Enol form of Beta Dicarbonyl compounds are more stable because it has a conjugated π-system and intramolecular H-bonding.
Sunday, January 19, 2020
Why enol form of Beta Dicarbonyl compound more stable than those of monocarbonyl compounds?
Enol form of Beta Dicarbonyl compounds are more stable because it has a conjugated π-system and intramolecular H-bonding.
Why enol form of phenol is more stable than its keto isomers?
Which of the Complex of the following pairs has the highest value of CFSE?
(1) [Co(CN)6]3- and [Co(NH3)6]3+
(1) Why all the tetrahedral Complexes are high spin Complexes?
(2) Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
(3) Why [Mn(H2O)6]+2 is colourless although in which Mn+2 ion had five unpaired electrons ?
(4) Why [FeF6]3– is colourless whereas [CoF6]3– is coloured?
(5) Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 is colour although both have +2 oxidation state and 3d8 configuration ?
(2) [Co(NH3)6]3+ and
[CoF6]3-
(3) [Co(H2O)6]3+ and [Rh(H2O)6]3+
(4) [Co(H2O)6]2+ and [Co(H2O)6]3+
(3) [Co(H2O)6]3+ and [Rh(H2O)6]3+
(4) [Co(H2O)6]2+ and [Co(H2O)6]3+
SOLUTION:
(1) CN is the stronger ligand than NH3 therefore CFSE of [Co(CN)6]3- will be more than [Co(NH3)6]3+
(2) NH3 is stronger ligand than F therefore CFSE of [Co(NH3)6]3+ will be more than [CoF6]3- .
(3) Co belong to 3d series whereas The Rh belong to 4d series. More the value of n more is CFSE therefore CFSE of [Rh(H2O)6]3+ is more than [Co(H2O)6]3+ .
(4) Oxidation number of Co in [Co(H2O)6]3+ is more than the Oxidation number of [Co(H2O)6]2+ therefore, CFSE of [Co(H2O)6]3+ is more than [Co(H2O)6]2+ .
(1) CN is the stronger ligand than NH3 therefore CFSE of [Co(CN)6]3- will be more than [Co(NH3)6]3+
(2) NH3 is stronger ligand than F therefore CFSE of [Co(NH3)6]3+ will be more than [CoF6]3- .
(3) Co belong to 3d series whereas The Rh belong to 4d series. More the value of n more is CFSE therefore CFSE of [Rh(H2O)6]3+ is more than [Co(H2O)6]3+ .
(4) Oxidation number of Co in [Co(H2O)6]3+ is more than the Oxidation number of [Co(H2O)6]2+ therefore, CFSE of [Co(H2O)6]3+ is more than [Co(H2O)6]2+ .
Related
Questions:
(1) Why all the tetrahedral Complexes are high spin Complexes?
(2) Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
(3) Why [Mn(H2O)6]+2 is colourless although in which Mn+2 ion had five unpaired electrons ?
(4) Why [FeF6]3– is colourless whereas [CoF6]3– is coloured?
(5) Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 is colour although both have +2 oxidation state and 3d8 configuration ?
Friday, January 17, 2020
How can explain resorcinol (1,3-Dihydroxybenzene) give positive Iodoform test?
Resorcinol (1,3-Dihydroxybenzene) give
positive iodoform test because it is in equilibrium with its tautomer cyclohex-2-en-1,5-dione.
It give 3, 5-dioxohexanoic acid (CH3-CO-CH=CH-CH2-COOH) on reaction with strong base (OH-) (I2+NaOH Iodoform reagent)
,which has a CH3-CO (acetyl) group, so it’ clear that it would be
give positive iodoform test.
How can explain that 1, 3, 5-Trihydroxybenzene give positive iodoform test?
How can explain that 1, 3, 5-Trihydroxybenzene give positive iodoform test?
1,3,5-trihydroxybenzene give positive
iodoform test because it is in equilibrium with its tautomer
cyclohexane-1,3,5-trione. It give 3,5-dioxohexanoic acid (CH3-CO-CH2-CO-CH2-COOH)
on reaction with strong base (OH-) (I2+NaOH Iodoform reagent)
,which has a CH3-CO (acetyl) group, so it’ clear that it would be
give positive iodoform test.
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