Concept explainers
Interpretation:
The compound that gives the given
Concept Introduction:
NMR spectroscopy:
It is the study of the interaction between
Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).
Integration value (I): The integration value at the bottom of the
Chemical shift values for protons in different electronic environment
Types of protons | Chemical shift |
Methyl | |
methylene | |
Methine | |
Allylic | |
Alkynyl | |
Aromatic methyl | |
Alkyl halide |
Alcohol | |
Vinyl |
Aryl | |
Aldehyde | |
carboxylic |
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Chapter 15 Solutions
ORGANIC CHEMISTRY 1 TERM ACCESS
- Highlight in red each acidic location on the organic molecule at left. Highlight in blue each basic location on the organic molecule at right. Note for advanced students: we mean acidic or basic in the Brønsted-Lowry sense only. Cl N شیخ x Garrow_forwardQ4: Draw the mirror image of the following molecules. Are the molecules chiral? C/ F LL CI CH3 CI CH3 0 CI CH3 CI CH3 CH3arrow_forwardComplete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?arrow_forward
- 4. C6H100 5 I peak 3 2 PPM Integration values: 1.79ppm (2), 4.43ppm (1.33) Ipeakarrow_forwardNonearrow_forward3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forward
- A chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forwardNonearrow_forward
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