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i cant see anything

Integral{sqrt[x^4 - x^2(a^2+b^2) + a^2*b^2]dx}, between some lim like 0 to a/2

can we say that since there r two phenyl grps, the +ve charge is delocalised more due to resonance in the biphenyl comp than in the comp with a single phenyl grp. so it wont be available for the attack of the OH ion. so the comp with the single phenyl grp will be easily hydrolysed right?

but isnt phemyl grp +R eff grp. so that will override the -I eff right?

no the comp is c6h5-ch(cl)-C6h5 and not c6h5-ch2-cl-c6h5

yep, and i use a mozilla firefox browser

ive asked the wrong ques, it shouldnt be ch2 but ch in C6H5Ch2(Cl)C6H5

with aq KOH, wat mech will be followed, SN1 or SN2.

wat abt the second case

none of these worked for me

which among the two given compounds is more easily hydrolysed, one is 1,1'-(chloromethanediyl)dibenzene, i.e C6H5Ch2(Cl)C6H5 and the other is chloromethyl benzene C6H5CH2Cl.

wat happenes when we compare 1-Chloropropane and C6H5Ch2(Cl)C6H5

but why the +M/+R effect, is there any reason for it to have +M/+R. (and thanks for the imediate replies.)

wats the resonace effect of the phenly group, +R or -R. and wat abt the nature of its inductive effect.

wat critical area

let the radius be R=D/2 then the field at a point close to the planet will be

field = GM/ ${R}^{2}$ tis is the g at that point but R=D/2 so in terms of D

field = 4GM/ ${D}^{2}$