• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Help with M7 Question (1 Viewer)

flowerp

Active Member
Joined
May 15, 2017
Messages
124
Gender
Undisclosed
HSC
N/A
Hi, I need help with the following question please:
How can one make 3-chloro-2-butanol ? How can one make 2,3-pentanediol?
Thank you :)
 
Last edited:

worldno17

Active Member
Joined
Oct 24, 2015
Messages
126
Gender
Female
HSC
N/A
For 3-chloro-2-butanol, you have but-2-ene --> addition reaction with Cl2 --> substitution of 2,3-dichloro-butane with hydroxide ion (from strong base e.g. NaOH)?

does that work?
 

jazz519

Moderator
Moderator
Joined
Feb 25, 2015
Messages
1,955
Location
Sydney
Gender
Male
HSC
2016
Uni Grad
2021
Hi, I need help with the following question please:
How can one make 3-chloro-2-butanol ? How can one make 2,3-pentanediol?
Thank you :)
So you have molecules where multiple functional groups are being added so the way you want to do these types of questions is think about how you could make each of them individually

For the 3-chlorobutan-2-ol you can use but-2-ene do a substitution reaction in presence of Uv light with chlorine gas. That will make 2-chlorobut-2-ene

Then react that with water and dilute Sulfuric acid catalyst and you make the 3-chlorobutan-2-ol
 

jazz519

Moderator
Moderator
Joined
Feb 25, 2015
Messages
1,955
Location
Sydney
Gender
Male
HSC
2016
Uni Grad
2021
For the other one Pentan-2,3-diol

use pent-2-ene do an addition reaction with chlorine gas. This will make 2,3-dichloropentane. Then add sodium hydroxide excess and a substitution reaction will occur giving you Pentan-2,3-diol
 

flowerp

Active Member
Joined
May 15, 2017
Messages
124
Gender
Undisclosed
HSC
N/A
For the other one Pentan-2,3-diol

use pent-2-ene do an addition reaction with chlorine gas. This will make 2,3-dichloropentane. Then add sodium hydroxide excess and a substitution reaction will occur giving you Pentan-2,3-diol
Thanks heaps jazz !
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top