Reactions Types - Organic Chemistry

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)

The correct ranking from most basic to least basic is:

The best bases are negatively charged, and the worst bases are positively charged (acidic). The stronger the base, the weaker (more stable) it's conjugate acid. An alkane is a very stable conjugate acid, which tells us that is the most basic of the set.

We know based on charge alone, that is more basic than but less basic than or .

We know that is more basic because the electronegativity of is less than that of . This means the lone pair of electrons on are held less tightly and more likely to pick up a proton.

Remember that the strongest acids have the weakest conjugate bases. is more acidic than because iodine has a larger atomic radius than bromine. , , and are strong acids and should be at the beginning of the list. Alkanes are not acidic. Acetic acid is a weak acid (pKa =4-5).

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.

Hydrogenation of a double bond involves the bond breaking and a hydrogen being added to each carbon of that double bond. You can tell the number of double bonds by taking the number of hydrogens added and dividing it by 2.

6 added hydrogen divided by 2 is 3 double bonds.

A hydrocarbon with zero degrees of unsaturation and carbons has hydrogens. For every two hydrogens less than , there is one degree of unsaturation. After hydrogenation, our final product has no double bonds. After calculation, we see that it has two degrees of unsaturation. This means that it has two rings.

In order to convert an alkene into an alkane, we need a catalyst.

We can reduce the alkene here by simply adding two hydrogens with as a reagent.

In this question, we're told that our starting material, 1-hexene, is being reacted with hydrobromic acid in the presence of ultraviolet (UV) light. To solve this, we need to consider how halogens add to alkenes, specifically in the presence of ultraviolet light.

First, it's important to note that UV light will cause the hydrogen and bromine atoms in to dissociate as free radicals. Because each of these atoms are electron deficient, they desperately want to react in order to fill their valence shells. And since the double bond in the alkene is electron dense, a reaction will occur between one of the electrons in this double bond and the free radical bromine.

The bromine radical adds to the alkene first at the 1-carbon in an anti-Markovnikov fashion. The reason it adds to this carbon (and not the 2-carbon) is because a secondary free radical is more stable than a primary free radical. Upon formation of this secondary radical, it will react with the hydrogen free radical in solution to generate the finished product, 1-bromohexane.

A catalyst is often required for an electrophilic aromatic substitution reaction. All reagents are paired with their correct catalysts except for . requires as a catalyst in order to react with benzene.

The reaction shown is a Markovnikov addition of a hydracid (HX) across a double bond. According to Markovnikov's rule, the hydrogen gets added to the lesser-substituted carbon in the double bond, and the halide (in this case, ) gets added to the more-substituted carbon. Thus, the correct answer is .

An addition reaction is a reaction in which the reactants react to combine and form one product. It is the opposite of an elimination reaction. In the reaction given, the reactants hydrochloric acid and ethylene combine to form the product 1-chloropropane.

An addition reaction is a reaction in which the reactants react to combine and form one product. It is the opposite of an elimination reaction. In the reaction given, the reactants 2-butene and molecular hydrogen combine to form the product butane.

An addition reaction is a reaction in which the reactants react to combine and form one product. It is the opposite of an elimination reaction. In the reaction given, the reactants hydrochloric acid and propyne combine to form the product 2-chloropropene.