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Question

Which of the following is not prepared by Kolbe's electrolytic process?

A
$$C_{4}H_{10}$$
B
$$C_{6}H_{14}$$
C
$$C_{3}H_{8}$$
D
$$C_{2}H_{6}$$
Solution
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Correct option is A. $$C_{3}H_{8}$$
Hydrocarbons prepared by Kolbe's Electrolysis contains even number of carbons.
Of the above given compounds, only $$\mathrm{C_3H_8}$$ has odd number of carbons in it. So, it can't be prepared by Kolbe's electrolysis process.

Hence, Option "A" is the correct answer.

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