Ketone and Aldehyde Reactions...?
829009
I won't answer this completely, because it looks very much like homework. \n\nI will say that an Aldehyde and a Ketone will yeild different degree alcohols. Reactions that yeild an primary alcohol from an Aldehyde would yeild a secondary alcohol from a ketone. Reactions that yeild an secondary alcohol from an Aldehyde would yeild a tertiary  alcohol from a ketone.
771541
Product 1: 3-methyl-2-pentene, HBr, and Water.\nReaction 1 is alkene formation by dehydration.\n\nProduct 2: 1-Chloro-2-ethylbutane and junk.\nRecation 2 is substition of OH with Cl.  I think this is an SN1 reaction but I'm not sure.  I'm sure about the product though.
827688
a) 2C2H6 + 7O2  --> 4CO2 + 6H2O\n\nb) C6H12 + 9O2  --> 6CO2 + 6H2O\n\nc) C5H12 + 8O2  --> 5CO2 + 6H2O\n\nYou must apply Markovnikov rule for the next addition reactions:\n\na) CH2=CHCH2CH3 + HCl  --> CH3CH(Cl)CH2CH3\n\nb) CH2=C(CH3)CH3 + HBr  --> CH3-CBr(CH3)CH3
319095
An alcohol could give a positive iodoform test after it has been oxidised to a methyl ketone. This of course implies that the OH group is on the 2nd carbon atom. I trust that you can work out the rest for yourself.
947223
Alcohol, aldehyde, ketone, or carboxylic acid.
412197
It is a nucleophilic substitution reaction, where the nucleophile -OH from KOH replaces -Cl from 1-chlorobutane; this forms KCl and 1-butanol as products.\n\nKOH + CH3-CH2-CH2-CH2Cl ---> KCl + CH3-CH2-CH2-CH2OH
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.
828259
Removing the OCH3 group leaves the framework: CH3CH2CHO and aldehyde! Since only one methoxy group is added the compound is a hemiacetal.
865969
Yes they react, for example:\n\nCH3COOCH2CH3 + HCOOH  <==> HCOOCH2CH3 + CH3COOH
787458
HCl interacts with pentoses producing furfural or furan-2-carboxaldehyde. The reaction causes a bright pink color. This distinguishes pentoses (sugars containing five carbons) and hexoses (sugars containing six carbons).
157082
Specifically: water, pyridine, squalene, urea, acetic acid, lactic acid, complex alcohols and glycols, ketones, and aldehydes.
130958
The equation is correct, if I perceive it right. You have 3 ethoxy groups on the boron atom, which does correspond to that formula. It looks like what might happen if you reacted borane (BH3), possibly tethered in a THF complex, with ethanol (EtOH) to get the product and H2 gas, but a lot of heat too. Such a substituted Borane would be rather reactive, as a Lewis acid.\n\nB(OC2H5)3 + 6O2 ---> H3B03 + 6CO2 + 6H2\n\nis my best guess
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.
933669
aldehyde\nthe name of the compound is propionaldehyde
833616
butanal - acrid odor\nheptanal - seen this as citrus and also as dry fish. go figure!\noctanal - seen this as citrus and also rancid.\nnonanal - plums\n\nI think that some of the aldehydes odour could be subjective and vary from person to person and perhaps with time, or they may be a multitude of receptors that react differently.\n\nKetones. yes.  Go and ask your wife, girlfriend or mother for the nail polish remover.  That is usually acetone.  Most of the ketones will have that odour, some milder.
66672
Possibly detoxing, but I agree with the other fellow who mentioned ketosis as well.  Sometimes you will see a white substance on the tongue which is supposed to be a type of eliminative waste trying to exit your body.  You are supposed to try to slough it off with a clean cloth or napkin, which might rid of any odor and speed up any detoxifying process.
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.