Answer on Organic Chemistry Question for naz
i) C2H6 into C2H5OH
ii) C2H5Br into C2H5COOH
iii) 1-bromopropane into 2-bromopropane
2. simple alkenes such as ethene undergo addition reactions with chlorine and bromine at room temperature whereas simple alkanes such as methane react only at elevated temperatures on in the presence of ultraviolet light. how do you account for the difference in reactivity in terms of structure and bonding?
C2H5Cl + KOH --> C2H5OH + KCl
ii. C2H5Br + KCN ———> C2H5CN + KBr
C2H5CN + 2H2O ———> C2H5COOH + NH3
iii. The Dehydrohaogenation of 1-bromo propane with alcoholic KOH gives propenewhich on again hydrohalogenation with HBr gives 2-bromo propane due to
Markonikove's rule for addition.
CH3CH2CH2Br + alc. KOH -------> CH3CH=CH2 + KBr
CH3CH=CH2 + HBr -------> CH3CHBrCH3
2. The only other reaction that an alkane will undergo is a reaction with ahalogen ( chlorine or bromine typically ) with UV light present as an initiator
of the reaction. The UV light causes the formation of free radical halogen atoms by providing
enough energy for the bond between the two halogen atoms to break. A halogen atom attacks the alkane, substituting itself for a hydrogen atom.
This substitution may occur many times in an alkane before the reaction is
finished. A similar process occurs high up in the earth's atmosphere when CFC's and other
organic solvents react with intense sunlight to produce free radicals, chlorine
atoms in this case. These attack molecules of ozone (O3) depleting ozone'sconcentration and leading to the "holes".
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