When organic molecules are coloured, this is typically due to conjugation. Conjugation refers to the path available to electron movement created through neighbouring pi orbitals with delocalised electrons, such as the alternate double bonds in aromatic rings. Electrons can move through the connected pi orbitals, after first being excited by ambient electromagnetic radiation. The longer the available path/conjugation, the higher the energy of light absorbed by the electrons. Thus, conjugation can cause the colour of the molecule to vary depending on the energy of the light being absorbed. The parent molecule (shown below) is yellow in colour due to the conjugation between the alternate double bonds. Generally speaking, conjugation can only exist in molecules that are planar – i.e. they exist in a flat conformation. In the case of the above structure (1), there is the potential to increase the conjugation by forcing the central section to remain planar by placing a hydrocarbon tether/ring at that site (shown below). Figure 1 - Ball-and-stick models of the geometry-optimized structures of 1, 5-7 showing the change in overall planarity caused by the additional hydrocarbon ring. (1) Considering the need for planarity over the molecule, rank the systems 5, 6 and 7 from the least to the most conjugated. 5 is Blank 1 Question 8[select: , the most conjugated, neither most nor least conjugated, least conjugated] . 6 is Blank 2 Question 8[select: , the most conjugated, neither most nor least conjugated, least conjugated] . 7 is Blank 3 Question 8[select: , the most conjugated, neither most nor least conjugated, least conjugated] . (1) Huck, L. A.; Leigh, W. J. Journal of Chemical Education 2010, 87 (12), 1384-1387.多重下拉选择题

登录即可查看完整答案
我们收录了全球超50000道真实原题与详细解析,现在登录,立即获得答案。
类似问题
When organic molecules are coloured, this is typically due to conjugation. Conjugation refers to the path available to electron movement created through neighbouring pi orbitals with delocalised electrons, such as the alternate double bonds in aromatic rings. Electrons can move through the connected pi orbitals, after first being excited by ambient electromagnetic radiation. The longer the available path/conjugation, the higher the energy of light absorbed by the electrons. Thus, conjugation can cause the colour of the molecule to vary depending on the energy of the light being absorbed. The parent molecule (shown below) is yellow in colour due to the conjugation between the alternate double bonds. Generally speaking, conjugation can only exist in molecules that are planar – i.e. they exist in a flat conformation. In the case of the above structure (1), there is the potential to increase the conjugation by forcing the central section to remain planar by placing a hydrocarbon tether/ring at that site (shown below). Figure 1 - Ball-and-stick models of the geometry-optimized structures of 1, 5-7 showing the change in overall planarity caused by the additional hydrocarbon ring. (1) Considering the need for planarity over the molecule, rank the systems 5, 6 and 7 from the least to the most conjugated. 5 is Blank 1 Question 8[select: , the most conjugated, neither most nor least conjugated, least conjugated] . 6 is Blank 2 Question 8[select: , the most conjugated, neither most nor least conjugated, least conjugated] . 7 is Blank 3 Question 8[select: , the most conjugated, neither most nor least conjugated, least conjugated] . (1) Huck, L. A.; Leigh, W. J. Journal of Chemical Education 2010, 87 (12), 1384-1387.
The implementation and loading phase of the Database Life Cycle (DBLC) involves _____.
Question textComplete the following protocol table for setting up your standard curve for protein. Remember, each of your standards, except the reagent blank, will be prepared in duplicate in the lab (i..e, you will make up each of these standards twice), giving you 11 standards. When filling in your answers, be careful with your formatting and note the following requirements: for all entries of concentrations, volumes and amounts in this table, include two decimal places (e.g. '1.5') for any decimal numbers less than one, make sure you precede the decimal point with the number zero (i.e. type in '0.2'; not '.2' without the zero) [table] Standard | Concentration of BSA standard | Volume of 20 mg/mL BSA stock solution required (mL) | Volume of 0.05 M NaOH diluent required (mL) | Amount of BSA present (mg) | Volume (mL) 1(reagent blank) | 0 | Answer 1 Question 6 | Answer 2 Question 6 | Answer 3 Question 6 | 0.8 2 | 4 | Answer 4 Question 6 | Answer 5 Question 6 | Answer 6 Question 6 | 0.8 3 | 8 | Answer 7 Question 6 | Answer 8 Question 6 | Answer 9 Question 6 | 0.8 4 | 12 | Answer 10 Question 6 | Answer 11 Question 6 | Answer 12 Question 6 | 0.8 5 | 16 | Answer 13 Question 6 | Answer 14 Question 6 | Answer 15 Question 6 | 0.8 6 | 20 | Answer 16 Question 6 | Answer 17 Question 6 | Answer 18 Question 6 | 0.8 [/table]
Which of the following parameters is used to determine the concentration of protein in a spectrophotometer?
更多留学生实用工具
希望你的学习变得更简单
加入我们,立即解锁 海量真题 与 独家解析,让复习快人一步!