Portmess Homework Solutions 47

Portmess Homework Solutions 47 - ( c ) The oxidation number...

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( b ) Converting a ketone to an ester increases the oxygen content of carbon and requires an oxidizing agent. ( c ) Reduction occurs when the hydrogen content increases, as in the conversion of a ketone to an alkane or to an alcohol. Reductions are carried out by using reagents that are reducing agents. 2.40 Methyl formate is an ester. ( a ) The oxidation numbers of the two carbon atoms in methyl formate and the carbon atoms in the reaction products can be determined by comparison with the entries in text Table 2.6. There has been no change in oxidation state in going from reactants to products, and the reac- tion is neither oxidation nor reduction. The number of carbon oxygen bonds does not change in this reaction. ( b ) As in part ( a ), the oxidation states of the carbon atoms in both the reactant and the products do not change in this reaction. The reaction is neither oxidation nor reduction.
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Unformatted text preview: ( c ) The oxidation number of one carbon of methyl formate has decreased in this reaction. This reaction is a reduction and requires a reagent that is a reducing agent. ( d ) The oxidation number of both carbon atoms of methyl formate has increased. This reaction is an oxidation and requires use of a reagent that is an oxidizing agent. HCOCH 3 O H 2 O 2 CO 2 1 Oxidation number 1 2 2 2 1 4 HCOCH 3 O 2 CH 3 OH Oxidation number 1 2 2 2 2 2 HCOCH 3 O 1 Oxidation number 1 2 2 2 HCONa O 1 2 CH 3 OH 2 2 HCOCH 3 O CH 3 OH 1 HCOH O Oxidation number 1 2 2 2 1 2 2 2 CH 3 CC(CH 3 ) 3 O oxidation reduction reduction CH 3 COC(CH 3 ) 3 O CH 3 CHC(CH 3 ) 3 OH CH 3 CH 2 C(CH 3 ) 3 Ester Alkane Alcohol Compound A ( d ) ALKANES 41...
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This note was uploaded on 09/20/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.

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