How many grams of ( S ) - 1 - chloro - 4 - ethyl - 2 - methylhexane and triphenylphosphine would you need to create 4 . 1 5g of ( S ) - 1 - chloro - 4 - ethyl - 2 - methyl triphenylphosphonium assuming an 8 1 % yield for the reaction?

Answer:

Explanation:

The standard molar mass is:

For (S )-1-chloro-4-ethyl-2-methylhexane = 162.5 g/mol

For triphenylphosphine  = 262 g/mol

For  ( S )-1-chloro-4-ethyl-2-methyl triphenylphosphonium = 424.5 g/mol

The mass required for 81% yield = [tex]\dfrac{81}{100} \times Theoretical \ yield = 4.15 g[/tex]

Theoretical yield = [tex]\dfrac{4.15}{0.81}[/tex]

= 5.1235 g

thus, since 424.5 g yield produce from 162.5 g

∴

5.1235 g yield will produce = [tex]\dfrac{162.5}{424.5}\times 5.1235 \ g[/tex]

= 1.9613 g of alkyl halide (-chloro) required.

Also, since 424.5 g yield produce from 262 g phosphine

∴

5.1235 g yield will produce = [tex]\dfrac{262}{424.5}\times 5.1235 \ g[/tex]

= 3.1622 g of triphenylphosphine required.