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Positional Isomerism

A hydrocarbon...

Question

A hydrocarbon $X (C_{6} H_{10})$ on reductive ozonolysis gives $C_{6} H_{10} O_{2}$ which gives positive Tollen's test. The compound below satisfying the criteria of X is:

A

Cyclopentane

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B

Cyclohexene

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C

Cyclohexane

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D

None of the above

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Solution

The correct option is B Cyclohexene
The ozonolysis product of X is a dicarbonyl which contains aldehyde group.
An aldehyde gives positive Tollen's test. $R C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to R C O O^{-} + 2 A g ↓ + 2 H_{2} O + 4 N H_{3} Aldehyde Tollen's Reagent Carboxylate ion Silver mirror$

Cyclohexane and cyclopentane do not undergoes ozonolysis as these do not have double bond.

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Similar questions

(A)

C_{5} H_{10} O

forms a phenylhydrazone and gives negative Tollen's and iodoform test. Compound (A) on reduction gives n-pentase. Compound

(A)

Q. A hydrocarbon X on treatement with

O_{3}

followed by the reduction of ozonide with

Z n - H_{2}

O gives Y. Y gives both Tollen's test as well as yellow precipitate with

N a O H / I_{2}

solution. Which is a possible structure of X?

(C_{5} H_{1} 0)

on ozonolysis gives a mixture of two compounds (Y) and (Z). Compound (Y) gives positive Fehling's test and iodoform test. Compound (Z) does not give Fehling's test but give iodoform test. Compounds (X), (Y) and (Z) are?

(X)

with a molecular formula

C_{5} H_{10} O

gives a positive 2,4-DNP test but a negative Tollen's test. On oxidation it gives a carboxylic acid

(Y)

with a molecular formula

C_{3} H_{6} O_{2}

. Potassium salt of

(Y)

undergoes Kolbe's reaction and gives a hydrocarbon

(Z)

(X) . (Y)

(Z)

.

(A), (B)

(C)

are three non-cyclic functional isomers of a carbonyl compound with molecular formula

C_{4} H_{8} O

(A)

(C)

give positive Tollen's test whereas isomer

(B)

does not give Tollen's test but gives positive Iodoform test. Isomers

(A)

(B)

on reduction with

Z n (H g) / c o n c

H C l

give the same product

(D)

Write the structures of

(A), (B), (C)

(D)

.

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