Question

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Answer 1

Explanation:

To Find -

• A, B & C

Solution:

If we treat 1-bromo butane with alcoholic KOH then it gives only 1 product i.e, but-1-ene.

And here it's mentioned that On treating C₄H₉Br with alcoholic KOH it gives two isomeric compounds.

So,

Now, It's cleared that C₄H₉Br is 2-bromo butane and not 1-bromo butane.

When 2-bromo butane reacts with alcoholic KOH then it forms but-1-ene and but-2-ene i.e, isomeric compounds with molecular formula C₄H₈

Reaction:

CH₃–CH₂–CH–CH₃ + KOH (alc.) →

|

Br

CH₃–CH₂–CH=CH₂ + CH₃–CH=CH–CH₃

Now,

On Ozonolysis B gives only 1 product CH₃CHO

Here, we see that but-2-ene gives only 1 product CH₃CHO

Reaction:

CH₃–CH=CH–CH₃ (ozonolysis) → 2CH₃CHO

And

but-1-ene gives two different product on ozonolysis

Reaction:

CH₃–CH₂–CH=CH₂ (ozonolysis) → CH₃–CH₂–CHO + HCHO

So,

Now, It's cleared that Compound B is but-2-ene

Therefore,

Compound A = 2-bromo butane

Compound B = but-2-ene

Compound C = but-1-ene

Answer 2

Answer:

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Explanation:

To Find -

A, B & C

Solution:

If we treat 1-bromo butane with alcoholic KOH then it gives only 1 product i.e, but-1-ene.

And here it's mentioned that On treating C₄H₉Br with alcoholic KOH it gives two isomeric compounds.

So,

Now, It's cleared that C₄H₉Br is 2-bromo butane and not 1-bromo butane.

When 2-bromo butane reacts with alcoholic KOH then it forms but-1-ene and but-2-ene i.e, isomeric compounds with molecular formula C₄H₈

Reaction:

CH₃–CH₂–CH–CH₃ + KOH (alc.) →

|

Br

CH₃–CH₂–CH=CH₂ + CH₃–CH=CH–CH₃

Now,

On Ozonolysis B gives only 1 product CH₃CHO

Here, we see that but-2-ene gives only 1 product CH₃CHO

Reaction:

CH₃–CH=CH–CH₃ (ozonolysis) → 2CH₃CHO

And

but-1-ene gives two different product on ozonolysis

Reaction:

CH₃–CH₂–CH=CH₂ (ozonolysis) → CH₃–CH₂–CHO + HCHO

So,

Now, It's cleared that Compound B is but-2-ene

Therefore,

Compound A = 2-bromo butane

Compound B = but-2-ene

Compound C = but-1-ene