Number of alkanes obtained on electrolysis of a mixture of CH3 COONa and C2 H5 COONa is:

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Given: Electrolysis of a mixture of CH3 COONa and C2 H5 COONa. Find: Number of alkanes formed. From CH3 COONa, the radical formed is CH3 COONa CH3 From C2 H5 COONa, the radical formed is C2 H5 COONa C2 H5 Now these radicals can combine as follows: 2 CH3 CH3 CH3 This gives ethane. 2 C2 H5 CH3 CH2 CH2 CH3 This gives butane. CH3 + C2 H5 CH3 CH2 CH3 This gives propane. The solution labels this as propene once, but the product formula shown is CH3 CH2 CH3, which is propane. Therefore, the alkanes obtained are ethane, propane, and butane. Hence, the number of alkanes formed is 3. The correct option is B.

Given: A mixture of acetate and propionate salts undergoes Kolbe electrolysis. Find: Count all distinct alkane products. The radicals formed are CH3 and C2 H5. Possible radical couplings are: 1. CH3 with CH3 2 CH3 C2 H6 2. C2 H5 with C2 H5 2 C2 H5 C4 H10 3. CH3 with C2 H5 CH3 + C2 H5 C3 H8 Thus, the distinct alkanes are C2 H6, C3 H8, and C4 H10. Therefore, the number of alkanes is 3.

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