Which carbonyl compounds do not undergo an aldol reaction when treated with −oh in h2o? select all that apply?
1 answer:
Answers:
Among many few compounds are shown below which can not undergo Aldol Condensation reactions.
Explanation:
Aldol Condensation reactions are given by those carbonyl compounds which are capable of forming conjugate bases called as Enolates. Enolates are formed when carbonyl compounds having hydrogen atoms at alpha position are treated with strong bases. This hydrogen atom at alpha position is mildly acidic in nature and readily donated to base as the upcoming conjugate base i.e. has stability due to resonance.
So those carbonyl compounds which lacks these acidic protons at alpha position are not able to form enolates, Hence, unable to undergo Aldol Condensation reactions. Few examples are attached below,
Among many few compounds are shown below which can not undergo Aldol Condensation reactions.
Explanation:
Aldol Condensation reactions are given by those carbonyl compounds which are capable of forming conjugate bases called as Enolates. Enolates are formed when carbonyl compounds having hydrogen atoms at alpha position are treated with strong bases. This hydrogen atom at alpha position is mildly acidic in nature and readily donated to base as the upcoming conjugate base i.e. has stability due to resonance.
So those carbonyl compounds which lacks these acidic protons at alpha position are not able to form enolates, Hence, unable to undergo Aldol Condensation reactions. Few examples are attached below,
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Using Hess's law we found:
1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)
2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).
The reactions of dissolution (10.1) and neutralization (10.2) are:
KOH(s) → KOH(aq) ΔHsoln (10.1)
KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)
1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.
Hence, to get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1
KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)
<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:
The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:
Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.
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Explanation:
(a) potassium oxide with water
According to reaction,1 mole of potassium oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.
(b) diphosphorus trioxide with water
According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water to give 2 moles of phosphorus acid.
(c) chromium(III) oxide with dilute hydrochloric acid,
According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.
(d) selenium dioxide with aqueous potassium hydroxide
According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.