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

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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 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)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 considerable amount 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 of dichromate (VI). Cr2O72¿+14H+¿+6e¿2Cr3+¿+7H2O¿¿¿¿
This half equation is then combined with the half-equation of the acidic oxidation of an aldehyde, 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 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.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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