[Gregorian]

Hi, I have 2 questions for you guys

Question 1:
If we have excess I-I (iodine) & butanone in basic conditions, do we replace all the hydrogens of both adjacent methyl groups?
If the iodine was not in excess which H gets replaced first !

Question 2:
What is the name of the molecule in the attached picture and how do you make it from benzene knowing that this synthesis can be done in three steps only!

I thought about using ozonolysis in reverse and using DMFDMA. I don't even know if that is possible. I never learned anything like that!

Thanks

 

[azmanam]

1) In butanone there's only 1 adjacent methyl group.

http://www.cem.msu.edu/~reusch/VirtTxtJml/suppmnt3.htm#halfm

2) If you don't know the Bergman, you'll have to 2 Friedel Crafts reactions.  But the 3 carbon chain can't be functionalized for both FC at the start, or you'll just tether 2 benzene rings together.  so the 2nd step will have to take some masked FC reagent and convert it into the FC reagent.

Ozonolysis does not run in reverse.

The compound is tetralin.

[Gregorian]

1) In butanone there's only 1 adjacent methyl group.

http://www.cem.msu.edu/~reusch/VirtTxtJml/suppmnt3.htm#halfm

2) If you don't know the Bergman, you'll have to 2 Friedel Crafts reactions.  But the 3 carbon chain can't be functionalized for both FC at the start, or you'll just tether 2 benzene rings together.  so the 2nd step will have to take some masked FC reagent and convert it into the FC reagent.

Ozonolysis does not run in reverse.

The compound is tetralin.

Thanks for your answer.
For #1 I meant by the two methyl groups these in the red boxes. I saw in the link that it says to only replace the hydrogens of the methyl, but aren't the hydrogens on the left also acidic!

supposing that the ketone is not a methyl ketone and is not symmetric, what do we use to determine which hydrogen of the alpha hydrogens will be replaced first

thanks

for #2
I don't get it 

[BlowUpEverything]

Look at it in terms of the conjugate base.  Which carbanion would be more stable - a primary or secondary?

[azmanam]

Useful information:

C-CH₃ is a methyl group.
C-CH₂-C is a methylene group.
C-CH(-C)-C (work with me, it's a carbon with one hydrogen and 3 other carbons attached) is a methine group.
Carbon with 4 carbons, no hydrogens, attached is a quaternary carbon.

There's good information here:
http://bio-che.mc.edu/valente/ch18.pdf

For #2, have you not learned the Friedel Craft's reaction?

[nox]

for alpha halogentions on asymmetrical ketones doesn't the reaction conditions used to generate the enolate control which side of the carbonyl group halogenation occurs?

if you use a strong, hindered base like LDA at -78 degrees in an aprotic solvent (eg THF) you obtain the kinetic enolate and halogenation occurs at the less substituted site (ie methyl group in your case)
whereas if you use a weaker base like potassium tert-butoxide with heating in a protic solvent (eg tert-butanol) you'd get the thermodynamic enolate and halogenation occurs at the more substituted site (the methylene group)

of course if you have a great excess of iodine I suppose halogenation would occur on both sides of the carbonyl group

[macman104]

for alpha halogentions on asymmetrical ketones doesn't the reaction conditions used to generate the enolate control which side of the carbonyl group halogenation occurs?

if you use a strong, hindered base like LDA at -78 degrees in an aprotic solvent (eg THF) you obtain the kinetic enolate and halogenation occurs at the less substituted site (ie methyl group in your case)
whereas if you use a weaker base like potassium tert-butoxide with heating in a protic solvent (eg tert-butanol) you'd get the thermodynamic enolate and halogenation occurs at the more substituted site (the methylene group)

of course if you have a great excess of iodine I suppose halogenation would occur on both sides of the carbonyl group

Correct, you can control it like that as long the you have different pkas of a little more than 1 or 2 (that would be 10 to 100 fold reactivity).

[waheed]

Hi, I have 2 questions for you guys

Question 1:
If we have excess I-I (iodine) & butanone in basic conditions, do we replace all the hydrogens of both adjacent methyl groups?
If the iodine was not in excess which H gets replaced first !

Question 2:
What is the name of the molecule in the attached picture and how do you make it from benzene knowing that this synthesis can be done in three steps only!

I thought about using ozonolysis in reverse and using DMFDMA. I don't even know if that is possible. I never learned anything like that!

Thanks

 

The name of the aromatic compound is tetraline(1,2,3,4-tetrahyronaphthalene). You can make it from benzene in 3 steps as follows: 1- Rxn of benzene w sucinic anhydride, and AlCl3. 2- reduction of the keto group, clemenson rdn or wolfkischner redn. 3- cyclization with PPA.