In a base-catalyzed elimination reaction, the base takes away a proton from the $β$ position of the leaving group (the group that goes away with the bonding pair of electrons) and a double bond is formed. When the base is bulky it prefers to take away the proton from the less hindered position.
In the above process, $Y$ and $Z$ are:

two molecules of ethanal

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propanal and ethanol

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propanoic acid and carbon dioxide

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propanal and methanal

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Solution

The correct option is C propanal and methanal
The dehydrobromination of 2-bromobutane in presence of bulky base gives 1-butene as major product. Reductive ozonolysis of 1-butene gives propanal and methanal.

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Similar questions

Q. A

13 f t

. ladder is leaning against a wall when its base starts to slide away At the instant when the base is

12 f t

. away from the wall, the base is moving away from the wall at the rate of

5 f t / s e c

. The rate at which the angle

θ

between the ladder and the ground is changing is

Conversion of Ketoximes to N- substituted amides is known as Beckmann rearrangement. In this rearrangement migration takes place from carbon to nitrogen. The reaction is catalyzed by acidic reagents. The main function of the acidic catalyst is to convert -OH of the oxime into a better leaving group by protonation.