[chalk5]

Hey All,

I'm new to the forum here and a struggling O-Chem student. I'm taking OChem 1, and although I do enjoy it, I'm still timid about my answers. I'm doing a test review problems, and I was wondering if I could get some help to verify my answers. I'm not hundred percent sure, and any inputs would be greatly appreciated. Anything that isn't clear, too vague, etc. Thank you in advance!


4. Explain why free radical halogenation produces racemic mixtures of products.
-Because the radical intermediate produces an SP2 like hybridization orbital with lone electron in the vacant 2P orbital, therefore attack of the radical electron on the C-H bond can take place from either side of the molecule.


5. Explain what it means to say that a one reaction reaction is more selective than another
-Transition state is the highest point in energy in a reaction's process. Once the transition state is reached, there is no barrier to go on to the product, but before that, the interacting molecules can go back to not interacting with each other. So, for a highly exothermic reaction with an early transition state, there is negligible energetic difference between the transition states to different products. This means once the molecules get to the transition state, there is no preference for which product they form, since the transition states for each product is the same. However, for a more endothermic/less exothermic reaction, with a later transition state, there is a large difference in transition state energies. So if the molecules are heading up to the higher energy transition state, they may not have enough energy and fall back down to products. But they may have enough energy, on average, to get to the lowest energy transition state. So the product that comes from the lower energy transition state gets formed selectively, while the barrier to get the product from the higher energy transition state is too high to overcome.
Note: I wasn’t sure if this is the way the question should be answered. Do I need to include regioselectivity/regiospecificity (constitutional isomers), stereoselectivity/stereospecificity (spacial isomers), enantioselectivity (selecting for a specific spacial arrangement), and Markovnakov/Anti-Markovnokov additions?



6. Explain why free radical chlorination is less selective than free radical bromination. Be sure your answer makes reference to bond energies, transition states, and the Hammond postulate (at a minimum).
-One outcome of the Hammond postulate is the placement of the transition state on a reaction coordinate diagram. More exothermic reactions have early transition states, which more closely resemble the reactants. More endothermic reactions have later transition states, and more closely resemble the product. Transition states that more resemble the products are more subject to energetic differences between possible products, since at this point in the reaction the product is almost formed. Radical chlorination is more exothermic than radical bromination, so it has an earlier transition state. This means the carbon-hydrogen bond is not very broken at the transition state, so all hydrogens are essentially equivalent and we don't see much selectivity. For bromination, with its later transition state, the C-H bond is almost entirely broken at the transition state, so all hydrogens are not equivalent here, and we do see selectivity for abstracting the hydrogen with the lowest bond dissociation energy.
Note: I need help in reference to bond energies!



9. In the reaction of HO- with iodomethane, the reaction rate increases when the concentration of HO- is increased. However, when HO- reacts with 2-iodo-2-methylpropane, the reaction is unaffected by the concentration of HO-. Give a complete explanation for this difference.
-So in reaction of HO- with iodomethane, with HO- being the nucleophile, the reaction of HO- with iodomethane is SN2 reaction. So therefore the rate equation for SN2 reaction is Rate= K<sub>[Nuc], so as the concentration of HO- increases then the rate of reaction also increases. On the other hand, the reaction of HO- and 2-iodo-2 methylpropane is SN1 with the rate equation being Rate= K[Substrate], so this indicates that concentration of nucleophile does not affect the rate of the reaction.


10. Is BF3 an electrophile or a nucleophile? Explain your answer.
-BF3 is an electrophile because the electrostatic potential map indicates that it is electron poor (blue). And the Lewis Structure shows that BF3 lacks a complete electron octet and can accept an electron pair from a nucleophile.


11. Which is the better nucleophile? CH3OH or CH3SH? Explain your answer
-CH3SH is a better nucleophile since Sulfur is larger and more polarizable than Oxygen so therefore it is a better nucleophile. CH3SH can do SN2 while CH3OH, being a weak nucleophile and a weak base, will typically result in both SN1 and E1 when added to alkyl halide.


13. Why does 4-iodo-4 methylpent 2-ene give two substitution products when reacted with a nucleophile, but 2-iodo-4-methylpentane only gives on product under the same reaction conditions? (Ignore stereochemistry for this question).
-Because with 4-iodo-4 methylpent 2-ene, there is resonance to the structure so the carbocation have different positions which gives two substitution products, while 2-iodo-4-methylpentane, the carbocation is always secondary and no hydride shift, so there is only one intermediate, which gives only one product.



17. OH- is a worse choice of nucleophile than t-butoxide [(CH3)3CO-] for converting 1-chloropentane to pent-1-ene. Explain why?
-OH- is a worse choice of nucleophile than t-butoxide for converting 1-chloropentane to pent-1-ene because [OH]- is a small base and [(CH3)3CO-] is a bulky base. [(CH3)3CO-] is the better base to make the 1-pentene, because the big bulky base will result in a less substituted alkene.This is due to hoffman. The smaller base will give the more substituted alkene.</sub>

[PhDoc]

It's really hard to imagine these questions not being part of a problem set assigned for a grade. Then again, perhaps it's just me.

You've got MAJOR issues with #9. There's a flaw in the question. Carbocations (pKa -10 to -15) form in strong base just as easily as HI (pKa -10) forms in strong base. If you replace HO- in part two with H2O, then you have a "legal" question.