what is the product of propene and hydrobromic acid?plz?
476736
=======\n\ngem is right.\n\nYou will get a mixture of two isomers of bromopropane.
858619
The easiest way to approach problems like these, my friend, is to look at your alkyl halide first and then look at your nucleophile or base.\n\nT-butyl bromide is a tertiary halide.  Tertiary halides undergo E1 and SN1 reactions because of their ability to form a stable carbocation intermediate.  If you recall, tertiary carbocations are more stable than primary and secondary.\n\nYour -OCH3 is a strong base known as methoxide, the conjugate base of methanol.  Strong bases favor elimination reactions and thus the elimination product is your answer.
128824
NaBH4 functions as BH4-, which donates a hydride (a hydrogen with a negative charge) to the aldehyde or ketone that you're working with. The product is a sodium salt/alkoxide. Water or HCl (acid) will protonate the alkoxide to generate the alcohol that you need. \n\nThe gas formed was probably hydrogen, from when the HCl reacted with the Sodium borohydride.
616657
i think you've named your compound incorrectly.  The phenol is already in the one position, so the propyl can't be.  I can only assume you mean 2-propylphenol.\n\nIf you mean 2-propyl phenol then there are a couple of ways.\n\nProtect your phenol with the MOM or MEM group by using phenol and MOM-Cl or MEM-Cl.  Treat with butyl lithium at -78oC to promote an ortho-directed lithiation.  MOM and MEM ethers work better than others as they allow additional co-ordination and stability points for your lithium through the additional oxygen lone-pairs, stabilising the phenyl lithium formed.  Treat with propanal.  The phenyl lithium will nucelophilically attack your aldehyde to give you a secondary alcohol.  De-hydrate the alcohol under acidic conditions to give the alkene, hydrogenate over Pd/C to give your alkane.  Alternatively you can de-hydrate by mesylating your alcohol and treating with a base such as TEA or DIPEA, reflux sometimes helps.  Finally remove your MEM or MOM protecting group using acid to release your desired 2-propylphenol.\n\nI hope this helps.
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
527310
As you know, benzene is rather unreactive. Its conjugate pi system is entirely impervious to conventional nucleophilic and electrophilic attack. For example, you can react HBr with it forever, and you won't get Bromobenzene.\n\nWhat you have to do is to find a way to "pump up" the electrophile that is being used to attack the ring. When you run a bromination, you do:\n\nPh + Br2 ---> (w/ FeBr3 catalyst) PhBr\n\nAn Iron bromide catalyst works as a Lewis Acid to create Br+ in situ. This Br+ is electrophilic enough to promote the benzene's pi electrons to attack, forming the sigma complex and leading to bromobenzene.\n\nNitration works the same way. NO3 is not a strong enough nucleophile to promote the reaction. The H2SO4 is a stronger acid than NO3 and it protonates it, yielding NO2+ (HSO4 + H20 as spectators). This NO2+, called nitronium, is enough to cause the nitration of the benzene ring.
216732
This might not be the simplest way possible, but here goes:\n\n(Starting with propanone)\n1. H3O+/I2 (to form 1-iodopropanone)\n2. (Ph)3P (triphenylphosphine forms the ylide)\n3. butanone* (Wittig reaction, now you should have hepta-3-ene-2,6-one)\n4. H2/Pd/C (hydrogenate the C=C; hopefully this'll leave the C=O's alone)\n5. H3O+ (intramolecular aldol condensation to close the ring, with dehydration to form the C=C double bond)\n\n*can be made from propylaldehyde and CH3MgBr\n\nNote: No guarantees this'll work as desired. :)
786324
i would guess D ethanol ch3ch2oh \nc is a gas propane\nand b diethyl ether is more volitile the d
986149
The Bromine test for alkenes works because bromine is electron deficient and alkenes have pi electrons forming a double bond that are open for attack.\n\nThe exact mechanism is shown here:\nhttp://en.wikipedia.org/wiki/Halogen_addition_reaction\n\nAlkanes can be brominated but the reaction is a lot slower and it requires a UV light catalyst (it's a radical bromination)
973000
C6H10 + Br2 ---> C6H10Br2.\n\nThe two Br atoms join at either end of the double bond.
892054
Is Hydrorogen chloride a solid? I've only ever thought of it as Hydrochloric Acid (get it out of a bottle!) which, of course, liberates See-oh-2 from See-ay-2-see-oh-3.\n\nPS. I asked about the solid 'cos on a practical level be careful of Toluene - it's nasty stuff. It's not used to make HCl, but I can't remember how you do make it. Sorry.
37368
You'll need to read (or cut and paste) some of the facts behind those carbon compounds.\n\nXav covered the pure carbon compounds.\n\nMethol seems to be the odd man out, a simple alcohol better known as methanol - CH3OH. \n\nC60Br24 is brominated buckminster fullerene rather than the pure carbon molecule. It will have brom atoms sticking out from the soccer ball shape.\n\nMolecule or crystal models usually come as stick (or stick and ball) models, or as calotte model with spherical balls intersecting. The stick model emphasizes the chemical bonds you know from structural formula of organic molecules and the location of the atom core, whereas the calotte model signifies the extend of the electron cloud surrounding the atom core.
933669
aldehyde\nthe name of the compound is propionaldehyde
583132
It shouldn't have anything significant.  Propylene glycol (1,2-propanediol) has no pi-bonds, so there wouldn't be any UV-absorbing electron structure.\n\nHO-CH2-CHOH-CH3
838542
The answer is B, butyric acid, or the ester of it.
516493
a) Toluene and Cl2 form p-chlorotoluene and HCl in presence of AlCl3\nb) nitrobenzene + Cl2 -(AlCl3)-> m-chloro-nitrobenzene\nc) chlorobenzene + Cl2 -(AlCl3)-> p-dichlorobenzene\nd) 1-(3-chlorophenyl)ethanone\ne) 3-chlorophenyl acetate\nf) methyl 3-chlorobenzoate\n\nOrganic Chem rocks!
137566
B/L acids HBr, NH4, H3O, HSO3, H20 respectively. the other reactant is a B/L base.