Doubt from goc

1571126414243-1430934300 15711264781151827662594 1571126518921-52420403
Plz help @Shwetanshu_2018

For 10 option A, D
For chlorine it is -I and +R effects, it increases electron density at ortho and para positions.
For No2 it pulls electrons from the ring, making electron deficient at ortho and para and indirectly have more negative charge at meta. So it is meta directing group.

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Ans 1015711292429602834463410205431093

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For Question 1 is the answer A ?

For 10th it’s AD
Remember that not only Cl but also other halogens are deactivating yet ortho, para directing groups

For 7th it should be D because acetic acid has polarised O-H bond and it being in non polar solvent ( benzene ) interacts with other molecule of acetic acid and forms H bonding and thus dimerisation happens

@Shwetanshu_2018 all the Q are multi correct answers are
1.ABD
6.ABC
7.ABD
10.AD

For question 10 i saw correct statement(s) and for others i say correct statement so did not check for other option :sweat_smile: My bad !

Okay just tell in which all options you are facing problem

@Shwetanshu_2018 I'm too weak in chemistry :pensive: .Can you plz explain all the options.

Okay so
Q7
Option A ) When water is in liquid state then it has intermolecular H bonding in it and hence bonds are stronger , molecules are closer but in solid state (ice) there is no H bonding and hence lesser attraction resulting in decrease in density and hence ice floats

Option B ) I’m a bit skeptical about it because using the term solvation of ion would be more appropriate as compared to H bonding but still , due to lesser hinderance in primary amine there is more stabilisation of the ion formed when N donate it’s electron pair . And thus the resulting ion is more stable and thus primary amine > tertiary amine in basic strength

  1. Option C
    Formic acid is more acidic than acetic acid because in acetic acid CH3 is there which destabilises the negative charge on it’s conjugate base and hence is less stable and thus weaker acid .
    It has no relation with H bonding
    Option D i have already explained

For q6)
OH being a +M group decides where the halogen must attack (ortho+para) so option D is correct
But due to steric hindrance , few positions like ortho(in between OH and tert butyl )and para isn’t available , this hindrance is due to both the attaching halogen as well as tert butyl
And hence option A and B are also correct

Option C is incorrect as tert butyl can only show +I and not +H or +M and hence it has negligible effect and doesn’t influence the formation of any product