Question 3.. PART A. Synthesis Map from Primary Alcohol to

Added on - 16 Sep 2019

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Question 3.PART A. Synthesis Map from Primary Alcohol to Carboxylic AcidRCH2OHRCHO[O]RCOOHPrimary alcohols and aldehydes can be oxidised to carboxylic acids with the use of acidifiedpotassium dichromate (VI) solution. In this reaction, the orange coloured solution of potassiumdichromate (VI) solution turns into green colour.Step 1. Conversion of alcohol to aldehyde (Jones Oxidation)RCH2OH+Cr2O72¿+8H+¿RCHO+2Cr3+¿+7H2O¿¿¿Alcohol can directly be converted to carboxylic acid using an oxidising agent. However, in orderto 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 considerableamount and distilling aldehyde off would make sure that it is not oxidised further.a.Conversion of aldehyde to AcidRCHO+Cr2O72¿+8H+¿RCOOH+2Cr3+¿+7H2O¿¿¿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 ofdichromate (VI).Cr2O72¿+14H+¿+6e¿2Cr3+¿+7H2O¿¿¿¿
This half equation is then combined with the half-equation of the acidic oxidation of analdehyde, as given below:RCHO+H2ORCOOH+2H+¿+2e¿¿¿Part B. Synthesis Map from Ethene to Halo AlkaneAlkenes usually react with hydrogen halides in cold where the double bond is seen to break andthe hydrogen atom is then attached to one carbon atom whereas the bromine atom is attached toadjacent 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 andconsequently forms a new C-H bond. In the first step, the attacking electrophile i.e. the polarisedhydrogen bromide moleculeHδ+¿Brδ¿¿¿, first splits in heterolytical manner and the alkene isprotonated 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 abromoalkance. A pair of electrons is then donated so as to form a new C-Br bond.CH2=CH2+HBrCH3CH2BrThe reaction is a perfect example of electrophilic addition.Step 1. Formation of corresponding carbocationCH2=CH2+Hδ+¿Brδ¿C(H)H2CH2+¿¿¿¿Step 2. Formation of halo alkaneC(H)H2CH2+¿+Br¿C(H)H2CH2+¿¿¿¿¿
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