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The following organic compounds cannot exist. Why? a. 2 -chloro-2-butyne b. 2 -methyl-2-propanone c. \(1,\) l-dimethylbenzene d. 2 -pentanal e. 3 -hexanoic acid f. \(5,5\) -dibromo- \(-\) cyclobutanol

Short Answer

Expert verified
The given organic compounds cannot exist for the following reasons: a. 2-chloro-2-butyne: No place for chlorine to attach to the second carbon atom due to the triple bond. b. 2-methyl-2-propanone: Attaching a methyl group to the second carbon violates the octet rule. c. 1,1-dimethylbenzene: Attaching two methyl groups to one carbon in the benzene ring breaks the octet rule. d. 2-pentanal: No hydrogen atom connected to the second carbon for the aldehyde group. e. 3-hexanoic acid: Carboxylic acid functional group only bonds at the terminal carbon atom, not an internal one. f. 5,5-dibromo-cyclobutanol: No 5th carbon in a cyclobutanol molecule for bromine atoms to attach.

Step by step solution

01

a. 2-chloro-2-butyne

2-chloro-2-butyne implies that a chlorine atom is connected to the second carbon atom of a butyne molecule. Butyne has a triple bond between two of the carbon atoms. In the case of 2-butyne, the triple bond should be between the second and third carbon atoms. However, there is no place for the chlorine atom to attach to the second carbon atom, as it already has four bonds (one with the neighboring carbon, and three as part of the triple bond with the third carbon). This results in an invalid structure, so 2-chloro-2-butyne cannot exist.
02

b. 2-methyl-2-propanone

2-methyl-2-propanone implies that there is a methyl group attached to the second carbon atom of a propanone molecule. However, propanone (known as acetone) has only three carbon atoms with a carbonyl group (C=O) at the second carbon, and it is already bonded with the other two carbons. Adding a methyl group would violate the octet rule as the second carbon atom would have five bonds instead of the maximum of four. Thus, this compound cannot exist.
03

c. 1,1-dimethylbenzene

The name 1,1-dimethylbenzene implies that there are two methyl groups attached to the same carbon atom in the benzene ring. However, each carbon in a benzene ring is already attached to two neighboring carbon atoms, and one hydrogen atom, which means that there is only one available bond for the addition of a methyl group. Attaching two methyl groups on a single carbon breaks the octet rule. Therefore, the compound cannot exist. The correct compound is 1,2-dimethylbenzene, where two methyl groups are attached to adjacent carbon atoms.
04

d. 2-pentanal

2-pentanal implies there is an aldehyde functional group (C=O) attached at the second carbon atom of a pentane chain. However, in an aldehyde, the carbonyl carbon is bonded to a hydrogen and an alkyl group. In the case of 2-pentanal, there is no hydrogen atom connected to the second carbon atom, as it is bonded to two other carbon atoms from the alkyl chain. A correct compound with an aldehyde group would be pentanal instead of 2-pentanal, where the aldehyde group is connected at the end of the chain.
05

e. 3-hexanoic acid

3-hexanoic acid implies that a carboxylic acid functional group (COOH) is attached at the third carbon atom of the hexane chain. However, COOH functional groups only bond at the terminal carbon atom of an alkyl chain, not an internal one. There cannot be a hexanoic acid attached to the third carbon. A correct compound would be hexanoic acid, where the carboxylic acid group is connected to the end carbon atom.
06

f. 5,5-dibromo-cyclobutanol

5,5-dibromo-cyclobutanol implies that two bromine atoms are attached to the 5th carbon atom in the cyclobutanol molecule. However, cyclobutanol is a four-membered ring with an alcohol group (OH) attached to one of the carbon atoms. There is no 5th carbon in a cyclobutanol molecule, which means that 5,5-dibromo-cyclobutanol cannot exist. A correct compound might be 1,1-dibromo-cyclobutanol, in which two bromine atoms are attached to the first carbon atom.

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Key Concepts

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

Invalid Molecular Structures
When we say a molecular structure is invalid in organic chemistry, we mean that the arrangement of atoms within the proposed molecule isn’t possible according to known chemical rules. Invalid structures often arise when naming or constructing molecules without adhering to basic chemical principles like the octet rule or the nature of functional groups.

For instance, in the exercises provided:
  • 2-chloro-2-butyne is invalid because there is no free valence on the second carbon to bond with chlorine due to a triple bond already connecting it to another carbon.

  • 1,1-dimethylbenzene is also impossible because a single benzene carbon cannot support two methyl groups while maintaining the benzene structure.

  • 5,5-dibromo-cyclobutanol is non-existent since cyclobutanol is a four-membered ring with no room for two bromines on a non-existent fifth carbon atom.
Recognizing invalid structures is critical in organic chemistry to avoid proposing unfeasible compounds and ensures the scientific accuracy of reported chemical species.
Bonding Rules in Organic Molecules
Bonding rules are essential for understanding how atoms connect in organic molecules. The fundamental concept here is the octet rule, which posits that atoms are usually most stable when they have eight electrons in their valence shell. In hydrocarbons, each carbon forms four bonds in total. These can be shared with carbon or other atoms like hydrogen, oxygen, or halogens. When naming or interpreting organic molecules, understanding how these bonds can be formed prevents creating impossible structures.

In the example of 2-methyl-2-propanone, we spot an error because the central carbon atom already bonds through the carbonyl group, leaving no room to add a methyl group without surpassing four bonds, thus violating the octet rule. Recognizing bonding constraints can serve as a quick check against molecular misconfigurations, providing a clearer pathway to valid structure formulation.
Functional Groups
Functional groups are specific groupings of atoms within molecules that have distinctive chemical properties. Common examples include hydroxyl groups in alcohols, carbonyl groups in ketones, and carboxyl groups in acids. Understanding functional groups is vital for accurately predicting the reactivity, polarity, phase, color, and even the biological activity of organic compounds. Errors often come up when function groups aren't placed correctly according to their defining rules.
  • For example, an aldehyde group should appear at the end position of a carbon chain, as seen with the corrected form of pentanal, but not as internal groups as attempted in 2-pentanal.
  • Similarly, a carboxyl group, typical in acids such as hexanoic acid, must bond with the terminal carbon to be valid, not internally connecting, which creates a non-existent 3-hexanoic acid structure.
Analyzing functional group placement helps chemists accurately predict and name compounds, ensuring their structural validity and expected chemical behavior.

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Most popular questions from this chapter

When toluene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\right)\) reacts with chlorine gas in the presence of iron(III) catalyst, the product is a mixture of the ortho and para isomers of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{ClCH}_{3}\) . However, when the reaction is light-catalyzed with no \(\mathrm{Fe}^{3+}\) catalyst present, the product is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}\) . Explain.

Identify each of the following compounds as a carboxylic acid, ester, ketone, aldehyde, or amine. a. Anthraquinone, an important starting material in the manufacture of dyes:

Give an example reaction that would yield the following products as major organic products. See Exercises 22.68 and 22.71 for some hints. For oxidation reactions, just write oxidation over the arrow and don’t worry about the actual reagent. a. primary alcohol b. secondary alcohol c. tertiary alcohol d. aldehyde e. ketone f. carboxylic acid g. ester

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 \(68 . )\) (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.) $$ \begin{array}{l}{\text { a. } 1 \text { -butanol } \longrightarrow \text { butane }} \\ {\text { b. } 1 \text { -butanol } \longrightarrow 2 \text { -butanone }}\end{array} $$

If one hydrogen in a hydrocarbon is replaced by a halogen atom, the number of isomers that exist for the substituted compound depends on the number of types of hydrogen in the original hydrocarbon. Thus there is only one form of chloroethane (all hydrogens in ethane are equivalent), but there are two isomers of propane that arise from the substitution of a methyl hydrogen or a methylene hydrogen. How many isomers can be obtained when one hydrogen in each of the compounds named below is replaced by a chlorine atom? $$ \begin{array}{ll}{\text { a. } n \text { -pentane }} & {\text { c. } 2,4 \text { -dimethylpentane }} \\ {\text { b. } 2 \text { -methylbutane }} & {\text { d. methylcyclobutane }}\end{array} $$

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