Organic Chemistry Synthesis Problems

This chemistry assignment presents solved problems focusing on organic synthesis. It covers the synthesis of carboxylic acids from primary alcohols using acidified potassium dichromate, the synthesis of haloalkanes from alkenes (specifically ethene to bromoethane), the synthesis of alcohols by reducing aldehydes, and the synthesis of esters from alcohols and acyl chlorides (ethyl ethanoate example). Each synthesis includes a step-by-step mechanism and explanation, detailing the reactions and the underlying chemical principles. The problems demonstrate electrophilic addition, oxidation, and reduction reactions, along with detailed reaction mechanisms.

Question 3.
PART A. Synthesis Map from Primary Alcohol to Carboxylic Acid
R−C H2 OH → R−CHO [ O ]
→
R−COOH
Primary alcohols and aldehydes can be oxidised to carboxylic acids with the use of acidified
potassium dichromate (VI) solution. In this reaction, the orange coloured solution of potassium
dichromate (VI) solution turns into green colour.
Step 1. Conversion of alcohol to aldehyde (Jones Oxidation)
R−C H2 OH +Cr2 O7
2−¿+8 H + ¿→R −CHO+ 2 Cr3+¿ +7H 2 O ¿
¿ ¿
Alcohol can directly be converted to carboxylic acid using an oxidising agent. However, in order
to synthesis an aldehyde, excess of alcohol is used where aldehyde is distilled off on forming.
The excess of alcohol would make sure that oxidising agent is not present in considerable
amount and distilling aldehyde off would make sure that it is not oxidised further.
a. Conversion of aldehyde to Acid
R−CH O+Cr2 O7
2−¿+8 H+ ¿→R−C O OH +2 Cr3+¿ +7H 2O ¿
¿ ¿
The aldehyde will reduce the orange colour of the dichromate (VI) ions to the green chromium
(III) ions. The electron-half equation can be represented as given below for the reduction of
dichromate (VI).
Cr2 O7
2−¿+14 H+ ¿+6 e−¿ → 2Cr
3+¿ +7H 2O ¿
¿ ¿¿

This half equation is then combined with the half-equation of the acidic oxidation of an
aldehyde, as given below:
R−CHO+ H2 O → R−COOH + 2 H+¿+2 e−¿¿ ¿
Part B. Synthesis Map from Ethene to Halo Alkane
Alkenes usually react with hydrogen halides in cold where the double bond is seen to break and
the hydrogen atom is then attached to one carbon atom whereas the bromine atom is attached to
adjacent carbon atom. Ethene on reacting with hydrogen bromide forms bromo ethane.
HR is an electrophile as it accepts a pair of electron in order to from the alkene bond and
consequently forms a new C-H bond. In the first step, the attacking electrophile i.e. the polarised
hydrogen bromide moleculeHδ +¿− Brδ−¿¿¿, first splits in heterolytical manner and the alkene is
protonated consequently, thereby forms the carbocation and a bromide ion.
Secondly, the bromide ion thus formed is then seen to combine with the carbocation and forms a
bromoalkance. A pair of electrons is then donated so as to form a new C-Br bond.
C H2=C H2+ HBr → C H3 C H2 Br
The reaction is a perfect example of electrophilic addition.
Step 1. Formation of corresponding carbocation
C H2=C H2+ H δ+¿−Brδ−¿→ C ( H ) H 2−C H 2
+¿¿ ¿ ¿
Step 2. Formation of halo alkane
C ( H ) H2−C H2
+¿+ Br−¿→ C ( H ) H 2−C H 2
+¿ ¿ ¿ ¿ ¿

Question 4
a. Synthesis of alcohol by reducing aldehyde
Step 1.
R−CH ¨O+ H−O H2
+¿⇌ R−CH =O+ ¿−H + H 2O ¿¿
Step 2.
R−CH =O+¿− H+ H2 ¨O slow
⇔
R−CH (OH )−O+ ¿¿¿ ¿
Step 3.
R−CH ( OH ) −O+¿¿¿
b. Synthesis of aldehyde
R−C H2 OH +Cr2 O7
2−¿+8 H + ¿→R−CHO+ 2 Cr3+¿ +7H 2O ¿
¿ ¿
C. Synthesis of esters from alcohol and acyl chloride
Ethyl ethanoate (an ester) is formed when ethanoyl chloride is reacted instantly with cold
ethanol. The reaction is highly exothermic in nature and a steamy acidic gas is seen to given off.
RCOCl+ R'−OH → R CO−O R'+ HCl
The mechanism
Step 1. In the first step, addition takes place. The nucleophilic reaction takes place from one of
the lone pairs present on oxygen atom of the alcohol molecule on the positive carbon atom.
RCOCl+ R'− ¨O H → R−C ¿

Step 2. In the second step, elimination takes place. Initially, the carbon-oxygen double bond is
seen to be reformed and then the chloride ion is pushed off.
Step 3. De protonation