A hydrocarbon C7H12 on reaction with H2 in presence of Lindlar’s catalyst yields X. On reductive ozonolysis X gives a mixture of two aldehydes Y and Z out of which only Y can give iodoform test. The structure of hydrocarbon C7H12 is
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Answer:
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in first step,only CIS isomer is formed.not trans isomer.due to Lindlers catalyst.
Answer:
The structure of hydrocarbon C7H12 is Cycloheptene .
Explanation:
Let's first understand that what is a Lindlar's catalyst?
- It is a combination of palladium and calcium carbonate (CaCO3).
- To obtain alkene from an alkyne, lindlar's catalyst can be used.
So when a hydrocarbon C7H12 ( Cycloheptene) reacts with
H2 in the presence of Lindlar's catalyst then reaction will not take place.
- So the X will be the same C7H12
- Then, on reductive ozonolysis X that is C7H12 gives the mixture of two products, one is ketone which is Y and the other one is aldehyde which is Z.
- Now between ketone and aldehyde, only ketone which is Y can give iodoform test, as iodoform test tell us about the presence of aldehyde or ketone in which a methyl group is one of the group which is directly attached to the carbonyl carbon.
- So in ketone, methyl group is directly attached to the carbonyl carbon and hence it gives us the iodoform test.
O
1. O3 ║
C7H12 --------------------> CH3-C- CH2-CH2-CH2-CH2-CHO
2. Zn/ H2O
- Here, the mixture of products are formed Y and Z
Here Y = CH3-CO-CH3 ( Acetone)
Z = CH3-CH2-CH2-CHO ( Butanal )
- So in this case only Y can give iodoform test which is acetone
- As in iodoform test ketone reacts with
and NaOH and CHI3 will produce. NaOH
- CH3-CO-CH3 + I2 ------------> CH3-CO-Na + CHI3 + NaI + H2O
(Chloroform )
And the structure of hydrocarbon is written on the attachment file completely.