The initially chiral carbon has an hybridization. Once treated with ,an oxidizing agent, the secondary alcohol in the compound is oxidized to a ketone. The central carbon is no longer a chiral center (a carbon with a double bond cannot be chiral), and the double bond (pi-bond) formed between carbon and oxygen gives the molecule an hybridization.

hybridization is formed with triple bonds, and an hybridization does not exist.

Benzylamine consists of a benzene ring with a -CH2NH2 substituent. The benzene ring contains three double bonds, and the substituent has none.

Each bond in a compound represents a sigma bond. Each additional bond represents a pi bond; thus, double binds result from one sigma and one pi bond, and triple bonds from one sigma and two pi bonds. In benzylamine, there are three double bonds and seventeen total bonds, three of which are double bonds. The compound will have a total of seventeen sigma bonds and three pi bonds.

Recall the formula for degrees of unsaturation as the sum of the number of pi bonds and the number of rings. This value is equivalent to the below formula.

C is the number of carbons, H is the number of hydrogens, X is the number of halogens, and N is the number of nitrogens. Using this formula, compound A has four degrees of unsaturation, while the product only has two.

The reaction described is a hydrogenation reaction, which will reduce double bonds. The reaction saturates two pi bonds of compound A, meaning the remaining degrees of unsaturation in that compound have to come from two rings. Essentially, due to the reaction given, we know that the product contains no double bonds but still has two degrees of unsaturation. These unsaturation points must correspond to rings.

The stability of a bond can be compared to others by determining the hybridization of the bond. It helps to remember that the bond with the most s character is considered the most stable.

Carbon dioxide has a carbon that is connected to two oxygens by two double bonds. This means that the carbon has sp hybridization. Since the bond has 50% s character, it is the most stable out of the available compounds.

Methane contains only single bonds and will have an sp3 hybridized carbon, with only 25% s character. Acetone and ethene each contain one double bond and will contain an sp2 hybridized carbon with 33% s character.

The answer is arrows A and C. The carbon that is pointed to by arrow C is hybridized. We see that its bond angles are at 120º (the full substituent points into the page) with a p-orbital that is involved in the pi bond of the carbonyl. Phenyl rings are also made up of carbons that are all hybridized. The nitrogen pointed to by arrow B has two electrons that are not shown, but cause the atom to be hybridized. The methyl groups denoted by arrow D are also hybridized.

We can quickly tell the hybridization of atoms by observing their double bonds and unbonded electrons. As a rule of thumb, any carbon, nitrogen, or oxygen involved in a double bond will be hybridized. Any of these atoms with no double bonds will be hybridized. Finally, nitrogens or carbons involved in triple bonds are hybridized.

For a compound to be considered aromatic, it must be flat, cyclic, and conjugated and it must obey Huckel's rule. Huckel's rule states that an aromatic compound must have pi electrons in the overlapping p orbitals in order to be aromatic (n in this formula represents any integer). Only compounds with 2, 6, 10, 14, . . . pi electrons can be considered aromatic. Compound A has 6 pi electrons, compound B has 4, and compound C has 8. This eliminates answers B and C. Answer D is not cyclic, and therefore cannot be aromatic. The only aromatic compound is answer choice A, which you should recognize as benzene.

The C–N bond is a single bond, and the carbonyl bond is a double bond. Double bonds are stronger and shorter than single bonds.

It helps to remember that bond energy is inversely proportional to bond length. In other words, the shorter the bond, the higher the bond energy.

Bond length is shortened when pi bonds are involved, so double and triple bonds are much shorter than simple sigma bonds. The bond length between the carbons in ethyne is the shortest out of all options because they are triple bonded to one another. This also makes it the most stable bond, and gives it the largest bond energy.

Benzene and 1,3-butadiene will have relatively high energy stored in double bonds, but will be unable to match a triple bond. Propane has only single bonds and will have the lowest bond energy of these answers.

A degree of unsaturation results from either a ring or a double bond. In the case of a benzene ring or phenyl substituent, four degrees of unsaturation are present: the ring itself and the three included double bonds.

Compounds A, B, and C all possess a benzene ring, but compounds A and C also have one additional degree of unsaturation. Compound A contains a carbon-oxygen double bond in the ketone group and compound C contains a carbon-carbon double bond in the alkene group. In addition to the four degrees from the ring, these additions bring their total degrees of unsaturation to five. Compound B has no additional double bond, and therefore has four degrees of unsaturation.

In order to find the hybridization of an atom, simply count the number of sigma bonds and lone pair electrons around the atom. The carbon in formic acid is double bonded to an oxygen, and has two single bonds. This means that it has sp2 hybridization. Upon being reduced to methanol, the carbon now has four single bonds surrounding it. As a result, the carbon now has sp3 hybridization.

Remember that a triple bond corresponds to sp hydrization, a double bond to sp2, and single bonds to sp3 for a carbon atom.