[icyhead]

Hi im having some trouble with nucleophilic subsitution/elimination, im doing a biology course but this is from a chemistry module, so im struggling... anyway.

Ive got one question that asks me for the product/s of a reaction, while explaining it fully.

The Reaction is an alkane with an -OMS group on the 2nd carbon, pointing backwards from the molecule. Its reacting with NaN3. I dont know anything about either of these molecules other than MS is a very good leaving group, and NaN3 is a weak base but a very strong nucleophile. I think it could be a SN1 reaction or maybe an elimination. Can anyone help me understand it and perhaps offer a solution?

Ive also got another question with a tertiary alcohol reacting with conc. h2so4, ive been told OH isnt a leaving group, so im not sure whats going on in this reaction.

[leaftye]

I'm a little groggy, and I'm doing this mentally which is always a bad idea, but I'll give it a shot.

Since it's a weak base, I don't see an elimination happening.  I'm think :N=N=N:. is going to do a concerted substitution (SN2) at the carbon OMS is attached to.

As to the second part, the acid could add a hydrogen proton to the OH group, making it H2O, which is a great leaving group.

[alphahydroxy]

Yup, assuming you mean "OMs", as in "mesylate" or "trifluoromethylsulfonate", the azide (very good nucleophile - shaped like a little missile no less!) is going to displace it an an S_N2 fashion.

I think I agree with leaftye with the second one also - while OH is not a good leaving group generally, +OH₂ is an excellent one, thus protonation of the hydroxyl leads to E₁ elimination and rearrangement to isobutene.

[icyhead]

Thanks for the help, and yes i did mean mesylate, and will the OMs all be displaced or just the Ms ?
Also im not sure how the OMs bonds to the alkane, can i check is it  -C=O-SCH3O2?

[Squirmy]

-OMs is displaced.  Can you see why it would be a good leaving group?

      O
      ||
R-O-S-CH₃
      ||
      O

[icyhead]

im really grasping at straws here but..
I think there is hypoconjugation which makes it more stable and a better leaving group.
Id also assume its a weak base (due to the single bonded O-S ??) and neutral - though thats from guessing, not knowing how.

[Squirmy]

Methane sulfonic acid is one of sulfuric acid's organic cousins.

     O
     ||
HO-S-CH₃   Methane sulfonic acid
     ||
     O


     O
     ||
HO-S-OH     Sulfuric acid
     ||
     O


Hopefully, you know sulfuric acid is a strong acid. Well, so is methane sulfonic acid. Strong acids make weak (stable) conjugate bases. ^-OMs = conj. base of methane sulfonic acid.

As for why those conjugate bases are stable, there's resonance stabilization and the oxygens attached to sulfur are inductively withdrawing, both lessening the localized charge.

[icyhead]

ah thanks a lot Ive never even seen sulphuric acids' structure before but that explains OMs's popularity as a leaving group.
Ive still got at least 1 question to go:

When the Molecule shows  a skeleton strucute of an alkane with the OMs as a wedge bond above it, does it mean there is a H behind the structure?
And ive also got the same reaction with the OMs leaving group, but the nucleophile as NaOC2H5, it seems like a bulky molecule probably a weak nuc, and a weak base, is that right?

[Squirmy]

When the Molecule shows  a skeleton strucute of an alkane with the OMs as a wedge bond above it, does it mean there is a H behind the structure?

Assuming the carbon skeleton is in the plane (regular lines, not dashes or wedges), then yes.

And ive also got the same reaction with the OMs leaving group, but the nucleophile as NaOC2H5, it seems like a bulky molecule probably a weak nuc, and a weak base, is that right?

NaOC2H5 a strong nucleophile and a strong base.

With a secondary mesylate, you'll likely get E2 (NaOC2H5 acts as a base), due to the steric requirements of backside attack in an Sn2. NaOC2H5 with a primary mesylate would give Sn2 (NaOC2H5 acts as a nucleophile).

[icyhead]

Ok 1 more question, moving on to adding br2 to alkenes, having some trouble with the stereoisomers that are formed. It says on the sheet some of the questions must give multiple stereoisomers, but I cant see how... i keep looking at the diagrams of the end products and it seems like they can be rotated.

1 . Pent-1-ene

2 . (cis) Hex-2-ene (non cyclic)

3 .  (trans) Hex-2-ene (non-cyclic)

[Squirmy]

check this page, and let us know if you still have questions:

http://en.wikipedia.org/wiki/Halogen_addition_reaction

[icyhead]

I have a question on the Halogen addition reaction, if the alkene is irregular, i.e one side of the double bond differs from the other, and the bond is Trans (E) is there still only 1 product formed? (the wikipedia example only shows a symettrical alkene)
I ask this because i checked and looking at my 2 diagrams they appear to be emantiomers.

[Squirmy]

The symmetrical trans-alkene only forms 1 product because the product is a meso compound. That is, even though the product has chiral carbons, the molecule is not chiral overall because of an internal mirror plane (which requires symmetry).

Since a non-symmetrical trans-alkene won't form a product with an internal mirror plane, 2 enantiomers will form.

[icyhead]

Thanks for the quick *delete me*
Next ive got reactions of alkenes with Osmium tetroxide followed by aqueous sodium sulfite. My lectures breezed over reactions of osmium tetroxide with a couple of examples with cycloalkenes.
I think its similar to reacting with bromine. I reckon it will add an O to each side of the double bond, creating carbonyls.
But then ive no idea what the Na2SO3 does... done a quick google search didnt yield any helpful results.

[Squirmy]

http://www.cem.msu.edu/~reusch/VirtualText/addene2.htm#add4b

The Na₂SO₃ is used to hydrolyze the cyclic intermediate containing osmium and to reduce the osmium.

Note that the stereochemistry of this reaction is quite different from halogenation. The OH groups add syn, whereas the halogenation is an anti addition... http://en.wikipedia.org/wiki/Syn_addition