Concept (explain why this is so..)

Remember the rules of what gives compounds a higher boiling point (and melting point):

Van der Waals Forces

More electrons = more Van der Waals interactions = higher boiling point

So in general, the more "stuff" in a molecule (the higher the molecular weight), the higher its b.p. Example: C₁₀H₂₂ will have a higher b.p. than C₅H₁₂.

How to remember this trend? What do you grill with? propane gas. (C₃H₈)

What's in a lighter? butane liquid (C₄H₁₀)

Hydrogen Bonding

Compounds that can hydrogen bond have higher b.p./m.p/ than those that don't.

Compounds can hydrogen bond is they have a N, O, or F bonded directly to an H.

In organic chemistry, the best examples are alcohols (ROH) and amines (RNH₂).

Branching

Branching makes it harder for molecules to pack together, which makes it harder to form Van der Waals interactions, and so tends to lower b.p./m.p.

a) (highest b.p.) n-octane > n-pentane > ethane (lowest b.p.)

Reason: octane has the most "stuff" (higher molecular weight).

b) 1-butanol > 1-chloroethane > n-butane

Reason: 1-butanol can hydrogen bond. 1-chloroethane has a higher molecular weight than n-butane.

c) n-octane > 2-methylheptane > 2,5-dimethylhexane

Reason: Branching. n-octane has no branching. 2,5-dimethylhexane has the most branching. Notice that each compound has the same molecular formua- C₈H₁₈. (and therefore the same weight).

The trick to this problem is to look for implicit hydrogens- hydrogens (protons) that are not drawn in, but assumed to be there because there are no charges.

Also remember these rules:

• If an atom has no multiple bonds, the total number of "things" (bonds and lone pairs) surrounding the atom is probably four, so it's sp³ hybridized (s¹ + p³ = 4).
• If an atom has one double bond, the total number of "things"surrounding the atom is probably three, so it's sp² hybridized (s¹ + p² = 3).
• If an atom has one triple bond or two double bonds, the total number of "things" surrounding the atom is probably two, so it's sp hybridized (s¹ + p¹ = 2).
•  

A sp² This oxygen has one double bond and two lone pairs. Three things total = sp².

B sp² This carbon has one double bond and two single bonds, to a carbon and an oxygen (so two bonds). Three things total = sp².

C sp³ This nitrogen has one long pair, two implicit hydrogens (two bonds), and one single bond to a carbo (another bond). Four things total = sp³.

D sp³ This carbon has one implicit hydrogen and three single bonds. Four things total = sp³.

E sp³ This oxygen has two lone pairs and two bonding pairs. Four things total = sp³.

F sp² This carbon has one double bond (one bond), one implicit hydrogen (another bond), and is bonded to one other carbon (a third bond). So three bonds in total. Three things total = sp².

G sp This carbon has a triple bond (one bond) and a single bond (another bond). So two things in total = sp.

H sp This nitrogen has a lone pair and a triple bond. Two things total = sp.

For angles, remember that electrons repel each other, and so will try to be as far apart from each other as possible:

• Best way to separate two things is a straight line (180°).
• Best way to separate three things is a "trifecta" (120°).
• Best way to separate four things in three dimensional space is a tetrahedron (109.5°).

X 120° The carbon between the two oxygens is sp² hybridized, so the shape is trigonal planar and the bond angle is 120°.

Y 180° The carbon bonded to the nitrogen is sp hybridized, so the shape is linear and the bond angle is 180°.

Z 109.5° The carbon is sp³ hybridized (the most common type of carbon). So the shape is tetrahedral and the bond angle is 109.5°.

There are separate explanations for SN1 and SN2.

S_N1: The S_N1 mechanism involves a carbocation intermediate, and both allylic and benzylic carbocations have resonance, which increases the stability of their carbocations, and speeds up the rate of S_N1 reaction.

S_N2: This explanation is less obvious, and is probably only mentioned in passing in your orgo textbook, if at all. 

The pi systems present in allylic and benzylic halides are able to overlap with the pi orbitals of the nucleophile and the leaving group in the 5-coordinate transition state of an S_N2 reaction. This orbital overlap lowers the energy of the transition state (which stabilizes it), and so increases the rate of S_N2 reaction.

There are other possible explanations as well (Which also vary by textbook). For one, an allylic halide is less sterically hindered than an alkyl halide (less hydrogens sticking out in 3D).

Also, the carbons on the double bond of allyl and benzyl compounds are sp² hybridized, and so are more electronegative than sp³ carbons. So the sp² carbons pull alway some electron density from the alpha carbon (the carbon attached to the leaving group), making it more electrophilic, and thus increasing S_N2 reactivity.

a) There are five different carbons in 2-methylbutane, so hypothetically, you should get five different products, 20% A, 20% B, 20% C, 20% D, and 20% E.

But notice that A and B are the same product! So only four products are possible.

b) Based on statistics alone, we would expect product A (or B) to be the major product, encompassing 40% of all the products (20% A + 20% B).

But this is not what happens in real life because A is primary, and for free-radical halogenation reactions, more substituted products are favored. So the major product will be C, which is tertiary.

This happens because the intermediate is a radical, and radical stability goes 3° > 2°  > 1°.

c) Chlorine reacts faster than bromine, and so is less selective than bromine.

For example, we know that the 3° halide will be the major product. But with chlorine, the proportion might turn out to be ~70% 3° , 20% 2°, and 10% 1°.

But since bromine reacts more slowly, you will get better selectivity. The proportion of products will might be something like ~95% 3° , 4% 2°, and 1% 1°.

While on this topic of reactivity, it's interesting to note that iodine reacts too slowly to be useful, and fluorine reacts so violently it's dangerous!

When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).

Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.

Reductive amination is a two step process. First, an imine or enamine is formed from a carbonyl and an imine.

Then, a the imine (or enamine) is reduced to an amine (using reducing agents such as H₂, NaBH₄, NaBH₃(CN), NaBH(OAc)₃, etc.). For practice on forming imine and enamines, see problem 712).

Electron withdrawing groups (EWG) reduce electron density and reduce basicity. Resonance delocalizes electron density and so also decreases basicity (the electrons are less available to pick up a proton).

Electron donating groups (EDG) add electron density and increase basicity.

Alkyl groups are weakly EDG (induction), and hydrogen is neither EDG nor EWG. So cyclohexylamine (compound X) is a stronger base than ammonia (compound W).

Aniline (compound Y) has resonance, so it is less basic than compound X, which does not have resonance.

N-acyl aniline (compound Z) has more resonance forms than aniline, so it will be the least basic.

So the overall over is:

(most basic) X > W > Y > Z (least basic)

sp² carbons are more electronegative (electron withdrawing) than sp³ carbons due to increased s-character. Electron withdrawing groups (EWG) make acids more acidic and bases less basic.

For this reason, piperidine (compound A) is the more basic than pyridine (compound B):

(more basic) A > B (less basic)

Nitro (-NO₂) is an EWG, so 4-nitropyridine (compound C) will be less basic than compound A.

Dimethylamine is an electron donating group (EDG), so it adds electron density to pyridine and increases basicity. So dimethylamino pyridine (DMAP, compound D) will be more basic than compound A:

(more basic) D > B > C (less basic)

So how does compound A compare with compound D? There is no way of predicting this based on EWG/EDG rules alone. After all, what should be more basic: piperidine, which has sp³ carbons (increases base strength), or DMAP, which has sp² carbons (decreases base strength) but also an EDG (increases base strength) ?

There's no way to tell. But it turns out that Piperidine is a little more basic than DMAP. So the overal order is:

(more basic) A > D > B > C (least basic)

a) Bromine is an electron withdrawing group (EWG), which makes nearby protons more acidic. So the carbonyl with two bromines is the most acidic.

b) To perform alpha bromination, the enolate (or enol) must first be formed, which then attacks Br₂.

The problem with doing this under basic conditions is that each successive bromination leads to a carbonyl that is more acidic, and so forms an enolate even more easily. So once this reaction starts, it's difficult to control.

Enols are less nucleophilic than enolates, so this react can be controlled under acidic conditions. (Br₂/H3O⁺ instead of Br₂/NaOH).

Hydrogens D and E are alpha to two carbonyls and so will be the more acidic than C, which is only alpha to one carbonyl. This is because the enolates that arise from deprotonate at D and E have more resonance forms than the enolate that arrises at C.

Because ketones are more electron withdrawing than esters, D will be more acidic than E. So far we have:

D > E > C > (other)

The carbanions that would arrise from deprotonation of carbons A and B both do not have any resonance forms, so we don't expect either to be acidic. But because A is next to a fluorine atom, which is an electron withdrawing group, A will be more acidic than B. So the overall order is:

(most acidic) D > E > C > A > B (least acidic)