Draw all the isomers that have molecular formula C 5 H 11 Br. ( Hint: There are eight.) b. Give a common name for each isomer that has a common name.

Hello, everyone. In this video, we have to draw all the members of C three H seven I and we also have to give them their common names. So first we have to draw the Iceman's which we need to first calculate for the degree of saturation. So the formula for our degree of saturation is to C plus two plus N minus H minus X all divided by two. So see here represents the number of carbons and represents the number of nitrogen, H represents the number of protons or hydrogen. And an X here represents a number of religions. So that's florine, chlorine, bromine, and iodine. So for the given formula of C three H seven I, we can go ahead and use this to calculate for the degree of saturation. So we see here that for carbons, we have a total of three atoms. So we'll plug in three for C and this all plus two plus N, we don't have any nitrogen here in our molecule. So plus zero for our hae Jin's here, we see we have seven. So we'll put minus seven then for our halogen here, that's just one that's iodine. So that's minus one. Again, all of this is going to be divided by two. So once I put all this into my calculator, I get that the degree of the situation here for this molecule is equal to zero. So zero degree of saturation means that it does not have a double bond or a ring. So the possible with the molecular formula being C three H seven, I is going to be where we have a linear structure. So we have this going on here and we have an iodine at the end, we're going to have a more branch here, some sort of chain here. So we have these three carbons. And then again, at the end, we have an iodine. So we have something like this. So to read the common names of alcohol highlights, we first read the name of the alcohol group, followed by the name of the halogen. So in the name of the halogen, the line, so I N E at the end becomes I D E. So for example, if we have roaming, so let's break this out roaming. So we have the I N E ending, this goes on to become bromide. So we have I D E ending here. So the out group in the first summer is proposal. So we'll write that out. So the out Gail group again, that is probable. So the name of this first one here is going to be pro pro I O died. So usually it's iodine when we switched out the suffix into I D E. Alright. So for our second one here again, the out group is going to be so appropriate. So this means that our nomenclature is going to be Isopropyl, I'll died. Alright. So go ahead and highlight those two names of their final names for our customers. Also go ahead and circle Our two structures of our C three H 7.

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

4. Alkanes and Cycloalkanes

Naming Alkyl Halides

Organic Chemistry

4. Alkanes and Cycloalkanes

Naming Alkyl Halides

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3:19 minutes

Problem 72b

Textbook Question

Draw all the isomers that have molecular formula C₅H₁₁Br. (Hint: There are eight.)
b. Give a common name for each isomer that has a common name.

Verified step by step guidance

Step 1: Understand the problem. The molecular formula C5H11Br indicates a brominated alkane with five carbon atoms. Isomers are compounds with the same molecular formula but different structural arrangements. We need to draw all possible structural isomers and name them appropriately.

Step 2: Begin by considering the straight-chain structure (n-pentane). Replace one hydrogen atom with a bromine atom at different positions along the chain to create isomers. For example, bromine can be attached to the first, second, or third carbon atom in the chain.

Step 3: Next, consider branched-chain structures. Create isomers by branching the carbon chain (e.g., isobutane or neopentane skeletons). Replace one hydrogen atom with bromine at different positions in these branched structures.

Step 4: For each structure, ensure that you are not duplicating isomers. Check for symmetry in the molecule to avoid counting the same isomer twice. For example, bromine on the first carbon of n-pentane is equivalent to bromine on the fifth carbon due to symmetry.

Step 5: Assign common names to each isomer where applicable. Common names are based on the alkyl group attached to bromine (e.g., n-butyl bromide, isobutyl bromide, etc.). Use IUPAC naming conventions for any isomers that do not have widely recognized common names.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Isomerism

Isomerism refers to the phenomenon where compounds with the same molecular formula exhibit different structural or spatial arrangements. In organic chemistry, isomers can be classified into structural isomers, which differ in the connectivity of atoms, and stereoisomers, which differ in the orientation of atoms in space. Understanding isomerism is crucial for identifying and drawing all possible isomers for a given molecular formula.

Branched vs. Straight-Chain Alkanes

Branched and straight-chain alkanes are types of hydrocarbons that differ in their structure. Straight-chain alkanes have a continuous chain of carbon atoms, while branched alkanes contain one or more branches off the main chain. Recognizing these differences is essential for drawing isomers, as the branching can lead to multiple structural variations for the same molecular formula.

Common Names of Organic Compounds

Common names are informal names used to identify organic compounds, often based on historical or traditional naming conventions rather than systematic IUPAC rules. For example, isomers of C5H11Br may include common names like 'pentyl bromide' or 'isobutyl bromide.' Familiarity with these common names helps in quickly identifying and communicating about specific isomers in organic chemistry.

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