The main product of the reaction would be
$2 - b u t e n e + c h l o r o f o r m N a O H - --- \to$

Butanoic acid

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2-methyl butanoic acid

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1,1,1-trichloro-2-methyl butane

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1,4-butane diol

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Solution

The correct option is B 2-methyl butanoic acid
$C H_{3} C H = C H - C H_{3} + C H C l_{3} ⟶ C H_{3} - C C l_{3} | C H - C H_{2} - C H_{3} N a O H - ----- \to h y d r o l y s i s$
$C H_{3} - C (O H)_{3} | C H - C H_{2} - C H_{3} - H_{2} O - --- \to C H_{3} - C O O H | C H - C H_{2} C H_{3} 2 - m e t h y l b u t a n o i c a c i d$
Answer is (b).

Suggest Corrections

Similar questions

{C H}_{3} M g B r

H C l

3

1

2

2

List -I (Products)	List -II (Reactants-Reaction)
A) Methane	(1) Sabatier - Senderens reaction of propylene
B) Ethane	(2) Wurtz reaction of methyl iodide
C) Propane	(3) Decarboxylation of sodium acetate
D) Butane	(4) Sodium butanoate by Kolbes electrolysis

A hydrocarbon `x' decolourises bromine water. On hydrogenation `x' gives

2-methyl butane. On ozonalysis it gives acetone and acetaldehyde as the

products. The hydrocarbon `x' is

B r_{2}

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