Draw all 12 acyclic (no rings) isomers of formula C 4 H 7 Br. Include stereoisomers.

Welcome back, everyone show all four a cyclic isomers with the formula C three H five cl note, if applicable include stereo isomers to begin with. We want to have a good idea if there are any pi bonds present in the structure. So we're going to calculate the hydrogen deficiency index 21 half, multiplied by two, multiplied by the number of carbon atoms, which is three, right plus two minus the number of hydrogens, which is five minus the number of halogens, which is a one. What do we get? Well, essentially we get a result of one, meaning there must be a one pi bond or one ring, but we are interested in a cyclic isomers. So we're not interested in those rings, right? And because we are not interested in the ranks, we're going to only focus on the structures that have double bonds because there must be one pi bond. So let's go ahead and draw them. If we have three carbon atoms, we can just draw a three member chain right now. What can we do? Well, we can essentially immediately assign a double bond between carbons one and two and then simply think where can we attach that chlorine, we can attach it at the end of the chain. So that's one of our structures. However, we can also attach chlorine to carbon number two, right. We have used up carbon number three. Now we can use carbon number two. So let's change the position of chlorine and let's bond it to carbon number two. Well done. So we have two structures and now we simply want to think of any additional structures. Can we add that chlorine to carbon? Number one? And the answer is yes, we can. So we are simply shifting our chlorine to every possible carbon so far. Now, it looks like that's it. But we also have to include stereo isomers. The first two structures cannot exhibit c trans isomers because we have two equivalent hydrogens bonded to one of our double bonded carbons. But the final structure is trans because we have two hydrogens on the opposite sides of the double bond. And we can draw its cis form where the two Hagens would be on the same side of the double bond, meaning the alki group and the chloral group will be on the same side of the double bond. That's our Z isomer because the previous one was an E isomer and that's it. We have all the four structures possible. Thank you for watching.

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1. A Review of General Chemistry

Constitutional Isomers

Organic Chemistry

1. A Review of General Chemistry

Constitutional Isomers

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Problem 44b

Textbook Question

Draw all 12 acyclic (no rings) isomers of formula C₄H₇Br. Include stereoisomers.

Verified step by step guidance

Step 1: Begin by identifying the possible carbon skeletons for the compound C4H7Br. Since the compound is acyclic, consider linear and branched structures. The simplest structure is a straight chain of four carbon atoms.

Step 2: For each carbon skeleton, determine the possible positions for the bromine atom. In a straight chain, the bromine can be attached to any of the four carbon atoms, leading to different structural isomers.

Step 3: Consider the presence of double bonds, as the formula C4H7Br suggests unsaturation. Explore the placement of a double bond between different pairs of carbon atoms in each skeleton, ensuring the correct number of hydrogen atoms are present.

Step 4: Identify stereoisomers by examining the configurations around any chiral centers or double bonds. For example, if a double bond is present, consider both E/Z configurations, and if a chiral center is present, consider R/S configurations.

Step 5: Compile all possible isomers, including structural and stereoisomers, ensuring that each isomer adheres to the molecular formula C4H7Br. Verify that you have accounted for all 12 isomers by systematically checking each configuration and structure.

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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 have the same molecular formula but different structural or spatial arrangements. In the context of C4H7Br, isomers can be structural, where the connectivity of atoms differs, or stereoisomers, where the spatial arrangement of atoms differs without changing the connectivity.

Stereoisomerism

Stereoisomerism involves isomers that have the same structural formula but differ in the spatial orientation of atoms. This includes geometric isomers (cis-trans) and optical isomers (enantiomers). For C4H7Br, identifying stereocenters or double bonds can help determine possible stereoisomers, which are crucial for drawing all isomers.

Alkyl Halides

Alkyl halides are organic compounds containing a halogen atom (like bromine) bonded to an alkyl group. Understanding the reactivity and structure of alkyl halides is essential for drawing isomers of C4H7Br, as the position of the bromine atom can significantly affect the compound's properties and the number of possible isomers.

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