1
$\begingroup$

As a general sense I know that cis isomers are less symmetrical and more polar in nature whereas trans isomers are less polar. Now given those information, would it be appropriate to assume that cis isomers are more water soluble owing to their polarity?

Also there was a discussion in our class where my teacher told us that between the cis and trans Butenedioic acids, the cis acid is not water soluble on account of the fact that it undergoes dehydration and the carbonyl groups are bonded by an oxygen atom in between eliminating hydrogen bonds (??).

For the cis Butenedioic acid

For the trans Butenedioic acid

I would like to know if there is any constant character when it comes to solubility of cis and trans isomers.

$\endgroup$
0

2 Answers 2

4
$\begingroup$

That's not quite correct and can be quite easily to be shown as wrong by looking up data tables and/or MSDS. Fumaric acid (the common name of the trans-isomer) is less soluble, with a solubility in water of 4.9 g/L whereas maleic acid (the cis-isomer) has a solubility of 478.8 g/L.

Maleic anhydride would also react in water to form maleic acid.

$\endgroup$
2
  • $\begingroup$ And regarding my last question? Is there any common rule governing it? Are either of cis or trans isomers more water soluble than the other? $\endgroup$ Commented May 21, 2016 at 15:17
  • 1
    $\begingroup$ @user30287 As with most things in chemistry, there is no hard and fast rule. I'm sure given a sufficiently complex molecule you can find an exception. $\endgroup$
    – IT Tsoi
    Commented May 21, 2016 at 15:30
0
$\begingroup$

Actually in the cis isomer there is a repulsive force acting between the molecules but when we see the trans structure there is a compact structure, there is less repulsive force... so due to this repulsive force the molecules in cis isomer will be farther from each other and hence will be more soluble.

$\endgroup$
1
  • $\begingroup$ What exactly do you mean by repulsive force. I'd say there's simply much less H-bonding to overcome during dissolution. $\endgroup$
    – Mithoron
    Commented Nov 15, 2016 at 22:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.