1) Esters have fairly strong intermolecular forces due to polarity and form separate layers from water since they cannot hydrogen bond with water molecules. 2) Amides are usually made by reaction of acyl halides with ammonia. 3) Amides have high boiling points and are soluble in water because they can hydrogen bond to themselves and water molecules.

1. CHEMISTRY 3.5 ANSWERS
WORKSHEET NINE ORGANIC CHEMISTRY
Carboxylic acids and their derivatives #2
1. As a liquid at the reaction temperature, esters must have high melting points.
This suggests that they have fairly strong intermolecular forces due to the
polarity of the molecules.
As sweet smelling liquids, the esters must also be moderately volatile.
By forming a separate layer on top of the aqueous phase, esters show that they
are not water soluble. This is because they are unable to hydrogen bond to
water molecules.
2. (a) HOCH2CH(OH)CH2OH
(b) 1,2,3-trihydroxypropane or propan-1,2,3-triol
(c) O
CH3(CH2)16C
O CH2
O
CH3(CH2)16C
O CH
O
CH3(CH2)16C
O CH2
(d) In an esterification reaction, the ester link between the acid and the
alcohol is made by the removal of an OH and an H which join to form
H2O. The formation of this water molecule each time an ester linkage is
made, allows the reaction to be classified as a condensation reaction.
3. (a) HCOOH + CH3CH2OH  HCOOCH2CH3 + H2O
(b) The reverse reaction is called hydrolysis because the ester is reacting
with water to become the acid and alcohol again.
(c) A dilute alkali solution such as sodium hydroxide.

2. (d) HCOOCH2CH3 + H2O  HCOOH + CH3CH2OH
The OH- in the alkali reacts with the methanoic acid molecule
producing the methanoate ion. This removal of product shifts the
equilibrium to the right favouring the hydrolysis reaction.
HCOOH + OH-(aq) → HCOO-(aq) + H2O(l)
4. Amides are usually made by reaction of acyl halides with ammonia.
5. (a) Propanamide
(b) Methanamide
6. Amides have high boiling points and are soluble in water because they have
the ability to hydrogen bond to themselves as well as to water molecules.
7. CH3COOCl + NH3  CH3COONH2 + HCl
8. (a) CH3CH2COONH2 + H3O+(aq)  CH3CH2COOH + NH4+
(b) CH3CH2COONH2 + OH-(aq)  CH3CH2COO- + NH3