Test: Covalent Bonds - MCAT MCQ

The sulfur atom undergoes excitation to promote two electrons, one from 3s and one from 3p, not two from 2s, to two empty 3d orbitals.

In the excited state, the intermixing of one s, three p, and two, not three d, orbitals gives 6 half-filled sp²d³, not sp²d³,, hybridized orbitals.

Sulfur forms 6 σ hybridized bonds with fluorine atoms , and each fluorine atom makes use of a half-filled 2p_z, not 3p_z, orbital for the bond formation.

The orbitals are all oriented such that all bond angles are 90° and have an octahedral configuration is the correct answer.

Alcohols, carboxylic acids, and only 1° and 2° amines and amides may serve as H-bond donors and acceptors. Ethers, ketones, 3° amines and amides can serve as acceptors.

Carboxylic acids generally have high boiling points, but amides can technically have boiling points almost 100 degrees greater than carboxylic acids

The dimeric H-bonded structure appears to be a good representation of acetic acid in both the gaseous and liquid state. If acetic acid dimerized, it would have a molecular weight of 120 and a boiling point similar to a molecule of that weight.

Tetramethoxymethane is a molecule with a molecular weight of 136, which is sufficiently close to 120, and a boiling point of 114. The similar boiling point with acetic acid confirms our suspicions that it dimerized in liquid and gaseous phases.

Amides indeed have high melting points, which makes most of them solids at room temperatures. Amides have higher boiling points than carboxylic acids due to its highly polar carbonyl group as indicated by its resonance structure, leaving carbon with a negative charge and nitrogen with a positive charge, and the H-bonding between primary and secondary amides.

This is an exercise in counting the total number of stereogenic centers. Stereogenic centers include chiral carbons and double-bonded carbons in cis-trans alkenes.

On the left half of the molecule, there are 9 asymmetric carbons and 2 double bonds:

On the right half of the molecule, there are 8 asymmetric carbons and 2 double bonds:

There are 21 stereogenic centers in total in this compound.

The acetylide ion (HCC‾ ) has 2 sp-hybridized carbons due to the triple bond since two unhybridized p orbitals are needed to form the two π bonds. The C-H bond is a sigma bond formed from the overlap of a sp hybridized and 1s orbital.

The nitronium ion (NO2⁺) has an sp-hybridized nitrogen, and the cation is formed from removing the lone electron from paramagnetic NO. The nitrogen is double-bonded to each of the oxygens necessitating two unhybridized p orbitals to form the two π bonds.

Carbon dioxide should be familiar as a linear molecule, in which the carbon has no lone pairs. The central carbon is sp-hybridized since it is double-bonded to each of the oxygens.

The nitrite ion (NO2‾) will be sp2-hybridized since nitrogen forms a π bond with one of the oxygens through an unhybridized p orbital. The other orbitals become hybridized into sp2, and nitrogen forms a hybridized bond to each of the oxygens and has a lone pair in the other hybridized orbital. One oxygen is single-bonded and has a negative charge, and the other is double-bonded and neutral.

The adamantane derivative features four asymmetric substituted carbon atoms, and it only exists as two stereoisomers (enantiomers) rather than the sixteen predicted by the 2ⁿ rule. However, the number of observed stereoisomers is only 2 since once the first stereocenter is defined, the remaining three are determined by their relative configuration.

The configurations of the asymmetric bridgehead carbon units in this structure are (R), (S), (R) and (R) respectively, for the Br, Cl, CO₂H and CH₃ substituents.

The central adamantyl residue may be regarded as an extended tetrahedron, and the stereogenic center is located in the center of the adamantyl cage, shown by the dotted lines below connecting at the center of the molecule, and the configuration shown here is (R).

In naming any compound systemically, we must determine the primary substituent group which will become the suffix. Here it is the carboxylic acid group. We are looking for some kind of -oic acid.

We can determine the stereochemistry of the compound, which is (S), and that is shown below:

Technically, we should have the answer with those two pieces of information, but there are two answers with To be thorough, the next step in IUPAC nomenclature is to find the longest carbon chain starting at the carbon of the carboxylic acid. It would seem it would be 4 carbons, but it will be only 3 carbons because of the amide group as a prefix.

The amine group at C-2 will become 2-amino, and the amide group attached at C-3, the substituent including the carbon, will become 3-carbamoyl and not 4-carbamoy

The name of the compound is (S)-2-amino-3-carbomylpropanoic acid.

When 2-butanol is placed in pyridine with 4-toluenesulfonyl chloride, the alcohol is converted into the tosylate, which makes it a better leaving group. The reaction is known to occur with retention of configuration,here is no change in the stereochemistry.

When the tosylate is then placed into a sodium azide and ethanol solution, the azide acts as a nucleophile in a SN2 fashion to attack from the backside and substitute the tosylate with an azide group. There is an inversion of stereochemistry due to the backside attack.

2-butanol has a configuration of (R), so then compound A will have an (R) configuration and compound B an (S) configuration due to the inversion. These are our relative configurations of A and B.

Since we know the sign of rotation and the designation of R or S for the parent, we have its absolute configuration.

While for compounds A and B, without further experimentation, we do not ascertain whether R corresponds to (+) or (-). But, because you know the configuration of the starting alcohol, then the absolute configuration of the product is also known.

The answer is that the relative configurations of A and B can be determined to be (R) and (S), while the absolute and relative configurations of the parent is already known.

Covalent bonds are formed between nonmetallic elements. Oxygen is a nonmetal and therefore tends to form covalent bonds.

Carbon and hydrogen are both nonmetals and tend to form covalent bonds when they combine.

In covalent bonds, electrons are shared between atoms, allowing them to achieve a more stable electron configuration.