Question

In which of the following methods, the number of carbon atoms in alkane (forms as a product) is less than that of the parent compound (reactant)?

• A. Reduction of alkyl halide
• B. Wurtz reaction
• C. Decarboxylation of alkali salt of a fatty acid
• D. Hydrolysis of Griganard reagent

Answer 1

Answer: (C)

Decarboxylation of alkali salt of a fatty acid

A) In reduction of alkyl halide, the number of carbon atoms in alkane formed as product is same as the reactant.
$R-X\stackrel{\left[2H\right]}{\to }R-H+H-X$
B) In Wurtz reaction, coupling of R takes place without any loss of carbon atoms.
$2R-X+2Na\to R-R+2NaX$
C) In decarboxylation of alkali salt of a fatty acid, the reaction occurs at anode is:
$2C{H}_{3}CO{O}^{\bullet }\to 2C{H}_3^{\bullet }+2C{O}_2$
$2C{H}_3^{\bullet }\to C_2{H}_6$
Here, one carbon escapes as carbon dioxide. So, alkane (product) has one carbon less than reactant.
D) In hydrolysis of Grignard reagent, the number of carbon atoms in reactant (Grignard reagent) remains same in product (alkane).
$RMgX+H_{2}O\to RH+Mg\left(OH\right)X$