[eureka123]

What is the final product ?

[nj_bartel]

What's LDA act as?

[orgoclear]

LDA is a hindered strong base. So basically you have to decide where it will attack i.e. which acidic H it will remove to form the carbanion.

[eureka123]

oh...i know all that ...i have doubts regarding final product....

i am posting two options..tell me which it will be

[nj_bartel]

Are alkyl chains electron donating or withdrawing?

[eureka123]

electron donating..

plz tell me the final answer....

[Dan]

Remember, you only have 1 eq of LDA, you can only deprotonate your starting material once

[nj_bartel]

Unless your product has a more acidic proton than your reagant.  Is that the case?

[eureka123]

thats what my doubt is....

plz all the experts here reply...

i feel ans should be 2nd option..but ans given in book is 1st option

[nj_bartel]

Well I'd say you're right and the book is wrong.

[lavoisier]

I agree.
It might be a typo in your book.

However, even if some dialkylated material were formed, it would be in mixture with some monoalkylated material.
Unless we think that a rather hindered 1,3-dicarbonyl compound is allylated by an excess of allyl bromide without any base? I don't think so.

[eureka123]

Thanx everyone

but

Unless your product has a more acidic proton than your reagant.  Is that the case?

plz explain why the above situtation with an example if possible

[nj_bartel]

If trifluoromethane was your substituent, that would increase the acidity of the remaining proton.

[eureka123]

ok ..tahnx everyone for your contributions...
such thpe of sicussion help a lot