Reactions Types - Organic Chemistry

The side of the reaction that is favored will have the acid with the higher , because the reaction goes (strong acid + strong base weak acid + weak base).

Addition reactions involve breaking one pi bond (double bond) and forming two sigma bonds in the product.

E1 reactions occur in two steps, forming a carbocation intermediate, which is most stable if there is a protic solvent present. Furthermore, the use of a weak base favors E1.

E2 reactions occur in one step; thus no carbocation intermediate is formed, and an aprotic solvent is favored. E2 reactions are favored by strong bases and higher temperatures.

An elimination reaction occurs when there is a release of atoms in a given compound to produce two or more products. In the reaction given a hydrogen and chloride atom are eliminated from the original compound to form one 2-butene, potassium chloride and water molecule.

Heterolytic bond breaking occurs in polar compounds to form to products of opposite charges. In these types of reaction two electrons from the original bond stays with one fragment upon cleavage. In the reaction given, the bond between the hydrogen and chlorine atom is broken with the two electrons from the original bond staying with the chlorine atom. The resulting products are hydrogen ion and chloride ion.

Homolytic bond breaking occurs when a molecule breaks up to form two or more new products. In the reaction given, molecular chlorine forms two radicals in which one electron stays with each fragment formed.

All of the given answers reflect SN1 reactions, except the claim that SN1 reactions are favored by weak nucleophiles.

SN1 reactions occur in two steps and involve a carbocation intermediate. The product demonstrates inverted stereochemistry (no racemic mixture). Tertiary substrates are preferred in this mechanism because they provide stabilization of the carbocation.

SN2 reaction mechanisms are favored by methyl/primary substrates because of reduced steric hindrance. No carbocation is formed via an SN2 mechanism since the mechanism is concerted; thus a strong nuclephile is used.

Substitution reactions—regardless of the mechanism—involve breaking one sigma bond, and forming another sigma bond (to another group).

This reaction would use an SN1 mechanism because the leaving group, bromine, is on a tertiary carbon, which is a carbon attached to three other carbon atoms. The bulk of these methyl groups would make SN2 impossible, but it would make the carbocation produced by an SN1 reaction very stable. The methyl group would lead to hyperconjugation, which is a type of resonance that stabilizes transition states.

This reaction would occur using an SN1 mechanism because the leaving group is attached to a tertiary carbon, a carbon atom with three of the carbon atoms attached to it. The rate laws for SN1 mechanisms do not depend on the nucleophile concentration. The slow-step occurs unimolecularly within the molecule with the leaving group.

The reaction shown is a nucleophilic substitution reaction. The molecule shown is a primary alkyl bromide, and the nucleophile that will be used is the iodide anion (). In this type of reaction the iodide will displace the bromide on the organic molecule, generating iodoethane ().

The pKa value indicates how strong an acid is, and acid strength increases as pKa decreases. The side of a reaction with a lower pKa is going to dissociate more, pushing the equilibrium over to the other side. The equilibrium will thus lie on the side with the HIGHER pKa.

Since the pKa of acetic acid (4.76) is higher than the pKa of trifluoroacetic acid (0), the reaction will shift to the left to reach equilibrium.

The governing principle regarding the prediction of values (relative to other compounds) is to assess the stability of the product formed by the release of a proton. The release of the alkyne hydrogen in compound III results in a carbanion, a highly unstable species, so it is expected that this compound is the least acidic and has the highest . Intuition serves well in this instance and we see that hydrogens bound to a triple bond have a value of around 25. The relative stabilities of the remaining compounds may be assessed in the same manner. Compound IV is the second weakest acid because the three methyl groups donate electron density such that if the oxygen is deprotonated, the resulting negative charge is destabilized. Methanol and water have a unique, non-intuitive relationship regarding their relative acidities. One would assume that water should be a stronger acid than other acids bound to alkyl groups (by the reasoning expressed for compound IV). This is the case for all alcohols except methanol, in which the delocalization of charge allowed by the increased molecular size outweighs the destabilization caused by electron donation. Thus methanol is a slightly stronger acid than water. This is evidenced in their values: 15.7 for water and 15.5 for methanol. The correct ordering of the given compounds is: III, IV, II, I.

An easy way to consider relative base strengths is to consider the strength of the compounds' conjugate acids.The stronger the conjugate acid, the weaker the base. Water (compound IV) is the least basic of the compounds because its conjugate acid, is the strongest of the given compounds' conjugate acids . Carboxylate ions (compound III) are highly stabilized by resonance and predominate at neutral pH. The conjugate carboxylic acids readily donate protons (acetic acid: ). Ammonia (compound I) has considerable basicity; binding a fourth hydrogen produces ammonium ion , which predominates at neutral pH Sodium propoxide (compound II) is a strong base, bearing a full negative charge on its oxygen. Its conjugate acid, 1-propanol, is a rather weak acid . Since its conjugate acid is the weakest (highest ), sodium propoxide is the strongest base. Based on the previous observations the correct ordering of the compounds is: II, I, III, IV.

Use the Henderson Hasselbalch equation:

Molecule B will have a more acidic alpha-carbon because once the alpha proton becomes dissociated, the conjugate base will have relatively more stability than the conjugate base of molecule A.

When the alpha-carbon on molecule A loses it's proton, the conjugate base is not as stable. The reason for this is because the oxygen that is involved in the ester bond can contribute its electrons towards a resonance structure. Therefore, after the alpha-proton is lost, the alpha-carbon will have a negative charge that will be destabilized by the delocalized negative charge of the resonance structures.

From the start, we know we can eliminate answer choice because it is the only answer choice that is not a strong acid. Now we have three strong acids, but we have to determine which is strongest. To do so, we take the conjugate base of each strong acid to see which conjugate base is the weakest acid. Remember: weaker conjugate base means a stronger acid. is the largest ion of the bunch. Its large size allows it to better stabilize the negative charge and so it is the weakest (most stable) conjugate base. Because the weakest conjugate base leads to the strongest acid, is our correct answer.

Recall that the stronger an acid, the lower the pKa.

II (two fluorine atoms really close to has largest inductive effect, so bond is most weakened, and pKa is lowest)

V (one fluorine really close to has strong inductive effect)

I (one fluorine a little further away from has weaker inductive effect)

IV (no inductive effect)

III (alcohols are much less acidic than carboxylic acids, and it has the highest pKa of all)