-1
$\begingroup$

Which of the following is the most stable?

(i) Conjugated alkadiene $(\ce{CH2=CH-CH=CH2})$
(ii) Isolated alkadiene $(\ce{CH2=CH-CH2-CH=CH2})$
(iii) Cumulated alkadiene $(\ce{CH2=C=CH2})$
(iv) All are equal.

I am not able to understand the meaning of "isolated" here. I am thinking the answer here to be (ii), as it will show more hyperconjugation, but answer is (i).

$\endgroup$
  • 6
    $\begingroup$ Well, it's simple: conjugated double bonds feel great, isolated double bonds feel normal, and cumulated bonds feel terrible. Hyperconjugation is a small and unsignificant thing compared to that. $\endgroup$ – Ivan Neretin Mar 12 at 10:41
  • 1
    $\begingroup$ "Isolated" means that the double bonds do not interact and it is, in this example, the opposite of "conjugated". The key here is to see if the double bonds and single bonds are alternating like double-single-double-etc. $\endgroup$ – SteffX Mar 12 at 12:40
  • $\begingroup$ KM: Make an effort to justify your answer of ii. That way the question will not be marked as homework and dumped. Otherwise you will not get help. $\endgroup$ – user55119 Mar 27 at 19:25
2
$\begingroup$

It’s better to compare dienes with the same total chain length and thus the same molecular formula, for example the following pentadienes, so that the differences in enthalpy of formation (or as well in enthalpy of hydrogenation) directly indicate the effects of the arrangements of the double bonds.

Isolated double bonds
– penta-1,4-diene ($\Delta_\mathrm f H^\circ=105.7\ \mathrm{kJ/mol}$)

Conjugated double bonds
– (3​Z)-penta-1,3-diene ($\Delta_\mathrm f H^\circ=81.4\ \mathrm{kJ/mol}$)
– (3​E)-penta-1,3-diene ($\Delta_\mathrm f H^\circ=76.1\ \mathrm{kJ/mol}$)

Cumulated double bonds
– penta-1,2-diene ($\Delta_\mathrm f H^\circ=140.7\ \mathrm{kJ/mol}$)
– penta-2,3-diene ($\Delta_\mathrm f H^\circ=133.1\ \mathrm{kJ/mol}$)

(All values taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)

Double bonds are isolated if they are separated by two or more single bonds so that they cannot interact with each other. The enthalpy of hydrogenation is close to the sum of the enthalpies of hydrogenation for the individual double bonds.

Double bonds are conjugated if they are separated by just one single bond. Because of the interaction between the double bonds, systems containing conjugated double bonds are more stable than similar systems with isolated double bonds.

Successive double bonds with no intervening single bonds are called cumulated double bonds. Systems containing cumulated double bonds are less stable than similar systems with isolated double bonds.

$\endgroup$
  • $\begingroup$ As a suggestion, a short sentence rationalizing why cumulated double bonds are less stable would make this an even better answer. $\endgroup$ – Zhe Mar 13 at 22:43
0
$\begingroup$

A graphic, which is not to scale, of @Loong's explanation and @Zhe's request is offered below. [All values are in Kcal/mol (NIST). Values adjacent to structures are heats of formation and heats of hydrogenation are adjacent to arrows.] Although the order of stability is conjugated>isolated>cumulative, the heats of formation of 1,4-pentadiene and 1,3-butadiene are in the wrong order although 1,2-propadiene (allene) is OK. Hydrogenation of these three dienes give different alkanes. The heats of hydrogenation show the least amount of heat is liberated from the most stable, conjugated diene and the most heat liberated from the least stable, cumulated diene. As Loong has pointed out, comparing C5 compounds is appropriate. Now (E)-1,3-pentadiene (in red) is more stable than 1,4-pentadiene which is more stable than 1,2-pentadiene (in red). Hydrogenation results in the formation of pentane in all three cases. The more stable the diene, the less heat is liberated on hydrogenation.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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