Hi Ahmed,
These questions are an excellent challenge! First let's look at the formula: C5H10O2. A completely saturated hydrocarbon would have 2n+2 hydrogens (n = the number of C's). So the completely saturated hydrocarbon with 5 carbons would have 12 hydrogens, and here we only have 10. (12-10)/2 = 1 degree of unsaturation, meaning we have a ring or a double bond somewhere in there (triple bonds would account for 2 degrees unsaturation).
Let's do the second question first:
A branched-chain molecule with two different functional groups that cannot react with either MnO4– / H+ or Na2CO3 solution.
Branched chain means we can toss out the ring idea, our one degree of unsaturation must be from a double bond. So what reacts with MnO4- followed by acidic workup?
KMnO4 followed by H+ is most often seen cleaving double bonds. The double bond to the other carbon is replaced by a double bond to an oxygen, so it becomes a carbonyl. Unlike ozonolysis, where a carbon of a double bond that has a hydrogen bonded to it will remain as just a hydrogen, oxidative cleavage by MnO4- will replace a hydrogen with an -OH group, forming a carboxylic acid rather than an aldehyde. If the carbon of the double bond has two hydrogens, you would form carbonic acid, which would immediately react further to form CO2 and water.
MnO4- reacting with an aldehyde would result in oxidation to the carboxylic acid. MnO4- can also oxidize primary or secondary alcohols (but not tertiary!). The primary ROH would become a carboxylic acid and the secondary ROH would become a ketone.
So the oxygen-containing functional groups that would NOT react with the MnO4- include ketones RCOR, carboxylic acids RCOOH, esters RCOOR, and ethers ROR. But if we have two different functional groups present, the carboxylic acid and ester don't make a whole lot of sense, we wouldn't have enough oxygens for a second functional group!
Further, the carboxylic acid is also ruled out by the other requirement, that we not react Na2CO3. A carboxylic acid would donate it's proton to the Na2CO3, which would further react to form CO2 and water. So we are limited to a ketone and an ether. Great! We've identified the 2 different functional groups contained in this isomer. If that's all we're considering we still have a couple different possibilities though... so we need to also keep in mind that they have specified this is branched chain molecule! This leaves us with just one possibility...
H3C-C=O-CH(CH3)-O-CH3 (hopefully you are able to decipher this condensed structural formula)
Now let's talk about the first question.
A straight-chain molecule that can decolorise bromine water, but cannot exist as geometric (cis-trans) isomers. It can react with Cr2O72– / H+, but the resulting organic product will not form an orange-red solid when heated with Benedict’s solution.
Again, we can rule out the ring, since they tell us this is a straight-chain molecule. Our one degree of unsaturation is again a double bond. So the possibilities include: alkene, ketone, aldehyde, ester, or carboxylic acid. The alkene would also have to have 2 oxygens in there somewhere, maybe as alcohols or ethers. The ketone or aldehyde would have to have an alcohol or ether present too to account for the second oxygen.
But wait a second, we are told we can't have cis-trans isomers of this molecule! There's no way to have an alkene that is a straight chain without cis-trans isomers. So not an alkene.
Next, we know it decolorizes bromine water. So what reacts with aqueous Br2? Anything with a double bond pretty much. This isn't really new info, we already knew we had a double bond, and further more that the double bond is not an alkene, so it must be a carbonyl.
What else? It reacts with Cr2O7-2/H+. This is chromic acid or Jones reagent, a strong oxidizing agent. Primary alcohols get oxidized to carboxylic acids, secondary alcohols get oxidized to ketones, aldehydes get oxidized to carboxylic acids, but ketones, carboxylic acids, and esters do not react. So our double bond is a ketone or an aldehyde, and we still have another oxygen in there somewhere, probably as an alcohol. And that alcohol could be primary or secondary. (Can't be tertiary, this is a straight chain molecule.)
So let's write out the possibilities of the Jones reagent reaction:
ketone and primary alcohol → ketone and carboxylic acid
ketone and secondary alcohol → two ketones
aldehyde and primary alcohol → two carboxylic acids
aldehyde and secondary alcohol → carboxylic acid and ketone
Finally the last piece of info we are given: The resulting product of the reaction with Jones will NOT form an orange-red solid when heated with Benedict's solution. This is where I step a little out of my comfort zone, I have not had to know much about Benedict's solution in the past, but I believe we can figure this out.
So what does Benedict's solution do? It is another oxidizing agent, typically used as a test for the presence of what are called reducing sugars. Reducing sugars are sugars that can interconvert between their ring and straight chain forms. Meaning, they are simple sugars where the ketone or aldehyde functional group is available to react, as opposed to a complex sugar made up of multiple sugar monomers linked together by the formation of a hemiacetal (can't go back to the straight chain form in that case).
So which of the above products does NOT have a ketone in it? The third one, the one with two carboxylic acid functional groups. This formed by the reaction of Jones with an aldehyde and primary alcohol. So the molecule we are looking for is a straight five carbon chain aldehyde and primary alcohol:
H-C=O-CH2-CH2-CH2-CH2-OH
Hope this all makes sense!
