Question

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 1

Step-by-step 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 =
`(81)/(100) * Theoretical \ yield = 4.15 g`

Theoretical yield =
`(4.15)/(0.81)`

= 5.1235 g

thus, since 424.5 g yield produce from 162.5 g

∴

5.1235 g yield will produce =
`(162.5)/(424.5)* 5.1235 \ g`

= 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 =
`(262)/(424.5)* 5.1235 \ g`

= 3.1622 g of triphenylphosphine required.