The product of addition of bromine to (R)-3-buten-2-ol will bea. a 50:50 mixture of enantiomers. b. a mixture of enantiomers formed in unequal amounts. c. a 50:50 mixture of diastereomers. d. a mixture of diastereomers formed in unequal amounts. e. optically inactive.

Option D is the answer

D. a mixture of diastereomers formed in unequal amounts.

Explanation:

Chiral intermediates give chiral products

- form unequal mixtures of diastereomers

- transition states are now diastereotopic

• Optically active reactants leads to optically active products.

find attached a diagram of the reaction.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.