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Standard XII

Chemistry

Physical Properties of Alkenes

An alkene A ...

Question

An alkene A $(C_{10} H_{20})$ on oxidative ozonolysis yields two molecules of $(C_{5} H_{10} O)$ B. B on treatment with $L i A I H_{4}$ yields C $(C_{5} H_{12} O)$ which can't be resolved into enantiomers. Which of following is correct?

A: 3,4-diethyl-3-hexene

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B: 3-pentanone

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C: 3-pentanol

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All of them are correct

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Solution

The correct option is D All of them are correct
The compound A is 3,4-diethyl-3-hexene.
On oxidative ozonolysis, it gives two molecules of 3-pentanone which is compound B.
The reduction of compound B with $L i A l H_{4}$ gives 3=pentanol which is compound C.

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Similar questions

Q. An organic compound

A (C_{5} H_{10} O_{2})

reacts with

B r_{2}

in presence of phosphorus to produce

B

which can be resolved into enantiomers.

B

on dehydrobromination yields

C (C_{5} H_{8} O_{2})

which does not show stereoisomerism.

C

on decarboxylation gives an alkene which on further ozonolysis gives

D

and

E

D

gives Schiff's test but

E

does not. The correct identity of these compounds are:

Q. An organic compound

X (C_{5} H_{8} O_{2})

liberated

C O_{2}

on treatment with

N a H C O_{3}

. X exist in two stereoisomeric forms but none of them is optically active. X on selective olefinic hydrogenation yields

Y (C_{5} H_{10} O_{2})

which can be resolved into enantiomers. The correct deduction is /are :

Q. A hydrocarbon

A

(C_{5} H_{10})

on addition of

H B r

in presence of a peroxide forms

(C_{5} H_{11} B r)

B

which is found to be resolvable into enantiomers.

A

on addition of

H B r

in aqueous medium produces

C

which is an isomer of

B

but can't be resolved.

A

on reductive ozonolysis produced

(C_{4} H_{8} O)

D

as one product.

D

on catalytic reduction with

N a B H_{4}

produced E

(C_{4} H_{10} O)

which is resolvable.

E

on dehydration with concentrated sulphuric acid produced

F

(C_{4} H_{8})

which on ozonolysis followed by work-up with

Z n - H_{2} O

produced two molecules of ethanol. Identify

A

F

Q. An organic compound has molecular formula A

(C_{7} H_{10})

and known to decolourized

B r_{2}

-water solution.A on reduction with

H_{2} / P t

produces B

(C_{7} H_{14})

. B on monochlorination with

C l_{2} / h v

produces four isomeric monochloro derivatives

C_{7} H_{13} C l

. A on ozonolysis followed by work-up with Zn-dimethyl sulphide affords two and only two products, one being methanal and other C

(C_{6} H_{8} O_{3})

. C is oxidised with acidic solution of dochromate to yields D

(C_{6} H_{8} O_{5})

which is soluble in

N a H C O_{3}

. D is reduced with

N a B H_{4}

to yield E

(C_{6} H_{10} O_{3})

which may be resolved into enantiomers. Identify A to E.

Q. An Organic compounds 'A'

C_{4} H_{8}

on treatment with

K M n O_{4} / H^{+}

yields compound 'B'

C_{3} H_{6} O

. Compound 'A' also yields compound 'B' an ozonolysis. Compound 'A' is

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An alkene A (C10H20) on oxidative ozonolysis yields two molecules of (C5H10O) B. B on treatment with LiAIH4 yields C (C5H12O) which can't be resolved into enantiomers. Which of following is correct?

The correct option is D All of them are correctThe compound A is 3,4-diethyl-3-hexene.On oxidative ozonolysis, it gives two molecules of 3-pentanone which is compound B.The reduction of compound B with LiAlH4 gives 3=pentanol which is compound C.

An alkene A $(C_{10} H_{20})$ on oxidative ozonolysis yields two molecules of $(C_{5} H_{10} O)$ B. B on treatment with $L i A I H_{4}$ yields C $(C_{5} H_{12} O)$ which can't be resolved into enantiomers. Which of following is correct?

The correct option is D All of them are correct
The compound A is 3,4-diethyl-3-hexene.
On oxidative ozonolysis, it gives two molecules of 3-pentanone which is compound B.
The reduction of compound B with $L i A l H_{4}$ gives 3=pentanol which is compound C.