According me i am getting both a and c correct , but not getting any idea about b and answer is given B .please explain
In cas of b2 there is +I effect of CH3 which decreases acidic character.
Hence b2 will be less acidic than b1
Usually -I effect increase acidity and +I effect decrease decreases acidity.
So answer should be D ?
How u are getting a and c? Attack of base is not steric sensitive
See it is cope elimination , so hoffman product will be formed now for hoffman that beta1 alkene is stable
Yield of Hoffman product is more than saytzeff in this mechanism
But saytzeff alkene is always more stable than Hoffman
But product ka formation hoffman se ho rha h na toh hoffman ke hisaaf se vo jada stable hoga and i have a doubt that here more than products can be formed ?
2 product will formed : saytzeff and Hoffman alkene
Ok but beta 2 is why not sterically hindered?
I think all options are correct but in question it is asked specific reason for attack of base and attack of base is not affected by stability of alkene and steric hindrance .
Attack of base completely dependent on acidic nature of H.
IF any other more acidic hydrogen present even in steric hindrance condition then also base attacks on that acidic hydrogen.
Hence most suitable answer is b not d.