(a)
Interpretation:
It is to be explained why the pKa values of all three isomers in which a methyl group is attached to the ring are higher than the pKa of phenol itself.
Concept introduction:
When comparing uncharged acids, the stronger acid is the one whose negatively charged conjugate base is more stable. Electron donating substituents attached to the ring decrease the acidity as compared to non-substituted rings. Alkyl group is an electron donating group, thus, it will destabilize the negatively charged conjugate base to a greater extent than non-substituted rings.
(b)
Interpretation:
Why the meta isomer has the lowest
Concept introduction:
When comparing uncharged acids, the stronger acid is the one whose negatively charged conjugate base is more stable. Electron donating substituents attached to the ring decrease the acidity as compared to non-substituted rings. Methyl groups is an electron donating group, thus, it will destabilize the negatively charged conjugate base to a greater extent than non-substituted rings. Greater the resonance stabilization of the conjugate base, stronger is the acid. In the resonance contributors, the structure in which an atom bearing a negative charge becomes destabilized as the number of electron-donating groups attached to it increases. This structure will not contribute significantly to the resonance hybrid.
Trending nowThis is a popular solution!
Chapter 6 Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
- Nonearrow_forward4. Draw and label all possible isomers for [M(py)3(DMSO)2(CI)] (py = pyridine, DMSO dimethylsulfoxide).arrow_forwardThe emission data in cps displayed in Table 1 is reported to two decimal places by the chemist. However, the instrument output is shown in Table 2. Table 2. Iron emission from ICP-AES Sample Blank Standard Emission, cps 579.503252562 9308340.13122 Unknown Sample 343.232365741 Did the chemist make the correct choice in how they choose to display the data up in Table 1? Choose the best explanation from the choices below. No. Since the instrument calculates 12 digits for all values, they should all be kept and not truncated. Doing so would eliminate significant information. No. Since the instrument calculates 5 decimal places for the standard, all of the values should be limited to the same number. The other decimal places are not significant for the blank and unknown sample. Yes. The way Saman made the standards was limited by the 250-mL volumetric flask. This glassware can report values to 2 decimal places, and this establishes our number of significant figures. Yes. Instrumental data…arrow_forward
- 7. Draw a curved arrow mechanism for the following reaction. HO cat. HCI OH in dioxane with 4A molecular sievesarrow_forwardTry: Convert the given 3D perspective structure to Newman projection about C2 - C3 bond (C2 carbon in the front). Also, show Newman projection of other possible staggered conformers and circle the most stable conformation. Use the template shown. F H3C Br Harrow_forwardNonearrow_forward