OH).
Explaining the mechanism , OH⁻ attacks the carbon bonded to Cl (electrophilic) , Cl⁻
leaves, its a one-step reaction.
M5, D4
The conditions needed for this reaction is firstly, a strong base like Sodium hydroxide
(NaOH) or potassium hydroxide (KOH) provides the hydroxide ion (OH⁻).
Reflux heating to make sure that the reaction mixture is properly heated to continue
without losing volatile reactants (such as the haloalkane). Maintaining the high
temperature allows the reaction to speed up without evaporating the reactants, without
the heat the reaction would be slow so the reactants would evaporate before reacting.
However, if the reaction is too hot then it would be at risk of decomposition. Although
there wouldn't be enough energy to cross the activation barrier, the reaction rate would
be much lower if it were carried out at room temperature or without reflux.
, It needs a solvent which is usually, aqueous ethanol, which is a mixture of ethanol and
water , Ethanol helps in the dissolution of the organic haloalkane; water aids in the
dissolution of NaOH. If only waters used then the haloalkane may not dissolve well and
cause a slow reaction , if only ethanol is used then is used then the NaOH may not
dissolve enough. While 1-chloropropane is mostly non-polar and dissolves better in
ethanol, sodium hydroxide is soluble in water. By making sure that both reactants are in
the same phase, this mixed solvent maximises molecular collisions and promotes an
effective SN2 reaction. The haloalkane may separate from the aqueous phase in the
absence of ethanol, decreasing interaction with hydroxide ions and delaying the
process. In a similar vein, NaOH would not dissolve efficiently to provide nucleophilic
OH⁻ ions in the absence of water.
Due to its low steric hindrance, SN2 functions well with primary halides such as
1-chloropropane. If tertiary halides or secondary halides were used SN2 wouldnt occur
and SN1 might take over. The SN2 mechanism is extremely sensitive to steric
hindrance, the reaction especially benefits from the use of a main halogenoalkane. The
nucleophile may reach and attack the electrophilic carbon directly because primary
carbon centres are less obstructed. The reaction would be slower or may proceed via
an SN1 pathway, which requires a carbocation intermediate and might result in side
reactions or rearrangements, if a secondary or tertiary halogenoalkane were
substituted.
This reaction needs a strong acid because it needs the OH- nucleophile, without the
strong acid (NaOH) there would be no OH- which would mean no substitution and no
alcohol formed.
Reaction 2 ( Preparation of a compound with a carbonyl functional group of how to
prepare ethanal from ethanol)
P5
Balanced chemical equation:
CH₃CH₂OH + [O] → CH₃CHO + H₂O
The function groups involved in the reaction is primary alcohol(-OH) which comes from
the compound Ethanol , and aldehyde (-CHO , with carbonyl group C=O) which comes
from the compound Ethanal. This reaction introduces a carbonyl functional group into
the product.
Explaining the reaction mechanism , the process involves a primary alcohol being
oxidised to an aldehyde , One hydrogen from the carbon bonded to the hydroxyl group
(-OH) and the hydrogen from the hydroxyl group itself are eliminated. As a result, a
Introduction:
In this assignment I will be explaining the chemical reactions carried out practically,
assessing the importance of the conditions chosen for the reactions carried out
practically and evaluating the importance of the reaction conditions for the reactions.
Researching methods for organic compounds for the following 3 types of organic
compounds:
● 1. Preparation of a compound without a carbonyl functional group of how to
prepare propan-1-ol from 1-chloropropane.
● 2. Preparation of a compound with a carbonyl functional group of how to prepare
ethanal from ethanol.
● 3. Preparation of an aromatic compound of how to prepare nitrobenzene from
benzene.
P5
Reaction 1 (Preparation of a compound without a carbonyl functional group of how to
prepare propan-1-ol from 1-chloropropane)
Balanced chemical equation:
CH₃CH₂CH₂Cl + NaOH → CH₃CH₂CH₂OH + NaCl
The functional groups involved are Haloalkane (-Cl) and Primary alcohol (-OH).
Explaining the mechanism , OH⁻ attacks the carbon bonded to Cl (electrophilic) , Cl⁻
leaves, its a one-step reaction.
M5, D4
The conditions needed for this reaction is firstly, a strong base like Sodium hydroxide
(NaOH) or potassium hydroxide (KOH) provides the hydroxide ion (OH⁻).
Reflux heating to make sure that the reaction mixture is properly heated to continue
without losing volatile reactants (such as the haloalkane). Maintaining the high
temperature allows the reaction to speed up without evaporating the reactants, without
the heat the reaction would be slow so the reactants would evaporate before reacting.
However, if the reaction is too hot then it would be at risk of decomposition. Although
there wouldn't be enough energy to cross the activation barrier, the reaction rate would
be much lower if it were carried out at room temperature or without reflux.
, It needs a solvent which is usually, aqueous ethanol, which is a mixture of ethanol and
water , Ethanol helps in the dissolution of the organic haloalkane; water aids in the
dissolution of NaOH. If only waters used then the haloalkane may not dissolve well and
cause a slow reaction , if only ethanol is used then is used then the NaOH may not
dissolve enough. While 1-chloropropane is mostly non-polar and dissolves better in
ethanol, sodium hydroxide is soluble in water. By making sure that both reactants are in
the same phase, this mixed solvent maximises molecular collisions and promotes an
effective SN2 reaction. The haloalkane may separate from the aqueous phase in the
absence of ethanol, decreasing interaction with hydroxide ions and delaying the
process. In a similar vein, NaOH would not dissolve efficiently to provide nucleophilic
OH⁻ ions in the absence of water.
Due to its low steric hindrance, SN2 functions well with primary halides such as
1-chloropropane. If tertiary halides or secondary halides were used SN2 wouldnt occur
and SN1 might take over. The SN2 mechanism is extremely sensitive to steric
hindrance, the reaction especially benefits from the use of a main halogenoalkane. The
nucleophile may reach and attack the electrophilic carbon directly because primary
carbon centres are less obstructed. The reaction would be slower or may proceed via
an SN1 pathway, which requires a carbocation intermediate and might result in side
reactions or rearrangements, if a secondary or tertiary halogenoalkane were
substituted.
This reaction needs a strong acid because it needs the OH- nucleophile, without the
strong acid (NaOH) there would be no OH- which would mean no substitution and no
alcohol formed.
Reaction 2 ( Preparation of a compound with a carbonyl functional group of how to
prepare ethanal from ethanol)
P5
Balanced chemical equation:
CH₃CH₂OH + [O] → CH₃CHO + H₂O
The function groups involved in the reaction is primary alcohol(-OH) which comes from
the compound Ethanol , and aldehyde (-CHO , with carbonyl group C=O) which comes
from the compound Ethanal. This reaction introduces a carbonyl functional group into
the product.
Explaining the reaction mechanism , the process involves a primary alcohol being
oxidised to an aldehyde , One hydrogen from the carbon bonded to the hydroxyl group
(-OH) and the hydrogen from the hydroxyl group itself are eliminated. As a result, a
