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[SchoolBoyDJ]

Second organic test this cominig monday

1. Which of the following is not an electrophile?
a) AlCl_4-[/sub]   b) BH₃     c)  NH₄⁺

I was able to eliminate c, but the other two gave me trouble.  My thinking was that:  If carbon can form 4 bonds (sp3 hybridized), why can't Boron?  So I was thinking Boron had an open p orbital (currently sp2 hybridized).  AlCl4- seemed to already be full.  I know that elements in period 3 or higher can form more than the expected number of bonds, but why the heck would something with full orbitals, and a negative charge still want more negative/electrons 

2.  The following is a ______ alkene



a)monosubstituted
b)disubstituted
c)trisubstituted
d)tertasubstituted


Now you'd name this 2-ethyl-1-butene right?  I thought this would be monosubstituted because it has one group coming off the main chain.  I guess substitution takes into account all R groups (everything but H right?) coming off the double bond?

3.  When 2-pentene is heated with a catalytic amount of I₂, an equilibrium concentration of two isomers is obtained.  The dominant isomer is:

c) an equal amount of each is present at equilibrium



I didn't have much of a guess on this, and I guess that's because I don't know what is happening to the molecule while reacting with I₂. A is the correct answer, but with the knowledge I have now I'd argue b as it seems like it would be more stable...the pi bond almost seems more enclosed.  Help >_<

Thanks everyone!

[chrisf]

That first one, it says which is "NOT an electrophile." That one wouldn't be an electrophile.. just as you said.

[valdorod]

1) as chris said, the question asked for the one who is not an electrophile.  The other three, b, c and d are all lewis acids, which qualifies them as electrophiles.  AlCl₃ is also a lewis acid but AlCl₄^- is not

2)  Substituted alkenes have the followind denominations.  Thus b is the correct answer.



3) while I am not exactly sure what I2 does, I can tell you that if you have a choice between a trans and a cis isomer of the same molecule, the trans version is typically the one in highest yield and the more stable.  Thus a being in a trans configuration would be my choice.

[SchoolBoyDJ]

OMG LOL!  I can't believe I typed that out and still didn't notice is says NOT XD!!!!  Wow let's hope I'm a little more on guard for the test tomorrow.

Yes valdro, you are correct about the cis and trans and I wasn't thinking of that while making my decision.  Thank you all very much!  I'll let you guys know how it goes tomorrow ^_^

[AceOrganic]

So, on #3, I2 can add reversibly to alkenes.  It adds to the alkene and forms a carbocation intermediate.  Instead of adding the other I- to form the dihalide, the iodine acts as a leaving group and eliminates.  The reason A is the answer is because once you form the cation intermediate, the trans is more thermodynmically favored b/c of sterics.

for more info on this, see http://books.google.com/books?id=txmp1aoCJp8C&pg=PA270&lpg=PA270&dq=alkene+equilibrium+iodine&source=web&ots=cFzAg8gxSI&sig=cJjQFm1FMBDWsIkoZAvfVyzgf1Y&hl=en&sa=X&oi=book_result&resnum=8&ct=result#PPA270,M1

[valdorod]

Ace

From the way the question is worded, I do not think it is an addition reaction.

First -  I₂ is not a reactant but a catalyst.

Second  - none of the answers are halides

Third - addition of a halogen to an alkene produces an alkane, once again none of the answers are alkanes nor halides.

[AceOrganic]

Hi valdorod, I have to disagree with you on this on, or maybe I did not explain myself well.  I2 is not a catalyst, but is present in catalytic amounts.  The overall reaction is addition-elimination which takes an alkene and gives an alkene.  You say that the addition of a halogen to an alkene gives an alkane (I can only assume you meant an alkyl halide), but that is not the case for iodine.  Below is the reaction (Attached as jpg), along with an excerpt from a textbook on it.

"Iodine intially reacts with alkenes to form a cationic intermediate"..."this is then eliminated to reform the alkene"

[valdorod]

Instead of adding the other I- to form the dihalide, the iodine acts as a leaving group and eliminates.

Ace, you explained it correctly the first time, I missed two very important words in your description, and focused on the adding and dihalide.  Your explanation makes sense.  After clicking on your first link, I focused on the bromine and chlorine additions, skimming throught the fluorine and iodine examples.

[SchoolBoyDJ]

Thanks everyone   We hadn't learned about the "halonium" ions yet so this question didn't make much sense to me at the time.