In Q4 Cu2Br2 catalyses 1,4-addition of grignard reagent. So answer will be a beta-methyl ketone.
4 Answers
Q3 Gilman reagent, for 1,4-addition to alpha-beta unsaturated ketone ((CH3)2CuLi).
Q1. RLi is most reactive due to polarising power of lithium. Negative charge on R is maximum in this type of organometallic reagent.
In Q4, since there is no steric hindrance in the substrate, grignard reagent adds to the more polar C=O bond (1,2-addition). Hydrolysis of this gives alcohol in place of =O and the double bond is also hydrated into alcohol(as per Markownikov addition).
In Q5, dibromocarbene is formed by alpha elimination and it adds to alkene to form a dibromocyclopropane(base of the triangle is where the alkene is, and on top most vertex two bromine atoms are present). SInce Gilman reagent is an alkylating reagent when it comes to halides, it will remove the two bromine atoms and replace them with ethyl groups.
@pritish
thx for Q1..i was looking for exact reason..
for Q3..where is 1-4 addn ??
Q4 ..i also did like that but ans given is through attack on double bond...is it due to Cu2Cl2..thats wat i wanted to know
Q5..ans is rite.....but i didnt knew that carbene could be formed from CHBr3..all i knew was that CHCl3 is common source.can u brief me all possible surece of carbene ??