Question

Organic help: @abb0t

Answer 1

Why doesnt acetylene react with NaOH?

Answer 2

NaOH is a base, and acetlyene is relatively acidic, but not acidic enough to donate a proton to NaOH. You'd need a real strong base like NaH or NaNH\(_2\)

Answer 3

Thanks.2. Why does 2,3 dicholoro butane exhibit optical isomerism even though it has a plane of symmetry?

Answer 4

If you want to get more into why, you can also look at it's hybdirization as well and see the stability of it vs ethene.

Answer 5

But can you speak of it in simple terms? It has a plane of symmetry means its achiral so how can we take it as a optical isomer?

Answer 6

Well, there are three stereoisomers, if im looking at it correctly, there's RR, RS, and SS...

Answer 7

remember that optical isomerism is simply their rotation in plane polarized light.

Answer 8

The R-R, and S-S are enantiomers.

Answer 9

hmm...But plane of symmetry means no optical activity isnt it?

Answer 10

Not necessarily, look at their enantiomers. they are nonsuperimposible mirror images of one another, and henceforeth, chiral. Am i right?

Answer 11

This might help clarify chirality and optical isomerism: http://info.piercecollege.edu/title3/aln/chem102/Chemistry102_01.html
as well as this site I found: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/stereochemie/chiralitaet.vlu/Page/vsc/en/ch/12/oc/stereochemie/optische_aktivitaet/optische_aktivitaet.vscml.html

The topic is a bit difficult to understand at first, and I may not be the best person to help explain, but those sites do a pretty good job.

Answer 12

okay I ttink I got it.

Answer 13

Think doesn't mean you fully understand it :P

Answer 14

3. On mixing a certain alkane with chlorine and irradiating it with UV light, it forms only 1 monochloroalkane. This alkane would be?

Answer 15

This is what is called a radical reaction involving 3 major steps: (a) initiation (b) propagation (c) termination

And the product formed would be the more stable one. Follows markovnikovs rule for most stable product.

Answer 16

although, you may find trace amounts of 1-chloro product, you are generally forming the 1st product i drew.

You can, however, get 1-chloroalkane if you added H\(_2\)O\(_2\), in which case, you'd have 2 radical reactions occuring and it gets a bit more complex, BUT not much.

Answer 17

But we are starting with an alkane here

Answer 18

Oh sorry. HAHA. Yes, i meant alkane, not alkene. SORRY. haha

Answer 19

2º > 1º

Answer 20

okay so how do we know which alkane it is?

Answer 21

the one that's more stable. Are you familiar with radical stability? It also follows markovnikov rule; formation of more stable product. tertiary > secondary > primary

Answer 22

Yes I know that. But why only monochlorination happen here?

Answer 23

Well, one thing you should note is that radical reactions occur repeatedly, meaning, you can form a copious amount of products! Radicals are very reactive and you COULD possibly form dichloro compounds, but you just want to focus on the monochloro product because it is the one that will be found more when subject to analysis. But certainly, you will have various products.

Answer 24

The answer given is neopentane and reason is all H atoms are same and of 1 degree :O

Answer 25

Oh. Lol. I guess i missread the question. I didn't see what they were asking for. I just thought they meant some alkane, as in ANY alkane. Lol. I didn't see it asked "what alkane" lol

Answer 26

lool. Now answer it >.>

Answer 27

It's very similar to what I explained. Since all the alkanes are the same, the only thing you are removing here is a hydrogen. In a radical reaction, basically what is happening is:

Br:Br \(\rightarrow\) 2 Br•
Then you have: