When reacting KMnO4, OH- and HEAT plus a H+ on Aldehydes and Ketones what products would be expected?
80529
I'd expect oxidation to form the carboxylic acid. The aldehyde will form a primary carboxylic acid and CO2 gas. The ketone will form 2 secondary carboxylic acids.
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
752218
2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O
422972
In the first reaction, the base is NH2-, and the conjugate acid is NH3 (ammonia).\n\nIn the second reaction, the acid is HClO4 (perchloric acid) and the conjugate base is ClO4- (perchlorate).\n\nIn the third reaction, the acid is HSO4- (hydrogen sulfate) and the base is CO3-2 (carbonate). The conjugate base of HSO4- is SO4-2 (sulfate) and the conjugate acid of CO3-2 is HCO3- (hydrogen carbonate).
867182
6HCl +3 I2 + KClO3 --->6 ICl+ KCl + 3H2O\n\n\nMnO4 - +2 NO2- +4 H+ ---> Mn2+2 NO3- +2 H2O
903801
Carbonyl groups are not alkenes.\n\nThat being said, ketones and aldehydes that have hydrogens on the alpha carbon will react with bromine. This reactin is fairly quick.\n\nEsters and carboxylic acids do not react as quickly, but will eventually react. This reaction is known as the Hell-Volhard-Zelinsky reaction.\n\nIf you got a positive bromine test, you should use another test such as dinitrophenylhydrazine or the Tollens reagent to see if you have a carbonyl.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
740149
example of ketone having four alpha hydrogen atoms-3-pentanone. \n  i don't think all symmetric ketones have four alpha hydrogen atoms e.g. 2,4-di methyl pentanone is symmetrical ketone but with only two alpha hydrogen atoms.
267385
The REAGENT is an ylide. The PRODUCT is an ALKENE (and triphenylphosphine oxide). The reaction is:\naldehyde/ketone + triphenylphosphonium ylide ----> (Z)-alkene + triphenylphosphine oxide.
766197
CO2\nCarbon Dioxide\nHere is the reaction:\nHC2H3O2(aq) + NaHCO3(aq) ----> NaC2H3O2(aq) + H2O(l) + CO2(g)
957626
It is because you have the wrong compound (HSO3)!\n\n2 HI + K2SO3 ---> 2KI + H2SO3
762929
CH3CONH2 + Na+-OH ---->  CH3COO -Na+ +  :NH3\n\nThis is a amide hydrolysis reaction, similar to the hydrolysis of an ester. Hydroxide attacks carbonyl to give tetrahedral intermediate. Explulsion of an amide ion gives a carboxylic acid, which is deprotonated quickly to give the carboxylate ion and ammonia.
464400
An aldehyde is either a functional group consisting of a terminal carbonyl group or a compound containing a terminal carbonyl group.\n\nCommon reactions\naldehydes react with alcohols and an acid or base to a hemiacetal, the hemiacetal and alcohol and more acid catalyst further react to an acetal and water \nSimple hemiacetals are usually unstable, although cyclic ones such as glucose do exist. Acetals are stable, but these revert to the aldehyde in the presence of aqueous acid. \nTreating aldehydes with oxidizing agents such as potassium permanganate, nitric acid, chromium(VI) oxide or acidified potassium dichromate), will yield a carboxylic acid. \nTreating aldehydes with Tollens' reagent (which is prepared by adding a drop of sodium hydroxide solution into silver nitrate solution to give a precipitate of silver(I) oxide, and then adding just enough dilute ammonia solution to redissolve the precipitate in aqueous ammonia to produce [Ag(NH3)2]+ complex) will convert aldehydes to carboxylic acids without attacking carbon-carbon double-bonds. This reaction is also known as the silver mirror test. \nAldehydes can react with water (under acidic or basic conditions) to form hydrates, R-C(H)(OH)(OH), although these are only stable when strong electron withdrawing groups are present, as in chloral hydrate. The mechanism is identical to hemiacetal formation. \nAldehydes can react with HCN to form cyanohydrins, R-C(H)(OH)(CN). \nTreating an aldehyde with a Grignard reagent can yield an alcohol with a substituted group from the Grignard reagent. \nTreating aldehydes with NH3 derivatives (like NH2OH, H2NNH2, 2,4-dinitrophenylhydrazine) will result in the formation of an oxime or hydrazone. This is classed as an addition-elimination reaction or addition-condensation reaction. If the simple hydrazone (RCH=NHNH2) of an aldehyde is heated with a base such as KOH, the terminal carbon is fully reduced via the Wolff-Kishner reaction to a methyl group. The Wolff-Kishner reaction may be performed as a one-pot reaction, giving the overall conversion RCH=O → RCH3. \nReaction of aldehydes with reducing agents such as magnesium gives diols in a Pinacol coupling reaction \nThe Wittig reaction takes aldehydes to alkenes and the Corey-Fuchs reaction takes aldehydes to alkynes both with a triphenylphosphine reagent. The Corey-Chaykovsky reagent is a sulfonium ylide which convert aldehydes to epoxides. \n\nExamples of Aldehydes:\n\nMethanal (Formaldehyde) \nEthanal (Acetaldehyde) \nPropanal \nButanal \nPentanal \nGlucose \nBenzaldehyde
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
369946
The reaction is generally thought to proceed through a nucleophilic addition mechanism.Grignard reagents such as methylmagnesium bromide are sources of a nucleophile that can attack the + end of the C=O double bond in aldehydes and ketones.If we treat the product of this reaction with water, we get an tertiary alcohol.Sterically hindered substrates may react according to an SET (single electron transfer) mechanism.