Interpretation: The Lewis structures of the small compounds have to be found by using their Lewis dot structures of involving atoms and by following the octet rule.
Concept introduction: The arrangement of atoms that are bonded together determines its constitution and molecular formula of that particular compound. This concept is referred as structural isomers or in more modern term constitutional isomers. Each atom has a typical valency or valence which is defined as the ability of an atom to form a chemical bond with other atoms. For example, carbon has four valence or tetravalent that means carbon has the capacity to form four bonds with other elements or other atoms. Nitrogen atom is trivalent. Hydrogen is monovalent in nature.
Lewis dot structures are also called electron dot structures. The diagrams are named after Gilbert N. Lewis, who described them in his 1916 article entitled The Atom and the Molecule. The number of electrons present in the outermost shell of an atom that participate in forming
The Lewis structure of small compounds can be drawn by combining the Lewis dot structures of the atoms involved in the formation of that compound. In order to achieve complete noble gas configuration, octet rule is followed to find bonding nature of atoms. Octet means 8 electrons in their valence shells. The compounds which obey the octet rule are called stable compounds.
To find: The four constitutional isomers of C3H9N, their Lewis structures and the number of lone pair of electrons on the nitrogen atom in each isomer
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Organic Chemistry
- Complete boxes in the flow chart. Draw the structure of the organic compound foundin each layer after adding 3M NaOH and extraction. Make sure to include any charges. Provide explanation on answers.arrow_forward== Vid4Q2 Unanswered ☑ Provide IUPAC name of product in the reaction below A 3,4-dimethylcyclohexene B 1,2-dimethylcyclohexane C 1,2-dimethylcyclohexene D 3,4-dimethylcyclohexane H₂ Pdarrow_forward5. Use the MS data to answer the questions on the next page. 14.0 1.4 15.0 8.1 100- MS-IW-5644 26.0 2.8 27.0 6.7 28.0 1.8 29.0 80 4.4 38.0 1.0 39.0 1.5 41.0 1.2 42.0 11.2 43.0 100.0 44.0 4.3 79.0 1.9 80.0 2.6 Relative Intensity 40 81.0 1.9 82.0 2.5 93.0 8.7 20- 95.0 8.2 121.0 2.0 123.0 2.0 136.0 11.8 0 138.0 11.5 20 40 8. 60 a. Br - 0 80 100 120 140 160 180 200 220 m/z Identify the m/z of the base peak and molecular ion. 2 b. Draw structures for each of the following fragments (include electrons and charges): 43.0, 93.0, 95.0, 136.0, and 138.0 m/z. C. Draw a reasonable a-fragmentation mechanism for the fragmentation of the molecular ion to fragment 43.0 m/z. Be sure to include all electrons and formal charges. 6. Using the values provided in Appendix E of your lab manual, calculate the monoisotopic mass for the pyridinium ion (CsH6N) and show your work.arrow_forward
- Nonearrow_forwardStereochemistry: Three possible answers- diastereomers, enantiomers OH CH₂OH I -c=0 21108 1101 41745 HOR CH₂OH IL Но CH₂OH TIL a. Compounds I and III have this relationship with each other: enantiomers b. Compounds II and IV have this relationship with each other: c. Compounds I and II have this relationship with each other: d. *Draw one structure that is a stereoisomer of II, but neither a diastereomer nor an enantiomer. (more than one correct answer)arrow_forwardNonearrow_forward
- Don't used Ai solutionarrow_forwardIn mass spectrometry, alpha cleavages are common in molecules with heteroatoms. Draw the two daughter ions that would be observed in the mass spectrum resulting from an alpha cleavage of this molecule. + NH2 Q Draw Fragment with m/z of 72arrow_forwardDon't used Ai solution and don't used hand raitingarrow_forward
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