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Which of the noncyclic isomers of bromochloropropene are optically active?

Short Answer

Expert verified
The optically active noncyclic isomer of bromochloropropene is 1-Bromo-2-chloropropene, as it has a stereocenter at the carbon in position \(2\) with four different groups attached to it.

Step by step solution

01

Identify the molecular formula for bromochloropropene

First, we need to find the molecular formula for bromochloropropene. Bromochloropropene contains bromine (Br), chlorine (Cl), carbon (C), and hydrogen (H). Propene, a 3-carbon molecule, has the formula C3H6. Since the molecule has one bromine and one chlorine atom, the overall molecular formula will be C3H5BrCl.
02

Identify noncyclic isomers

Now, we will generate all the possible noncyclic isomers for the given molecular formula. There are two possibilities for placing Br and Cl on propene: 1. 1-Bromo-1-chloropropene 2. 1-Bromo-2-chloropropene So we have two possible noncyclic isomers.
03

Identify the stereocenters

Next, we will identify any stereocenters in each isomer. A stereocenter is an atom with four different groups attached to it. 1. In 1-Bromo-1-chloropropene, there are no carbons with four distinct groups attached, so there are no stereocenters. 2. In 1-Bromo-2-chloropropene, the carbon in position 2 is a stereocenter as it has four different groups attached to it: H, Br, Cl, and the rest of the propene.
04

Determine the optical activity of the isomers

An optically active compound must have a chiral center and a non-superimposable mirror image (enantiomers). We have found that only 1-Bromo-2-chloropropene has a stereocenter. As a result, only this isomer can have two non-superimposable mirror images, making it an optically active compound. Thus, the optically active noncyclic isomer of bromochloropropene is 1-Bromo-2-chloropropene.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Noncyclic Isomers
Understanding noncyclic isomers can be crucial when studying organic chemistry, specifically when analyzing optically active compounds. Noncyclic isomers are different molecules that have the same molecular formula but differ in the arrangement of their atoms without forming any rings. With bromochloropropene, for example, we look at how bromine (Br) and chlorine (Cl) can be attached to a propene backbone to create different structures.

When creating noncyclic isomers, we must consider each unique position that substituents like Br and Cl can occupy on the carbon chain. For instance, in 1-Bromo-1-chloropropene, both halogens are attached to the first carbon on the propene, whereas in 1-Bromo-2-chloropropene, they are on separate carbons. The former arrangement does not allow for optical activity because there is symmetry; hence, no molecule can have a non-superimposable mirror image. The latter arrangement, however, shows diversity in position leading to potential optical activity due to the presence of a stereocenter.
Stereocenters
A stereocenter, also known as a chiral center, is an atom that bears groups of atoms in such a way that an interchange of any two groups leads to a stereoisomer. In simpler terms, it's like a junction from which four different paths diverge. For a compound to have optical activity, it must have at least one stereocenter. The presence of a stereocenter correlates with the molecule's ability to rotate plane-polarized light, a characteristic we associate with chiral substances.

In our problem, 1-Bromo-1-chloropropene lacks a stereocenter because it does not have a carbon atom with four unique substituents. In contrast, 1-Bromo-2-chloropropene does have a carbon (the second carbon in the chain) bonded to four different groups - a hydrogen atom, a bromine atom, a chlorine atom, and an ethyl group. This asymmetry is essential for creating two non-identical mirror images, known as enantiomers, which leads us to conclude that 1-Bromo-2-chloropropene is the optically active isomer of the two.
Enantiomers
Enantiomers can be thought of as chemical mirror images of each other, just as our left hand is a mirror image of our right hand. They are a pair of molecules that have the same molecular formula, the same connectivity of atoms, but are non-superimposable. This means you cannot place one enantiomer on top of the other and have all aspects of the molecules perfectly align.

These isomers are important because they can have vastly different biological activities. In the context of bromochloropropene isomers, 1-Bromo-2-chloropropene contains a stereocenter which means it can exist as a pair of enantiomers. Each enantiomer will rotate plane-polarized light in opposite directions, one to the left (levorotatory or 'L') and one to the right (dextrorotatory or 'D'). It's this rotation that signifies the optical activity of the compound. Pharmaceuticals, in particular, are heavily impacted by enantiomers, as one form of a drug might be beneficial while its counterpart could be inactive or even harmful.

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Most popular questions from this chapter

Cis-trans isomerism is also possible in molecules with rings. Draw the cis and trans isomers of 1,2 -dimethylcyclohexane. In Exercise 41, you drew all of the noncyclic structural and geometric isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~F}\). Now draw the cyclic structural and geometric isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~F}\).

Alcohols are very useful starting materials for the production of many different compounds. The following conversions, starting with 1 -butanol, can be carried out in two or more steps. Show the steps (reactants/catalysts) you would follow to carry out the conversions, drawing the formula for the organic product in each step. For each step, a major product must be produced. (See Exercise 62.) (Hint: In the presence of \(\mathrm{H}^{+}\), an alcohol is converted into an alkene and water. This is the exact reverse of the reaction of adding water to an alkene to form an alcohol.) a. 1 -butanol \(\longrightarrow\) butane b. 1-butanol \(\longrightarrow 2\) -butanone

Draw all the structural and geometrical (cis-trans) isomers of bromochloropropene.

Stretch a rubber band while holding it gently to your lips. Then slowly let it relax while still in contact with your lips. a. What happens to the temperature of the rubber band on stretching? b. Is the stretching an exothermic or endothermic process? c. Explain the above result in terms of intermolecular forces. d. What is the sign of \(\Delta S\) and \(\Delta G\) for stretching the rubber band? e. Give the molecular explanation for the sign of \(\Delta S\) for stretching.

Consider the compounds butanoic acid, pentanal, \(n\) -hexane, and 1 -pentanol. The boiling points of these compounds (in no specific order) are \(69^{\circ} \mathrm{C}, 103^{\circ} \mathrm{C}, 137^{\circ} \mathrm{C}\), and \(164^{\circ} \mathrm{C}\). Match the boiling points to the correct compound.

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