Substitution reaction<span> monitors second-order kinetics; that is, the degree
of the reaction hinge on on
the concentration of two first-order reactants. In the circumstance of bimolecular nucleophilic </span>substitution, these two
reactants are the nucleophile and the haloalkane. So for this problem, the
answer is ch3br.
Answer:
2.6 ×10^-42
Explanation:
From
∆G= -RTlnK
∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1
R= 8.314 Jmol-1K-1
T= 25°C + 273= 298K
-237.2×10^3= 8.314 × 298 × ln K
ln K= -237.2×10^3/2477.572
K = 2.6 ×10^-42
Answer:
Here's what I get
Explanation:
(a) Intermediates
The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).
(b) Relative Stabilities
The relative stabilities decrease in the order shown.
N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.
(c) Relative reactivities
The relative reactivities would be
C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃
Answer: The correct answer is option (A).
Explanation:
Polar molecules are molecules in which formation of partial charges takes place due to which dipole moment gets created in a molecule. Molecules with polar bonds that s bond with partly ionic character. And water is of the example of polar molecule.
Electronegative oxygen atom in water molecule attracts the electron bond pair towards itself which generates partial negative charge on oxygen atom and partial positive charge on both hydrogen atoms.
Where as water has higher value surface tension due to strong intermolecular association of the water molecule due to presence of hydrogen bonding.And it is more denser is liquid state than in its solid state.
Hence,the correct answer is option (A).
C. C and Pb; Carbon and Lead being in the same group.
