Question

CH —C ≡ CH□(→┴40%H₂SO₄)A
□(→┴Isomerisation )CH₃ —C—CH₃
||
O
structure of 'A' and type of isomerism in the above reaction
are respectively
(a) Prop-1-en-2-ol, metamerism
(b) Prop-1-en-1-ol, tautomerism
(c) Prop-2-en-2-ol, geometrical isomerism
(d) Prop -1-en-2-ol, tautomerism

Answer 1

CH —C ≡ CH□(→┴40%H₂SO₄)A

□(→┴Isomerisation )CH₃ —C—CH₃

||

O

structure of 'A' and type of isomerism in the above reaction

are respectively

(a) Prop-1-en-2-ol, metamerism

(b) Prop-1-en-1-ol, tautomerism

(c) Prop-2-en-2-ol, geometrical isomerism

(d) Prop -1-en-2-ol, tautomerism

Answer 2

Prop -1-en-2-ol, tautomerism is seen.

Explanation:

• The reactant is the propyne.

• The reagent is Hg2+ in H2SO4.

• The reaction going in here is the oxymercuration demercuration reaction.

• This reaction converts triple bonds to double bonds by means of reduction.

• This reaction doesn't convert to alkanes.

• And the reaction is electrophilic water addition.

• The product formed will be Prop -1-en-2-ol.

• Now this compound has conjugated double bond and OH group.

• This will show tautomerism.

• Keto enol tautomerism will convert this product to acetone as shown.

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