Mehak wanted to prepare an alkane, she added soda lime to sodium ethanoate. The reaction that took place would be: ${CH}_{3} COONa + NaOH + CaO Δ \to {Na}_{2} {CO}_{3} +$

{CH}_{4}

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{CH}_{3} - {CH}_{3}

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{CH}_{3} - COOH

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{CH}_{2} = {CH}_{2}

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Solution

The correct option is A ${CH}_{4}$
Sodium salts of carboxylic acids on heating with soda lime give alkanes containing one carbon atom less than the carboxylic acid. This reaction is known as decarboxylation. Decarboxylation of sodium ethanoate leads to formation of methane.

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A g N O_{3}

An alkane with molecular formula $C_{6} H_{14}$ gives two monochloro alkanes on chlorination. What is the structure of alkane?

S_{N} 2

c o n c . H C l :

(a) C H_{3} - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{2} - C l

(b) C H_{3} - C H_{3} | C | C H_{3} - C H_{2} O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{1} | C | C H_{3} - C H_{2} - C H_{3}

(c) C H_{3} - O H | C | C H_{3} - C H_{3} A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C l | C | C H_{3} - C H_{3}

(d) C H_{2} = C H - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{2} = C H - C H_{2} - C l

I_{2} / N a O H

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